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Solution_manual_to_accompany_boyce_elementary_differential_equations_8e_mathematic87.blogfa.com

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Unformatted text preview: —————————————————————————— CHAPTER 1. —— Chapter One Section 1.1 1. For C "Þ& , the slopes are negative, and hence the solutions decrease. For C "Þ& , the slopes are positive, and hence the solutions increase. The equilibrium solution appears to be Ca>b œ "Þ& , to which all other solutions converge. 3. For C "Þ& , the slopes are :9=3tive, and hence the solutions increase. For C "Þ& , the slopes are negative, and hence the solutions decrease. All solutions appear to diverge away from the equilibrium solution Ca>b œ "Þ& . 5. For C "Î# , the slopes are :9=3tive, and hence the solutions increase. For C "Î# , the slopes are negative, and hence the solutions decrease. All solutions diverge away from ________________________________________________________________________ page 1 —————————————————————————— CHAPTER 1. —— the equilibrium solution Ca>b œ "Î# . 6. For C # , the slopes are :9=3tive, and hence the solutions increase. For C # , the slopes are negative, and hence the solutions decrease. All solutions diverge away from the equilibrium solution Ca>b œ # . 8. For all solutions to approach the equilibrium solution Ca>b œ #Î$ , we must have C w ! for C #Î$ , and C w ! for C #Î$ . The required rates are satisfied by the differential equation C w œ # $C . 9. For solutions other than Ca>b œ # to diverge from C œ # , C a>b must be an increasing function for C # , and a decreasing function for C # . The simplest differential equation whose solutions satisfy these criteria is C w œ C # . 10. For solutions other than Ca>b œ "Î$ to diverge from C œ "Î$ , we must have C w ! for C "Î$ , and C w ! for C "Î$ . The required rates are satisfied by the differential equation C w œ $C " . 12. Note that C w œ ! for C œ ! and C œ & . The two equilibrium solutions are C a>b œ ! and Ca>b œ & . Based on the direction field, C w ! for C & ; thus solutions with initial values greater than & diverge from the solution Ca>b œ & . For ! C &, the slopes are negative, and hence solutions with initial values between ! and & all decrease toward the ________________________________________________________________________ page 2 —————————————————————————— CHAPTER 1. —— solution Ca>b œ ! . For C ! , the slopes are all positive; thus solutions with initial values less than ! approach the solution Ca>b œ ! . 14. Observe that C w œ ! for C œ ! and C œ # . The two equilibrium solutions are C a>b œ ! and Ca>b œ # . Based on the direction field, C w ! for C # ; thus solutions with initial values greater than # diverge from Ca>b œ # . For ! C #, the slopes are also positive, and hence solutions with initial values between ! and # all increase toward the solution Ca>b œ # . For C ! , the slopes are all negative; thus solutions with initial values less than ! diverge from the solution Ca>b œ ! . 16. a+b Let Q a>b be the total amount of the drug ain milligramsb in the patient's body at any given time > a2<=b . The drug is administered into the body at a constant rate of &!! 71Î2<Þ The rate at which the drug leaves the bloodstream is given by !Þ%Q a>b Þ Hence the accumulation rate of the drug is described by the differential equation .Q œ &!! !Þ% Q .> a71Î2<b Þ a, b Based on the direction field, the amount of drug in the bloodstream approaches the equilibrium level of "#&! 71 aA3>238 + 0 /A 29?<=bÞ 18. a+b Following the discussion in the text, the differential equation is ________________________________________________________________________ page 3 —————————————————————————— CHAPTER 1. —— 7 or equivalently, .@ œ 71 # @# .> a,b After a long time, .@ # œ 1 @# Þ .> 7 .@ .> ¸ ! Þ Hence the object attains a terminal velocity given by @_ œ Ê 71 Þ # # a- b Using the relation # @_ œ 71 , the required drag coefficient is # œ !Þ!%!) 51Î=/- Þ a. b 19. All solutions appear to approach a linear asymptote aA3>2 =69:/ /;?+6 >9 "b. It is easy to verify that Ca>b œ > $ is a solution. 20. ________________________________________________________________________ page 4 —————————————————————————— CHAPTER 1. —— All solutions approach the equilibrium solution Ca>b œ ! Þ 23. All solutions appear to diverge from the sinusoid Ca>b œ 25. which is also a solution corresponding to the initial value Ca!b œ &Î# . $ È# =38Ð> 1 Ñ " , % All solutions appear to converge to Ca>b œ ! . First, the rate of change is small. The slopes eventually increase very rapidly in magnitude. 26. ________________________________________________________________________ page 5 —————————————————————————— CHAPTER 1. —— The direction field is rather complicated. Nevertheless, the collection of points at which the slope field is zero, is given by the implicit equation C$ 'C œ #># Þ The graph of these points is shown below: The y-intercepts of these curves are at C œ ! , „È' . It follows that for solutions with initial values C È' , all solutions increase without bound. For solutions with initial values in the range C È' and ! C È' , the slopes remain negative, and hence these solutions decrease without bound. Solutions with initial conditions in the range È' C ! initially increase. Once the solutions reach the critical value, given by the equation C$ 'C œ #># , the slopes become negative and remain negative. These solutions eventually decrease without bound. ________________________________________________________________________ page 6 —————————————————————————— CHAPTER 1. —— Section 1.2 1a+b The differential equation can be rewritten as Integrating both sides of this equation results in 68k& C k œ > -" , or equivalently, & C œ - /> . Applying the initial condition C a!b œ C! results in the specification of the constant as - œ & C! . Hence the solution is C a>b œ & aC! &b/> Þ .C œ .> Þ &C All solutions appear to converge to the equilibrium solution Ca>b œ & Þ 1a- bÞ Rewrite the differential equation as .C œ .> Þ "! #C Integrating both sides of this equation results in " 68k"! #C k œ > -" , or # equivalently, & C œ - /#> . Applying the initial condition C a!b œ C! results in the specification of the constant as - œ & C! . Hence the solution is C a>b œ & aC! &b/#> Þ All solutions appear to converge to the equilibrium solution Ca>b œ & , but at a faster rate than in Problem 1a Þ 2a+bÞ The differential equation can be rewritten as ________________________________________________________________________ page 7 —————————————————————————— CHAPTER 1. —— Integrating both sides of this equation results in 68kC &k œ > -" , or equivalently, C & œ - /> . Applying the initial condition C a!b œ C! results in the specification of the constant as - œ C! & . Hence the solution is C a>b œ & aC! &b/> Þ .C œ .> Þ C& All solutions appear to diverge from the equilibrium solution Ca>b œ & . 2a,bÞ Rewrite the differential equation as .C œ .> Þ #C & Integrating both sides of this equation results in " 68k#C &k œ > -" , or equivalently, # #C & œ - /#> . Applying the initial condition C a!b œ C! results in the specification of the constant as - œ #C! & . Hence the solution is C a>b œ #Þ& aC! #Þ&b/#> Þ All solutions appear to diverge from the equilibrium solution Ca>b œ #Þ& . 2a- b. The differential equation can be rewritten as .C œ .> Þ #C "! Integrating both sides of this equation results in " 68k#C "!k œ > -" , or equivalently, # C & œ - /#> . Applying the initial condition C a!b œ C! results in the specification of the constant as - œ C! & . Hence the solution is C a>b œ & aC! &b/#> Þ ________________________________________________________________________ page 8 —————————————————————————— CHAPTER 1. —— All solutions appear to diverge from the equilibrium solution Ca>b œ & . 3a+b. Rewrite the differential equation as .C œ .> , , +C " which is valid for C Á , Î+. Integrating both sides results in + 68k, +C k œ > -" , or equivalently, , +C œ - /+> . Hence the general solution is C a>b œ a, - /+> bÎ+ Þ Note that if C œ ,Î+ , then .CÎ.> œ ! , and C a>b œ ,Î+ is an equilibrium solution. a, b a3b As + increases, the equilibrium solution gets closer to Ca>b œ ! , from above. Furthermore, the convergence rate of all solutions, that is, + , also increases. a33b As , increases, then the equilibrium solution C a>b œ ,Î+ also becomes larger. In this case, the convergence rate remains the same. a333b If + and , both increase abut ,Î+ œ constantb, then the equilibrium solution Ca>b œ ,Î+ remains the same, but the convergence rate of all solutions increases. 5a+b. Consider the simpler equation .C" Î.> œ +C" . As in the previous solutions, rewrite the equation as .C" œ + .> Þ C" Integrating both sides results in C" a>b œ - /+> Þ a,bÞ Now set Ca>b œ C" a>b 5 , and substitute into the original differential equation. We find that ________________________________________________________________________ page 9 —————————————————————————— CHAPTER 1. —— +C" ! œ +aC" 5 b , . That is, +5 , œ ! , and hence 5 œ ,Î+ . a- b. The general solution of the differential equation is Ca>b œ - /+> ,Î+ Þ This is exactly the form given by Eq. a"(b in the text. Invoking an initial condition Ca!b œ C! , the solution may also be expressed as Ca>b œ ,Î+ aC! ,Î+b/+> Þ 6a+b. The general solution is :a>b œ *!! - />Î# , that is, :a>b œ *!! a:! *!!b/>Î# . With :! œ )&! , the specific solution becomes :a>b œ *!! &!/>Î# . This solution is a decreasing exponential, and hence the time of extinction is equal to the number of months it takes, say >0 , for the population to reach zero. Solving *!! &!/>0 Î# œ ! , we find that >0 œ # 68a*!!Î&!b œ &Þ() months. a,b The solution, :a>b œ *!! a:! *!!b/>Î# , is a decreasing exponential as long as :! *!! . Hence *!! a:! *!!b/>0 Î# œ ! has only one root, given by a- b. The answer in part a,b is a general equation relating time of extinction to the value of the initial population. Setting >0 œ "# months , the equation may be written as *!! œ /' , *!! :! which has solution :! œ )*(Þ('*" . Since :! is the initial population, the appropriate answer is :! œ )*) mice . >0 œ # 68Œ *!! Þ *!! :! a,b. R days œ R Î$! months . The hypothesis is stated mathematically as :! /<N/30 œ #:! . It follows that <R Î$! œ 68a#b , and hence the rate constant is given by < œ $! 68a#bÎR Þ The units are understood to be per month . 9a+b. Assuming no air resistance, with the positive direction taken as downward, Newton's Second Law can be expressed as 7 .@ œ 71 .> 7a+b. The general solution is :a>b œ :! /<> . Based on the discussion in the text, time > is measured in months . Assuming " month œ $! days , the hypothesis can be expressed as :! /<†" œ #:! . Solving for the rate constant, < œ 68a#b , with units of per month . in which 1 is the gravitational constant measured in appropriate units. The equation can be ________________________________________________________________________ page 10 —————————————————————————— CHAPTER 1. —— written as .@Î.> œ 1 , with solution @a>b œ 1> @! Þ The object is released with an initial velocity @! . a,b. Suppose that the object is released from a height of 2 units above the ground. Using the fact that @ œ .BÎ.> , in which B is the downward displacement of the object, we obtain the differential equation for the displacement as .BÎ.> œ 1> @! Þ With the origin placed at the point of release, direct integration results in Ba>b œ 1># Î# @! > . Based on the chosen coordinate system, the object reaches the ground when Ba>b œ 2 . Let > œ X be the time that it takes the object to reach the ground. Then 1X # Î# @! X œ 2 . Using the quadratic formula to solve for X , Xœ @! „È@! #12 Þ 1 The positive answer corresponds to the time it takes for the object to fall to the ground. The negative answer represents a previous instant at which the object could have been launched upward awith the same impact speed b, only to ultimately fall downward with speed @! , from a height of 2 units above the ground. a- b. The impact speed is calculated by substituting > œ X into @a>b in part a+bÞ That is, @aX b œ È@! #12 . 10a+,bb. The general solution of the differential equation is Ua>b œ - /<> Þ Given that Ua!b œ "!! mg , the value of the constant is given by - œ "!! . Hence the amount of thorium-234 present at any time is given by Ua>b œ "!! /<> . Furthermore, based on the hypothesis, setting > œ " results in )#Þ!% œ "!! /< Þ Solving for the rate constant, we find that < œ 68a)#Þ!%Î"!!b œ Þ"*(*'/week or < œ Þ!#)#)/day . a- b. Let X be the time that it takes the isotope to decay to one-half of its original amount. From part a+b, it follows that &! œ "!! /<X , in which < œ Þ"*(*'/week . Taking the natural logarithm of both sides, we find that X œ $Þ&!"% weeks or X œ #%Þ&" .+C s . 11. The general solution of the differential equation .UÎ.> œ < U is Ua>b œ U! /<> , in which U! œ Ua!b is the initial amount of the substance. Let 7 be the time that it takes the substance to decay to one-half of its original amount , U! . Setting > œ 7 in the solution, we have !Þ& U! œ U! /<7 . Taking the natural logarithm of both sides, it follows that <7 œ 68a!Þ&b or <7 œ 68 # Þ ________________________________________________________________________ page 11 —————————————————————————— CHAPTER 1. —— 12. The differential equation governing the amount of radium-226 is .UÎ.> œ < U , with solution Ua>b œ Ua!b/<> Þ Using the result in Problem 11, and the fact that the half-life 7 œ "'#! years , the decay rate is given by < œ 68a#bÎ"'#! per year . The amount of radium-226, after > years, is therefore Ua>b œ Ua!b/!Þ!!!%#()'> Þ Let X be the time that it takes the isotope to decay to $Î% of its original amount. Then setting > œ X, and UaX b œ $ Ua!b , we obtain $ Ua!b œ Ua!b/!Þ!!!%#()'X Þ Solving for the decay % % time, it follows that !Þ!!!%#()' X œ 68a$Î%b or X œ '(#Þ$' years . 13. The solution of the differential equation, with Ua!b œ !, is Ua>b œ GZ a" />ÎGV bÞ As > p _ , the exponential term vanishes, and hence the limiting value is UP œ GZ . 14a+b. The accumulation rate of the chemical is Ð!Þ!"Ña$!!b grams per hour . At any given time > , the concentration of the chemical in the pond is Ua>bÎ"!' grams per gallon . Consequently, the chemical leaves the pond at a rate of a$ ‚ "!% bUa>b grams per hour . Hence, the rate of change of the chemical is given by .U œ $ !Þ!!!$ Ua>b gm/hr . .> a,b. The differential equation can be rewritten as Since the pond is initially free of the chemical, Ua!b œ ! . .U œ !Þ!!!$ .> Þ "!!!! U Integrating both sides of the equation results in 68k"!!!! Uk œ !Þ!!!$> G . Taking the natural logarithm of both sides gives "!!!! U œ - /!Þ!!!$> . Since Ua!b œ ! , the value of the constant is - œ "!!!! . Hence the amount of chemical in the pond at any time is Ua>b œ "!!!!a" /!Þ!!!$> b grams . Note that " year œ )('! hours . Setting > œ )('! , the amount of chemical present after one year is Ua)('!b œ *#((Þ(( grams , that is, *Þ#(((( kilograms . a- b. With the accumulation rate now equal to zero, the governing equation becomes .UÎ.> œ !Þ!!!$ Ua>b gm/hr . Resetting the time variable, we now assign the new initial value as Ua!b œ *#((Þ(( grams . a. b. The solution of the differential equation in Part a- b is Ua>b œ *#((Þ(( /!Þ!!!$> Þ Hence, one year after the source is removed, the amount of chemical in the pond is Ua)('!b œ '(!Þ" grams . ________________________________________________________________________ page 12 —————————————————————————— CHAPTER 1. —— a/b. Letting > be the amount of time after the source is removed, we obtain the equation "! œ *#((Þ(( /!Þ!!!$> Þ Taking the natural logarithm of both sides, !Þ!!!$ > œ œ 68a"!Î*#((Þ((b or > œ ##ß ((' hours œ #Þ' years . a0 b 15a+b. It is assumed that dye is no longer entering the pool. In fact, the rate at which the dye leaves the pool is #!! † c; a>bÎ'!!!!d kg/min œ #!!a'!Î"!!!bc; a>bÎ'!d gm per hour . Hence the equation that governs the amount of dye in the pool is .; œ !Þ# ; .> a gm/hrb . The initial amount of dye in the pool is ; a!b œ &!!! grams . a- b. The amount of dye in the pool after four hours is obtained by setting > œ % . That is, ; a%b œ &!!! /!Þ) œ ##%'Þ'% grams . Since size of the pool is '!ß !!! gallons , the concentration of the dye is !Þ!$(% grams/gallon . a. b. Let X be the time that it takes to reduce the concentration level of the dye to !Þ!# grams/gallon . At that time, the amount of dye in the pool is "ß #!! grams . Using the answer in part a,b, we have &!!! /!Þ# X œ "#!! . Taking the natural logarithm of both sides of the equation results in the required time X œ (Þ"% hours . a/b. Note that !Þ# œ #!!Î"!!! . Consider the differential equation .; < œ ;. .> "!!! Here the parameter < corresponds to the flow rate, measured in gallons per minute . Using the same initial value, the solution is given by ; a>b œ &!!! /< >Î"!!! Þ In order to determine the appropriate flow rate, set > œ % and ; œ "#!! . (Recall that "#!! gm of ________________________________________________________________________ page 13 a,b. The solution of the governing differential equation, with the specified initial value, is ; a>b œ &!!! /!Þ# > Þ —————————————————————————— CHAPTER 1. —— dye has a concentration of !Þ!# gm/gal ). We obtain the equation "#!! œ &!!! /< Î#&! Þ Taking the natural logarithm of both sides of the equation results in the required flow rate < œ $&( gallons per minute . ________________________________________________________________________ page 14 —————————————————————————— CHAPTER 1. —— Section 1.3 1. The differential equation is second order, since the highest derivative in the equation is of order two. The equation is linear, since the left hand side is a linear function of C and its derivatives. 3. The differential equation is fourth order, since the highest derivative of the function C is of order four. The equation is also linear, since the terms containing the dependent variable is linear in C and its derivatives. 4. The differential equation is first order, since the only derivative is of order one. The dependent variable is squared, hence the equation is nonlinear. 5. The differential equation is second order. Furthermore, the equation is nonlinear, since the dependent variable C is an argument of the sine function, which is not a linear function. 7. C" a>b œ /> Ê C"w a>b œ C"ww a>b œ /> . Hence C"ww C" œ ! Þ Also, C# a>b œ -9=2 > Ê C"w a>b œ =382 > and C#ww a>b œ -9=2 > . Thus C#ww C# œ ! Þ 10. C" a>b œ >Î$ Ê C"w a>b œ "Î$ and C"ww a>b œ C"www a>b œ C"wwww a>b œ ! Þ Clearly, C" a>b is a solution. Likewise, C# a>b œ /> >Î$ Ê C#w a>b œ /> "Î$ , C#ww a>b /> , C#www a>b œ /> , C#wwww a>b œ /> . Substituting into the left hand side of the equation, we find that /> %a /> b $a/> >Î$b œ /> %/> $/> > œ > . Hence both functions are solutions of the differential equation. 11. C" a>b œ >"Î# Ê C"w a>b œ >"Î# Î# and C"ww a>b œ >$Î# Î% . Substituting into the left hand side of the equation, we have #># ˆ >$Î# Î% ‰ $>ˆ>"Î# Î# ‰ >"Î# œ >"Î# Î# $ >"Î# Î# >"Î# œ! 9. Ca>b œ $> ># Ê C w a>b œ $ #> . Substituting into the differential equation, we have >a$ #>b a$> ># b œ $> #># $> ># œ ># . Hence the given function is a solution. Likewise, C# a>b œ >" Ê C#w a>b œ ># and C#ww a>b œ # >$ . Substituting into the left hand side of the differential equation, we have #># a# >$ b $>a ># b >" œ % >" $ >" >" œ ! . Hence both functions are solutions of the differential equation. 12. C" a>b œ ># Ê C"w a>b œ #>$ and C"ww a>b œ ' >% . Substituting into the left hand side of the differential equation, we have ># a' >% b &>a #>$ b % ># œ ' ># "! ># % ># œ ! . Likewise, C# a>b œ >2 68 > Ê C#w a>b œ >$ #>$ 68 > and C#ww a>b œ & >% ' >% 68 >. Substituting into the left hand side of the equation, we have ># a & >% ' >% 68 >b &>a>$ #>$ 68 >b %a>2 68 >b œ & >2 ' >2 68 > ________________________________________________________________________ page 15 —————————————————————————— CHAPTER 1. —— & >2 "! >2 68 > % >2 68 > œ ! Þ Hence both functions are solutions of the differential equation. 13. Ca>b œ a-9= >b68 -9= > > =38 > Ê C w a>b œ a=38 >b68 -9= > > -9= > and C ww a>b œ a-9= >b68 -9= > > =38 > =/- > . Substituting into the left hand side of the differential equation, we have a a-9= >b68 -9= > > =38 > =/- >b a-9= >b68 -9= > > =38 > œ a-9= >b68 -9= > > =38 > =/- > a-9= >b68 -9= > > =38 > œ =/- > . Hence the function Ca>b is a solution of the differential equation. 15. Let Ca>b œ /<> . Then C ww a>b œ <# /<> , and substitution into the differential equation results in <# /<> # /<> œ !. Since /<> Á ! , we obtain the algebraic equation <# # œ !Þ The roots of this equation are <"ß# œ „ 3È# Þ 17. Ca>b œ /<> Ê C w a>b œ < /<> and C ww a>b œ <# /<> . Substituting into the differential equation, we have <# /<> </<> ' /<> œ ! . Since /<> Á ! , we obtain the algebraic equation <# < ' œ ! , that is, a< #ba< $b œ ! . The roots are <"ß# œ $ , # . 18. Let Ca>b œ /<> . Then C w a>b œ </<> , C ww a>b œ <# /<> and C www a>b œ <$ /<> . Substituting the derivatives into the differential equation, we have <$ /<> $<# /<> #</<> œ ! . Since /<> Á ! , we obtain the algebraic equation <$ $<# #< œ ! Þ By inspection, it follows that <a< "ba< #b œ ! . Clearly, the roots are <" œ ! , <# œ " and <$ œ # Þ 20. Ca>b œ >< Ê C w a>b œ < ><" and C ww a>b œ <a< "b><# . Substituting the derivatives into the differential equation, we have ># c<a< "b><# d %>a< ><" b % >< œ ! . After some algebra, it follows that <a< "b>< %< >< % >< œ ! . For > Á ! , we obtain the algebraic equation <# &< % œ ! Þ The roots of this equation are <" œ " and <# œ % Þ 21. The order of the partial differential equation is two, since the highest derivative, in fact each one of the derivatives, is of second order. The equation is linear, since the left hand side is a linear function of the partial derivatives. 23. The partial differential equation is fourth order, since the highest derivative, and in fact each of the derivatives, is of order four. The equation is linear, since the left hand side is a linear function of the partial derivatives. 24. The partial differential equation is second order, since the highest derivative of the function ?aBß Cb is of order two. The equation is nonlinear, due to the product ? † ?B on the left hand side of the equation. 25. ?" aBß Cb œ -9= B -9=2 C Ê It is evident that derivatives are ` ?" `B# # ` ?" `C# # œ ! Þ Likewise, given ?# aBß C b œ 68aB# C # b, the second ` # ?" `B# œ -9= B -9=2 C and ` # ?" `C# œ -9= B -9=2 C Þ ________________________________________________________________________ page 16 —————————————————————————— CHAPTER 1. —— ` # ?2 # %B# œ# # `B# B C# aB C # b # ` # ?2 # %C# œ# # `C# B C# aB C # b # Adding the partial derivatives, ` # ?2 ` # ?2 # %B# # %C # œ# # # # `B# `C# B C# B C# aB C # b # aB C # b # % %aB# C# b œ# # B C# aB C # b # œ !. Hence ?# aBß Cb is also a solution of the differential equation. ` # ?" œ -# =38 -B =38 -+> # `B ` # ?" œ -# +# =38 -B =38 -+> `># It is easy to see that +# ` ?#" œ `B # 27. Let ?" aBß >b œ =38 -B =38 -+> . Then the second derivatives are ` # ?" `># Clearly, ?# aBß >b is also a solution of the partial differential equation. # # ` # ?# œ =38aB +>b `B# ` # ?# œ +# =38aB +>b # `> . Likewise, given ?# aBß >b œ =38aB +>b , we have 28. Given the function ?aBß >b œ È1Î> /B Î%! > , the partial derivatives are È1Î> /B# Î%!# > È1Î> B# /B# Î%!# > ?BB œ #! # > %! % > # # B# Î%!# > B# Î%!# > È1> / È1 B / ?> œ #># %!# ># È> È1 ˆ#!# >B# ‰/B# Î%!# > %! # > # È > It follows that !# ?BB œ ?> œ Hence ?aBß >b is a solution of the partial differential equation. . ________________________________________________________________________ page 17 —————————————————————————— CHAPTER 1. —— 29a+b. a,bÞ The path of the particle is a circle, therefore polar coordinates are intrinsic to the problem. The variable < is radial distance and the angle ) is measured from the vertical. Newton's Second Law states that ! F œ 7a Þ In the tangential direction, the equation of motion may be expressed as ! J) œ 7 +) , in which the tangential acceleration, that is, the linear acceleration along the path is +) œ P . # )Î.># Þ Ð +) is positive in the direction of increasing ) Ñ. Since the only force acting in the tangential direction is the component of weight, the equation of motion is 71 =38 ) œ 7P .# ) Þ .># ÐNote that the equation of motion in the radial direction will include the tension in the rodÑ. a- b. Rearranging the terms results in the differential equation .# ) 1 =38 ) œ ! Þ # .> P ________________________________________________________________________ page 18 —————————————————————————— CHAPTER 2. —— Chapter Two Section 2.1 1a+bÞ a,bÞ Based on the direction field, all solutions seem to converge to a specific increasing function. a- bÞ The integrating factor is .a>b œ /$> , and hence Ca>b œ >Î$ "Î* /#> - /$> Þ It follows that all solutions converge to the function C" a>b œ >Î$ "Î* Þ 2a+bÞ a,b. All slopes eventually become positive, hence all solutions will increase without bound. a- bÞ The integrating factor is .a>b œ /#> , and hence Ca>b œ >$ /#> Î$ - /#> Þ It is evident that all solutions increase at an exponential rate. 3a+b ________________________________________________________________________ page 18 —————————————————————————— CHAPTER 2. —— a,b. All solutions seem to converge to the function C! a>b œ " Þ 4a+b. a- bÞ The integrating factor is .a>b œ /#> , and hence Ca>b œ ># /> Î# " - /> Þ It is clear that all solutions converge to the specific solution C! a>b œ " . a,b. Based on the direction field, the solutions eventually become oscillatory. a- bÞ The integrating factor is .a>b œ > , and hence the general solution is C a >b œ $-9=a#>b $ =38a#>b %> # > 5a+b. in which - is an arbitrary constant. As > becomes large, all solutions converge to the function C" a>b œ $=38a#>bÎ# Þ ________________________________________________________________________ page 19 —————————————————————————— CHAPTER 2. —— a,b. All slopes eventually become positive, hence all solutions will increase without bound. a- bÞ The integrating factor is .a>b œ /B:a ' #.>b œ /#> Þ The differential equation can w be written as /#> C w #/#> C œ $/> , that is, a/#> C b œ $/> Þ Integration of both sides of the equation results in the general solution Ca>b œ $/> - /#> Þ It follows that all solutions will increase exponentially. 6a+b a,bÞ All solutions seem to converge to the function C! a>b œ ! Þ Ca>b œ -9=a>b =38a#>b # # > > > a- bÞ The integrating factor is .a>b œ ># , and hence the general solution is in which - is an arbitrary constant. As > becomes large, all solutions converge to the function C! a>b œ ! Þ 7a+b. ________________________________________________________________________ page 20 —————————————————————————— CHAPTER 2. —— a,bÞ All solutions seem to converge to the function C! a>b œ ! Þ a- bÞ The integrating factor is .a>b œ /B:a># b, and hence C a>b œ ># /> - /> Þ It is clear that all solutions converge to the function C! a>b œ ! . # # 8a+b a,bÞ All solutions seem to converge to the function C! a>b œ ! Þ a- bÞ Since .a>b œ a" ># b# , the general solution is Ca>b œ c>+8" a>b G dÎa" ># b# Þ It follows that all solutions converge to the function C! a>b œ ! . 9a+bÞ ________________________________________________________________________ page 21 —————————————————————————— CHAPTER 2. —— a,b. All slopes eventually become positive, hence all solutions will increase without bound. 10a+b. a- bÞ The integrating factor is .a>b œ /B:ˆ' " .>‰ œ />Î# . The differential equation can # w be written as />Î# C w />Î# CÎ# œ $> />Î# Î# , that is, ˆ/>Î# CÎ#‰ œ $> />Î# Î#Þ Integration of both sides of the equation results in the general solution Ca>b œ $> ' - />Î# Þ All solutions approach the specific solution C! a>b œ $> ' Þ a,b. For C ! , the slopes are all positive, and hence the corresponding solutions increase without bound. For C ! , almost all solutions have negative slopes, and hence solutions tend to decrease without bound. a- bÞ First divide both sides of the equation by > . From the resulting standard form, the integrating factor is .a>b œ /B:ˆ ' " .>‰ œ "Î> . The differential equation can be > written as C w Î> CÎ># œ > /> , that is, a CÎ>bw œ > /> Þ Integration leads to the general solution Ca>b œ >/> - > Þ For - Á ! , solutions diverge, as implied by the direction field. For the case - œ ! , the specific solution is Ca>b œ >/> , which evidently approaches zero as > p _ . 11a+b. a,bÞ The solutions appear to be oscillatory. ________________________________________________________________________ page 22 —————————————————————————— CHAPTER 2. —— a- bÞ The integrating factor is .a>b œ /> , and hence Ca>b œ =38a#>b # -9=a#>b - /> Þ It is evident that all solutions converge to the specific solution C! a>b œ =38a#>b # -9=a#>b . 12a+bÞ a- bÞ The integrating factor is .a>b œ /#> . The differential equation can be w written as />Î# C w />Î# CÎ# œ $># Î# , that is, ˆ/>Î# CÎ#‰ œ $># Î#Þ Integration of both sides of the equation results in the general solution Ca>b œ $># "#> #% - />Î# Þ It follows that all solutions converge to the specific solution C! a>b œ $># "#> #% . a,b. All solutions eventually have positive slopes, and hence increase without bound. 14. The integrating factor is .a>b œ /#> . After multiplying both sides by .a>b, the w equation can be written as ˆ/2> C‰ œ > Þ Integrating both sides of the equation results in the general solution Ca>b œ ># /#> Î# - /#> Þ Invoking the specified condition, we require that /# Î# - /# œ ! . Hence - œ "Î# , and the solution to the initial value problem is Ca>b œ a># "b/#> Î# Þ 16. The integrating factor is .a>b œ /B:ˆ' # .>‰ œ ># . Multiplying both sides by .a>b, > w # the equation can be written as a> Cb œ -9=a>b Þ Integrating both sides of the equation results in the general solution Ca>b œ =38a>bÎ># - ># Þ Substituting > œ 1 and setting the value equal to zero gives - œ ! . Hence the specific solution is Ca>b œ =38a>bÎ># Þ 17. The integrating factor is .a>b œ /#> , and the differential equation can be written as ˆ/2> C‰w œ " Þ Integrating, we obtain /2> C a>b œ > - Þ Invoking the specified initial condition results in the solution Ca>b œ a> #b/#> Þ 19. After writing the equation in standard 0 orm, we find that the integrating factor is .a>b œ /B:ˆ' % .>‰ œ >% . Multiplying both sides by .a>b, the equation can be written as > ˆ>% C‰w œ > /> Þ Integrating both sides results in >% C a>b œ a> "b/> - Þ Letting > œ " and setting the value equal to zero gives - œ ! Þ Hence the specific solution of the initial value problem is Ca>b œ ˆ>$ >% ‰/> Þ 21a+b. ________________________________________________________________________ page 23 —————————————————————————— CHAPTER 2. —— The solutions appear to diverge from an apparent oscillatory solution. From the direction field, the critical value of the initial condition seems to be +! œ " . For + " , the solutions increase without bound. For + " , solutions decrease without bound. a,bÞ The integrating factor is .a>b œ />Î# . The general solution of the differential equation is Ca>b œ a)=38a>b %-9=a>bbÎ& - />Î# . The solution is sinusoidal as long as - œ ! . The initial value of this sinusoidal solution is +! œ a)=38a!b %-9=a!bbÎ& œ %Î& Þ a- bÞ See part a,b. 22a+bÞ All solutions appear to eventually increase without bound. The solutions initially increase or decrease, depending on the initial value + . The critical value seems to be +! œ " Þ a,bÞ The integrating factor is .a>b œ />Î# , and the general solution of the differential equation is Ca>b œ $/>Î$ - />Î# Þ Invoking the initial condition C a!b œ + , the solution may also be expressed as Ca>b œ $/>Î$ a+ $b />Î# Þ Differentiating, follows that C w a!b œ " a+ $bÎ# œ a+ "bÎ# Þ The critical value is evidently +! œ " Þ ________________________________________________________________________ page 24 —————————————————————————— CHAPTER 2. —— a- b. For +! œ " , the solution is Ca>b œ $/>Î$ # />Î# , which afor large >b is dominated by the term containing />Î# Þ is Ca>b œ a)=38a>b %-9=a>bbÎ& - />Î# . 23a+bÞ As > p ! , solutions increase without bound if C a"b œ + Þ% , and solutions decrease without bound if Ca"b œ + Þ% Þ a,b. The integrating factor is .a>b œ /B:ˆ' >" .>‰ œ > /> Þ The general solution of the > differential equation is Ca>b œ > /> - /> Î> . Invoking the specified value C a"b œ + , we have " - œ + / . That is, - œ + / " . Hence the solution can also be expressed as Ca>b œ > /> a+ / "b /> Î> . For small values of > , the second term is dominant. Setting + / " œ ! , critical value of the parameter is +! œ "Î/ Þ a- b. For + "Î/ , solutions increase without bound. For + "Î/ , solutions decrease without bound. When + œ "Î/ , the solution is C a>b œ > /> , which approaches ! as > p ! . 24a+b. As > p ! , solutions increase without bound if C a"b œ + Þ% , and solutions decrease without bound if Ca"b œ + Þ% Þ ________________________________________________________________________ page 25 —————————————————————————— CHAPTER 2. —— a,b. Given the initial condition, Ca 1Î#b œ + , the solution is Ca>b œ a+ 1# Î% -9= >bÎ> Þ Since lim -9= > œ " , solutions increase without bound if + %Î1# , and solutions decrease without bound if + %Î1# Þ Hence the critical value is +! œ %Î1# œ !Þ%&#)%(ÞÞÞ. >Ä! a- bÞ For + œ %Î1# , the solution is C a>b œ a" -9= >bÎ> , and lim C a>b œ "Î# . Hence the solution is bounded. >Ä! 25. The integrating factor is .a>b œ /B:ˆ' " .>‰ œ />Î# Þ Therefore general solution is # Ca>b œ c%-9=a>b )=38a>bdÎ& - />Î# Þ Invoking the initial condition, the specific solution is Ca>b œ c%-9=a>b )=38a>b * />Î# dÎ& . Differentiating, it follows that Setting C w a>b œ ! , the first solution is >" œ "Þ$'%$ , which gives the location of the first stationary point. Since C ww a>" b ! , the first stationary point in a local maximum. The coordinates of the point are a"Þ$'%$ ß Þ)#!!)b. C w a>b œ %=38a>b )-9=a>b %Þ& />Î# ‘Î& C ww a>b œ %-9=a>b )=38a>b #Þ#& />Î# ‘Î& 26. The integrating factor is .a>b œ /B:ˆ' # .>‰ œ /#>Î$ , and the differential equation $ can w be written as a/#>Î$ Cb œ /#>Î$ > /#>Î$ Î# Þ The general solution is C a>b œ Ð#" '>ÑÎ) - /#>Î$ . Imposing the initial condition, we have C a>b œ Ð#" '>ÑÎ) aC! #"Î)b/#>Î$ . Since the solution is smooth, the desired intersection will be a point of tangency. Taking the derivative, C w a>b œ $Î% a#C! #"Î%b/#>Î$ Î$ Þ Setting C w a>b œ ! , the solution is >" œ $ 68ca#" )C! bÎ*dÞ Substituting into the solution, the respective value at the # stationary point is Ca>" b œ $ * 68 $ * 68a#" )C! b. Setting this result equal to zero, # % ) we obtain the required initial value C! œ a#" * /%Î$ bÎ) œ "Þ'%$ Þ 27. The integrating factor is .a>b œ />Î4 , and the differential equation can be written as w a/>Î4 Cb œ $ />Î4 # />Î4 -9=a#>bÞ The general solution is Invoking the initial condition, Ca!b œ ! , the specific solution is Ca>b œ "# c)-9=a#>b '%=38a#>bdÎ'& - />Î4 Þ Ca>b œ "# )-9=a#>b '%=38a#>b ()) />Î4 ‘Î'& Þ As > p _ , the exponential term will decay, and the solution will oscillate about an average value of "# , with an amplitude of )ÎÈ'& Þ ________________________________________________________________________ page 26 —————————————————————————— CHAPTER 2. —— 29. The integrating factor is .a>b œ /$>Î# , and the differential equation can be written w as a/$>Î# Cb œ $> /$>Î# # />Î# Þ The general solution is C a>b œ #> %Î$ % /> - /$>Î# Þ Imposing the initial condition, C a>b œ #> %Î$ % /> aC! "'Î$b /$>Î# Þ As > p _ , the term containing /$>Î# will dominate the solution. Its sign will determine the divergence properties. Hence the critical value of the initial condition is C! œ "'Î$Þ The corresponding solution, Ca>b œ #> %Î$ % /> , will also decrease without bound. Note on Problems 31-34 : Let 1a>b be given, and consider the function Ca>b œ C" a>b 1a>b , in which C" a>b p _ as > p _ . Differentiating, C w a>b œ C"w a>b 1 w a>b . Letting + be a constant, it follows that C w a>b +C a>b œ C"w a>b +C" a>b 1 w a>b +1a>bÞ Note that the hypothesis on the function C" a>b will be satisfied, if C"w a>b +C" a>b œ ! . That is, C" a>b œ - /+> Þ Hence Ca>b œ - /+> 1a>b, which is a solution of the equation C w +C œ 1 w a>b +1a>bÞ For convenience, choose + œ " . 31. Here 1a>b œ $ , and we consider the linear equation C w C œ $ Þ The integrating w factor is .a>b œ /> , and the differential equation can be written as a/> C b œ $/> Þ The general solution is Ca>b œ $ - /> Þ 33. 1a>b œ $ > Þ Consider the linear equation C w C œ " $ > ÞThe integrating w factor is .a>b œ /> , and the differential equation can be written as a/> C b œ a# >b/> Þ The general solution is Ca>b œ $ > - /> Þ 34. 1a>b œ % ># Þ Consider the linear equation C w C œ % #> ># ÞThe integrating w factor is .a>b œ /> , and the equation can be written as a/> C b œ a% #> ># b/> Þ The general solution is Ca>b œ % ># - /> Þ ________________________________________________________________________ page 27 —————————————————————————— CHAPTER 2. —— Section 2.2 2. For B Á " , the differential equation may be written as C .C œ cB# Îa" B$ bd.B Þ Integrating both sides, with respect to the appropriate variables, we obtain the relation C# Î# œ " 68k" B$ k - Þ That is, C aBb œ „É # 68k" B$ k - Þ $ $ 3. The differential equation may be written as C# .C œ =38 B .B Þ Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation C " œ -9= B - Þ That is, aG -9= BbC œ ", in which G is an arbitrary constant. Solving for the dependent variable, explicitly, CaBb œ "ÎaG -9= Bb . 5. Write the differential equation as - 9=# #C .C œ -9=# B .B, or =/- # #C .C œ -9=# B .BÞ Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation >+8 #C œ =38 B -9= B B - Þ 7. The differential equation may be written as aC /C b.C œ aB /B b.B Þ Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation C # # / C œ B # # / B - Þ 8. Write the differential equation as a" C# b.C œ B# .B Þ Integrating both sides of the equation, we obtain the relation C C $ Î$ œ B$ Î$ - , that is, $C C $ œ B$ GÞ 9a+b. The differential equation is separable, with C# .C œ a" #Bb.B Þ Integration yields C" œ B B# - Þ Substituting B œ ! and C œ "Î' , we find that - œ ' Þ Hence the specific solution is C" œ B# B ' . The explicit form is CaBb œ "ÎaB# B 'bÞ a, b a- b. Note that B# B ' œ aB #baB $b . Hence the solution becomes singular at B œ # and B œ $ Þ 10a+bÞ CaBb œ È#B #B# % Þ ________________________________________________________________________ page 28 —————————————————————————— CHAPTER 2. —— 10a,bÞ 11a+bÞ Rewrite the differential equation as B /B .B œ C .C Þ Integrating both sides of the equation results in B /B /B œ C # Î# - Þ Invoking the initial condition, we obtain - œ "Î# Þ Hence C # œ #/B #B /B "Þ The explicit form of the solution is CaBb œ È#/B #B /B " Þ The positive sign is chosen, since C a!b œ "Þ a, bÞ a- bÞ The function under the radical becomes negative near B œ "Þ( and B œ !Þ(' Þ 11a+bÞ Write the differential equation as <# .< œ ) " . ) Þ Integrating both sides of the equation results in the relation <" œ 68 ) - Þ Imposing the condition <a"b œ # , we obtain - œ "Î# . The explicit form of the solution is <a) b œ #Îa" # 68 )bÞ ________________________________________________________________________ page 29 —————————————————————————— CHAPTER 2. —— a, bÞ a- bÞ Clearly, the solution makes sense only if ) ! Þ Furthermore, the solution becomes singular when 68 ) œ "Î# , that is, ) œ È/ Þ 13a+bÞ CaBb œ È# 68a" B# b % Þ a, bÞ 14a+b. Write the differential equation as C$ .C œ Ba" B# b .B Þ Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation C# Î# œ È" B# - Þ Imposing the initial condition, we obtain - œ $Î# Þ Hence the specific solution can be expressed as C# œ $ #È" B# Þ The explicit form of the solution is CaBb œ "ÎÉ$ #È" B# Þ The positive sign is chosen to "Î# satisfy the initial condition. ________________________________________________________________________ page 30 —————————————————————————— CHAPTER 2. —— a, bÞ a- bÞ The solution becomes singular when #È" B# œ $ . That is, at B œ „È& Î# Þ 15a+bÞ CaBb œ "Î# ÈB# "&Î% Þ a, bÞ 16a+b. Rewrite the differential equation as %C$ .C œ BaB# "b.B Þ Integrating both sides of the equation results in C% œ aB# "b# Î% - Þ Imposing the initial condition, we obtain - œ ! Þ Hence the solution may be expressed as aB# "b# %C % œ ! Þ The explicit form of the solution is CaBb œ ÈaB# "bÎ# Þ The sign is chosen based on C a!b œ "ÎÈ# Þ ________________________________________________________________________ page 31 —————————————————————————— CHAPTER 2. —— a, bÞ a- bÞ The solution is valid for all B − ‘ . a, bÞ 17a+bÞ CaBb œ &Î# ÈB$ /B "$Î% Þ a- b. The solution is valid for B "Þ%& Þ This value is found by estimating the root of %B$ %/B "$ œ ! Þ 18a+b. Write the differential equation as a$ %Cb.C œ a/B /B b.B Þ Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation $C #C # œ a/B /B b - Þ Imposing the initial condition, C a!b œ " , we obtain - œ (Þ Thus, the solution can be expressed as $C #C # œ a/B /B b (Þ Now by completing the square on the left hand side, #aC $Î%b# œ a/B /B b '&Î) . Hence the explicit form of the solution is CaBb œ $Î% È'&Î"' -9=2 B Þ ________________________________________________________________________ page 32 —————————————————————————— CHAPTER 2. —— a, bÞ a- b. Note the '& "' -9=2 B ! , as long as kBk #Þ" Þ Hence the solution is valid on the interval #Þ" B #Þ" . 19a+bÞ C aBb œ 1Î$ " =38" a$ -9=# BbÞ $ a, bÞ 20a+bÞ Rewrite the differential equation as C# .C œ +<-=38 BÎÈ" B# .B Þ Integrating # both sides of the equation results in C$ Î$ œ a+<-=38 Bb Î# - Þ Imposing the condition $$ Ca!b œ !, we obtain - œ !Þ The explicit form of the solution is CaBb œ É # a+<-=38 Bb#Î$ Þ ________________________________________________________________________ page 33 —————————————————————————— CHAPTER 2. —— a, b . a- b. Evidently, the solution is defined for " Ÿ B Ÿ " Þ 22. The differential equation can be written as a$C# %b.C œ $B# .B Þ Integrating both sides, we obtain C$ %C œ B$ - Þ Imposing the initial condition, the specific solution is C$ %C œ B$ " Þ Referring back to the differential equation, we find that C w p _ as C p „#ÎÈ$ Þ The respective values of the abscissas are B œ "Þ#(' , "Þ&*) Þ Hence the solution is valid for "Þ#(' B "Þ&*) Þ 24. Write the differential equation as a$ #Cb.C œ a# /B b.B Þ Integrating both sides, we obtain $C C # œ #B /B - Þ Based on the specified initial condition, the solution can be written as $C C # œ #B /B " Þ Completing the square, it follows that CaBb œ $Î# È#B /B "$Î% Þ The solution is defined if #B /B "$Î% ! , that is, "Þ& Ÿ B Ÿ # aapproximatelyb. In that interval, C w œ ! , for B œ 68 # Þ It can be verified that C ww a68 #b ! . In fact, C ww aBb ! on the interval of definition. Hence the solution attains a global maximum at B œ 68 # Þ 26. The differential equation can be written as a" C# b" .C œ #a" Bb.B Þ Integrating both sides of the equation, we obtain +<->+8C œ #B B# - Þ Imposing the given initial condition, the specific solution is +<->+8C œ #B B# Þ Therefore, C aBb œ >+8a#B B# bÞ Observe that the solution is defined as long as 1Î# #B B# 1Î# Þ It is easy to see that #B B# "Þ Furthermore, #B B# œ 1Î# for B œ #Þ' and !Þ' Þ Hence the solution is valid on the interval #Þ' B !Þ' Þ Referring back to the differential ________________________________________________________________________ page 34 —————————————————————————— CHAPTER 2. —— equation, the solution is stationary at B œ "Þ Since C ww aBb ! on the entire interval of definition, the solution attains a global minimum at B œ " Þ 28a+b. Write the differential equation as C" a% Cb" .C œ >a" >b" .> Þ Integrating both sides of the equation, we obtain 68 kCk 68kC %k œ %> %68k" >k - Þ Taking the exponential of both sides, it follows that kCÎaC %bk œ G /%> Îa" >b% Þ It follows that as > p _ , kCÎaC %bk œ k" %ÎaC %bkp _ . That is, C a>b p % Þ a,bÞ Setting Ca!b œ # , we obtain that G œ " . Based on the initial condition, the solution may be expressed as CÎaC %b œ /%> Îa" >b% Þ Note that CÎaC %b ! , for all > !Þ Hence C % for all > !Þ Referring back to the differential equation, it follows that C w is always positive. This means that the solution is monotone increasing. We find that the root of the equation /%> Îa" >b% œ $** is near > œ #Þ)%% Þ a- bÞ Note the Ca>b œ % is an equilibrium solution. Examining the local direction field, we see that if Ca!b ! , then the corresponding solutions converge to C œ % . Referring back to part a+b, we have CÎaC %b œ cC! ÎaC! %bd/%> Îa" >b% , for C! Á % Þ Setting % > œ # , we obtain C! ÎaC! %b œ a$Î/# b C a#bÎaC a#b %bÞ Now since the function 0 aCb œ CÎaC %b is monotone for C % and C % , we need only solve the equations C! ÎaC! %b œ $**a$Î/# b% and C! ÎaC! %b œ %!"a$Î/# b% Þ The respective solutions are C! œ $Þ''## and C! œ %Þ%!%# Þ 30a0 bÞ ________________________________________________________________________ page 35 —————————————————————————— CHAPTER 2. —— 31a- b C " 32a+b. Observe that aB# $C # bÎ#BC œ " ˆ B ‰ # is homogeneous. $C #B Þ Hence the differential equation a,b. The substitution C œ B @ results in @ B @ w œ aB# $B# @# bÎ#B# @ . The transformed equation is @ w œ a" @# bÎ#B@ Þ This equation is separable, with general solution @# " œ - B Þ In terms of the original dependent variable, the solution is B# C # œ - B$ Þ a- bÞ 33a- bÞ ________________________________________________________________________ page 36 —————————————————————————— CHAPTER 2. —— C C " 34a+bÞ Observe that a%B $C bÎÐ#B CÑ œ # B # B ‘ Þ Hence the differential equation is homogeneous. a,b. The substitution C œ B @ results in @ B @ w œ # @Îa# @b. The transformed equation is @ w œ a@# &@ %bÎÐ# @ÑB Þ This equation is separable, with general solution a@%b# k@"k œ GÎB$ Þ In terms of the original dependent variable, the solution is a%B Cb# kBCk œ GÞ a- bÞ 35a- b. 36a+bÞ Divide by B# to see that the equation is homogeneous. Substituting C œ B @ , we obtain B @ w œ a" @b# Þ The resulting differential equation is separable. a,bÞ Write the equation as a" @b# .@ œ B" .B Þ Integrating both sides of the equation, we obtain the general solution "Îa" @b œ 68kBk - Þ In terms of the original dependent variable, the solution is C œ B cG 68kBkd" B Þ ________________________________________________________________________ page 37 —————————————————————————— CHAPTER 2. —— a- bÞ a,bÞ Integrating both sides of the transformed equation yields " & 68k" &@# k œ 68kBk - , that is, " &@# œ GÎkBk& Þ In terms of the original dependent variable, the general solution is &C# œ B# GÎkBk$ Þ a- bÞ C " $C 37a+bÞ The differential equation can be expressed as C w œ " ˆ B ‰ # B . Hence the # equation is homogeneous. The substitution C œ B @ results in B @ w œ a" &@# bÎ#@ . #@ " Separating variables, we have "&@# .@ œ B .B Þ C " C " 38a+bÞ The differential equation can be expressed as C w œ $ B # ˆ B ‰ . Hence the # equation is homogeneous. The substitution C œ B @ results in B @ w œ a@# "bÎ#@, that @ " is, @##" .@ œ B .B Þ a,bÞ Integrating both sides of the transformed equation yields 68k@# "k œ 68kBk - , that is, @# " œ G kBkÞ In terms of the original dependent variable, the general solution is C# œ G B# kBk B# Þ ________________________________________________________________________ page 38 —————————————————————————— CHAPTER 2. —— a- bÞ ________________________________________________________________________ page 39 —————————————————————————— CHAPTER 2. —— Section 2.3 5a+b. Let U be the amount of salt in the tank. Salt enters the tank of water at a rate of # " ˆ" " =38 >‰ œ " " =38 > 9DÎ738 Þ It leaves the tank at a rate of # UÎ"!! 9DÎ738Þ % # # % Hence the differential equation governing the amount of salt at any time is .U "" œ =38 > UÎ&! Þ .> #% The initial amount of salt is U! œ &! 9D Þ The governing ODE is linear, with integrating w factor .a>b œ />Î&! Þ Write the equation as ˆ/>Î&! U‰ œ />Î&! ˆ " " =38 >‰Þ The # % specific solution is Ua>b œ #& "#Þ&=38 > '#&-9= > '$"&! />Î&! ‘Î#&!" 9D Þ a, bÞ a- bÞ The amount of salt approaches a steady state, which is an oscillation of amplitude "Î% about a level of #& 9D Þ 6a+b. The equation governing the value of the investment is .WÎ.> œ < W . The value of the investment, at any time, is given by W a>b œ W! /<> Þ Setting W aX b œ #W! , the required time is X œ 68a#bÎ< Þ a- bÞ Referring to Parta+b, < œ 68a#bÎX . Setting X œ ) , the required interest rate is to be approximately < œ )Þ'' % Þ 8a+b. Based on the solution in Eq.a"'b , with W! œ ! , the value of the investments with contributions is given by W a>b œ #&ß !!!a/<> "bÞ After ten years, person A has WE œ $#&ß !!!a"Þ##'b œ $ $!ß '%! Þ Beginning at age $& , the investments can now be analyzed using the equations WE œ $!ß '%! /Þ!)> and WF œ #&ß !!!a/Þ!)> "bÞ After thirty years, the balances are WE œ $ $$(ß ($% and WF œ $ #&!ß &(*Þ a,bÞ For the case < œ (% œ Þ!( , X ¸ *Þ* C<= Þ a,bÞ For an unspecified rate < , the balances after thirty years are WE œ $!ß '%! /$!< and WF œ #&ß !!!a/$!< "bÞ ________________________________________________________________________ page 40 —————————————————————————— CHAPTER 2. —— a- b . a. bÞ The two balances can never be equal. 11a+b. Let W be the value of the mortgage. The debt accumulates at a rate of <W , in which < œ Þ!* is the annual interest rate. Monthly payments of $ )!! are equivalent to $ *ß '!! per yearÞ The differential equation governing the value of the mortgage is .WÎ.> œ Þ!* W *ß '!! Þ Given that W! is the original amount borrowed, the debt is W a>b œ W! /Þ!*> "!'ß ''(a/Þ!*> "bÞ Setting W a$!b œ ! , it follows that W! œ $ **ß &!! . a,bÞ The total payment, over $! years, becomes $ #))ß !!! . The interest paid on this purchase is $ "))ß &!! . 13a+b. The balance increases at a rate of < W $/yr, and decreases at a constant rate of 5 $ per year. Hence the balance is modeled by the differential equation .WÎ.> œ <W 5 . The balance at any time is given by W a>b œ W! /<> 5 a/<> "bÞ < a,b. The solution may also be expressed as W a>b œ ÐW! 5 Ñ/<> 5 Þ Note that if the < < withdrawal rate is 5! œ < W! , the balance will remain at a constant level W! Þ 5 a- b. Assuming that 5 5! , W aX! b œ ! for X! œ " 68’ 55! “Þ < a. b. If < œ Þ!) and 5 œ #5! , then X! œ )Þ'' years . a/b. Setting W a>b œ ! and solving for /<> in Parta,b, /<> œ results in 5 œ <W! /<X Îa/<X "bÞ 5 5<W! Þ Now setting > œ X a0 bÞ In parta/b, let 5 œ "#ß !!! , < œ Þ!) , and X œ #! . The required investment becomes W! œ $ ""*ß ("& . 14a+bÞ Let U w œ < U Þ The general solution is Ua>b œ U! /<> Þ Based on the definition of half-life, consider the equation U! Î# œ U! /&($! < Þ It follows that ________________________________________________________________________ page 41 —————————————————————————— CHAPTER 2. —— &($! < œ 68Ð"Î#Ñ, that is, < œ "Þ#!*( ‚ "!% per year. a,b. Hence the amount of carbon-14 is given by Ua>b œ U! /"Þ#!*(‚"! > Þ % % a- bÞ Given that UaX b œ U! Î& , we have the equation "Î& œ /"Þ#!*(‚"! X Þ Solving for the decay time, the apparent age of the remains is approximately X œ "$ß $!%Þ'& years . 15. Let T a>b be the population of mosquitoes at any time > . The rate of increase of the mosquito population is <T Þ The population decreases by #!ß !!! per day. Hence the equation that models the population is given by .T Î.> œ <T #!ß !!! . Note that the variable > represents days . The solution is T a>b œ T! /<> #!ß!!! a/<> "bÞ In the < absence of predators, the governing equation is .T" Î.> œ <T" , with solution T" a>b œ T! /<> Þ Based on the data, set T" a(b œ #T! , that is, #T! œ T! /(< Þ The growth rate is determined as < œ 68a#bÎ( œ Þ!**!# per dayÞ Therefore the population, including the predation by birds, is T a>b œ # ‚ "!& /Þ!**> #!"ß **(a/Þ!**> "b œ œ #!"ß **(Þ$ "*((Þ$ /Þ!**> Þ 16a+b. Ca>b œ /B:c#Î"! >Î"! #-9=Ð>ÑÎ"!dÞ The doubling-time is 7 ¸ #Þ*'$# Þ a,bÞ The differential equation is .CÎ.> œ CÎ"! , with solution C a>b œ C a!b/>Î"! Þ The doubling-time is given by 7 œ "!68a#b ¸ 'Þ*$"& Þ a- b. Consider the differential equation .CÎ.> œ a!Þ& =38Ð#1>Ñb CÎ& Þ The equation is " " separable, with C .C œ ˆ!Þ" & =38Ð#1>щ.> Þ Integrating both sides, with respect to the appropriate variable, we obtain 68 C œ a1> -9=Ð#1>ÑbÎ"!1 - Þ Invoking the initial condition, the solution is Ca>b œ /B:ca" 1> -9=Ð#1>ÑbÎ"!1dÞ The doubling-time is 7 ¸ 'Þ$)!% Þ The doubling-time approaches the value found in parta,b. a. b . 17a+b. The differential equation .CÎ.> œ <a>b C 5 is linear, with integrating factor .a>b œ /B: ' <a>b.>‘Þ Write the equation as a. C bw œ 5 .a>b Þ Integration of both ________________________________________________________________________ page 42 —————————————————————————— CHAPTER 2. —— sides yields the general solution C œ 5 ' .a7 b. 7 C! .a!b‘Î.a>b . In this problem, the integrating factor is .a>b œ /B:ca-9= > >bÎ&dÞ a,bÞ The population becomes extinct, if Ca>‡ b œ ! , for some > œ >‡ . Referring to parta+b, we find that Ca>‡ b œ ! Ê ( >‡ ! /B:ca-9= 7 7 bÎ&d. 7 œ & /"Î& C- Þ It can be shown that the integral on the left hand side increases monotonically, from zero to a limiting value of approximately &Þ!)*$ . Hence extinction can happen only if & /"Î& C- &Þ!)*$ , that is, C- !Þ)$$$ Þ a- b. Repeating the argument in parta,b, it follows that Ca>‡ b œ ! Ê ( >‡ ! /B:ca-9= 7 7 bÎ&d. 7 œ " "Î& / C- Þ 5 19a+b. Let Ua>b be the volume of carbon monoxide in the room. The rate of increase of CO is aÞ!%ba!Þ"b œ !Þ!!% 0 >$ Î738 Þ The amount of CO leaves the room at a rate of a!Þ"bUa>bÎ"#!! œ Ua>bÎ"#!!! 0 >$ Î738 Þ Hence the total rate of change is given by the differential equation .UÎ.> œ !Þ!!% Ua>bÎ"#!!! Þ This equation is linear and separable, with solution Ua>b œ %) %) /B:a >Î"#!!!b 0 >$ Þ Note that U! œ ! 0 >$ Þ Hence the concentration at any time is given by Ba>b œ Ua>bÎ"#!! œ Ua>bÎ"# % . a,b. The concentration of CO in the room is Ba>b œ % %/B:a >Î"#!!!b %Þ A level of !Þ!!!"# corresponds to !Þ!"# % . Setting Ba7 b œ !Þ!"# , the solution of the equation % %/B:a >Î"#!!!b œ !Þ!"# is 7 ¸ $' minutes . page 43 a. b. Evidently, C- is a linear function of the parameter 5 . Hence extinction can happen only if /"Î& C- Î5 &Þ!)*$ , that is, C- %Þ"''( 5 Þ ________________________________________________________________________ —————————————————————————— CHAPTER 2. —— 20a+bÞ The concentration is - a>b œ 5 T Î< a-! 5 T Î<b/<>ÎZ Þ It is easy to see that - a>p_b œ 5 T Î< Þ a,bÞ - a>b œ -! /<>ÎZ . The reduction times are X&! œ 68Ð#ÑZ Î< and X"! œ 68Ð"!ÑZ Î<Þ a- bÞ The reduction times, in years, are XW œ 68a"!ba'&Þ#bÎ"#ß #!! œ %$!Þ)& XQ œ 68a"!ba"&)bÎ%ß *!! œ ("Þ% à XI œ 68a"!ba"(&bÎ%'! œ 'Þ!& XS œ 68a"!ba#!*bÎ"'ß !!! œ "(Þ'$ Þ 21a- bÞ 22a+b. The differential equation for the motion is 7 .@Î.> œ @Î$! 71 Þ Given the initial condition @a!b œ #! m/s , the solution is @a>b œ %%Þ" '%Þ" /B:a >Î%Þ&b . Setting @a>" b œ ! , the ball reaches the maximum height at >" œ "Þ')$ sec . Integrating @a>b , the position is given by Ba>b œ $")Þ%& %%Þ" > #))Þ%& /B:a >Î%Þ&bÞ Hence the maximum height is Ba>" b œ %&Þ() m . a- bÞ a,bÞ Setting Ba># b œ ! , the ball hits the ground at ># œ &Þ"#) sec . 23a+bÞ The differential equation for the upward motion is 7 .@Î.> œ . @# 71 , in which . œ "Î"$#& . This equation is separable, with . @#7 .@ œ .> Þ Integrating 71 ________________________________________________________________________ page 44 —————————————————————————— CHAPTER 2. —— both sides and invoking the initial condition, @a>b œ %%Þ"$$ >+8aÞ%#& Þ### >bÞ Setting @a>" b œ ! , the ball reaches the maximum height at >" œ "Þ*"' sec . Integrating @a>b , the position is given by Ba>b œ "*)Þ(& 68c-9=a!Þ### > !Þ%#&bd %)Þ&( Þ Therefore the maximum height is Ba>" b œ %)Þ&' m . a,bÞ The differential equation for the downward motion is 7 .@Î.> œ .@# 71 Þ 7 This equation is also separable, with 71. @# .@ œ .> Þ For convenience, set > œ ! at the top of the trajectory. The new initial condition becomes @a!b œ ! . Integrating both sides and invoking the initial condition, we obtain 68ca%%Þ"$ @bÎa%%Þ"$ @bd œ >Î#Þ#& Þ Solving for the velocity, @a>b œ %%Þ"$a" />Î#Þ#& bÎa" />Î#Þ#& b Þ Integrating @a>b , the # position is given by Ba>b œ **Þ#* 68’/>Î#Þ#& Îa" />Î#Þ#& b “ ")'Þ# Þ To estimate the a- bÞ duration of the downward motion, set Ba># b œ ! , resulting in ># œ $Þ#(' sec . Hence the total time that the ball remains in the air is >1 ># œ &Þ"*# sec . 24a+bÞ Measure the positive direction of motion downward . Based on Newton's #nd law, the equation of motion is given by 7 .@ !Þ(& @ 71 , ! > "! œœ "# @ 71 , > "! .> Þ Note that gravity acts in the positive direction, and the drag force is resistive. During the first ten seconds of fall, the initial value problem is .@Î.> œ @Î(Þ& $# , with initial velocity @a!b œ ! fps Þ This differential equation is separable and linear, with solution @a>b œ #%!a" />Î(Þ& b. Hence @a"!b œ "('Þ( fps Þ a,b. Integrating the velocity, with Ba>b œ ! , the distance fallen is given by Hence Ba"!b œ "!(%Þ& ft . Ba>b œ #%! > ")!! />Î(Þ& ")!! . ________________________________________________________________________ page 45 —————————————————————————— CHAPTER 2. —— a- bÞ For computational purposes, reset time to > œ ! . For the remainder of the motion, the initial value problem is .@Î.> œ $#@Î"& $# , with specified initial velocity @a!b œ "('Þ( fps Þ The solution is given by @a>b œ "& "'"Þ( /$# >Î"& Þ As > p _ , @a>b p @P œ "& fps Þ Integrating the velocity, with Ba!b œ "!(%Þ& , the distance fallen after the parachute is open is given by Ba>b œ "& > (&Þ) /$# >Î"& ""&!Þ$ Þ To find the duration of the second part of the motion, estimate the root of the transcendental equation "& X (&Þ) /$# X Î"& ""&!Þ$ œ &!!! Þ The result is X œ #&'Þ' sec Þ a. bÞ 25a+b. Measure the positive direction of motion upward . The equation of motion is given by 7.@Î.> œ 5 @ 71 . The initial value problem is .@Î.> œ 5@Î7 1 , with @a!b œ @! . The solution is @a>b œ 71Î5 a@! 71Î5 b/5>Î7 Þ Setting @a>7 b œ !, the maximum height is reached at time >7 œ a7Î5 b68ca71 5 @! bÎ71dÞ Integrating the velocity, the position of the body is Ba>b œ 71 >Î5 ”Š Hence the maximum height reached is B7 œ Ba>7 b œ 7 @! 7# 71 5 @! 1Š ‹ 68” •Þ 5 5 71 7# 7 @! 5>Î7 ÑÞ ‹ 1 •Ð" / 5 5 a,bÞ Recall that for $ ¥ " , 68a" $ b œ $ " $ # " $ $ " $ % á # $ % 26a,b. a- bÞ 5Ä ! 71 lim 71a5 @!5 b/ 71 5 5>Î7 œ lim 5Ä ! 28a+b. In terms of displacement, the differential equation is 7@ .@Î.B œ 5 @ 71 Þ .@ This follows from the chain rule : .@ œ .B .B œ @ .@ . The differential equation is .> .> .> separable, with 7Ä! lim ˆ 71 @! ‰/5>Î7 ‘ œ ! , since 5 > 7 a5 @ ! 71b/5>Î7 œ 1> Þ 7Ä! lim /5>Î7 œ ! Þ ________________________________________________________________________ page 46 —————————————————————————— CHAPTER 2. —— Ba@b œ 7@ 7# 1 71 5 @ # 68º ºÞ 5 5 71 The inverse exists, since both B and @ are monotone increasing. In terms of the given parameters, Ba@b œ "Þ#& @ "&Þ$" 68k!Þ!)"' @ "kÞ a,bÞ Ba"!b œ "$Þ%& meters . The required value is 5 œ !Þ#% . a- bÞ In parta+b, set @ œ "! m/s and B œ "! meters . 29a+bÞ Let B represent the height above the earth's surface. The equation of motion is Q7 given by 7 .@ œ K aVBb# , in which K is the universal gravitational constant. The .> 7@ .@ Q7 œ K . .B aV Bb# symbols Q and V are the mass and radius of the earth, respectively. By the chain rule, This equation is separable, with @ .@ œ KQ aV Bb# .B Þ Integrating both sides, and invoking the initial condition @a!b œ È#1V , the solution is @# œ #KQ aV Bb" #1V #KQ ÎV Þ From elementary physics, it follows that 1 œ KQ ÎV # . Therefore @aBb œ È#1 ’VÎÈV B “Þ a Note that 1 œ ()ß &%& mi/hr# .b a,bÞ We now consider .BÎ.> œ È#1 ’VÎÈV B “. This equation is also separable, with ÈV B .B œ È#1 V .> Þ By definition of the variable B, the initial condition is #Î$ Ba!b œ !Þ Integrating both sides, we obtain Ba>b œ $ ˆÈ#1 V > # V $Î# ‰‘ V Þ # $ Setting the distance BaX b V œ #%!ß !!! , and solving for X , the duration of such a flight would be X ¸ %* hours . 32a+bÞ Both equations are linear and separable. The initial conditions are @a!b œ ? -9= E and Aa!b œ ? =38 E . The two solutions are @a>b œ ? -9= E /<> and Aa>b œ 1Î< a? =38 E 1Î<b/<> Þ ________________________________________________________________________ page 47 —————————————————————————— CHAPTER 2. —— a,bÞ Integrating the solutions in parta+b, and invoking the initial conditions, the coordinates are Ba>b œ ? -9= Ea" /<> b and < a- bÞ ? Ca>b œ 1>Î< ˆ1 ?< =38 E 2<# ‰Î<# Š =38 E 1Î<# ‹/<> Þ < a. bÞ Let X be the time that it takes the ball to go $&! ft horizontally. Then from above, /X Î& œ a? -9= E (!bÎ? -9= E Þ At the same time, the height of the ball is given by CaX b œ "'! X #'( "#&?=38 E Ð)!! &? =38 EÑca? -9= E (!bÎ? -9= E dÞ Hence E and ? must satisfy the inequality )!!68” ? -9= E (! • #'( "#&?=38 E Ð)!! &? =38 EÑca? -9= E (!bÎ? -9= E d "! Þ ? -9= E "Î# 33a+bÞ Solving equation a3b, C w aBb œ ca5 # C bÎCd chosen, since C is an increasing function of B . . The positive answer is a,b. Let C œ 5 # =38# > . Then .C œ #5 # =38 > -9= > .> Þ Substituting into the equation in parta+b, we find that #5 # =38 > -9= > .> -9= > œ Þ .B =38 > Hence #5 # =38# > .> œ .B Þ a- bÞ Letting ) œ #> , we further obtain 5 # =38# ) . ) œ .B Þ Integrating both sides of the # equation and noting that > œ ) œ ! corresponds to the origin, we obtain the solutions Ba) b œ 5 # a) =38 )bÎ# and cfrom parta,bd Ca)b œ 5 # a" -9= )bÎ# Þ a. b. Note that CÎB œ a" -9= )bÎa) =38 )bÞ Setting B œ " , C œ # , the solution of the equation a" -9= ) bÎa) =38 )b œ # is ) ¸ "Þ%!" . Substitution into either of the expressions yields 5 ¸ #Þ"*$ Þ ________________________________________________________________________ page 48 —————————————————————————— CHAPTER 2. —— Section 2.4 2. Considering the roots of the coefficient of the leading term, the ODE has unique solutions on intervals not containing ! or % . Since # − a! ß %b , the initial value problem has a unique solution on the interval a! ß %b Þ 3. The function >+8 > is discontinuous at odd multiples of 1 . Since # 1 initial value problem has a unique solution on the interval ˆ 1 ß $# ‰Þ # 1 # 1 $1 # , the 5. :a>b œ #>Îa% ># b and 1a>b œ $># Îa% ># b. These functions are discontinuous at B œ „# . The initial value problem has a unique solution on the interval a # ß #bÞ 6. The function 68 > is defined and continuous on the interval a! ß _b . Therefore the initial value problem has a unique solution on the interval a! ß _b. 7. The function 0 a> ß C b is continuous everywhere on the plane, except along the straight line C œ #>Î& Þ The partial derivative `0 Î`C œ (>Îa#> &C b# has the same region of continuity. 9. The function 0 a> ß C b is discontinuous along the coordinate axes, and on the hyperbola ># C # œ " . Furthermore, `0 „" C 68k>C k œ # # C# b `C C a" > a " ># C # b # has the same points of discontinuity. 10. 0 a> ß C b is continuous everywhere on the plane. The partial derivative `0 Î`C is also continuous everywhere. 12. The function 0 a> ß C b is discontinuous along the lines > œ „5 1 and C œ " . The partial derivative `0 Î`C œ -9>a>bÎa" C b# has the same region of continuity. 14. The equation is separable, with .CÎC # œ #> .> Þ Integrating both sides, the solution is given by Ca>b œ C! ÎÐ" C! ># ÑÞ For C! ! , solutions exist as long as ># "ÎC! . For C! Ÿ ! , solutions are defined for all > . 15. The equation is separable, with .CÎC $ œ .> Þ Integrating both sides and invoking the initial condition, Ca>b œ C! ÎÈ#C! > " Þ Solutions exist as long as #C! > " ! , that is, #C! > " . If C! ! , solutions exist for > "Î#C! . If C! œ ! , then the solution Ca>b œ ! exists for all > . If C! ! , solutions exist for > "Î#C! . 16. The function 0 a> ß C b is discontinuous along the straight lines > œ " and C œ ! . The partial derivative `0 Î`C is discontinuous along the same lines. The equation is ________________________________________________________________________ page 49 ————————————————————————— CHAPTER 2. —— separable, with C .C œ ># .>Îa" >$ bÞ Integrating and invoking the initial condition, the # "Î# solution is Ca>b œ # 68k" >$ k C! ‘ Þ Solutions exist as long as $ # that is, C! # 68k" >$ kÞ For all C! Ðit can be verified that C! œ ! yields a valid $ solution, even though Theorem #Þ%Þ# does not guarantee oneÑ , solutions exists as long as # k" >$ k /B:a $C! Î#bÞ From above, we must have > " . Hence the inequality $ # may be written as > /B:a $C! Î#b " Þ It follows that the solutions are valid for "Î$ # c/B:a $C! Î#b "d > _ . # # 68¸" >$ ¸ C! ! , $ 17. 18. Based on the direction field, and the differential equation, for C! ! , the slopes eventually become negative, and hence solutions tend to _ . For C! ! , solutions increase without bound if >! ! Þ Otherwise, the slopes eventually become negative, and solutions tend to zero . Furthermore, C! œ ! is an equilibrium solution. Note that slopes are zero along the curves C œ ! and >C œ $ . 19. ________________________________________________________________________ page 50 —————————————————————————— CHAPTER 2. —— For initial conditions a>! ß C! b satisfying >C $ , the respective solutions all tend to zero . Solutions with initial conditions above or below the hyperbola >C œ $ eventually tend to „_ . Also, C! œ ! is an equilibrium solution. 20. Solutions with >! ! all tend to _ . Solutions with initial conditions a>! ß C! b to the right of the parabola > œ " C # asymptotically approach the parabola as > p _ . Integral curves with initial conditions above the parabola aand C! !b also approach the curve. The slopes for solutions with initial conditions below the parabola aand C! !b are all negative. These solutions tend to _ Þ 21. Define C- a>b œ # a> - b$Î# ?a> - b, in which ?a>b is the Heaviside step function. $ Note that C- a- b œ C- a!b œ ! and C- ˆ- a$Î#b#Î$ ‰ œ "Þ a,bÞ Let - œ # a$Î#b#Î$ Þ a+bÞ Let - œ " a$Î#b#Î$ Þ a- bÞ Observe that C! a#b œ # a#b$Î# , C- a>b # a#b$Î# for ! - #, and that C- a#b œ ! for $ $ - #Þ So for any - !, „C- a#b − c #ß #dÞ 26a+bÞ Recalling Eq. a$&b in Section #Þ", ________________________________________________________________________ page 51 —————————————————————————— CHAPTER 2. —— " Þ ( .a=b1a=b .= .a>b .a>b "' .a>b Cœ It is evident that C" a>b œ " .a>b " a,b. By definition, .a>b œ /B:ˆ ' :a>b.>‰. Hence C"w œ :a>b That is, C"w :a>bC" œ !Þ and C# a>b œ .a=b1a=b .=. " .a>b œ :a>bC" Þ a- bÞ C#w œ Š :a>b That is, C#w :a>bC# œ 1a>bÞ " '> .a>b ‹ ! .a=b1a=b .= " Š .a>b ‹.a>b1a>b œ :a>bC# 1a>bÞ 30. Since 8 œ $, set @ œ C # . It follows that .@ œ #C $ .C and .C œ C# .@ Þ .> .> .> .> $ Substitution into the differential equation yields C# .@ &C œ 5C $ , which further .> results in @ w #&@ œ #5 Þ The latter differential equation is linear, and can be written as w a/#&> b œ #5 Þ The solution is given by @a>b œ #5> /#&> -/#&> Þ Converting back to the original dependent variable, C œ „@"Î# Þ 31. Since 8 œ $, set @ œ C # . It follows that .@ œ #C $ .C and .C œ C# .@ Þ .> .> .> .> C$ .@ $ The differential equation is written as # .> a>-9= > X bC œ 5C , which upon further substitution is @ w #a>-9= > X b@ œ #Þ This ODE is linear, with integrating factor .a>b œ /B:ˆ#' a>-9= > X b.>‰ œ /B:a #>=38 > #X >bÞ The solution is > ! $ $ @a>b œ #/B:a#>=38 > #X >b( /B:a #>=38 7 #X 7 b. 7 - /B:a #>=38 > #X >bÞ Converting back to the original dependent variable, C œ „@"Î# Þ 33. The solution of the initial value problem C"w #C" œ !, C" a!b œ " is C" a>b œ /#> Þ Therefore ya" b œ C" a"b œ /# Þ On the interval a"ß _bß the differential equation is C#w C# œ !, with C# a>b œ -/> Þ Therefore C a" b œ C# a"b œ -/" Þ Equating the limits Ca" b œ Ca" b, we require that - œ /" Þ Hence the global solution of the initial value problem is Ca>b œ œ Note the discontinuity of the derivative C a >b œ œ #/#> , ! > " Þ /"> , > " /#> , ! Ÿ > Ÿ " Þ /"> , > " ________________________________________________________________________ page 52 —————————————————————————— CHAPTER 2. —— Section 2.5 1. For C! ! , the only equilibrium point is C ‡ œ ! . 0 w a!b œ + ! , hence the equilibrium solution 9a>b œ ! is unstable. 2. The equilibrium points are C‡ œ +Î, and C ‡ œ ! . 0 w a +Î, b ! , therefore the equilibrium solution 9a>b œ +Î, is asymptotically stable. 3. ________________________________________________________________________ page 53 —————————————————————————— CHAPTER 2. —— 4. The only equilibrium point is C‡ œ ! . 0 w a!b ! , hence the equilibrium solution 9a>b œ ! is unstable. 5. The only equilibrium point is C‡ œ ! . 0 w a!b ! , hence the equilibrium solution 9a>b œ ! is +=C7:>9>3-+66C stable. 6. ________________________________________________________________________ page 54 —————————————————————————— CHAPTER 2. —— 7a,b. 8. The only equilibrium point is C‡ œ " . Note that 0 w a"b œ ! , and that C w ! for C Á " . As long as C! Á " , the corresponding solution is monotone decreasing. Hence the equilibrium solution 9a>b œ " is semistable. 9. ________________________________________________________________________ page 55 —————————————————————————— CHAPTER 2. —— 10. The equilibrium points are C‡ œ ! , „" . 0 w aC b œ " $C # . The equilibrium solution 9a>b œ ! is unstable, and the remaining two are asymptotically stable. 11. 12. The equilibrium points are C‡ œ ! , „# . 0 w aC b œ )C %C $ . The equilibrium solutions 9a>b œ # and 9a>b œ # are unstable and asymptotically stable, respectively. The equilibrium solution 9a>b œ ! is semistable. ________________________________________________________________________ page 56 —————————————————————————— CHAPTER 2. —— 13. The equilibrium points are C‡ œ ! and " . 0 w aC b œ #C 'C # %C $ . Both equilibrium solutions are semistable. 15a+b. Inverting the Solution a""b, Eq. a"$b shows > as a function of the population C and the carrying capacity O . With C! œ OÎ$, " a"Î$bc" aCÎO bd > œ 68º ºÞ < aCÎO bc" a"Î$bd " a"Î$bc" a#Î$bd 7 œ 68º ºÞ < a#Î$bc" a"Î$bd Setting C œ #C! , That is, 7 œ " 68 % Þ If < œ !Þ!#& per year, 7 œ &&Þ%& years. < " ! c" " d X œ 68º ºÞ < " c" ! d a,bÞ In Eq. a"$b, set C! ÎO œ ! and CÎO œ " . As a result, we obtain Given ! œ !Þ", " œ !Þ* and < œ !Þ!#& per year, 7 œ "(&Þ() years. 16a+b. ________________________________________________________________________ page 57 —————————————————————————— CHAPTER 2. —— 17. Consider the change of variable ? œ 68aCÎO bÞ Differentiating both sides with respect to >, ? w œ C w ÎCÞ Substitution into the Gompertz equation yields ? w œ <?, with solution ? œ ?! /<> Þ It follows that 68aCÎO b œ 68aC! ÎO b/<> Þ That is, C œ /B:68aC! ÎO b/<> ‘Þ O a,bÞ Solving for >, a+b. Given O œ )!Þ& ‚ "!' , C! ÎO œ !Þ#& and < œ !Þ(" per year, C a#b œ &(Þ&) ‚ "!' . " 68aCÎO b > œ 68” •Þ < 68aC! ÎO b Setting Ca7 b œ !Þ(&O , the corresponding time is 7 œ #Þ#" years. 19a+b. The rate of increase of the volume is given by rate of flow in rate of flow out. That is, . Z Î.> œ 5 !+È#12 Þ Since the cross section is constant, .Z Î.> œ E.2Î.>Þ Hence the governing equation is . 2Î.> œ ˆ5 !+È#12 ‰ÎEÞ " 5# a,bÞ Setting .2Î.> œ !, the equilibrium height is 2/ œ #1 ˆ !+ ‰ Þ Furthermore, since 0 w a2/ b !, it follows that the equilibrium height is asymptotically stable. a- b. Based on the answer in parta,b, the water level will intrinsically tend to approach 2/ . Therefore the height of the tank must be greater than 2/ ; that is, 2/ Z ÎE. 22a+b. The equilibrium points are at C‡ œ ! and C ‡ œ ". Since 0 w aC b œ ! #!C , the equilibrium solution 9 œ ! is unstable and the equilibrium solution 9 œ " is asymptotically stable. a,b. The ODE is separable, with cCa" C bd" .C œ ! .> . Integrating both sides and invoking the initial condition, the solution is Ca>b œ C! /!> Þ " C! C! /!> It is evident that aindependent of C! b 23a+b. Ca>b œ C! /"> Þ > Ä _ lim Ca>b œ ! and >Ä_ lim Ca>b œ " . a,b. From parta+b, .BÎ.> œ ! B C! /"> Þ Separating variables, .BÎB œ ! C! /"> .> Þ Integrating both sides, the solution is Ba>b œ B! /B:! C! Î" ˆ" /"> ‰‘Þ a- b. As > p _ , Ca>b p ! and Ba>b p B! /B:a! C! Î" bÞ Over a long period of time, the ________________________________________________________________________ page 58 —————————————————————————— CHAPTER 2. —— proportion of carriers vanishes. Therefore the proportion of the population that escapes the epidemic is the proportion of susceptibles left at that time, B! /B:a! C! Î" bÞ 25a+b. Note that 0 aBb œ BcaV V- b + B# d, and 0 w aBb œ aV V- b $+ B# . So if aV V- b ! , the only equilibrium point is B‡ œ ! . 0 w a!b ! , and hence the solution 9a>b œ ! is asymptotically stable. a,b. If aV V- b ! , there are three equilibrium points B‡ œ ! , „ÈaV V- bÎ+ Þ Now 0 w a!b !, and 0 w ˆ„ÈaV V- bÎ+ ‰ !. Hence the solution 9 œ ! is unstable, and the solutions 9 œ „ÈaV V- bÎ+ are asymptotically stable. a- b . ________________________________________________________________________ page 59 —————————————————————————— CHAPTER 2. —— Section 2.6 1. Q aBß Cb œ #B $ and R aBß C b œ #C # . Since QC œ RB œ ! , the equation is exact. Integrating Q with respect to B , while holding C constant, yields <aBß Cb œ œ B# $B 2aC b . Now <C œ 2 w aC b , and equating with R results in the possible function 2aCb œ C # #C . Hence <aBß C b œ B# $B C # #C , and the solution is defined implicitly as B# $B C # #C œ - . 2. Q aBß Cb œ #B %C and R aBß C b œ #B #C . Note that QC Á RB , and hence the differential equation is not exact. 4. First divide both sides by a#BC #bÞ We now have Q aBß C b œ C and R aBß C b œ B . Since QC œ RB œ ! , the resulting equation is exact. Integrating Q with respect to B , while holding C constant, results in <aBß Cb œ BC 2aC b . Differentiating with respect to C , <C œ B 2 w aC b . Setting <C œ R , we find that 2 w aC b œ ! , and hence 2 aC b œ ! is acceptable. Therefore the solution is defined implicitly as BC œ - . Note that if BC " œ ! , the equation is trivially satisfied. 6. Write the given equation as a+B ,C b.B a,B -C b.C . Now Q aBß C b œ +B ,C and R aBß C b œ ,B -C . Since QC Á RB , the differential equation is not exact. 8. Q aBß Cb œ /B =38 C $C and R aBß C b œ $B /B =38 C . Note that QC Á RB , and hence the differential equation is not exact. 10. Q aBß Cb œ CÎB 'B and R aBß C b œ 68 B #. Since QC œ RB œ "ÎB, the given equation is exact. Integrating R with respect to C , while holding B constant, results in <aBß Cb œ C 68 B #C 2 aBb Þ Differentiating with respect to B, <B œ CÎB 2 w aBb. Setting <B œ Q , we find that 2w aBb œ 'B , and hence 2aBb œ $B# . Therefore the solution is defined implicitly as $B# C 68 B #C œ - . 11. Q aBß Cb œ B 68 C BC and R aBß C b œ C 68 B BC . Note that QC Á RB , and hence the differential equation is not exact. 13. Q aBß Cb œ #B C and R aBß C b œ #C B. Since QC œ RB œ ", the equation is exact. Integrating Q with respect to B , while holding C constant, yields <aBß Cb œ œ B# BC 2aC b. Now <C œ B 2 w aC b. Equating <C with R results in 2 w aC b œ #C , and hence 2aCb œ C # . Thus <aBß C b œ B# BC C # , and the solution is given implicitly as B# BC C # œ - . Invoking the initial condition C a"b œ $ , the specific solution is B# BC C # œ (. The explicit form of the solution is CaBb œ " ’B È#) $B# “Þ # Hence the solution is valid as long as $B# Ÿ #) Þ 16. Q aBß Cb œ C /#BC B and R aBß C b œ ,B /#BC . Note that QC œ /#BC #BC /#BC , and RB œ , /#BC #,BC /#BC Þ The given equation is exact, as long as , œ " . Integrating ________________________________________________________________________ page 60 —————————————————————————— CHAPTER 2. —— R with respect to C , while holding B constant, results in <aBß Cb œ /#BC Î# 2aBb Þ Now differentiating with respect to B, <B œ C /#BC 2w aBb. Setting <B œ Q , we find that 2w aBb œ B , and hence 2aBb œ B# Î# . Conclude that <aBß C b œ /#BC Î# B# Î# . Hence the solution is given implicitly as /#BC B# œ - . 17. Integrating <C œ R , while holding B constant, yields <aBß Cb œ ' R aBß C b.C 2 aBbÞ ` Taking the partial derivative with respect to B , <B œ ' `B R aBß C b.C 2 w aBbÞ Now ` set <B œ Q aBß C b and therefore 2w aBb œ Q aBß C b ' `B R aBß C b.C . Based on the fact ` that QC œ RB , it follows that `C c2w aBbd œ ! . Hence the expression for 2w aBb can be integrated to obtain 2aBb œ ( Q aBß C b.B ( ”( 18. Observe that ` `C cQ aBbd ` R aBß C b.C •.B . `B œ 20. QC œ C " -9= C C # =38 C and RB œ # /B a-9= B =38 BbÎC . Multiplying both sides by the integrating factor .aBß Cb œ C /B , the given equation can be written as a/B =38 C #C =38 Bb.B a/B -9= C #-9= Bb.C œ ! . Let Q œ .Q and R œ .R . Observe that Q C œ R B , and hence the latter ODE is exact. Integrating R with respect to C , while holding B constant, results in <aBß Cb œ /B =38 C #C -9= B 2 aBb Þ Now differentiating with respect to B, <B œ /B =38 C #C =38 B 2 w aBb. Setting <B œ Q , we find that 2w aBb œ ! , and hence 2aBb œ ! is feasible. Hence the solution of the given equation is defined implicitly by /B =38 C #C -9= B œ " . 21. QC œ " and RB œ # . Multiply both sides by the integrating factor .aBß C b œ C to obtain C# .B a#BC C # /C b.C œ !. Let Q œ CQ and R œ CR . It is easy to see that Q C œ R B , and hence the latter ODE is exact. Integrating Q with respect to B yields <aBß Cb œ BC # 2aC b . Equating <C with R results in 2 w aC b œ C # /C , and hence 2aCb œ /C aC # #C #b. Thus <aBß C b œ BC # /C aC # #C #b, and the solution is defined implicitly by BC# /C aC # #C #b œ - . ` `B cR aC bd œ !Þ 24. The equation .Q .R C w œ ! has an integrating factor if a.Q bC œ a.R bB , that is, .C Q .B R œ .RB .QC . Suppose that RB QC œ V aBQ CR b, in which V is some function depending only on the quantity D œ BC . It follows that the modified form of the equation is exact, if .C Q .B R œ .V aBQ CR b œ V a. BQ . CR b. This relation is satisfied if .C œ a. BbV and .B œ a. CbV . Now consider . œ .aBCbÞ Then the partial derivatives are .B œ .w C and .C œ .w B . Note that .w œ . .Î.D . Thus . must satisfy .w aD b œ V aD bÞ The latter equation is separable, with . . œ V aD b.D , and .aD b œ ' V aD b.D Þ Therefore, given V œ V aBC b, it is possible to determine . œ .aBC b which becomes an integrating factor of the differential equation. ________________________________________________________________________ page 61 —————————————————————————— CHAPTER 2. —— 28. The equation is not exact, since RB QC œ #C " . However, aRB QC bÎQ œ œ a#C "bÎC is a function of C alone. Hence there exists . œ .aC b , which is a solution of the differential equation .w œ a# "ÎCb. . The latter equation is separable, with . .Î. œ # "ÎC . One solution is .aC b œ /B:a#C 68 C b œ /#C ÎC . Now rewrite the given ODE as /#C .B a#B /#C "ÎC b.C œ ! . This equation is exact, and it is easy to see that <aBß Cb œ B /#C 68 C . Therefore the solution of the given equation is defined implicitly by B /#C 68 C œ - . 30. The given equation is not exact, since RB QC œ )B$ ÎC $ 'ÎC # . But note that aRB QC bÎQ œ #ÎC is a function of C alone, and hence there is an integrating factor . œ .aCb. Solving the equation .w œ a#ÎCb. , an integrating factor is .aCb œ C # Þ Now rewrite the differential equation as a%B$ $Cb.B a$B %C $ b.C œ !. By inspection, <aBß Cb œ B% $BC C % , and the solution of the given equation is defined implicitly by B% $BC C % œ - . 32. Multiplying both sides of the ODE by . œ cBC a#B C bd" , the given equation is equivalent to ca$B CbÎa#B# BC bd.B caB C bÎa#BC C # bd.C œ ! Þ Rewrite the differential equation as ” # # " " •.B ” •.C œ ! . B #B C C #B C It is easy to see that QC œ RB Þ Integrating Q with respect to B, while keeping C constant, results in <aBß Cb œ #68kBk 68k#B C k 2aC b . Now taking the partial derivative with respect to C , <C œ a#B C b" 2 w aC b . Setting <C œ R , we find that 2 w aCb œ "ÎC , and hence 2aC b œ 68 kC k . Therefore <aBß Cb œ #68kBk 68k#B C k 68 kC k , and the solution of the given equation is defined implicitly by #B$ C B# C # œ - . ________________________________________________________________________ page 62 —————————————————————————— CHAPTER 2. —— Section 2.7 2a+b. The Euler formula is C8" œ C8 2a#C8 "b œ a" #2 bC8 2 . 4a+b. The Euler formula is C8" œ a" #2bC8 $2 -9= >8 . a. bÞ The differential equation is linear, with solution Ca>b œ a" /#> bÎ# Þ a. b. The exact solution is Ca>b œ a'-9= > $=38 > ' /#> bÎ& . 5. All solutions seem to converge to 9a>b œ #&Î* . 6. Solutions with positive initial conditions seem to converge to a specific function. On the other hand, solutions with negative coefficients decrease without bound. 9a>b œ ! is an equilibrium solution. 7. ________________________________________________________________________ page 63 —————————————————————————— CHAPTER 2. —— All solutions seem to converge to a specific function. 8. Solutions with initial conditions to the 'left' of the curve > œ !Þ"C # seem to diverge. On the other hand, solutions to the 'right' of the curve seem to converge to zero. Also, 9a>b is an equilibrium solution. 9. All solutions seem to diverge. 10. ________________________________________________________________________ page 64 —————————————————————————— CHAPTER 2. —— Solutions with positive initial conditions increase without bound. Solutions with negative initial conditions decrease without bound. Note that 9a>b œ ! is an equilibrium solution. # 12. The iteration formula is C8" œ a" $2bC8 2 >8 C8 . a>! ß C! b œ a! ß !Þ&bÞ 11. The Euler formula is C8" œ C8 $2ÈC8 &2 . The initial value is C! œ # . $ 14. The iteration formula is C8" œ a" 2 >8 bC8 2 C8 Î"! Þ a>! ß C! b œ a! ß "bÞ 17. The Euler formula is C8" œ C8 The initial point is a>! ß C! b œ a" ß #bÞ 18a+b. See Problem 8. 19a+bÞ # 2aC8 #>8 C8 b . $ ># 8 # # a,b. The iteration formula is C8" œ C8 2 C8 2 >8 . The critical value of ! appears to be near !! ¸ !Þ')"& . For C! !! , the iterations diverge. ________________________________________________________________________ page 65 —————————————————————————— CHAPTER 2. —— 20a+b. The ODE is linear, with general solution Ca>b œ > - /> . Invoking the specified initial condition, Ca>! b œ C! , we have C! œ >! - />! Þ Hence - œ aC! >! b/>! Þ Thus the solution is given by 9a>b œ aC! >! b/>>! > . a,b. The Euler formula is C8" œ a" 2bC8 2 2 >8 . Now set 5 œ 8 " . a- b. We have C" œ a" 2bC! 2 2 >! œ a" 2 bC! a>" >! b 2 >! . Rearranging the terms, C" œ a" 2baC! >! b >" . Now suppose that C5 œ a" 2b5 aC! >! b >5 , for some 5 " . Then C5" œ a" 2bC5 2 2 >5 . Substituting for C5 , we find that C5" œ a" 2b5" aC! >! b a" 2b >5 2 2 >5 œ a" 2 b5" aC! >! b >5 2 . Noting that >5" œ >5 5 , the result is verified. a. b. Substituting 2 œ a> >! bÎ8 , with >8 œ > , C8 œ Œ" > >! 8 aC! >! b > . 8 pointwise convergence is proved. Taking the limit of both sides, as 8 p _ , and using the fact that lim a" +Î8b8 œ /+ , 8Ä_ 23. The exact solution is 9a>b œ >Î# /#> . The Euler formula is C8" œ a" #2bC8 2Î# 2 >8 . Since C! œ " , C" œ a" #2 b 2Î# œ a" #2 b >" Î# Þ It is easy to show by mathematical induction, that C8 œ a" #2b8 >8 Î# . For > ! , set 2 œ >Î8 and thus >8 œ > . Taking the limit, we find that lim C8 œ lim ca" #>Î8b8 >Î#d œ œ /#> >Î# . Hence pointwise convergence is proved. 8Ä_ 8Ä_ 21. The exact solution is 9a>b œ /> . The Euler formula is C8" œ a" 2bC8 . It is easy to see that C8 œ a" 2b8 C! œ a" 2b8 Þ Given > ! , set 2 œ >Î8 . Taking the limit, we find that lim C8 œ lim a" >Î8b8 œ /> Þ 8Ä_ 8Ä_ ________________________________________________________________________ page 66 —————————————————————————— CHAPTER 2. —— Section 2.8 2. Let D œ C $ and 7 œ > " . It follows that .DÎ. 7 œ a.DÎ.>ba.>Î. 7 b œ .DÎ.> . Furthermore, .DÎ.> œ .CÎ.> œ " C $ . Hence .DÎ. 7 œ " aD $b$ . The new initial condition is D a7 œ !b œ ! Þ 3. The approximating functions are defined recursively by 98" a>b œ '! #c98 a=b "d.= . Setting 9! a>b œ ! , 9" a>b œ #> . Continuing, 9# a>b œ #># #> , 9$ a>b œ % >$ #># #> , $ 9% a>b œ # >% % >$ #># #> , â . Given convergence, set $ $ > 9a>b œ 9" a>b "c95" a>b 95 a>bd _ œ #> " _ 5œ# 5œ" +5 5 >. 5x Comparing coefficients, +$ Î$x œ %Î$ , +% Î%x œ #Î$ , â . It follows that +$ œ ) , +% œ "', and so on. We find that in general, that +5 œ #5 . Hence 9a>b œ " _ 5œ" #> 5x #5 5 > œ / ". 5. The approximating functions are defined recursively by 98" a>b œ ( c 98 a=bÎ# =d.= . > ! Setting 9! a>b œ ! , 9" a>b œ ># Î# . Continuing, 9# a>b œ ># Î# >$ Î"# ,9$ a>b œ ># Î# >$ Î"# >% Î*' , 9% a>b œ ># Î# >$ Î"# >% Î*' >& Î*'! , â . Given convergence, set ________________________________________________________________________ page 67 —————————————————————————— CHAPTER 2. —— 9a>b œ 9" a>b "c95" a>b 95 a>bd _ œ ># Î# " 5œ$ 5œ" _ +5 5 >. 5x Comparing coefficients, +$ Î$x œ "Î"# , +% Î%x œ "Î*' , +& Î&x œ "Î*'! , â . We find that +$ œ "Î# , +% œ "Î%, +& œ "Î) , â Þ In general, +5 œ #5" . Hence 9a>b œ " _ #5# a >b 5 5x 5œ# œ % />Î# #> % . 6. The approximating functions are defined recursively by 98" a>b œ ( c98 a=b " =d.= . > ! Setting 9! a>b œ ! , 9" a>b œ > ># Î# , 92 a>b œ > >$ Î' , 9$ a>b œ > >% Î#% , 9% a>b œ œ > >& Î"#! , â . Given convergence, set 9a>b œ 9" a>b "c95" a>b 95 a>bd _ 5œ" # œ > > Î# ># Î# >$ Î'‘ >$ Î' >% Î#%‘ â œ>!!â. Note that the terms can be rearranged, as long as the series converges uniformly. ________________________________________________________________________ page 68 —————————————————————————— CHAPTER 2. —— 8a+b. The approximating functions are defined recursively by Set 9! a>b œ !. The iterates are given by 9" a>b œ ># Î# , 92 a>b œ ># Î# >& Î"! , 9$ a>b œ ># Î# >& Î"! >) Î)! , 9% a>b œ ># Î# >& Î"! >) Î)! >"" Î))! ,â . Upon inspection, it becomes apparent that 8" " >$ >' a >$ b 98 a>b œ > ” â • # #†& #†&†) # † & † )âc# $a8 "bd 98" a>b œ ( =# 98 a=b =‘.= . > ! # œ ># " 8 5œ" a, b . # † & † )âc# $a5 "bd a >$ b 5" Þ The iterates appear to be converging. 9a+b. The approximating functions are defined recursively by Set 9! a>b œ !. The first three iterates are given by 9" a>b œ >$ Î$ , 92 a>b œ >$ Î$ >( Î'$ , 9$ a>b œ >$ Î$ >( Î'$ #>"" Î#!(* >"& Î&*&$& . ________________________________________________________________________ page 69 # 98" a>b œ ( =# 98 a=b‘.= . > ! —————————————————————————— CHAPTER 2. —— a, b . The iterates appear to be converging. 10a+b. The approximating functions are defined recursively by Set 9! a>b œ !. The first three iterates are given by 9" a>b œ > , 92 a>b œ > >% Î% , 9$ a>b œ > >% Î% $>( Î#) $>"! Î"'! >"$ Î)$$ . a, b . $ 98" a>b œ ( " 98 a=b‘.= . > ! The approximations appear to be diverging. 12a+b. The approximating functions are defined recursively by 98" a>b œ ( ” > ! Note that "Îa#C #b œ above iteration formula by " # 5œ! ! C 5 S ˆC( ‰. For computational purposes, replace the ' $=# %= # •.= . #a98 a=b "b ________________________________________________________________________ page 70 —————————————————————————— CHAPTER 2. —— ' "> 5 98" a>b œ ( – ˆ$=# %= #‰" 98 a=b—.= . #! 5œ! Set 9! a>b œ !. The first four approximations are given by 9" a>b œ > ># >$ Î# , 92 a>b œ > ># Î# >$ Î' >% Î% >& Î& >' Î#% â, 9$ a>b œ > ># Î# >% Î"# $>& Î#! %>' Î%& â, 9% a>b œ > ># Î# >% Î) (>& Î'! >' Î"& â a, b . The approximations appear to be converging to the exact solution, 9a>b œ " È" #> #># >$ Þ 13. Note that 98 a!b œ ! and 98 a"b œ " , a 8 " . Let + − a! ß "b . Then 98 a+b œ +8 . Clearly, lim +8 œ ! . Hence the assertion is true. 8Ä_ 14a+b. 98 a!b œ ! , a 8 " . Let + − Ð! ß "Ó. Then 98 a+b œ #8+ /8+ œ #8+Î/8+ Þ # # Using l'Hospital's rule, lim #+DÎ/+D œ lim "ÎD/+D œ ! . Hence lim 98 a+b œ ! Þ # # # #" " a,b. '! #8B /8B .B œ /8B ¸! œ " /8 Þ Therefore, DÄ_ DÄ_ 8Ä_ 8Ä_ ! lim ( 98 aBb.B Á ( lim 98 aBb.B . " " ! 8Ä_ 15. Let > be fixed, such that a> ß C" bß a> ß C# b − H . Without loss of generality, assume that C" C# . Since 0 is differentiable with respect to C , the mean value theorem asserts that b 0 − aC" ß C# b such that 0 a> ß C" b 0 a> ß C# b œ 0C a> ß 0baC" C# bÞ Taking the absolute value of both sides, k0 a> ß C" b 0 a> ß C# bk œ k0C a> ß 0bk kC" C# kÞ Since, by assumption, `0 Î`C is continuous in H, 0C attains a maximum on any closed and bounded subset of H . ________________________________________________________________________ page 71 ————————————————————————— CHAPTER 2. —— Hence k0 a> ß C" b 0 a> ß C# bk Ÿ O kC" C# kÞ 16. For a sufficiently small interval of >, 98" a>b , 98 a>b − H . Since 0 satisfies a Lipschitz condition, k0 a> ß 98 a>bb 0 a> ß 98" a>bbk Ÿ O k98 a>b 98" a>bkÞ Here O œ 7+B k0C k . k> k k>k > 17a+bÞ 9" a>b œ '! 0 a= ß !b.= . Hence k9" a>bk Ÿ '! k0 a= ß !bk.= Ÿ '! Q .= œ Q k>k , in which Q is the maximum value of k0 a> ß C bk on H . > a,b. By definition, 9# a>b 9" a>b œ '! c0 a= ß 9" a=bb 0 a= ß !bd.= . Taking the absolute k>k value of both sides, k9# a>b 9" a>bk Ÿ '! kc0 a= ß 9" a=bb 0 a= ß !bdk.= . Based on the k>k k>k results in Problems 16 and 17, k9# a>b 9" a>bk Ÿ '! O k9" a=b !k.= Ÿ OQ '! k=k.= . Evaluating the last integral, we obtain k9# a>b 9" a>bk Ÿ Q O k>k# Î# . a- b. Suppose that for some 3 " . By definition, 93" a>b 93 a>b œ '! c0 a> ß 93 a=bb 0 a= ß 93" a=bbd.= . It follows that > k93 a>b 93" a>bk Ÿ Q O 3" k>k3 3x k93" a>b 93 a>bk Ÿ ( Ÿ( Ÿ( k> k k> k k> k ! k0 a= ß 93 a=bb 0 a= ß 93" a=bbk.= O k93 a=b 93" a=bk.= O Q O 3" k=k3 .= 3x ! Q O 3 k>k3" Q O 3 23" œ Ÿ Þ a 3 "b x a 3 "b x ! Hence, by mathematical induction, the assertion is true. 18a+b. Use the triangle inequality, k+ ,k Ÿ k+k k,k . k98 a>bk Ÿ Q " 8 a,b. For k>k Ÿ 2 , k9" a>bk Ÿ Q 2 , and k98 a>b 98" a>bk Ÿ Q O 8" 2 8 Îa8 xb . Hence Q 8 aO2b3 " Þ O 3œ" 3 x O 3" 23 3x 3œ" œ ________________________________________________________________________ page 72 —————————————————————————— CHAPTER 2. —— a- b. The sequence of partial sums in a,b converges to Q ˆ/O2 "‰Þ By the comparison O test, the sums in a+b also converge. Furthermore, the sequence k98 a>bk is bounded, and hence has a convergent subsequence. Finally, since individual terms of the series must tend to zero, k98 a>b 98" a>bk p ! , and it follows that the sequence k98 a>bk is convergent. > > 19a+b. Let 9a>b œ '! 0 a= ß 9a=bb.= and <a>b œ '! 0 a= ß <a=bb.= Þ Then by linearity of > the integral, 9a>b <a>b œ '! c0 a= ß 9a=bb 0 a= ß <a=bbd.= Þ a,b. It follows that k9a>b <a>bk Ÿ '! k0 a= ß 9a=bb 0 a= ß <a=bbk.= Þ > a- b. We know that 0 satisfies a Lipschitz condition, based on k`0 Î`Ck Ÿ O in H. Therefore, > ! k0 a> ß C" b 0 a> ß C# bk Ÿ O kC" C# k , k9a>b <a>bk Ÿ ( k0 a= ß 9a=bb 0 a= ß <a=bbk.= Ÿ ( O k9a=b <a=bk.= . > ! ________________________________________________________________________ page 73 —————————————————————————— CHAPTER 2. —— Section 2.9 1. Writing the equation for each 8 ! , C" œ !Þ* C! , C# œ !Þ* C" , C$ œ !Þ* C# and so on, it is apparent that C8 œ a !Þ*b8 C! . The terms constitute an alternating series, which converge to zero, regardless of C! . 3. Write the equation for each 8 ! , C" œ È$ C! , C# œ È%Î# C" , C$ œ È&Î$ C# ,â. Upon substitution, we find that C# œ Èa% † $bÎ# C" , C$ œ Èa& † % † $bÎa$ † #b C! , â Þ It can be proved by mathematical induction, that C8 œ " a 8 #b x Ê C! È# 8x "È œ a8 "ba8 #b C! . È# This sequence is divergent, except for C! œ ! . 4. Writing the equation for each 8 ! , C" œ C! , C# œ C" , C$ œ C# , C% œ C$ , and so on, it can be shown that C! C8 œ œ C! , for 8 œ %5 or 8 œ %5 " , for 8 œ %5 # or 8 œ %5 $ The sequence is convergent only for C! œ ! . 6. Writing the equation for each 8 ! , C" œ !Þ& C! ' C# œ !Þ& C" ' œ !Þ&a!Þ& C! 'b ' œ a!Þ&b# C! ' a!Þ&b' C$ œ !Þ& C# ' œ !Þ&a!Þ& C" 'b ' œ a!Þ&b$ C! 'c" a!Þ&b Ð!Þ&Ñ# d ã C8 œ a!Þ&b8 C! "#c" a!Þ&b8 d which can be verified by mathematical induction. The sequence is convergent for all C! , and in fact C8 p "# . 7. Let C8 be the balance at the end of the 8-th day. Then C8" œ a" <Î$&'b C8 . The solution of this difference equation is C8 œ a" <Î$'&b8 C! , in which C! is the initial balance. At the end of one year, the balance is C$'& œ a" <Î$'&b$'& C! . Given that < œ Þ!(, C$'& œ a" <Î$'&b$'& C! œ "Þ!(#& C! . Hence the effective annual yield is a"Þ!(#& C! C! bÎC! œ (Þ#& % . 8. Let C8 be the balance at the end of the 8-th month. Then C8" œ a" <Î"#b C8 #& . As in the previous solutions, we have ________________________________________________________________________ page 74 —————————————————————————— CHAPTER 2. —— C8 œ 38 ”C! a"#b#& < in which 3 œ a" <Î"#b. Here < is the annual interest rate, given as ) % . Therefore C$' œ a"Þ!!''b$' ’"!!! a"#b#& <“ #& #& , • "3 "3 œ #ß #)$Þ'$ dollars. 9. Let C8 be the balance due at the end of the 8-th month. The appropriate difference equation is C8" œ a" <Î"#b C8 T . Here < is the annual interest rate and T is the monthly payment. The solution, in terms of the amount borrowed, is given by C8 œ 38 ”C! T T , • "3 "3 in which 3 œ a" <Î"#b and C! œ )ß !!! Þ To figure out the monthly payment, T , we require that C$' œ !Þ That is, 336 ”C! T T . •œ "3 "3 After the specified amounts are substituted, we find the T œ $ #&)Þ"% . 11. Let C8 be the balance due at the end of the 8-th month. The appropriate difference equation is C8" œ a" <Î"#b C8 T , in which < œ Þ!* and T is the monthly payment. The initial value of the mortgage is C! œ "!!ß !!! dollars. Then the balance due at the end of the 8-th month is C8 œ 38 ”C! T T Þ • "3 "3 "#T "#T . • < < where 3 œ a" <Î"#b. In terms of the specified values, C8 œ a!Þ!!(&b8 ”"!& Setting 8 œ $!a"#b œ $'! , and C$'! œ ! , we find that T œ )!%Þ'# dollarsÞ For the monthly payment corresponding to a #! year mortgage, set 8 œ #%! and C#%! œ ! . 12. Let C8 be the balance due at the end of the 8-th month, with C! the initial value of the mortgage. The appropriate difference equation is C8" œ a" <Î"#b C8 T , in which < œ !Þ" and T œ *!! dollars is the maximum monthly payment. Given that the life of the mortgage is #! years, we require that C#%! œ !. The balance due at the end of the 8th month is C8 œ 38 ”C! T T Þ • "3 "3 In terms of the specified values for the parameters, the solution of ________________________________________________________________________ page 75 —————————————————————————— CHAPTER 2. —— aÞ!!)$$b#%! ”C! is C! œ "!$ß '#%Þ'# dollars. 15. "#a"!!!b "#a"!!!b •œ !Þ" !Þ" 16. For example, take 3 œ $Þ& and ?! œ "Þ" : ________________________________________________________________________ page 76 —————————————————————————— CHAPTER 2. —— 19a+b. $# œ a3# 3" bÎa3$ 3# b œ a$Þ%%* $bÎa$Þ&%% $Þ%%*b œ %Þ(#'$ . a,b. % diff œ a- b. Assuming a3$ 3# bÎa3% 3$ b œ $ , 3% ¸ $Þ&'%$ k$ $# k $ ‚ "!! œ k%Þ''*#%Þ($'$k %Þ''*# ‚ "!! ¸ "Þ## % Þ a. b. A period "' solutions appears near 3 ¸ $Þ&'& . " a/b. Note that a38" 38 b œ $8 a38 38" bÞ With the assumption that $8 œ $ , we have a38" 38 b œ $ " a38 38" b, which is of the form C8" œ ! C8 , 8 $ . It follows that a35 35" b œ $ $5 a3$ 3# b for 5 % . Then 38 œ 3" a3# 3" b a3$ 3# b a3% 3$ b â a38 38" b œ 3" a3# 3" b a3$ 3# bc" $ " $ # â $ $8 d " $ %8 œ 3" a3# 3" b a3$ 3# b” •Þ " $ " $ Hence lim 38 œ 3# a3$ 3# b $ " ‘Þ Substitution of the appropriate values yields 8Ä_ 8Ä_ lim 38 œ $Þ&'** ________________________________________________________________________ page 77 —————————————————————————— CHAPTER 2. —— Miscellaneous Problems 1. Linear Ò C œ -ÎB# B$ Î& ÓÞ 2. Homogeneous Ò+<->+8aCÎBb 68ÈB# C # œ - Ó. 3. Exact Ò B# BC $C C $ œ ! Ó. 4. Linear in BaCb Ò B œ - / C C / C Ó. 5. Exact Ò B# C BC # B œ - Ó. 6. Linear Ò C œ B" a" /"B b Ó. # 7. Let ? œ B# Ò B# C # " œ - /C Ó. 8. Linear Ò C œ a% -9= # -9= BbÎB# Ó. 9. Exact Ò B# C B C # œ - Ó . 10. . œ .aBb Ò C # ÎB$ CÎB# œ - Ó. 11. Exact Ò B$ Î$ BC /C œ - Ó. 12. Linear Ò C œ - /B /B 68a" /B b Ó. 13. Homogeneous Ò #ÈCÎB 68 kBk œ - Ó. 14. Exact/Homogeneous Ò B# #BC #C # œ $% Ó. 15. Separable Ò C œ -Î-9=2# aBÎ#b Ó. 16. Homogeneous Ò Š#ÎÈ$ ‹+<->+8’Ð#C BÑÎÈ$ B“ 68 kBk œ - Ó. 17. 18. 19. 20. 21. 22. 23. Linear Ò C œ - /$B /#B Ó. Linear/Homogeneous Ò C œ - B# B Ó. . œ .aBb Ò $C #BC $ "!B œ ! Ó. Separable Ò /B /C œ - Ó. Homogeneous Ò /CÎB 68 kBk œ - Ó. Separable Ò C$ $C B$ $B œ # Ó. Bernoulli Ò "ÎC œ B' B# /#B .B -B Ó. 24. Separable Ò =38# B =38 C œ - Ó. 25. Exact Ò B# ÎC +<->+8aCÎBb œ - Ó. 26. . œ .aBb Ò B# #B# C C # œ - Ó. 27. . œ .aBb Ò =38 B -9= #C " =38# B œ - Ó. # 28. Exact Ò #BC BC $ B$ œ - Ó. 29. Homogeneous Ò +<-=38aCÎBb 68 kBk œ - Ó. 30. Linear in BaCb Ò BC # 68 kC k œ ! Ó. 31. Separable 32. . œ .aCb Ò B 68 kBk B" C # 68 kC k œ - Ó. Ò B$ C # BC $ œ % Ó. ________________________________________________________________________ page 78 —————————————————————————— CHAPTER 3. —— Chapter Three Section 3.1 1. Let C œ /<> , so that C w œ < /<> and C ww œ < /<> . Direct substitution into the differential equation yields a<# #< $b/<> œ ! . Canceling the exponential, the characteristic equation is <# #< $ œ ! Þ The roots of the equation are < œ $ , " . Hence the general solution is C œ -" /> -# /$> . 2. Let C œ /<> . Substitution of the assumed solution results in the characteristic equation <# $< # œ ! Þ The roots of the equation are < œ # , " . Hence the general solution is C œ -" /> -# /#> . 4. Substitution of the assumed solution C œ /<> results in the characteristic equation #<# $< " œ ! Þ The roots of the equation are < œ "Î# , " . Hence the general solution is C œ -" />Î# -# /> . 6. The characteristic equation is %<# * œ ! , with roots < œ „$Î# . Therefore the general solution is C œ -" /$>Î# -# /$>Î# . 8. The characteristic equation is <# #< # œ ! , with roots < œ "„È$ . Hence the general solution is C œ -" /B:Š" È$‹> -# /B:Š" È$‹>. 9. Substitution of the assumed solution C œ /<> results in the characteristic equation <# < # œ ! Þ The roots of the equation are < œ # , " . Hence the general solution is C œ -" /#> -# /> . Its derivative is C w œ #-" /#> -# /> . Based on the first condition, Ca!b œ " , we require that -" -# œ " . In order to satisfy C w a!b œ " , we find that #-" -# œ " . Solving for the constants, -" œ ! and -# œ " . Hence the specific solution is Ca>b œ /> . 11. Substitution of the assumed solution C œ /<> results in the characteristic equation '<# &< " œ ! Þ The roots of the equation are < œ "Î$ , "Î# . Hence the general solution is C œ -" />Î$ -# />Î# . Its derivative is C w œ -" />Î$ Î$ -# />Î# Î# . Based on the first condition, Ca!b œ " , we require that -" -# œ % . In order to satisfy the condition C w a!b œ " , we find that -" Î$ -# Î# œ ! . Solving for the constants, -" œ "# and -# œ ) . Hence the specific solution is C a>b œ "# />Î$ ) />Î# . 12. The characteristic equation is <# $< œ ! , with roots < œ $ , ! . Therefore the general solution is C œ -" -# /$> , with derivative C w œ $ -# /$> . In order to satisfy the initial conditions, we find that -" -# œ # , and $ -# œ $ . Hence the specific solution is Ca>b œ " /$> . 13. The characteristic equation is <# &< $ œ ! , with roots ________________________________________________________________________ page 83 —————————————————————————— CHAPTER 3. —— & È"$ <"ß# œ „ . # # The general solution is C œ -" /B:Š & È"$‹>Î# -# /B:Š & È"$‹>Î#, with derivative Cw œ & È"$ & È"$ -" /B:Š & È"$‹>Î# -# /B:Š & È"$‹>Î# . # # In order to satisfy the initial conditions, we require that -" -# œ " , and È &È"$ -" &# "$ -# œ ! . Solving for the coefficients, -" œ Š" &ÎÈ"$ ‹Î# and # -# œ Š" &ÎÈ"$ ‹Î# Þ 14. The characteristic equation is #<# < % œ ! , with roots " È$$ <"ß# œ „ . % % The general solution is C œ -" /B:Š " È$$‹>Î% -# /B:Š " È$$‹>Î%, with derivative Cw œ " È$$ " È$$ -" /B:Š " È$$‹>Î% -# /B:Š " È$$‹>Î% . % % In order to satisfy the initial conditions, we require that -" -# œ ! , and È "È$$ -" "% $$ -# œ " . Solving for the coefficients, -" œ #ÎÈ$$ and % -# œ #ÎÈ$$ Þ The specific solution is Ca>b œ #’/B:Š " È$$‹>Î% /B:Š " È$$‹>Î%“ÎÈ$$ . ________________________________________________________________________ page 84 —————————————————————————— CHAPTER 3. —— 16. The characteristic equation is %<# " œ ! , with roots < œ „"Î# . Therefore the general solution is C œ -" />Î# -# />Î# . Since the initial conditions are specified at > œ # , is more convenient to write C œ ." /a>#bÎ# .# /a>#bÎ# Þ The derivative is given by C w œ ." /a>#bÎ# ‘Î# .# /a>#bÎ# ‘Î# . In order to satisfy the initial conditions, we find that ." .# œ " , and ." Î# .# Î# œ " . Solving for the coefficients, ." œ $Î# , and .# œ "Î# . The specific solution is Ca>b œ $ a>#bÎ# " a>#bÎ# / / # # $ >Î# / >Î# œ/ /Þ #/ # 18. An algebraic equation with roots # and "Î# is #<# &< # œ ! . This is the characteristic equation for the ODE #C ww &C w # C œ ! . 20. The characteristic equation is #<# $< " œ ! , with roots < œ "Î# , " . Therefore the general solution is C œ -" />Î# -# /> , with derivative C w œ -" />Î# Î# -# /> . In order to satisfy the initial conditions, we require -" -# œ # and -" Î# -# œ "Î# . Solving for the coefficients, -" œ $ , and -# œ " . The specific solution is Ca>b œ $/>Î# /> Þ To find the stationary point, set C w œ $/>Î# Î# /> œ ! . There is a unique solution, with >" œ 68a*Î%b. The maximum value is then C a>" b œ *Î%Þ To find ________________________________________________________________________ page 85 —————————————————————————— CHAPTER 3. —— the x-intercept, solve the equation $/>Î# /> œ ! Þ The solution is readily found to be ># œ 68* ¸ #Þ"*(# . 22. The characteristic equation is %<# " œ ! , with roots < œ „"Î# . Hence the general solution is C œ -" />Î# -# />Î# , with derivative C w œ -" />Î# Î# -# />Î# Î# . Invoking the initial conditions, we require that -" -# œ # and -" -# œ " . The specific solution is Ca>b œ a" " b/>Î# a" " b/>Î# Þ Based on the form of the solution, it is evident that as > p _ , C a>b p ! as long as " œ " . 23. The characteristic equation is <# a#! "b< !a! "b œ ! . Examining the coefficients, the roots are < œ ! , ! " . Hence the general solution of the differential equation is Ca>b œ -" /!> -# /a!"b> Þ Assuming ! − ‘ , all solutions will tend to zero as long as ! ! . On the other hand, all solutions will become unbounded as long as ! " ! , that is, ! " . 25. Ca>b œ # />Î# Î& $ /#> Î& . The minimum occurs at a>! , C! b œ a!Þ("'( ß !Þ("&&b. 26a+b. The characteristic roots are < œ $ , # . The solution of the initial value problem is Ca>b œ a' " b/#> a% " b/$> . a,b. The maximum point has coordinates >! œ 68’ $a%"b “, C! œ #a'" b a- b. C! œ " Ä_ %a'" b$ #(a%" b# % , as long as " ' 'È$ . " Ä_ %a'" b$ #(a%" b# . a. b. lim >! œ 68 $ . # lim C! œ _ . 29. Set @ œ C w and @ w œ C ww . Substitution into the ODE results in the first order equation > @ w @ œ " . The equation is linear, and can be written as a> @bw œ " . Hence the general solution is @ œ " -" Î> . Hence C w œ " -" Î> , and C œ > -" 68 > -# . 31. Setting @ œ C w and @ w œ C ww , the transformed equation is #># @ w @$ œ #> @ . This ________________________________________________________________________ page 86 —————————————————————————— CHAPTER 3. —— is a Bernoulli equation, with 8 œ $ . Let A œ @# . Substitution of the new dependent variable yields ># A w " œ #> A , or ># A w #> A œ œ " . Integrating, we find that A œ a> -" bÎ># . Hence @ œ „>ÎÈ> -" , that is, C w œ „>ÎÈ> -" . Integrating one more time results in Ca>b œ „ # a> #-" bÈ> -" -# Þ Ð Note that @ œ ! is also a $ solution of the transformed equationÑ. 32. Setting @ œ C w and @ w œ C ww , the transformed equation is @ w @ œ /> . This ODE is linear, with integrating factor .a>b œ /> . Hence @ œ C w œ a> -" b/> . Integrating, we obtain Ca>b œ a> -" b/> -# Þ 33. Set @ œ C w and @ w œ C ww . The resulting equation is ># @ w œ @# . This equation is separable, with solution @ œ C w œ >Îa" -" >bÞ Integrating, the general solution is as long as -" Á ! Þ For -" œ ! , the solution is C a>b œ ># Î# -# . Note that @ œ ! is also a solution of the transformed equation. 35. Let C w œ @ and C ww œ @ .@Î.C . Then @ .@Î.C C œ ! is the transformed equation for @ œ @aC b . This equation is separable, with @ .@ œ C .CÞ The solution is given by @# œ C # -" . Substituting for @ , we find that C w œ „È-" C # Þ This equation is also separable, with solution +<-=38ˆCÎÈ-" ‰ œ „ > -# , or C a>b œ ." =38a> .# b . 36. Let C w œ @ and C ww œ @ .@Î.C . It follows that @.@Î.C C@$ œ ! is the differential equation for @ œ @aC b . This equation is separable, with @# .@ œ C .CÞ The solution " " is given by @ œ cC # Î# -" d . Substituting for @ , we find that C w œ cC # Î# -" d . This equation is also separable, with aC# Î# -" b.C œ .> . The solution is defined implicitly by C$ Î' -" C -# œ > . 38. Setting C w œ @ and C ww œ @.@Î.C , the transformed equation is C @.@Î.C @$ œ !. This equation is separable, with @# .@ œ .CÎC Þ The solution is @aC b œ c-" 68kC kd" Þ Substituting for @ , we obtain a separable equation, a-" 68kC kb.C œ .B . The solution is given implicitly by -# C C 68kC k -$ œ > . 39. Let C w œ @ and C ww œ @.@Î.C . It follows that @.@Î.C @# œ #/C is the equation for @ œ @aC b . Inspection of the left hand side suggests a substitution A œ @# . The resulting equation is .AÎ.C #A œ %/C . This equation is linear, with integrating factor . œ /#C Þ We obtain . a/#C AbÎ.C œ % /C , which upon integration yields AaC b œ % /C -" /#C Þ Converting back to the original dependent variable, C w œ „/C È% /C -" Þ Separating variables, /C a% /C -" b"Î# .C œ „ .> . Integration yields È% /C -" œ „ #> -# . 41. Setting C w œ @ and C ww œ @.@Î.C , the transformed equation is @.@Î.C $C # œ !. ________________________________________________________________________ page 87 # Ca>b œ >Î-" -" 68k" -" >k -# , —————————————————————————— CHAPTER 3. —— This equation is separable, with @.@ œ $C # .C Þ The solution is C w œ @ œ È#C $ -" Þ The positive root is chosen based on the initial conditions. Furthermore, when > œ ! , C œ # , and C w œ @ œ % . The initial conditions require that -" œ ! . It follows that C w œ È#C $ Þ Separating variables and integrating, "ÎÈC œ >ÎÈ# -# . Hence the solution is Ca>b œ #Îa" > b# Þ 42. Setting @ œ C w and @ w œ C ww , the transformed equation is a" ># b@ w #> @ œ œ $># Þ Rewrite the equation as @ w #> @Îa" ># b œ $># Îa" ># bÞ This equation is linear, with integrating factor . œ " ># . Hence we have ˆ" ># ‰@‘w œ $># Þ Integrating both sides, @ œ $>" Îa" ># b -" Îa" ># b. Invoking the initial condition @a"b œ " , we require that -" œ & . Hence C w œ a$ &>bÎa> >$ b. Integrating, we obtain Ca>b œ $ 68c># Îa" ># bd & +<->+8a>b -# . Based on the initial condition # Ca"b œ # , we find that -# œ $ 68 # & 1 # . # % ________________________________________________________________________ page 88 —————————————————————————— CHAPTER 3. —— Section 3.2 1. [ ˆ/#> ß /$>Î# ‰ œ º 3. [ ˆ/#> ß > /#> ‰ œ º 5. [ ˆ/> =38 > ß /> -9= >‰ œ º 6. [ ˆ-9=# ) ß " -9= #) ‰ œ º -9=# ) #=38 ) -9= ) " -9= #) œ !. # =38 #) º /> =38 > /> -9= > œ /#> . /> a=38 > -9= >b /> a-9= > =38 >b º /#> #/#> >/#> %> ºœ/ . a" #>b/#> /#> #/#> ( >Î# /$>Î# $ $>Î# º œ / Þ # #/ 7. Write the equation as C ww a$Î>bC w œ " . :a>b œ $Î> is continuous for all > ! . Since >! ! , the IVP has a unique solution for all > ! . $ % # 9. Write the equation as C ww >% C w >a>%b C œ >a>%b . The coefficients are not continuous at > œ ! and > œ % . Since >! − a! ß %b , the largest interval is ! > % . 10. The coefficient $68k>k is discontinuous at > œ ! . Since >! ! , the largest interval of existence is ! > _ . B 11. Write the equation as C ww B$ C w 68kBk C œ ! . The coefficients are discontinuous B$ at B œ ! and B œ $ . Since B! − a! ß $b , the largest interval is ! B $ . 13. C"ww œ # . We see that ># a#b #a># b œ ! . C#ww œ # >$ , with ># aC#ww b #aC# b œ ! . Let C$ œ -" ># -# >" , then C$ww œ #-" #-# >$ . It is evident that C$ is also a solution. 16. No. Substituting C œ =38a># b into the differential equation, %># =38ˆ># ‰ #-9=ˆ># ‰ #> -9=ˆ># ‰:a>b =38ˆ># ‰; a>b œ ! . For the equation to be valid, we must have :a>b œ "Î> , which is not continuous, or even defined, at > œ ! . ________________________________________________________________________ page 89 —————————————————————————— CHAPTER 3. —— 17. [ a/#> ß 1a>bb œ /#> 1 w a>b #/#> 1a>b œ $/%> Þ Dividing both sides by /#> , we find that 1 must satisfy the ODE 1 w #1 œ $/#> Þ Hence 1a>b œ $> /#> - /#> Þ 22. The general solution is C œ -" /$> -# /> . [ a/$> ß /> b œ #/%> , and hence the exponentials form a fundamental set of solutions. On the other hand, the fundamental solutions must also satisfy the conditions C" a"b œ " , C"w a"b œ ! ; C# a"b œ ! , C#w a"b œ " Þ For C" , the initial conditions require -" -# œ / , $-" -# œ ! . The coefficients are -" œ /$ Î# , -# œ $/Î# . For the solution, C# , the initial conditions require -" -# œ ! , $-" -# œ / . The coefficients are -" œ /$ Î# , -# œ /Î# . Hence the fundamental solutions are ˜C" œ " /$a>"b $ /a>"b ß C# œ " /$a>"b " /a>"b ™Þ # # # # 23. Yes. C"ww œ % -9= #> ; C#ww œ % =38 #> . [ a-9= #> ß =38 #>b œ # Þ 24. Clearly, C" œ /> is a solution. C#w œ a" >b/> , C#ww œ a# >b/> Þ Substitution into the ODE results in a# >b/> #a" >b/> > /> œ ! . Furthermore, [ a/> ß >/> b œ /#> Þ Hence the solutions form a fundamental set of solutions. 26. Clearly, C" œ B is a solution. C#w œ -9= B , C#ww œ =38 B Þ Substitution into the ODE results in a" B -9> Bba =38 Bb Ba-9= Bb =38 B œ !. [ aC" , C# b œ B -9= B =38 B, which is nonzero for ! B 1 . Hence eB ß =38 Bf is a fundamental set of solutions. 28. T œ " , U œ B , V œ " . We have T ww U w V œ ! . The equation is exact. Note that aC w bw aBCbw œ ! . Hence C w BC œ -" . This equation is linear, with integrating # factor . œ /B Î# Þ Therefore the general solution is CaBb œ -" /B:ˆ B# Î#‰( /B:ˆ?# Î#‰.? -# /B:ˆ B# Î#‰Þ B B! 20. [ a0 ß 1b œ 0 1 w 0 w 1 œ > -9= > =38 > , and [ a? ß @b œ %0 1 w %0 w 1 . Hence [ a? ß @b œ %> -9= > %=38 > . 19. [ a0 ß 1b œ 0 1 w 0 w 1 . Also, [ a? ß @b œ [ a#0 1 ß 0 #1b . Upon evaluation, [ a? ß @b œ &0 1 w &0 w 1 œ &[ a0 ß 1b. 29. T œ " , U œ $B# , V œ B . Note that T ww U w V œ &B , and therefore the differential equation is not exact. 31. T œ B# , U œ B , V œ " . We have T ww U w V œ ! . The equation is exact. w Write the equation as aB# C w b aBCbw œ ! . Integrating, we find that B# C w B C œ - . Divide both sides of the ODE by B# . The resulting equation is linear, with integrating factor . œ "ÎB . Hence aCÎBbw œ - B$ . The solution is C a>b œ -" B" -# B . ________________________________________________________________________ page 90 —————————————————————————— CHAPTER 3. —— 33. T œ B# , U œ B , V œ B# / # . Hence the coefficients are #T w U œ $B and T ww U w V œ B# " / # . The adjoint of the original differential equation is given by B# . ww $B . w aB# " / # b. œ ! . 35. T œ " , U œ ! , V œ B . Hence the coefficients are given by #T w U œ ! and T ww U w V œ B . Therefore the adjoint of the original equation is . ww B . œ ! . ________________________________________________________________________ page 91 —————————————————————————— CHAPTER 3. —— Section 3.3 1. Suppose that ! 0 a>b " 1a>b œ ! , that is, !a># &>b " a># &>b œ ! on some interval M . Then a! " b># &a! " b> œ ! , a > − M . Since a quadratic .has at most two roots, we must have ! " œ ! and ! " œ ! . The only solution is ! œ " œ ! . Hence the two functions are linearly independent. 3. Suppose that /-> -9= .> œ E /-> =38 .> , for some E Á ! , on an interval M . Since the function =38 .> Á ! on some subinterval M! § M , we conclude that >+8 .> œ E on M! . This is clearly a contradiction, hence the functions are linearly independent. 4. Obviously, 0 aBb œ / 1aBb for all real numbers B . Hence the functions are linearly dependent. )Þ Note that 0 aBb œ 1aBb for B − Ò ! ß _Ñ, and 0 aBb œ 1aBb for B − Ð _ ß ! Ó. It follows that the functions are linearly dependent on ‘ and ‘ . Nevertheless, they are linearly independent on any open interval containing zero. 5. Here 0 aBb œ $1aBb for all real numbers. Hence the functions are linearly dependent. 9. Since [ a>b œ > =38# > has only isolated zeros, [ a>b cannot identically vanish on any open interval. Hence the functions are linearly independent. 10. Same argument as in Prob. 9. 11. By linearity of the differential operator, -" C" and -# C# are also solutions. Calculating the Wronskian, [ a-" C" ß -# C# b œ a-" C" ba-# C# bw a-" C" bw a-# C# b œ -" -# [ aC" ß C# b . Since [ aC" ß C# b is not identically zero, neither is [ a-" C" ß -# C# b . 13. Direct calculation results in [ a+" C" +# C# ß ," C" ,# C# b œ +" ,# [ aC" ß C# b ," +# [ aC" ß C# b œ a+" ,# +# ," b[ aC" ß C# b Þ 14. Let !ai jb " ai jb œ ! i ! j . Then ! " œ ! and ! " œ ! . The only solution is ! œ " œ ! . Hence the given vectors are linearly independent. Furthermore, any vector +" i +# j œ ˆ +" +# ‰ai jb ˆ +" +# ‰ai jb . # # # # 16. Writing the equation in standard form, we find that T a>b œ =38 >Î-9= > . Hence the =38 Wronskian is [ a>b œ , /B:ˆ ' -9= > .>‰ œ , /B:a68k-9= >kb œ , -9= > , in which , is > some constant. Hence the combinations are also linearly independent as long as +" ,# +# ," Á ! . ________________________________________________________________________ page 92 —————————————————————————— CHAPTER 3. —— 17. After writing the equation in standard form, we have T aBb œ "ÎB . The Wronskian " is [ a>b œ - /B:ˆ ' B .B‰ œ - /B:a 68kBkb œ -ÎkBk, in which - is some constant. 18. Writing the equation in standard form, we find that T aBb œ #BÎa" B# b. The " #B Wronskian is [ a>b œ - /B:ˆ ' "B# .B‰ œ - /B:a 68k" B# kb œ - k" B# k , in which - is some constant. 19. Rewrite the equation as :a>bC ww : w a>bC w ; a>bC œ ! Þ After writing the equation in standard form, we have T a>b œ : w a>bÎ:a>b . Hence the Wronskian is [ a>b œ - /B:Œ ( : w a >b .> œ - /B:a 68 :a>bb œ -Î:a>b . : a >b 21. The Wronskian associated with the solutions of the differential equation is given by [ a>b œ - /B:ˆ ' # .>‰ œ - /B:a #Î>b. Since [ a#b œ $ , it follows that for the ># hypothesized set of solutions, - œ $ / . Hence [ a%b œ $È/ . 22. For the given differential equation, the Wronskian satisfies the first order differential equation [ w :a>b[ œ ! . Given that [ is constant, it is necessary that :a>b ´ ! . 23. Direct calculation shows that [ a0 1 ß 0 2b œ a0 1ba0 2bw a0 1bw a0 2 b œ a0 1ba0 w 2 0 2 w b a0 w 1 0 1 w ba0 2 b œ 0 # [ a1 ß 2 b Þ 25. Since C" and C# are solutions, they are differentiable. The hypothesis can thus be restated as C"w a>! b œ C#w a>! b œ ! at some point >! in the interval of definition. This implies that [ aC" ß C# ba>! b œ ! . But [ aC" ß C# ba>! b œ - /B:ˆ ' :a>b.>‰ , which cannot be equal to zero, unless - œ ! . Hence [ aC" ß C# b ´ ! , which is ruled out for a fundamental set of solutions. ________________________________________________________________________ page 93 —————————————————————————— CHAPTER 3. —— Section 3.4 2. /B:a# $3b œ /# /$3 œ /# a-9= $ 3 =38 $b. 3. /31 œ -9= 1 3 =38 1 œ " . 4. /B:ˆ# 1 3‰ œ /# ˆ-9= 1 3 =38 1 ‰ œ /# 3 . # # # 6. 1"#3 œ /B:ca " #3b68 1d œ /B:a 68 1b/B:a# 68 1 3b œ " œ 1 c-9= a# 68 1b 3 =38 a# 68 1bdÞ " 1 /B:a# 68 1 3b œ 8. The characteristic equation is <# #< ' œ ! , with roots < œ " „ 3È& . Hence the general solution is C œ -" /> -9= È& > -# /> =38 È& >. 9. The characteristic equation is <# #< ) œ ! , with roots < œ % ß # . The roots are real and different, hence the general solution is C œ -" /%> -# /#> . 10. The characteristic equation is <# #< # œ ! , with roots < œ " „ 3. Hence the general solution is C œ -" /> -9= > -# /> =38 >. 12. The characteristic equation is %<# * œ ! , with roots < œ „ $ 3. Hence the # general solution is C œ -" -9= $ > -# =38 $ > . # # 13. The characteristic equation is <# #< "Þ#& œ !, with roots < œ " „ " 3. Hence # the general solution is C œ -" /> -9= " > -# /> =38 " > . # # 15. The characteristic equation is <# < "Þ#& œ ! , with roots < œ " „ 3. Hence # the general solution is C œ -" />Î# -9= > -# />Î# =38 > . 16. The characteristic equation is <# %< 'Þ#& œ ! , with roots < œ # „ $ 3. Hence # the general solution is C œ -" /#> -9= $ > -# /#> =38 $ > . # # 17. The characteristic equation is <# % œ ! , with roots < œ „ #3. Hence the general solution is C œ -" -9= #> -# =38 #> . Its derivative is C w œ #-" =38 #> #-# -9= #> . Based on the first condition, Ca!b œ ! , we require that -" œ ! . In order to satisfy the condition C w a!b œ " , we find that #-# œ " . The constants are -" œ ! and -# œ "Î# . Hence the specific solution is Ca>b œ " =38 #> . # 19. The characteristic equation is <# #< & œ ! , with roots < œ "„ #3. Hence the general solution is C œ -" /> -9= #> -# /> =38 #> . Based on the condition, C a1Î#b œ ! , we require that -" œ ! . It follows that C œ -# /> =38 #> , and so the first derivative is C w œ -# /> =38 #> #-# /> -9= #> . In order to satisfy the condition C w a1Î#b œ #, we find that #/1Î# -# œ # . Hence we have -# œ /1Î# . Therefore the specific solution is Ca>b œ />1Î# =38 #> . ________________________________________________________________________ page 94 —————————————————————————— CHAPTER 3. —— 20. The characteristic equation is <# " œ ! , with roots < œ „ 3. Hence the general solution is C œ -" -9= > -# =38 > . Its derivative is C w œ -" =38 > -# -9= > . Based on the first condition, Ca1Î$b œ # , we require that -" È$ -# œ % . In order to satisfy the condition C w a1Î$b œ % , we find that È$ -" -# œ ) . Solving these for the constants, -" œ " #È$ and -# œ È$ # . Hence the specific solution is a steady oscillation, given by Ca>b œ Š" #È$‹-9= > ŠÈ$ #‹=38 > . 21. From Prob. 15, the general solution is C œ -" />Î# -9= > -# />Î# =38 > . Invoking the first initial condition, Ca!b œ $ , which implies that -" œ $ . Substituting, it follows that C œ $/>Î# -9= > -# />Î# =38 > , and so the first derivative is Invoking the initial condition, C w a!b œ " , we find that $ -# œ " , and so -# œ # Hence the specific solution is Ca>b œ $/>Î# -9= > & />Î# =38 > Þ $ $ -# C w œ />Î# -9= > $/>Î# =38 > -# />Î# -9= > />Î# =38 > . # # & # . 24a+b. The characteristic equation is &<# #< ( œ ! , with roots < œ " „ 3 & È È È$% &. The solution is ? œ -" />Î& -9= &$% > -# />Î& =38 &$% > . Invoking the given initial conditions, we obtain the equations for the coefficients : -" œ # , # È$% -# œ & . That is, -" œ # , -# œ (ÎÈ$% . Hence the specific solution is ________________________________________________________________________ page 95 —————————————————————————— CHAPTER 3. —— ?a>b œ #/>Î& -9= È$% & È$% & > ( />Î& =38 È$% >. a,b. Based on the graph of ?a>b, X is in the interval "% > "' . A numerical solution on that interval yields X ¸ "%Þ&""& . 26a+b. The characteristic equation is <# #+ < a+# "b œ ! , with roots < œ + „ 3 Þ Hence the general solution is Ca>b œ -" /+> -9= > -# /+> =38 > . Based on the initial conditions, we find that -" œ " and -# œ + Þ Therefore the specific solution is given by Ca>b œ /+> -9= > + /+> =38 > œ È" +# /+> -9= a> 9b, in which 9 œ >+8" a+bÞ a,b. For estimation, note that kCa>bk Ÿ È" +# /+> . Now consider the inequality È" +# /+> Ÿ "Î"! . The inequality holds for > " 68’"!È" +# “Þ Therefore a- b. Similarly, X"Î% ¸ (Þ%#)% , X"Î# ¸ %Þ$!!$ , X# ¸ "Þ&""' , X$ ¸ "Þ"%*' . a. b . " X Ÿ + 68’"!È" +# “. Setting + œ " , numerical analysis gives X ¸ "Þ)('$ . + ________________________________________________________________________ page 96 —————————————————————————— CHAPTER 3. —— Note that the estimates X+ approach the graph of " È + 68’"! " +# “ as + gets large. 27. Direct calculation gives the result. On the other hand, it was shown in Prob. 3.3.23 that [ a0 1 ß 0 2b œ 0 # [ a1 ß 2 bÞ Hence [ ˆ/-> -9= .> ß /-> =38 .>‰ œ /#-> [ a-9= .> ß =38 .>b œ /#-> -9= .>a=38 .>bw a-9= .>bw =38 .>‘ œ . /#-> Þ a,b. C w œ 3 /3> , C ww œ 3# /3> œ /3> Þ Evidently, C is a solution and so C œ -" C" -# C# . 28a+b. Clearly, C" and C# are solutions. Also, [ a-9= >ß =38 >b œ -9=# > =38# > œ "Þ a- b. Setting > œ ! , " œ -" -9= ! -# =38 ! , and -" œ ! . Differentiating, 3 /3> œ -# -9= > . Setting > œ ! , 3 œ -# -9= ! and hence -# œ 3 . Therefore /3> œ -9= > 3 =38 > . 29. Euler's formula is /3> œ -9= > 3 =38 > . It follows that /3> œ -9= > 3 =38 > . Adding these equation, /3> /3> œ # -9= > . Subtracting the two equations results in /3> /3> œ #3 =38 > . 30. Let <" œ -" 3." , and <# œ -# 3.# . Then /B:a<" <# b> œ /B:ca-" -# b> 3a." .# b>d œ /a-" -# b> c-9=a." .# b> 3 =38a." .# b>d œ /a-" -# b> ca-9= ." > 3=38 ." >ba-9= .# > 3=38 .# >bd œ /-" > a-9= ." > 3=38 ." >b † /-# > a-9= ." > 3=38 ." >b Hence /a<" <# b> œ /<" > /<# > Þ 32. If 9a>b œ ?a>b 3 @a>b is a solution, then and a? ww 3@ ww b :a>ba? w 3@ w b ; a>ba? 3@b œ ! . After expanding the equation and separating the real and imaginary parts, ?ww :a>b? w ; a>b? œ ! @ww :a>b@ w ; a>b@ œ ! a? 3@bww :a>ba? 3@bw ; a>ba? 3@b œ ! , Hence both ?a>b and @a>b are solutions. .C .D 34a+b. By the chain rule, CaBbw œ .B B w Þ In general, .D œ .B .B . Setting D œ .C , .> .> .> .#C . # C .B .B .C . .B ‘ .B .D .B . .C .B ‘ .B we have .># œ .B .> œ .B .B .> .> œ ’ .B# .> “ .> .B .B .> .> Þ However, . .B ‘ .B .B .> .> .> œ ’ . B “ .B † .># # .B .> œ .#B .># . Hence .# C .># œ . # C .B ‘# .B# .> .C . # B .B .># . ________________________________________________________________________ page 97 —————————————————————————— CHAPTER 3. —— a,b. Substituting the results in Parta+b into the general ODE, C ww :a>bC w ; a>bC œ ! , `we find that . # C .B # .C . # B .C .B :a>b ; a>bC œ ! . # ” .> • # .B .B .> .B .> Collecting the terms, .B . # C .#B .B .C ” # :a>b • ; a>bC œ ! Þ ”• # .> .B .> .> .B # # a- b. Assuming .B ‘ œ 5 ; a>b , and ; a>b ! , we find that .> be integrated. That is, B œ 0a>b œ ' È5 ; a>b .> . .B .> œ È5 ; a>b , which can ;w #È ; a. b. Let 5 œ " . It follows that .# B .B .B # ; w a>b #:a>b; a>b . ” # : a> b • Î ” • œ .> .> .> #c; a>bd$Î# .# C ; w a>b #:a>b; a>b .C ” C œ !Þ • .B# .B #c; a>bd$Î# .#B .># :a>b .B œ .> .0 .> :a>b0a>b œ : È; Þ Hence As long as .BÎ.> Á ! , the differential equation can be expressed as # * For the case ; a>b ! , write ; a>b œ c ; a>bd , and set .B ‘ œ ; a>b . .> 36. :a>b œ $> and ; a>b œ ># . We have B œ ' > .> œ ># Î# . Furthermore, ; w a>b #:a>b; a>b œ ˆ" $># ‰Î># . #c; a>bd$Î# The ratio is not constant, and therefore the equation cannot be transformed. ; w a>b #:a>b; a>b œ ". #c; a>bd$Î# 37. :a>b œ > "Î> and ; a>b œ ># . We have B œ ' > .> œ ># Î# . Furthermore, The ratio is constant, and therefore the equation can be transformed. From Prob. 35, the transformed equation is .# C .C C œ !Þ # .B .B Based on the methods in this section, the characteristic equation is <# < " œ ! , with È roots < œ " „ 3 #$ . The general solution is # ________________________________________________________________________ page 98 —————————————————————————— CHAPTER 3. —— CaBb œ -" /BÎ# -9= È$ BÎ# -# /BÎ# =38 È$ BÎ# . Since B œ ># Î# , the solution in the original variable > is # Ca>b œ /> Î% ’-" -9= ŠÈ$ ># Î%‹ -# =38 ŠÈ$ ># Î%‹“ . ; w a>b #:a>b; a>b $ . œ $Î# È# #c; a>bd .# C $ .C C œ !Þ # È# .B .B 40. :a>b œ %Î> and ; a>b œ #Î># . We have B œ È# ' >" .> œ È# 68 > . Furthermore, The ratio is constant, and therefore the equation can be transformed. In fact, we obtain Based on the methods in this section, the characteristic equation is È# <# $< È# œ ! , with roots < œ È# , "ÎÈ# . The general solution is Since B œ È# 68 > , the solution in the original variable > is Ca>b œ -" /# 68 > -# /68 > œ -" ># -# >" Þ CaBb œ -" / È# B -# /BÎ È# . 41. :a>b œ $Î> and ; a>b œ "Þ#&Î># . We have B œ È"Þ#& ' >" .> œ È"Þ#& 68 > . Checking the feasibility of the transformation, ; w a>b #:a>b; a>b % œ . $Î# È& #c; a>bd .# C % .C C œ !Þ È& .B .B# Based on the methods in this section, the characteristic equation is È& <# %< È& œ ! , with roots < œ # „3 " . The general solution is È& È& Since #BÎÈ& œ 68 > , the solution in the original variable > is CaBb œ -" /#BÎ & -9= BÎÈ& -# /#BÎ & =38 BÎÈ& . È È The ratio is constant, and therefore the equation can be transformed. In fact, we obtain ________________________________________________________________________ page 99 —————————————————————————— CHAPTER 3. —— Ca>b œ -" /68 > -9=Š68È> ‹ -# /68 > =38Š68È> ‹ œ >" ’-" -9=Š68È> ‹ -# =38Š68È> ‹“. 42. :a>b œ %Î> and ; a>b œ 'Î># . Set B œ È' ' >" .> œ È' 68 > . Checking the feasibility of the transformation a‡ see Prob. 34 d , with ; !b, ; w a>b #:a>b; a>b & œ . $Î# È' #c ; a>bd .# C & .C C œ !Þ # È' .B .B The ratio is constant, and therefore the equation can be transformed. In fact, we obtain Based on the methods in this section, the characteristic equation is È' <# & < È' œ ! , with roots < œ È' , "ÎÈ' . The general solution is Since B œ È' 68 > , the solution in the original variable > is Ca>b œ -" /'68 > -# /68 > œ -" > ' -# >" Þ CaBb œ -" / È' B -# /BÎ È' . ________________________________________________________________________ page 100 —————————————————————————— CHAPTER 3. —— Section 3.5 2. The characteristic equation is *<# '< " œ ! , with the double root < œ "Î$ Þ Based on the discussion in this section, the general solution is Ca>b œ -" />Î$ -# > />Î$ . 3. The characteristic equation is %<# %< $ œ ! , with roots < œ "Î# , $Î# Þ The general solution is Ca>b œ -" />Î# -# /$>Î# . 4. The characteristic equation is %<# "#< * œ ! , with the double root < œ $Î# Þ Based on the discussion in this section, the general solution is Ca>b œ a-" -# >b/$>Î# . 5. The characteristic equation is <# #< "! œ ! , with complex roots < œ " „ $3Þ The general solution is Ca>b œ -" /> -9= $> -# /> =38 $> . 6. The characteristic equation is <# '< * œ ! , with the double root < œ $ Þ The general solution is Ca>b œ -" /$> -# > /$> . 7. The characteristic equation is %<# "(< % œ ! , with roots < œ "Î% , % Þ The general solution is Ca>b œ -" />Î% -# /%> . 8. The characteristic equation is "'<# #%< * œ ! , with the double root < œ $Î% Þ The general solution is Ca>b œ -" /$>Î% -# > /$>Î% . 10. The characteristic equation is #<# #< " œ ! , with complex roots < œ " „ " 3Þ # # The general solution is Ca>b œ -" />Î# -9= >Î# -# />Î# =38 >Î# . 11. The characteristic equation is *<# "#< % œ ! , with the double root < œ #Î$ Þ The general solution is Ca>b œ -" /#>Î$ -# > /#>Î$ . Invoking the first initial condition, it follows that -" œ # Þ Now C w a>b œ a%Î$ -# b/#>Î$ #-# > /#>Î$ Î$ . Invoking the second initial condition, %Î$ -# œ " , or -# œ (Î$ . Hence C a>b œ #/#>Î$ ( > /#>Î$ . $ Since the second term dominates for large > , Ca>b p _ . 13. The characteristic equation is *<# '< )# œ !, with complex roots < œ " „$ 3Þ $ The general solution is Ca>b œ -" />Î$ -9= $> -# />Î$ =38 $> . Based on the first initial condition, -" œ " Þ Invoking the second initial condition, "Î$ $-# œ # , or -# œ & . * & >Î$ >Î$ Hence Ca>b œ / -9= $> * / =38 $> . ________________________________________________________________________ page 101 —————————————————————————— CHAPTER 3. —— 15a+b. The characteristic equation is %<# "#< * œ ! , with the double root < œ $ Þ # The general solution is Ca>b œ -" /$>Î# -# > /$>Î# . Invoking the first initial condition, it follows that -" œ " Þ Now C w a>b œ a $Î# -# b/#>Î$ $ -# > /#>Î$ . The second # initial condition requires that $Î# -# œ % , or -# œ &Î# . Hence the specific solution is Ca>b œ /$>Î# & > /$>Î# . # a,b. The solution crosses the x-axis at > œ !Þ% . a- b. The solution has a minimum at the point a"'Î"& ß &/)Î& Î$b. a. b. Given that C w a!b œ , , we have $Î# -# œ , , or -# œ , $Î# . Hence the solution is Ca>b œ /$>Î# ˆ, $ ‰> /$>Î# . Since the second term dominates, the long# term solution depends on the sign of the coefficient , $ . The critical value is , œ $ Þ # # 16. The characteristic roots are <" œ <# œ "Î# . Hence the general solution is given by Ca>b œ -" />Î# -# > />Î# . Invoking the initial conditions, we require that -" œ # , and that " -# œ ,. The specific solution is C a>b œ #/>Î# a, "b> />Î# . Since the second term dominates, the long-term solution depends on the sign of the coefficient , ". The critical value is , œ "Þ ________________________________________________________________________ page 102 —————————————————————————— CHAPTER 3. —— 18a+b. The characteristic roots are <" œ <# œ #Î$ . Therefore the general solution is given by Ca>b œ -" /#>Î$ -# > /#>Î$ . Invoking the initial conditions, we require that -" œ + , and that #+Î$ -# œ " . After solving for the coefficients, the specific solution is Ca>b œ +/#>Î$ ˆ #+ "‰> /#>Î$ . $ 20a+b. The characteristic equation is <# #+< +# œ ! , with double root < œ +Þ Hence one solution is C" a>b œ -" /+> . a,b. Since the second term dominates, the long-term solution depends on the sign of the coefficient #$+ " . The critical value is + œ $Î# Þ a,b. Recall that the Wronskian satisfies the differential equation [ w #+[ œ ! . The solution of this equation is [ a>b œ - /#+> . a- b. By definition, [ œ C" C#w C"w C# . Hence -" /+> C#w +-" /+> C# œ - /#+> . That is, C#w + C# œ -# /+> . This equation is first order linear, with general solution C# a>b œ -# >/+> -$ /+> . Setting -# œ " and -$ œ ! , we obtain C# a>b œ >/+> Þ , , # 22a+b. Write +<# ,< - œ +ˆ<# + < + ‰. It follows that + œ #<" and + œ <" Þ # # Hence +<# ,< - œ +<# #+<" < +<" œ +a<# #<" < <" b œ +a< <" b# Þ We find that Pc/<> d œ a+<# ,< - b/<> œ +a< <" b# /<> Þ Setting < œ <" , Pc/<" > d œ ! . a,b. Differentiating Eq.a3b with respect to <, ` P/<> ‘ œ +>/<> a< <" b# #+/<> a< <" b. `< Now observe that ` ` `# ` P/<> ‘ œ + # ˆ/<> ‰ , ˆ/<> ‰- ˆ/<> ‰• ” `< `< `> `> # ` ` `` ` œ ”+ # Œ /<> , Œ /<> - Œ /<> • `> `< `> `< `< # ` ` œ + # ˆ>/<> ‰ , ˆ>/<> ‰- ˆ>/<> ‰. `> `> Hence Pc>/<> d œ +>/<> a< <" b# #+/<> a< <" b. Setting < œ <" , Pc>/<" > d œ ! . 23. Set C# a>b œ ># @a>b . Substitution into the ODE results in ># ˆ># @ ww %>@ w #@‰ %>ˆ># @ w #>@‰ '># @ œ ! . After collecting terms, we end up with >% @ ww œ ! . Hence @a>b œ -" -# > , and thus C# a>b œ -" ># -# >$ . Setting -" œ ! and -# œ " , we obtain C# a>b œ >$ Þ 24. Set C# a>b œ > @a>b . Substitution into the ODE results in ________________________________________________________________________ page 103 —————————————————————————— CHAPTER 3. —— ># a>@ ww #@ w b #>a>@ w @b #>@ œ ! . After collecting terms, we end up with >$ @ ww %># @ w œ ! . This equation is linear in the variable A œ @ w . It follows that @ w a>b œ - >% , and @a>b œ -" >$ -# . Thus C# a>b œ -" ># -# > . Setting -" œ " and -# œ ! , we obtain C# a>b œ ># Þ 26. Set C# a>b œ > @a>b . Substitution into the ODE results in @ ww @ w œ ! . This ODE is linear in the variable A œ @ w . It follows that @ w a>b œ -" /> , and @a>b œ -" /> -# . Thus C# a>b œ -" >/> -# > . Setting -" œ " and -# œ ! , we obtain C# a>b œ >/> Þ 28. Set C# aBb œ /B @aBb . Substitution into the ODE results in @ ww B# w @ œ !. B" B# .B B" This ODE is linear in the variable A œ @ w . An integrating factor is . œ /B:Œ( 29. Set C# aBb œ C1 aBb @aBb , in which C1 aBb œ B"Î% /B:ˆ#ÈB‰. It can be verified that C1 is a solution of the ODE, that is, B# C"ww aB !Þ")(&bC" œ ! . Substitution of the given form of C# results in the differential equation #B*Î% @ ww ˆ%B(Î% B&Î% ‰@ w œ ! . /@ Rewrite the equation as B" ‘ œ ! , from which it follows that @ w aBb œ - aB "b/B . Hence @aBb œ -" B/B -# and C# aBb œ -" B -# /B . Setting -" œ " and -# œ ! , we obtain C# aBb œ BÞ Bw /B œ Þ B" w This ODE is linear in the variable A œ @ w . An integrating factor is . œ /B:Œ( ”#B"Î# œ ÈB /B:ˆ%ÈB‰Þ w " •.B #B Rewrite the equation as ÈB /B:ˆ%ÈB‰ @ w ‘ œ ! , from which it follows that Integrating, @aBb œ -" /B:ˆ %ÈB‰ -# and as a result, @ w aBb œ - /B:ˆ %ÈB‰ÎÈB . Setting -" œ " and -# œ ! , we obtain C# aBb œ B"Î% /B:ˆ #ÈB‰Þ C# aBb œ -" B"Î% /B:ˆ #ÈB‰ -# B"Î% /B:ˆ#ÈB‰. ________________________________________________________________________ page 104 —————————————————————————— CHAPTER 3. —— 32. Direct substitution verifies that C" a>b œ /B:a $ B# Î#b is a solution of the ODE. Now set C# aBb œ C1 aBb @aBb. Substitution of C# into the ODE results in This ODE is linear in the variable A œ @ w . An integrating factor is . œ /B:a $ B# Î#b. Rewrite the equation as c /B:a $ B# Î#b@ w dw œ ! , from which it follows that @ w aBb œ -" /B:ˆ$ B# Î#‰ . B @ ww $ B@ w œ ! . Integrating, we obtain @aBb œ -" ( /B:ˆ$ ?# Î#‰.? @aB! b. B! Hence C# aBb œ -" /B:ˆ $ B# Î#‰( /B:ˆ$ ?# Î#‰.? -# /B:ˆ $B# Î#‰. B B! Setting -# œ ! , we obtain a second independent solution. 34. After writing the ODE in standard form, we have :a>b œ $Î> . Based on Abel's identity, [ aC" ß C# b œ -" /B:ˆ ' $ .>‰ œ -" >$ Þ As shown in Prob. 33, two solutions > of a second order linear equation satisfy In the given problem, C" a>b œ >" . Hence a> C# bw œ -" >" Þ Integrating both sides of the equation, C# a>b œ -" >" 68 > -# >" Þ 36. After writing the ODE in standard form, we have :aBb œ BÎaB "b. Based on B Abel's identity, [ aC" ß C# b œ - /B:ˆ' B" .B‰ œ - /B aB "bÞ Two solutions of a second order linear equation satisfy In the given problem, C" aBb œ /B . Hence a/B C# bw œ - /B aB "bÞ Integrating both sides of the equation, C# aBb œ -" B -# /B Þ Setting -" œ " and -# œ ! , we obtain C# aBb œ B Þ 37. Write the ODE in standard form to find :aBb œ "ÎB. Based on Abel's identity, [ aC" ß C# b œ - /B:ˆ ' " .B‰ œ - B" Þ Two solutions of a second order linear ODE B # satisfy aC# ÎC" bw œ [ aC" ß C# bÎC" . In the given problem, C" aBb œ B"Î# =38 B . Hence " =38 B C# œ - =38# B Þ ÈB w # aC# ÎC" bw œ [ aC" ß C# bÎC" . # aC# ÎC" bw œ [ aC" ß C# bÎC" . ________________________________________________________________________ page 105 —————————————————————————— CHAPTER 3. —— Integrating both sides of the equation, C# aBb œ -" B"Î# -9= B -# B"Î# =38 BÞ Setting -" œ " and -# œ ! , we obtain C# aBb œ B"Î# -9= B Þ 39a+b. The characteristic equation is +<# - œ ! . If + , - ! , then the roots are <"ß# œ „ 3È-Î+ . The general solution is Ca>b œ -" -9=Ê > -# =38Ê > , + + which is bounded. a,bÞ The characteristic equation is +<# ,< œ ! . The roots are <"ß# œ ! , ,Î+ , and hence the general solution is Ca>b œ -" -# /B:a ,>Î+bÞ Clearly, C a>b p -" . 40. Note that -9= > =38 > œ " =38 #> . So that " 5 -9= > =38 > œ " 5 =38 #> . If # # ! 5 # , then 5 =38 #> k=38 #>k and 5 =38 #> k=38 #>k. Hence # # " 5 -9= > =38 > œ " 5 =38 #> # " k=38 #>k !Þ 41. :a>b œ $Î> and ; a>b œ %Î># . We have B œ #' >" .> œ # 68 > , and > œ /BÎ# Þ Furthermore, ; w a>b #:a>b; a>b œ #. #c; a>bd$Î# .# C .C # C œ !Þ # .B .B The ratio is constant, and therefore the equation can be transformed. In fact, we obtain The general solution of this ODE is CaBb œ -" /B -# B/B . In terms of the original independent variable, Ca>b œ -" ># -# ># 68 > . ________________________________________________________________________ page 106 —————————————————————————— CHAPTER 3. —— Section 3.6 2. The characteristic equation for the homogeneous problem is <# #< & œ ! , with complex roots < œ "„#3 . Hence C- a>b œ -" /> -9= #> -# /> =38 #> . Since the function 1a>b œ $ =38 #> is not proportional to the solutions of the homogeneous equation, set ] œ E -9= #> F =38 #> . Substitution into the given ODE, and comparing the coefficients, results in the system of equations F %E œ $ and E %F œ ! . Hence $ ] œ "# -9= #> "( =38 #> . The general solution is C a>b œ C- a>b ] Þ "( 3. The characteristic equation for the homogeneous problem is <# #< $ œ ! , with roots < œ " , $ . Hence C- a>b œ -" /> -# /$> . Note that the assignment ] œ E>/> is not sufficient to match the coefficients. Try ] œ E>/> F># /> . Substitution into the differential equation, and comparing the coefficients, results in the system of $ $ equations %E #F œ ! and )F œ $ . Hence ] œ "' >/> ) ># /> . The general solution is Ca>b œ C- a>b ] Þ 5. The characteristic equation for the homogeneous problem is <# * œ ! , with complex roots < œ „$3. Hence C- a>b œ -" -9= $> -# =38 $> . To simplify the analysis, set 1" a>b œ ' and 1# a>b œ ># /$> . By inspection, we have ]" œ #Î$ . Based on the form of 1# , set ]# œ E/$> F>/$> G># /$> . Substitution into the differential equation, and comparing the coefficients, results in the system of equations ")E 'F #G œ ! , ")F "#G œ ! , and ")G œ " . Hence ]# œ " $> " " / >/$> ># /$> . "'# #( ") The general solution is Ca>b œ C- a>b ]" ]# Þ 7. The characteristic equation for the homogeneous problem is #<# $< " œ ! , with roots < œ " , "Î# . Hence C- a>b œ -" /> -# />Î# . To simplify the analysis, set 1" a>b œ ># and 1# a>b œ $ =38 > . Based on the form of 1" , set ]" œ E F> G># Þ Substitution into the differential equation, and comparing the coefficients, results in the system of equations E $F %G œ ! , F 'G œ ! , and G œ " . Hence we obtain ]" œ "% '> ># Þ On the other hand, set ]# œ H -9= > I =38 > . After substitution into the ODE, we find that H œ *Î"! and I œ $Î"! . The general solution is C a >b œ C - a > b ] " ] # Þ 9. The characteristic equation for the homogeneous problem is <# =# œ ! , with ! complex roots < œ „ =! 3. Hence C- a>b œ -" -9= =! > -# =38 =! > . Since = Á =! , set ] œ E -9= => F =38 => . Substitution into the ODE and comparing the coefficients results in the system of equations a=# =# bE œ " and a=# =# bF œ ! . Hence ! ! ]œ The general solution is Ca>b œ C- a>b ] . =# ! " -9= => Þ =# ________________________________________________________________________ page 107 —————————————————————————— CHAPTER 3. —— 10. From Prob. 9, C- a>b œ - . Since -9= =! > is a solution of the homogeneous problem, set ] œ E> -9= =! > F> =38 =! > . Substitution into the given ODE and comparing the coefficients results in E œ ! and F œ #" ! Þ Hence the general solution is = > Ca>b œ -" -9= =! > -# =38 =! > #=! =38 =! > . 12. The characteristic equation for the homogeneous problem is <# < # œ ! , with roots < œ " , # . Hence C- a>b œ -" /> -# /#> . Based on the form of the right hand side, that is, -9=2Ð#>Ñ œ a/#> /#> bÎ# , set ] œ E> /#> F/#> . Substitution into the given ODE and comparing the coefficients results in E œ "Î' and F œ "Î) Þ Hence the general solution is C a>b œ -" /> -# /#> > /#> Î' /#> Î) . 14. The characteristic equation for the homogeneous problem is <# % œ ! , with roots < œ „ #3. Hence C- a>b œ -" -9= #> -# =38 #> . Set ]" œ E F> G># Þ Comparing the coefficients of the respective terms, we find that E œ "Î) , F œ ! , G œ "Î% . Now set ]# œ H /> , and obtain H œ $Î& . Hence the general solution is Ca>b œ -" -9= #> -# =38 #> "Î) ># Î% $ /> Î& . Invoking the initial conditions, we require that "*Î%! -" œ ! and $Î& #-# œ # . Hence -" œ "*Î%! and -# œ (Î"! . 15. The characteristic equation for the homogeneous problem is <# #< " œ ! , with a double root < œ " . Hence C- a>b œ -" /> -# > /> . Consider 1" a>b œ > /> . Note that 1" is a solution of the homogeneous problem. Set ]" œ E># /> F>$ /> Ðthe first term is not sufficient for a matchÑ. Upon substitution, we obtain ]" œ >$ /> Î' . By inspection, ]# œ % . Hence the general solution is C a>b œ -" /> -# > /> >$ /> Î' % . Invoking the initial conditions, we require that -" % œ " and -" -# œ " . Hence -" œ $ and -# œ % . 17. The characteristic equation for the homogeneous problem is <# % œ ! , with roots < œ „#3. Hence C- a>b œ -" -9= #> -# =38 #> . Since the function =38 #> is a solution of the homogeneous problem, set ] œ E> -9= #> F> =38 #> . Upon substitution, we obtain ] œ $ > -9= #> . Hence the general solution is C a>b œ -" -9= #> -# =38 #> " > -9= #> . % % Invoking the initial conditions, we require that -" œ # and #-# $ œ " . Hence % -" œ # and -# œ "Î) . 18. The characteristic equation for the homogeneous problem is <# #< & œ ! , with complex roots < œ "„ #3. Hence C- a>b œ -" /> -9= #> -# /> =38 #> . Based on the form of 1a>b, set ] œ E> /> -9= #> F> /> =38 #> . After comparing coefficients, we obtain ] œ > /> =38 #> . Hence the general solution is Ca>b œ -" /> -9= #> -# /> =38 #> > /> =38 #> . Invoking the initial conditions, we require that -" œ " and -" #-# œ ! . Hence -" œ " and -# œ "Î# . ________________________________________________________________________ page 108 —————————————————————————— CHAPTER 3. —— 20. The characteristic equation for the homogeneous problem is <# " œ ! , with complex roots < œ „ 3. Hence C- a>b œ -" -9= > -# =38 >. Let 1" a>b œ > =38 > and 1# a>b œ > . By inspection, it is easy to see that ]# a>b œ " . Based on the form of 1" a>b, set ]" a>b œ E> -9= > F> =38 > G># -9= > H># =38 > Þ Substitution into the equation and comparing the coefficients results in E œ ! , F œ "Î% , G œ "Î% , and H œ ! Þ Hence ] a>b œ " " > =38 > " ># -9= > . % % 21. The characteristic equation for the homogeneous problem is <# &< ' œ ! , with roots < œ # , $. Hence C- a>b œ -" /#> -# /$> . Consider 1" a>b œ /#> a$> %b=38 > , and 1# a>b œ /> -9= #>. Based on the form of these functions on the right hand side of the ODE, set ]# a>b œ /> aE" -9= #> E# =38 #>b, ]" a>b œ aF" F# > b/#> =38 > aG" G#>b/#> -9= >. Substitution into the equation and comparing the coefficients results in ] a>b œ 23. The characteristic roots are < œ # , #. Hence C- a>b œ -" /#> -# >/#> . Consider the functions 1" a>b œ #># , 1# a>b œ %>/#> , and 1$ a>b œ > =38 #> . The corresponding forms of the respective parts of the particular solution are ]" a>b œ E! E" > E# ># , ]# a>b œ œ /#> aF# ># F$ >$ b, and ]$ a>b œ >aG" -9= #> G# =38 #>b aH" -9=#> H# =38#>b. Substitution into the equation and comparing the coefficients results in ] a>b œ " # " " ˆ$ %> #># ‰ >$ /#> > -9= #> a-9= #> =38 #>b . % $ ) "' "> $ ˆ/ -9= #> $/> =38 #>‰ >/#> a-9= > =38 >b /#> Š " -9= > &=38 >‹. # #! # 24. The homogeneous solution is C- a>b œ -" -9= #> -# =38 #>. Since -9= #> and =38 #> are both solutions of the homogeneous equation, set ] a>b œ >ˆE! E" > E# ># ‰-9= #> >ˆF! F" > F# ># ‰=38 #> . ] a>b œ Œ "$ " " > >$ -9= #> ˆ#)> "$># ‰=38 #> . $# "# "' Substitution into the equation and comparing the coefficients results in 25. The homogeneous solution is C- a>b œ -" /> -# >/#> . None of the functions on the right hand side are solutions of the homogenous equation. In order to include all possible combinations of the derivatives, consider ] a>b œ /> aE! E" > E# ># b-9= #> /> aF! F" > F# ># b=38 #> /> aG" -9= > G# =38 >b H/> . Substitution into the differential equation and comparing the coefficients results in ] a>b œ /> ˆE! E" > E# ># ‰-9= #> /> ˆF! F" > F# ># ‰=38 #> /> Š -9= > =38 >‹ #/> Î$ , # $ # $ ________________________________________________________________________ page 109 —————————————————————————— CHAPTER 3. —— in which E! œ %"!&Î$&"&# ß E" œ ($Î'(' ß E# œ &Î&# ß F! œ "#$$Î$&"&# ß F" œ "!Î"'* ß F# œ "Î&# Þ 26. The homogeneous solution is C- a>b œ -" /> -9= #> -# /> =38 #>. None of the terms on the right hand side are solutions of the homogenous equation. In order to include the appropriate combinations of derivatives, consider ] a>b œ /> aE" > E# ># b-9= #> /> aF" > F# ># b=38 #> /#> aG! G" >b-9= #> /#> aH! H" >b=38 #>. Substitution into the differential equation and comparing the coefficients results in ] a>b œ $ > $ " >/ -9= #> ># /> =38 #> /#> a( "!>b-9= #> "' ) #& " #> / a" &>b=38 #> . #& 27. The homogeneous solution is C- a>b œ -" -9= -> -# =38 ->. Since the differential operator does not contain a first derivative aand - Á 71b, we can set ] a>b œ "G7 =38 71> . R 7œ" Substitution into the ODE yields " 7# 1# G7 =38 71> -# "G7 =38 71> œ "+7 =38 71> . R R R 7œ" 7œ" 7œ" Equating coefficients of the individual terms, we obtain +7 G7 œ # , 7 œ "ß # â R . - 7# 1# 29. The homogeneous solution is C- a>b œ -" /> -9= #> -# /> =38 #>. The input function is independent of the homogeneous solutions, on any interval. Since the right hand side is piecewise constant, it follows by inspection that ] a>b œ œ "Î& , !, ! Ÿ > Ÿ 1Î# Þ > 1Î# For ! Ÿ > Ÿ 1Î# , the general solution is C a>b œ -" /> -9= #> -# /> =38 #> "Î& . Invoking the initial conditions Ca!b œ C w a!b œ ! , we require that -" œ "Î& , and that -# œ "Î"! . Hence on the interval ! Ÿ > Ÿ 1Î# . We now have the values C a1Î#b œ ˆ" /1Î# ‰Î& , and C w a1Î#b œ ! . For > 1Î# , the general solution is C a>b œ ." /> -9= #> .# /> =38 #> . It follows that Ca1Î#b œ /1Î# ." and C w a1Î#b œ /1Î# ." #/1Î# .# . Since the ________________________________________________________________________ page 110 C a >b œ " " ˆ#/> -9= #> /> =38 #>‰ & "! —————————————————————————— CHAPTER 3. —— solution is continuously differentiable, we require that Solving for the coefficients, ." œ #.# œ ˆ/1Î# "‰Î& . /1Î# ." #/1Î# .# œ ! . /1Î# ." œ ˆ" /1Î# ‰Î& 31. Since + ß ,ß - ! , the roots of the characteristic equation has negative real parts. That is, < œ !„" 3 , where ! ! . Hence the homogeneous solution is If 1a>b œ . , then the general solution is C- a>b œ -" /!> -9= " > -# /!> =38 " > . Since ! ! , Ca>b p .Î- as > p _ . If - œ !, then that characteristic roots are < œ ! and < œ ,Î+ . The ODE becomes +C ww ,C w œ . . Integrating both sides, we find that +C w ,C œ . > -" . The general solution can be expressed as Ca>b œ . >Î, -" -# /,>Î+ Þ Ca>b œ .Î- -" /!> -9= " > -# /!> =38 " > . In this case, the solution grows without bound. If , œ !, also, then the differential equation can be written as C ww œ .Î+ , which has general solution C a>b œ . ># Î#+ -" -# . Hence the assertion is true only if the coefficients are positive. 32a+b. Since H is a linear operator, H# C ,HC -C œ H# C a<" <# bHC <" <# C œ H# C <# HC <" HC <" <# C œ HaHC <# C b <" aHC <# C b œ aH <" baH <# bC . a,b. Let ? œ aH <# bC . Then the ODE a3b can be written as aH <" b? œ 1a>b, that is, ________________________________________________________________________ page 111 —————————————————————————— CHAPTER 3. —— ? w <" ? œ 1a>b. The latter is a linear first order equation in ? . Its general solution is From above, we have C w <# C œ ?a>b. This equation is also a first order ODE. Hence the general solution of the original second order equation is Ca>b œ / ( /<# 7 ?a7 b. 7 -# /<# > . <# > > >! ?a>b œ /<" > ( /<" 7 1a7 b. 7 -" /<" > . > >! Note that the solution Ca>b contains two arbitrary constants. 34. Note that a#H# $H "bC œ a#H "baH "bC . Let ? œ aH "bC , and solve the ODE #? w ? œ ># $=38 > Þ This equation is a linear first order ODE, with solution ?a>b œ />Î# ( /7 Î# ’7 # Î# =38 7 “. 7 - />Î# > >! $ # Now consider the ODE C w C œ ?a>b. The general solution of this first order ODE is in which ?a>b is given above. Substituting for ?a>b and performing the integration, Ca>b œ ># '> "% * $ -9= > =38 > -" />Î# -# /> Þ "! "! Ca>b œ /> ( /7 ?a7 b. 7 -# /> , > >! ' $ œ ># %> ) -9= > =38 > - />Î# Þ & & 35. We have aH# #H "bC œ aH "baH "bC . Let ? œ aH "bC , and consider the ODE ? w ? œ #/> Þ The general solution is ?a>b œ #> /> - /> Þ We therefore have the first order equation ? w ? œ #> /> -" /> Þ The general solution of the latter differential equation is Ca>b œ / ( c#7 -" d. 7 -# /> > > >! > ˆ # œ/ > -" > -# ‰Þ 36. We have aH# #HbC œ HaH #bC . Let ? œ aH #bC , and consider the equation ? w œ $ %=38 #> Þ Direct integration results in ?a>b œ $> #-9= #> - Þ The problem is reduced to solving the ODE C w #C œ $> #-9= #> - Þ The general solution of this first order differential equation is ________________________________________________________________________ page 112 —————————————————————————— CHAPTER 3. —— Ca>b œ /#> ( /#7 c$7 #-9= #7 - d. 7 -# /#> > $ " œ > a-9= #> =38 #>b -" -# /#> Þ # # >! ________________________________________________________________________ page 113 —————————————————————————— CHAPTER 3. —— Section 3.7 1. The solution of the homogeneous equation is C- a>b œ -" /#> -# /$> . The functions C" a>b œ /#> and C# a>b œ /$> form a fundamental set of solutions. The Wronskian of these functions is [ aC" ß C# b œ /&> . Using the method of variation of parameters, the particular solution is given by ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which ?" a>b œ ( œ #/> ?# a>b œ ( /$> a#/> b .> [ a>b 3. The solution of the homogeneous equation is C- a>b œ -" /> -# >/> . The functions C" a>b œ /> and C# a>b œ >/> form a fundamental set of solutions. The Wronskian of these functions is [ aC" ß C# b œ /#> . Using the method of variation of parameters, the particular solution is given by ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which ?" a>b œ ( ?# a>b œ ( œ $> >/> a$/> b .> [ a>b œ $># Î# /> a$/> b .> [ a>b Hence the particular solution is ] a>b œ #/> /> œ /> . /#> a#/> b .> [ a>b œ /#> Hence the particular solution is ] a>b œ $># /> Î# $># /> œ $># /> Î# . 4. The functions C" a>b œ />Î# and C# a>b œ >/>Î# form a fundamental set of solutions. The Wronskian of these functions is [ aC" ß C# b œ /> . First write the equation in standard form, so that 1a>b œ %/>Î# . Using the method of variation of parameters, the particular solution is given by ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which ?" a>b œ ( >/>Î# ˆ%/>Î# ‰ .> [ a>b œ #># ________________________________________________________________________ page 114 —————————————————————————— CHAPTER 3. —— />Î# ˆ%/>Î# ‰ ?# a>b œ ( .> [ a>b œ %> Hence the particular solution is ] a>b œ #># />Î# %># />Î# œ #># />Î# . 6. The solution of the homogeneous equation is C- a>b œ -" -9= $> -# =38 $>. The two functions C" a>b œ -9= $> and C# a>b œ =38 $> form a fundamental set of solutions, with [ aC" ß C# b œ $ . The particular solution is given by ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which ?" a>b œ ( ?# a>b œ ( =38 $>a* =/- # $>b .> [ a>b œ -=- $> -9= $>a* =/- # $>b .> [ a>b œ 68k=/- $> >+8 $>k Hence the particular solution is ] a>b œ " a=38 $>b68k=/- $> >+8 $>k. The general solution is given by Ca>b œ -" -9= $> -# =38 $> a=38 $>b68k=/- $> >+8 $>k " . 7. The functions C" a>b œ /#> and C# a>b œ >/#> form a fundamental set of solutions. The Wronskian of these functions is [ aC" ß C# b œ /%> . The particular solution is given by ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which >/#> a># /#> b ?" a>b œ ( .> [ a>b œ 68 > ?# a>b œ ( /#> a># /#> b .> [ a>b œ "Î> Hence the particular solution is ] a>b œ /#> 68 > /#> . Since the second term is a solution of the homogeneous equation, the general solution is given by Ca>b œ -" /#> -# >/#> /#> 68 > . 8. The solution of the homogeneous equation is C- a>b œ -" -9= #> -# =38 #>. The two functions C" a>b œ -9= #> and C# a>b œ =38 #> form a fundamental set of solutions, with [ aC" ß C# b œ # . The particular solution is given by ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which ________________________________________________________________________ page 115 —————————————————————————— CHAPTER 3. —— ?" a>b œ ( ?# a>b œ ( $ % =38 #>a$ -=- #>b .> [ a>b œ $>Î# -9= #>a$ -=- #>b .> [ a>b Hence the particular solution is ] a>b œ $ >-9= #> $ a=38 $>b68k=38 #>k. The general # % solution is given by Ca>b œ -" -9= #> -# =38 #> $ >-9= #> $ a=38 $>b68k=38 #>k. # % 9. The functions C" a>b œ -9= a>Î#b and C# a>b œ =38a>Î#b form a fundamental set of solutions. The Wronskian of these functions is [ aC" ß C# b œ "Î# . First write the ODE in standard form, so that 1a>b œ =/- a>Î#bÎ#. The particular solution is given by ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which ?" a>b œ ( ?# a>b œ ( œ> -9= a>Î#bc=/- a>Î#bd .> #[ a>b œ # 68c-9= a>Î#bd =38a>Î#bc=/- a>Î#bd .> #[ a>b œ 68k=38 #>k The particular solution is ] a>b œ #-9=a>Î#b68c-9= a>Î#bd > =38a>Î#b. The general solution is given by Ca>b œ -" -9= a>Î#b -# =38a>Î#b # -9=a>Î#b 68c-9= a>Î#bd > =38a>Î#b. 10. The solution of the homogeneous equation is C- a>b œ -" /> -# >/> . The functions C" a>b œ /> and C# a>b œ >/> form a fundamental set of solutions, with [ aC" ß C# b œ /#> . The particular solution is given by ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which ?" a>b œ ( " œ 68ˆ" ># ‰ # >/> a/> b .> [ a>ba" ># b ?# a>b œ ( The particular solution is ] a>b œ " /> 68a" ># b >/> +<->+8a>b. Hence the general # ________________________________________________________________________ page 116 / > a/ > b .> [ a>ba" ># b œ +<->+8 > —————————————————————————— CHAPTER 3. —— solution is given by Ca>b œ -" /> -# >/> " /> 68a" ># b >/> +<->+8a>b. # 12. The functions C" a>b œ -9= #> and C# a>b œ =38 #> form a fundamental set of solutions, with [ aC" ß C# b œ # . The particular solution is given by ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which "> ?" a>b œ ( 1a=b =38 #= .= # ?# a>b œ Hence the particular solution is > > " " ] a>b œ -9= #>( 1a=b =38 #= .= =38 #>( 1a=b -9= #= .= . # # "> ( 1a=b -9= #= .= # Note that =38 #> -9= #= -9= #> =38 #= œ =38a#> #=b. It follows that ] a>b œ "> ( 1a=b=38a#> #=b.= . # The general solution of the differential equation is given by "> Ca>b œ -" -9= #> -# =38 #> ( 1a=b=38a#> #=b.= . # 13. Note first that :a>b œ ! , ; a>b œ #Î># and 1a>b œ a$># "bÎ># . The functions C" a>b and C# a>b are solutions of the homogeneous equation, verified by substitution. The Wronskian of these two functions is [ aC" ,C# b œ $ . Using the method of variation of parameters, the particular solution is ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which >" a$># "b ?" a>b œ ( .> ># [ a > b œ ># Î' 68 > ?# a>b œ ( ># a$># "b .> ># [ a>b œ >$ Î$ >Î$ Therefore ] a>b œ "Î' ># 68 > ># Î$ "Î$ Þ Hence the general solution is Ca>b œ -" ># -# >" ># 68 > "Î# . ________________________________________________________________________ page 117 —————————————————————————— CHAPTER 3. —— 15. Observe that 1a>b œ > /#> . The functions C" a>b and C# a>b are a fundamental set of solutions. The Wronskian of these two functions is [ aC" ,C# b œ > /> . Using the method of variation of parameters, the particular solution is ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which /> a> /#> b ?" a>b œ ( .> [ a>b œ /#> Î# ?# a>b œ ( œ > /> a" >ba> /#> b .> [ a>b 16. Observe that 1a>b œ #a" >b /> . Direct substitution of C" a>b œ /> and C# a>b œ > verifies that they are solutions of the homogeneous equation. The Wronskian of the two solutions is [ aC" ,C# b œ a" >b /> . Using the method of variation of parameters, the particular solution is ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which #>a" >b/> ?" a>b œ ( .> [ a>b œ >/#> /#> Î# ?# a>b œ ( #a" >b .> [ a>b œ # /> Therefore ] a>b œ a" >b/#> Î# > /#> œ /#> Î# > /#> Î# Þ Therefore ] a>b œ >/> /> Î# #> /> œ >/> /> Î# Þ 17. Note that 1aBb œ 68 B . The functions C" aBb œ B# and C# aBb œ B# 68 B are solutions of the homogeneous equation, as verified by substitution. The Wronskian of the solutions is [ aC" ,C# b œ B$ . Using the method of variation of parameters, the particular solution is ] aBb œ ?" aBb C" aBb ?# aBb C# aBb, ?" aBb œ ( B# 68 Ba68 Bb .B [ aBb œ a68 Bb$ Î$ in which ________________________________________________________________________ page 118 —————————————————————————— CHAPTER 3. —— ?# aBb œ ( B# a68 Bb .B [ aBb œ a68 Bb# Î# Therefore ] aBb œ B# a68 Bb$ Î$ B# a68 Bb$ Î# œ B# a68 Bb$ Î' Þ 19. First write the equation in standard form. Note that the forcing function becomes 1aBbÎa" Bb . The functions C" aBb œ /B and C# aBb œ B are a fundamental set of solutions, as verified by substitution. The Wronskian of the solutions is [ aC" ,C# b œ a" Bb/B . Using the method of variation of parameters, the particular solution is ] aBb œ ?" aBb C" aBb ?# aBb C# aBb, ?" aBb œ ( ?# aBb œ ( Therefore ] aBb œ /B ( œ( B B B B in which / 7 a1 a 7 b b .7 a" 7 b[ a 7 b B 7 a1 a7 bb .7 a" 7 b[ a 7 b 7 a1 a7 bb . 7 B( a" 7 b[ a 7 b aB/7 /B 7 b1a7 b .7 Þ a" 7 b # / 7 / 7 a1 a 7 bb .7 a" 7 b[ a 7 b 20. First write the equation in standard form. The forcing function becomes 1aBbÎB# . The functions C" aBb œ B"Î# =38 B and C# aBb œ B"Î# -9= B are a fundamental set of solutions. The Wronskian of the solutions is [ aC" ,C# b œ "ÎB . Using the method of variation of parameters, the particular solution is ] aBb œ ?" aBb C" aBb ?# aBb C# aBb, ?" aBb œ ( B in which -9= 7 a1a7 bb .7 7 È7 B ?# aBb œ ( Therefore =38 7 a1a7 bb .7 7 È7 ________________________________________________________________________ page 119 —————————————————————————— CHAPTER 3. —— ] aBb œ =38 B B -9= 7 a1a7 bb -9= B .> ( ÈB ÈB ( 7 È7 B " =38aB 7 b 1a7 b œ .7 Þ ÈB ( 7 È7 B =38 7 a1a7 bb .7 7 È7 21. Let C" a>b and C# a>b be a fundamental set of solutions, and [ a>b œ [ aC" ß C# b be the corresponding Wronskian. Any solution, ?a>b, of the homogeneous equation is a linear combination ?a>b œ !" C" a>b !# C# a>b. Invoking the initial conditions, we require that Note that this system of equations has a unique solution, since [ a>! b Á !. Now consider the nonhomogeneous problem, Pc@d œ 1a>b ,with homogeneous initial conditions. Using the method of variation of parameters, the particular solution is given by ] a>b œ C" a>b( > >! > C# a=b 1a=b C" a=b 1a=b .= C# a>b( .= Þ [ a =b [ a=b >! C! œ !" C" a>! b !# C# a>! b C!w œ !" C"w a>! b !# C#w a>! b The general solution of the IVP a333b is @a>b œ "" C" a>b "# C# a>b ] a>b œ "" C" a>b "# C# a>b C" a>b?" a>b C# a>b?# a>b ! œ "" C" a>! b "# C# a>! b ] a>! b ! œ "" C"w a>! b "# C#w a>! b ] w a>! b in which ?" and ?# are defined above. Invoking the initial conditions, we require that Based on the definition of ?" and ?# , ] a>! b œ ! . Furthermore, since C" ?"w C# ?#w œ ! , it follows that ] w a>! b œ ! . Hence the only solution of the above system of equations is the trivial solution. Therefore @a>b œ ] a>b . Now consider the function C œ ?@ . Then PcCd œ Pc? @d œ Pc?d Pc@d œ 1a>b. That is, C a>b is a solution of the nonhomogeneous w problem. Further, Ca>! b œ ?a>! b @a>! b œ C! , and similarly, C w a>! b œ C! . By the uniqueness theorems, Ca>b is the unique solution of the initial value problem. 23. A fundamental set of solutions is C" a>b œ -9= > and C# a>b œ =38 > . The Wronskian [ a>b œ C" C#w C"w C# œ " . By the result in Prob. ##, ] a>b œ ( > >! > >! Finally, we have - 9=a=b =38a>b -9=a>b =38a=b œ =38a> =b. œ ( c-9=a=b =38a>b -9=a>b =38a=bd1a=b.= Þ -9=a=b =38a>b -9=a>b =38a=b 1a=b.= [ a =b ________________________________________________________________________ page 120 —————————————————————————— CHAPTER 3. —— 24. A fundamental set of solutions is C" a>b œ /+> and C# a>b œ /,> . The Wronskian [ a>b œ C" C#w C"w C# œ a, +b/B:ca+ , b>d . By the result in Prob. ##, ] a>b œ ( / / /+> /,= 1a=b.= [ a =b >! > += ,> " / / /+> /,= œ 1a=b.= Þ ( , + >! /B:ca+ , b=d > += ,> Hence the particular solution is > " ,a>=b ] a>b œ /+a>=b ‘1a=b.= Þ ( / , + >! 26. A fundamental set of solutions is C" a>b œ /+> and C# a>b œ >/+> . The Wronskian [ a>b œ C" C#w C"w C# œ /#+> . By the result in Prob. ##, / / /+> /,= ] a>b œ ( 1a=b.= [ a =b >! > += ,> " / / /+> /,= œ 1a=b.= Þ ( , + >! /B:ca+ , b=d > += ,> Hence the particular solution is ] a>b œ > " ,a>=b /+a>=b ‘1a=b.= Þ ( / , + >! 26. A fundamental set of solutions is C" a>b œ /+> and C# a>b œ >/+> . The Wronskian [ a>b œ C" C#w C"w C# œ /#+> . By the result in Prob. ##, ] a>b œ ( >/+=+> = /+>+= 1a=b.= [ a =b >! > a> =b/+=+> œ( 1a=b.= Þ /#+= >! > Hence the particular solution is ] a>b œ ( a> =b/+a>=b 1a=b.= Þ > >! 27. Depending on the values of +, , and - , the operator +H# ,H - can have three types of fundamental solutions. a 3b The characteristic roots <"ß# œ ! , " ; ! Á " . C" a>b œ /!> and C# a>b œ /"> . page 121 ________________________________________________________________________ —————————————————————————— CHAPTER 3. —— O a>b œ " /"> /!> ‘Þ "! a33b The characteristic roots <"ß# œ ! , " ; ! œ " . C" a>b œ /!> and C# a>b œ >/!> . O a>b œ > /!> Þ a333b The characteristic roots <"ß# œ -„ 3 . . C" a>b œ /-> -9= .> and C# a>b œ /-> =38 .>. O a>b œ " -> / =38 .> Þ . 28. Let Ca>b œ @a>bC" a>b, in which C" a>b is a solution of the homogeneous equation. Substitution into the given ODE results in By assumption, C"ww :a>bC" ; a>bC" œ ! , hence @a>b must be a solution of the ODE Setting A œ @ w , we also have A w C" c#C"w :a>bC" dA œ 1a>b . 30. First write the equation as C ww (>" C &># C œ >" . As shown in Prob. #), the function Ca>b œ >" @a>b is a solution of the given ODE as long as @ is a solution of >" @ ww c #># (># d@ w œ >" , that is, @ ww &>" @ w œ " . This ODE is linear and first order in @ w . The integrating factor is . œ >& . The solution is @ w œ >Î' - >& . Direct integration now results in @a>b œ ># Î"# -" >% -# . Hence C a>b œ >Î"# -" >& -# >" . a" >b @ ww # >" a" >b# ‘@ w œ > /#> , @ ww C" c#C"w :a>bC" d@ w œ 1a>b . @ ww C" #@ w C"w @C"ww :a>bc@ w C" @C"w d ; a>b@C" œ 1a>b . 31. Write the equation as C ww >" a" >bC >" C œ > /#> . As shown in Prob. #), the function Ca>b œ a" >b@a>b is a solution of the given ODE as long as @ is a solution of that is, @ ww "># >a>"b factor . œ >" a" >b / . The solution is @ w œ a># /#> -" >/> bÎa" >b# . Integrating, we obtain @a>b œ /#> Î# /#> Îa> "b -" /> Îa> "b -# Þ Hence the solution of the original ODE is Ca>b œ a> "b/#> Î# -" /> -# a> "b . " @w œ > #> >" / . # > This equation is first order linear in @ w , with integrating 32. Write the equation as C ww >a" >b" C a" >b C œ #a" >b /> . The function Ca>b œ /> @a>b is a solution to the given ODE as long as @ is a solution of ________________________________________________________________________ page 122 —————————————————————————— CHAPTER 3. —— /> @ ww #/> >a" >b" /> ‘@ w œ #a" >b /> , @ w œ a> "bˆ#/#> -" /> ‰. that is, @ ww ca# >bÎa" >bd@ w œ #a" >b /#> . This equation is first order linear in @ w , with integrating factor . œ /> Îa> "b. The solution is Integrating, we obtain @a>b œ a"Î# >b/#> -" >/> -# Þ Hence the solution of the original ODE is Ca>b œ a"Î# >b/> -" > -# /> . Section 3.8 1. V-9= $ œ $ and V=38 $ œ % Ê V œ È#& œ & and $ œ +<->+8Ð%Î$ÑÞ Hence ? œ & -9=a#> !Þ*#($bÞ 3. V -9= $ œ % and V=38 $ œ # Ê V œ È#! œ #È& and $ œ +<->+8Ð"Î#ÑÞ Hence ? œ #È& -9=a$> !Þ%'$'bÞ 4. V-9= $ œ # and V=38 $ œ $ Ê V œ È"$ and $ œ 1 +<->+8Ð$Î#Ñ. Hence ? œ È"$ -9=a1> %Þ"#%%bÞ 5. The spring constant is 5 œ #Îa"Î#b œ % lb/ft . Mass 7 œ #Î$# œ "Î"' lb-s# /ft . Since there is no damping, the equation of motion is " ww ? %? œ ! , "' that is, ? ww '%? œ ! . The initial conditions are ?a!b œ "Î% ft , ?w a!b œ ! fps . The general solution is ?a>b œ E -9= )> F =38 )> . Invoking the initial conditions, we have ?a>b œ " -9= )> . V œ $ inches, $ œ ! rad , =! œ ) rad/s , and X œ 1Î% sec . % 7. The spring constant is 5 œ $Îa"Î%b œ "# lb/ft . Mass 7 œ $Î$# lb-s# /ft . Since there is no damping, the equation of motion is $ ww ? "#? œ ! , $# that is, ? ww "#)? œ ! . The initial conditions are ?a!b œ "Î"# ft , ?w a!b œ # fps . The general solution is ?a>b œ E -9= )È# > F =38 )È# > . Invoking the initial conditions, we have ________________________________________________________________________ page 123 —————————————————————————— CHAPTER 3. —— ?a>b œ " " -9= )È# > =38 )È# > . È# "# % V œ È""Î"# ft , $ œ 1 +>+8Š$ÎÈ# ‹ rad , =! œ )È# rad/s, and X œ 1Ί%È# ‹ sec. 10. The spring constant is 5 œ "'Îa"Î%b œ '% lb/ft . Mass 7 œ "Î# lb-s# /ft . The damping coefficient is # œ # lb-sec/ft . Hence the equation of motion is " ww ? #? w '%? œ ! , # that is, ? ww %? w "#)? œ ! . The initial conditions are ?a!b œ ! ft, ?w a!b œ "Î% fps. The general solution is ?a>b œ E -9= #È$" > F =38 #È$" > . Invoking the initial conditions, we have ?a>b œ " /#> =38 #È$" > . È$" ) Solving ?a>b œ ! , on the interval c!Þ# , !Þ%d, we obtain > œ 1Î#È$" œ !Þ#)#" sec . Based on the graph, and the solution of ?a>b œ !Þ!" , we have k?a>bk Ÿ !Þ!" for > 7 œ !Þ#"%& . 11. The spring constant is 5 œ $ÎaÞ"b œ $! N/m . The damping coefficient is given as # œ $Î& N-sec/m . Hence the equation of motion is $ #? ww ? w $!? œ ! , & that is, ? ww !Þ$? w "&? œ ! . The initial conditions are ?a!b œ !Þ!& 7 and ?w a!b œ !Þ!" 7Î= . The general solution is ?a>b œ E -9= .> F =38 .> , in which . œ $Þ)(!!) rad/s . Invoking the initial conditions, we have Also, .Î=! œ $Þ)(!!)ÎÈ"& ¸ !Þ***#& . ?a>b œ /0.15> a!Þ!&-9= .> !Þ!!%&#=38 .>b . ________________________________________________________________________ page 124 —————————————————————————— CHAPTER 3. —— 13. The frequency of the undamped motion is =! œ " . The quasi frequency of the damped motion is . œ " È% # # Þ Setting . œ # =! , we obtain # œ # È& Þ # $ $ 14. The spring constant is 5 œ 71ÎP . The equation of motion for an undamped system is 71 7? ww ? œ !. P 1 Hence the natural frequency of the system is =! œ É P . The period is X œ #1Î=! . 15. The general solution of the system is ?a>b œ E -9= # a> >! b F =38 # a> >! b . Invoking the initial conditions, we have ?a>b œ ?! -9=# a> >! b a?!w Î# b=38# a> >! b. Clearly, the functions @ œ ?! -9=# a> >! b and A œ a?!w Î# b=38# a> >! b satisfy the given criteria. 16. Note that < =38a =! > ) b œ < =38=! > -9= ) < -9==! > =38) . Comparing the given expressions, we have E œ < =38 ) and F œ < -9= ) . That is, < œ V œ ÈE# F # , and >+8 ) œ EÎF œ "Î>+8 $ . The latter relation is also >+8 ) -9> $ œ " . 18. The system is critically damped, when V œ #ÈPÎG Þ Here V œ "!!! ohms . 21a+b. Let ? œ V/# >Î#7 -9=a.> $ b . Then attains a maximum when .>5 $ œ #5 1. Hence X. œ >5" >5 œ #1Î. . a,b. ?a>5 bÎ?a>5" b œ /B:a # >5 Î#7bÎ/B:a # >5" Î#7b œ /B:ca# >5" # >5 bÎ#7dÞ Hence ?a>5 bÎ?a>5" b œ /B:c# a#1Î.bÎ#7d œ /B:a# X. Î#7bÞ a- b. ? œ 68c?a>5 bÎ?a>5" bd œ # a#1Î.bÎ#7 œ 1# Î.7 . 22. The spring constant is 5 œ "'Îa"Î%b œ '% lb/ft . Mass 7 œ "Î# lb-s# /ft . The damping coefficient is # œ # lb-sec/ft . The quasi frequency is . œ #È$" rad/s . #1 Hence ? œ È$" ¸ "Þ"#)& Þ 25a+b. The solution of the IVP is ?a>b œ />Î) Š# -9= $ È( > !Þ#&#=38 $ È( >‹. ) ) ________________________________________________________________________ page 125 —————————————————————————— CHAPTER 3. —— Using the plot, and numerical analysis, 7 ¸ %"Þ("& . a,b. For # œ !Þ& , 7 ¸ #!Þ%!# à for # œ "Þ! , 7 ¸ *Þ"') à for # œ "Þ& , 7 ¸ (Þ")% . a- bÞ a. b. For # œ "Þ' , 7 ¸ (Þ#") à for # œ "Þ( , 7 ¸ 'Þ('( à for # œ "Þ) , 7 ¸ &Þ%($ à for # œ "Þ* , 7 ¸ 'Þ%'! . 7 steadily decreases to about 7738 ¸ %Þ)($ , corresponding to the critical value #! ¸ "Þ($ . a/b. We have ?a>b œ # $ œ >+8" È%# # . Hence k?a>bk Ÿ %/# >Î# È%# # -9=a.> $ b , in which . œ " È% # # , and # %/# >Î# È%# # . 26a+b. The characteristic equation is 7<# # < 5 œ ! . Since # # %57 , the roots È%75# # # are <"ß# œ #7 „3 #7 . The general solution is ?a>b œ /# >Î#7 –E -9= È%75 # # #7 > F =38 È%75 # # #7 > —Þ Invoking the initial conditions, E œ ?! and Fœ a#7@! # ?! b . È%75 # # ________________________________________________________________________ page 126 —————————————————————————— CHAPTER 3. —— a,b. We can write ?a>b œ V /# >Î#7 -9=a.> $ b , in which V œ Ë?# ! and $ œ +<->+8– a- b. V œ É?# ! a#7@! # ?! b# %75# # # a#7@! # ?! b# , %75 # # a#7@! # ?! b Þ ?! È%75 # # — # It is evident that V increases amonotonicallyb without bound as # p Š#È75 ‹ . # ?! 7@ +, œ #É 7a5?!%75@#!# ! b œ É %75# # . # 28a+b. The general solution is ?a>b œ E-9= È# > F=38 È# > . Invoking the initial conditions, we have ?a>b œ È# =38 È# > . a, b . a- bÞ The condition ? w a!b œ # implies that ?a>b initially increases. Hence the phase point travels clockwise. ________________________________________________________________________ page 127 —————————————————————————— CHAPTER 3. —— 29. ?a>b œ È"#( ) >. "' È"#( />Î) =38 31. Based on Newton's second law, with the positive direction to the right, "J œ 7? ww where "J œ 5? # ? w . Hence the equation of motion is 7? ww # ? w 5? œ ! . The only difference in this problem is that the equilibrium position is located at the unstretched configuration of the spring. 32a+b. The restoring force exerted by the spring is J= œ a5? &?$ bÞ The opposing viscous force is J. œ # ? w . Based on Newton's second law, with the positive direction to the right, J= J. œ 7? ww . Hence the equation of motion is 7? ww # ? w 5? &?$ œ ! . a,b. With the specified parameter values, the equation of motion is ? ww ? œ ! . The general solution of this ODE is ?a>b œ E -9= > F =38 > . Invoking the initial conditions, the specific solution is ?a>b œ =38 > . Clearly, the amplitude is V œ ", and the period of the motion is X œ #1 . a- b. Given & œ !Þ" , the equation of motion is ? ww ? !Þ" ?$ œ ! . A solution of the ________________________________________________________________________ page 128 —————————————————————————— CHAPTER 3. —— IVP can be generated numerically: a. b . a/bÞ The amplitude and period both seem to decrease. a0 b . ________________________________________________________________________ page 129 —————————————————————————— CHAPTER 3. —— ________________________________________________________________________ page 130 —————————————————————————— CHAPTER 3. —— Section 3.9 2. We have =38a!„" b œ =38 ! -9= " „ -9= ! =38 " . Subtracting the two identities, we obtain =38a! " b =38a! " b œ # -9= ! =38 " . Setting ! " œ (> and ! " œ '>, > ! œ 'Þ&> and " œ !Þ&> . Hence =38 (> =38 '> œ # =38 # -9= "$> . # 3. Consider the trigonometric identity -9=a!„" b œ -9= ! -9= " … =38 ! =38 " . Adding the two identities, we obtain -9=a! " b -9=a! " b œ # -9= ! -9= " . Comparing the expressions, set ! " œ #1> and ! " œ 1>. Hence ! œ $1>Î# and " œ 1>Î# . Upon substitution, we have -9=a1>b -9=a#1>b œ # -9=a$1>Î#b -9=a1>Î#b . 4. Adding the two identities =38a!„" b œ =38 ! -9= " „ -9= ! =38 " , it follows that =38a! " b =38a! " b œ #=38 ! -9= " . Setting ! " œ %> and ! " œ $>, we have ! œ (>Î# and " œ >Î# . Hence =38 $> =38 %> œ # =38a(>Î#b -9=a>Î#b . &? ww &!? w %*!? œ "! =38a>Î#b . 6. Using mks units, the spring constant is 5 œ &a*Þ)bÎ!Þ" œ %*! N/m , and the damping coefficient is # œ #Î!Þ!% œ &! N-sec/m . The equation of motion is The initial conditions are ?a!b œ ! m and ? w a!b œ !Þ!$ m/s . 8a+b. The homogeneous solution is ?- a>b œ E/&> -9=È($ > F/&> =38È($ > Þ Based on the method of undetermined coefficients, the particular solution is Y a>b œ " c "'! -9=a>Î#b $"#) =38a>Î#bdÞ "&$#)" Hence the general solution of the ODE is ?a>b œ ?- a>b Y a>b. Invoking the initial conditions, we find that E œ "'!Î"&$#)" and F œ $)$%%$È($ Î""")*&"$!! Þ Hence the response is ?a>b œ " $)$%%$È($ &> "'! /&> -9= È($ > / =38 È($ >— Y a>b. "&$#)" – ($!! a,b. ?- a>b is the transient part and Y a>b is the steady state part of the response. ________________________________________________________________________ page 131 —————————————————————————— CHAPTER 3. —— a- b . a. b. Based on Eqs. a*b and a"!b, the amplitude of the forced response is given by V œ #Î?, in which ? œ É#&a*) =# b# #&!! =# Þ The maximum amplitude is attained when ? is a minimum. Hence the amplitude is maximum at = œ %È$ rad/s . 9. The spring constant is 5 œ "# lb/ft and hence the equation of motion is ' ww ? "#? œ % -9= (> , $# that is, ? ww '%? œ '% -9= (> . The initial conditions are ?a!b œ ! ft, ?w a!b œ ! fps. $ The general solution is ?a>b œ E-9= )> F=38 )> '% -9= (> Þ Invoking the initial %& conditions, we have ?a>b œ '% -9= )> '% -9= (> œ "#) =38a>Î#b=38a"&>Î#b Þ %& %& %& 12. The equation of motion is #? ww ? w $? œ $-9= $> #=38 $> . Since the system is damped, the steady state response is equal to the particular solution. Using the method of undetermined coefficients, we obtain ________________________________________________________________________ page 132 —————————————————————————— CHAPTER 3. —— ?== a>b œ " a=38 $> -9= $>b. ' Further, we find that V œ È# Î' and $ œ +<->+8a "b œ $1Î% . Hence we can write È ?== a>b œ '# -9=a$> $1Î%b. 13. The amplitude of the steady-state response is given by Vœ # É7# a=! =# b# # # =# J! . Since J! is constant, the amplitude is maximum when the denominator of V is minimum . # # Let D œ =# , and consider the function 0 aD b œ 7# a=! D b # # D . Note that 0 aD b is a quadratic, with minimum at D œ =# # # Î#7# . Hence the amplitude V attains a ! # # maximum at =# œ =! # # Î#7# . Furthermore, since =! œ 5Î7 , and therefore 7+B # =# œ =! ”" 7+B ## •. #57 Substituting =# œ =# into the expression for the amplitude, 7+B Vœ J! # È# % Î%7# # # a=! # # Î#7# b J! œ È=# # # # % Î%7# ! J! œ Þ #=! È" # # Î%75 14a+b. The forced response is ?== a>b œ E-9= => F=38 => Þ The constants are obtain by the method of undetermined coefficients. That is, comparing the coefficients of -9= => and =38 =>, we find that 7=# E #=F 5E œ J! , and 7=# F #=E 5F œ ! . Solving this system results in E œ 7ˆ=# =# ‰Î? and F œ #=Î? , ! # # in which ? œ É7# a=! =# b # # =# Þ It follows that >+8 $ œ FÎE œ #= . 7a=# =# b ! a,b. Here 7 œ " , # œ !Þ"#& , =! œ " . Hence >+8 $ œ !Þ"#&=Îa" =# bÞ ________________________________________________________________________ page 133 —————————————————————————— CHAPTER 3. —— # 17a+b. Here 7 œ " , # œ !Þ#& , =# œ # , J! œ # . Hence ?== a>b œ ? -9=a=> $ b , ! = where ? œ Éa# =# b# =# Î"' œ " È'% '$=# "' =% , and >+8 $ œ %a#=# b Þ % a,bÞ The amplitude is a- b . Vœ È'% '$=# "' =% ) . a. b. See Prob. 13. The amplitude is maximum when the denominator of V is minimum. That is, when = œ =7+B œ $È"% Î) ¸ "Þ%!$" . Hence V a= œ =7+B b œ '%ÎÈ"#( . 18a+b. The homogeneous solution is ?- a>b œ E-9= > F=38 > Þ Based on the method of undetermined coefficients, the particular solution is Y a>b œ $ -9= => Þ " =# Hence the general solution of the ODE is ?a>b œ ?- a>b Y a>b. Invoking the initial conditions, we find that E œ $Îa=# "b and F œ ! Þ Hence the response is ________________________________________________________________________ page 134 —————————————————————————— CHAPTER 3. —— ?a>b œ $ c -9= => -9= > d. " =# a, b . ________________________________________________________________________ page 135 —————————————————————————— CHAPTER 3. —— Note that ?a>b œ ' a" =b> a = "b > =38” •=38” •. # "= # # 19a+b. The homogeneous solution is ?- a>b œ E-9= > F=38 > Þ Based on the method of undetermined coefficients, the particular solution is Y a>b œ $ -9= => Þ " =# Hence the general solution is ?a>b œ ?- a>b Y a>b. Invoking the initial conditions, we find that E œ a=# #bÎa=# "b and F œ " Þ Hence the response is a,.b ?a>b œ " $ -9= => ˆ=# #‰-9= > ‘ =38 > . # "= ________________________________________________________________________ page 136 —————————————————————————— CHAPTER 3. —— ________________________________________________________________________ page 137 —————————————————————————— CHAPTER 3. —— Note that ?a>b œ 20. ' a" =b> a = "b > =38” •=38” • -9= > =38 > . # "= # # ________________________________________________________________________ page 138 —————————————————————————— CHAPTER 3. —— 21. The general solution is ?a>b œ ?- a>b Y a>b, in which ?- a>b œ />Î"' – Y a>b œ È#&& È#&& "("$&) #&((&) -9= > =38 > "$#(#" "' "' — "$#(#"È#&& and " c%$')!! -9=aÞ$>b ")!!! =38aÞ$>bd Þ "$#(#" ________________________________________________________________________ page 139 —————————————————————————— CHAPTER 3. —— a+ b . a, b . 23. The general solution is ?a>b œ ?- a>b Y a>b, in which ?- a>b œ />Î"' – Y a>b œ È#&& È#&& *(%' "#&) -9= > =38 > %"!& "' "' — )#"È#&& and " c "&$' -9=a$>b (# =38a$>bd Þ %"!& ________________________________________________________________________ page 140 —————————————————————————— CHAPTER 3. —— a+ b . a, b . 24. 25a+b. ________________________________________________________________________ page 141 ————————————————————————— CHAPTER 3. —— ________________________________________________________________________ page 142 —————————————————————————— CHAPTER 3. —— ________________________________________________________________________ page 143 —————————————————————————— CHAPTER 3. —— a, b . a- b. The amplitude for a similar system with a linear spring is given by Vœ È#& %*=# #&=% & . ________________________________________________________________________ page 144 —————————————————————————— CHAPTER 3. —— ________________________________________________________________________ page 145 —————————————————————————— CHAPTER 4. —— Chapter Four Section 4.1 1. The differential equation is in standard form. Its coefficients, as well as the function 1a>b œ > , are continuous everywhere. Hence solutions are valid on the entire real line. 3. Writing the equation in standard form, the coefficients are rational functions with singularities at > œ ! and > œ " . Hence the solutions are valid on the intervals a _,!b, a! , "b , and a" , _b . 4. The coefficients are continuous everywhere, but the function 1a>b œ 68 > is defined and continuous only on the interval a! , _b. Hence solutions are defined for positive reals. 5. Writing the equation in standard form, the coefficients are rational functions with a singularity at B! œ " . Furthermore, :% aBb œ >+8 BÎaB "b is undefined, and hence not continuous, at B5 œ „a#5 "b1Î# , 5 œ !ß "ß #ß â Þ Hence solutions are defined on any interval that does not contain B! or B5 . 6. Writing the equation in standard form, the coefficients are rational functions with singularities at B œ „ # . Hence the solutions are valid on the intervals a _, #b, a # , #b , and a# , _b . 7. Evaluating the Wronskian of the three functions, [ a0" ß 0# ß 0$ b œ "% . Hence the functions are linearly independent. 9. Evaluating the Wronskian of the four functions, [ a0" ß 0# ß 0$ ß 0% b œ ! . Hence the functions are linearly dependent. To find a linear relation among the functions, we need to find constants -" ß -# ß -$ ß -% , not all zero, such that -" 0" a>b -# 0# a>b -$ 0$ a>b -% 0% a>b œ ! . Collecting the common terms, we obtain a-# #-$ -% b># a#-" -$ -% b> a $-" -# -% b œ ! , which results in three equations in four unknowns. Arbitrarily setting -% œ " , we can solve the equations -# #-$ œ " , #-" -$ œ " , $-" -# œ " , to find that -" œ #Î(, -# œ "$Î(, -$ œ $Î( . Hence #0" a>b "$0# a>b $0$ a>b (0% a>b œ ! . 10. Evaluating the Wronskian of the three functions, [ a0" ß 0# ß 0$ b œ "&' . Hence the functions are linearly independent. ________________________________________________________________________ page 146 —————————————————————————— CHAPTER 4. —— 11. Substitution verifies that the functions are solutions of the ODE. Furthermore, we have [ a"ß -9= >ß =38 >b œ " . 12. Substitution verifies that the functions are solutions of the ODE. Furthermore, we have [ a"ß >ß -9= >ß =38 >b œ " . 14. Substitution verifies that the functions are solutions of the ODE. Furthermore, we have [ a"ß >ß /> ß > /> b œ /#> . 15. Substitution verifies that the functions are solutions of the ODE. Furthermore, we have [ a"ß Bß B$ b œ 'B . 16. Substitution verifies that the functions are solutions of the ODE. Furthermore, we have [ aBß B# ß "ÎBb œ 'ÎB . 18. The operation of taking a derivative is linear, and hence a5 b a-" C" -# C# ba5 b œ -" C" -# C# . a5 b It follows that Rearranging the terms, we obtain Pc-" C" -# C# d œ -" PcC" d -# PcC# dÞ Since C" and C# are solutions, Pc-" C" -# C# d œ ! . The rest follows by induction. 19a+b. Note that . 5 a>8 bÎ.>5 œ 8a8 "bâa8 5 "b>85 , for 5 œ "ß #ß âß 8 . Hence Pc>8 d œ +! 8x +" c8a8 "bâ#d> â +8" 8 >8" +8 >8 Þ P/<> ‘ œ +! <8 /<> +" <8" /<> â +8" < /<> +8 /<> œ +! <8 +" <8" â +8" < +8 ‘/<> Þ a a a a Pc-" C" -# C# d œ -" C"8b -# C#8b :" -" C"8"b -# C#8"b ‘ â :8 c-" C" -# C# dÞ a,b. We have . 5 a/<> bÎ.>5 œ <5 /<> , for 5 œ !ß "ß #ß â . Hence a- b. Set C œ /<> , and substitute into the ODE. It follows that <% &<# % œ ! , with < œ „ "ß „ # . Furthermore, [ a/> ß /> ß /#> ß /#> b œ (# . 20a+b. Let 0 a>b and 1a>b be arbitrary functions. Then [ a0 ß 1b œ 0 1 w 0 w 1 . Hence [ w a0 ß 1b œ 0 w 1 w 0 1 ww 0 ww 1 0 w 1 w œ 0 1 ww 0 ww 1 . That is, [ w a0 ß 1 b œ º 0 0 ww 1 Þ 1 ww º Now expand the $-by-$ determinant as ________________________________________________________________________ page 147 —————————————————————————— CHAPTER 4. —— [ aC" ß C# ß C$ b œ C" º Differentiating, we obtain [ w aC" ß C# ß C$ b œ C"w º C#w C#ww C#w C" º www C# w w C$w C$w C#w w C" w C" ww º C# º ww ww º C$ º ww C$ C" C $ C" C#ww º C$w C"w C$w C"w C#w Þ www º C# º www www º C$ º www C$ C" C$ C" C#www º C#w C#ww C$w Cw C# º " C$ww º C"ww C$w Cw C$ º " C$ww º C"ww C#w Þ C#ww º The second line follows from the observation above. Now we find that âw ââ â C# C$ â â C" C#w C$w â â C" â ââ â [ w aC" ß C# ß C$ b œ â C"w C#w C$w â â C"w C#w C$w âÞ â ww â â www â C#www C3www â â C" C#ww C3ww â â C" Hence the assertion is true, since the first determinant is equal to zero. a,b. Based on the properties of determinants, â â :$ C" â w :# a>b:$ a>b[ œ â :# C"w â www â C" â : $ C$ â â :# C$w â â C3www â : $ C# :# C#w C#www Adding the first two rows to the third row does not change the value of the determinant. Since the functions are assumed to be solutions of the given ODE, addition of the rows results in â â : $ C# : $ C$ â â :$ C" â â :# C#w :# C$w âÞ :# a>b:$ a>b[ w œ â :# C"w â â â :" C"ww :" C#ww :" C3ww â a- b. The first order equation [ w œ :" a>b[ is linear, with integrating factor .a>b œ œ /B:ˆ' :" a>b.>‰ . Hence [ a>b œ - /B:ˆ ' :" a>b.>‰ . Furthermore, [ a>b is zero only if - œ ! . a. b. It can be shown, by mathematical induction, that âC C# â â" â Cw C#w â â" âã w [ aC" ß C# ß âß C8 b œ â â a82b a C#82b â â C" â a8b a â C" C#8b â â â â â â âÞ â a C882b â â a C88b â C8 w C8 ã It follows that :# a>b:$ a>b[ w œ :" a>b:# a>b:$ a>b[ . As long as the coefficients are not zero, we obtain [ w œ :" a>b[ . C8" w C8" a8 C8"2b a8b C8" ________________________________________________________________________ page 148 —————————————————————————— CHAPTER 4. —— Based on the reasoning in Parta,b, it follows that and hence [ w œ :" a>b[ . :# a>b:$ a>bâ:8 a>b[ w œ :" a>b:# a>b:$ a>bâ:8 a>b[ , 22. Inspection of the coefficients reveals that :" a>b œ ! . Based on Prob. #!, we find that [ w œ ! , and hence [ œ - . 23. After writing the equation in standard form, observe that :" a>b œ #Î> . Based on the results in Prob. #!, we find that [ w œ a #Î>b[ , and hence [ œ -Î># . 24. Writing the equation in standard form, we find that :" a>b œ "Î>. Using Abel's formula, the Wronskian has the form [ a>b œ - /B:ˆ ' " .>‰ œ -Î> . > 25a+b. Assuming that -" C" a>b -# C# a>b â -8 C8 a>b œ ! , then taking the first 8 " derivatives of this equation results in a -" C" a>b -# C# a>b â -8 C85 b a>b œ ! a5 b a5 b for 5 œ !ß "ß âß 8 " . Setting > œ >! , we obtain a system of 8 algebraic equations with unknowns -" ß -# ß âß -8 . The Wronskian, [ aC" ß C# ß âß C8 ba>! b, is the determinant of the coefficient matrix. Since system of equations is homogeneous, [ aC" ß C# ß âß C8 ba>! b Á ! implies that the only solution is the trivial solution, -" œ -# œ â œ -8 œ ! . a,b. Suppose that [ aC" ß C# ß âß C8 ba>! b œ ! for some >! . Consider the system of algebraic equations a -" C" a>! b -# C# a>! b â -8 C85 b a>! b œ ! , a5 b a5 b 5 œ !ß "ß âß 8 " , with unknowns -" ß -# ß âß -8 . Vanishing of the Wronskian, which is the determinant of the coefficient matrix, implies that there is a nontrivial solution of the system of homogeneous equations. That is, there exist constants -" ß -# ß âß -8 , not all zero, which satisfy the above equations. Now let Since the ODE is linear, Ca>b is also a nonzero solution. Based on the system of algebraic equations above, Ca>! b œ C w a>! b œ â œ C a8"b a>! b œ ! . This contradicts the uniqueness of the identically zero solution. 26. Let Ca>b œ C" a>b@a>b. Then C w œ C"w @ C" @ w , C ww œ C"ww @ #C"w @ w C" @ ww , and C www œ C"www @ $C"ww @ w $C"w @ ww C" @ www Þ Substitution into the ODE results in Ca>b œ -" C" a>b -# C# a>b â -8 C8 a>b. Since C" is assumed to be a solution, all terms containing the factor @a>b vanish. Hence ________________________________________________________________________ page 149 www ww w ww w w C" @ $C" @ w $C" @ ww C" @ www :" cC" @ #C" @ w C" @ ww d :# cC" @ C"@ w d :$C"@ œ !. —————————————————————————— CHAPTER 4. —— C1 @ www c:" C1 $C1w d@ ww c$C1ww #:1 C1w :2 C1 d@ w œ ! , which is a second order ODE in the variable ? œ @ w . 28. First write the equation in standard form: C www $ > # ww >" ' C ' # Cw # C œ !. >a > $ b > a> $ b > a> $b >a> %b ww @ œ !. >$ Let Ca>b œ ># @a>bÞ Substitution into the given ODE results in ># @ www $ Set A œ @ ww . Then A is a solution of the first order differential equation Aw $ >% A œ !Þ >a> $b This equation is linear, with integrating factor .a>b œ >% Îa> $b. The general solution is A œ - a> $bÎ>% . Integrating twice, it follows that @a>b œ -" >" -" ># -# > -$ Þ Hence Ca>b œ -" > -" -# >$ -$ ># Þ Finally, since C" a>b œ ># and C# a>b œ >$ are given solutions, the third independent solution is C$ a>b œ -" > -" Þ ________________________________________________________________________ page 150 —————————————————————————— CHAPTER 4. —— Section 4.2 1. The magnitude of " 3 is V œ È# and the polar angle is 1Î% . Hence the polar form is given by " 3 œ È# /31Î% Þ 3. The magnitude of $ is V œ $ and the polar angle is 1 Þ Hence $ œ $ /31 Þ 4. The magnitude of 3 is V œ " and the polar angle is $1Î# Þ Hence 3 œ /$13Î# Þ 5. The magnitude of È$ 3 is V œ # and the polar angle is 1Î' œ ""1Î' . Hence the polar form is given by È$ 3 œ # /""13Î' Þ 6. The magnitude of " 3 is V œ È# and the polar angle is &1Î% . Hence the polar form is given by " 3 œ È# /&13Î% Þ 7. Writing the complex number in polar form, " œ /#713 , where 7 may be any integer. Thus ""Î$ œ /#713Î$ Þ Setting 7 œ !ß "ß # successively, we obtain the three roots as ""Î$ œ " , ""Î$ œ /#13Î$ , ""Î$ œ /%13Î$ . Equivalently, the roots can also be written as " , -9=a#1Î$b 3 =38a#1Î$b œ " Š " È$ ‹, -9=a%1Î$b 3=38a%1Î$b œ " Š " È$ ‹. # # 9. Writing the complex number in polar form, " œ /#713 , where 7 may be any integer. Thus ""Î% œ /#713Î% Þ Setting 7 œ !ß "ß #ß $ successively, we obtain the three roots as ""Î% œ " , ""Î% œ /13Î# , ""Î% œ /13 , ""Î% œ /$13Î# . Equivalently, the roots can also be written as " , -9=a1Î#b 3 =38a1Î#b œ 3, -9=a1b 3=38a1b œ " , -9=a$1Î#b 3=38a$1Î#b œ 3. 10. In polar form, #a-9= 1Î$ 3 =38 1Î$b œ # /31Î$#71 , in which 7 is any integer. Thus c#a-9= 1Î$ 3 =38 1Î$bd"Î# œ #"Î# /31Î'71 . With 7 œ ! , one square root is given by #"Î# /31Î' œ ŠÈ$ 3‹ÎÈ# . With 7 œ " , the other root is given by #"Î# /3(1Î' œ Š È$ 3‹ÎÈ# . 11. The characteristic equation is <$ <# < " œ !Þ The roots are < œ "ß "ß " . One root is repeated, hence the general solution is C œ -" /> -# /> -$ >/> Þ 13. The characteristic equation is <$ #<# < # œ ! , with roots < œ "ß "ß # Þ The roots are real and distinct, hence the general solution is C œ -" /> -# /> -$ /#> Þ 14. The characteristic equation can be written as <# a<# %< %b œ ! Þ The roots are < œ !ß !ß #ß # . There are two repeated roots, and hence the general solution is given by C œ -" -# > -$ /#> -% >/#> Þ 15. The characteristic equation is <' " œ ! . The roots are given by < œ a "b"Î' , that is, the six sixth roots of " . They are /13Î'713Î$ , 7 œ !ß "ß âß & . Explicitly, ________________________________________________________________________ page 151 —————————————————————————— CHAPTER 4. —— < œ ŠÈ$ 3‹Î# , ŠÈ$ 3‹Î# , 3 , 3 , Š È$ 3‹Î# , Š È$ 3‹Î# . Hence the general solution is given by C œ / $ >Î# c-" -9= a>Î#b -# =38 a>Î#bd -$ -9= > È -% =38 > / $ >Î# c-& -9= a>Î#b -' =38 a>Î#bdÞ È 16. The characteristic equation can be written as a<# "ba<# %b œ ! . The roots are given by < œ „ "ß „ # . The roots are real and distinct, hence the general solution is C œ -" /> -# /> -$ /#> -% /#> Þ 17. The characteristic equation can be written as a<# "b œ ! . The roots are given by < œ „ " , each with multiplicity three. Hence the general solution is $ C œ -" /> -# >/> -$ ># /> -% /> -& >/> -' ># /> Þ 18. The characteristic equation can be written as <# ˆ<% "‰ œ ! . The roots are given by < œ !ß !ß „ "ß „ 3 . The general solution is C œ -" -# > -$ /> -% /> -& -9= > -' =38 > . 19. The characteristic equation can be written as <ˆ<% $<$ $<# $< #‰ œ ! . Examining the coefficients, it follows that <% $<$ $<# $< # œ a< "ba< #b ‚ a<# "bÞ Hence the roots are < œ !ß "ß #ß „ 3 . The general solution of the ODE is given by C œ -" -# /> -$ /#> -% -9= > -& =38 > . 20. The characteristic equation can be written as <a<$ )b œ ! , with roots < œ ! , # /#713Î$ , 7 œ !ß "ß # . That is, < œ !ß #ß " „ 3È$ . Hence the general solution is C œ -" -# /#> /> ’-$ -9=È$ > -% =38È$ > “Þ 22. The characteristic equation can be written as a<# "b œ ! . The roots are given by < œ „ 3 , each with 7?6>3:63-3>C >A9Þ The general solution is C œ -" -9= > -# =38 > >c-$ -9= > -% =38 > dÞ # # 21. The characteristic equation can be written as ˆ<% %‰ œ ! . The roots of the equation <% % œ ! are < œ " „ 3 , " „ 3 . Each of these roots has 7?6>3:63-3>C >A9Þ The general solution is C œ /> c-" -9= > -# =38 > d >/> c-$ -9= > -% =38 > d /> c-& -9= > -' =38 > d >/> c-( -9= > -) =38 > d. 24. The characteristic equation is <$ &<# '< # œ !Þ Examining the coefficients, we find that <$ &<# '< # œ a< "ba<# %< #b. Hence the roots are deduced as È È < œ " , # „È# . The general solution is C œ -" /> -# /Š# #‹> -$ /Š# #‹> Þ 25. The characteristic equation is ")<$ #"<# "%< % œ !Þ By examining the first and last coefficients, we find that ")<$ #"<# "%< % œ a#< "ba*<# '< %b. ________________________________________________________________________ page 152 —————————————————————————— CHAPTER 4. —— Hence the roots are < œ "Î# , Š "„È$ ‹Î$ . The general solution of the ODE is given by C œ -" />Î# />Î$ ’-# -9=Š>ÎÈ$ ‹ -$ =38Š>ÎÈ$ ‹ “Þ 26. The characteristic equation is <% (<$ '<# $!< $' œ !Þ By examining the first and last coefficients, we find that The roots are < œ #ß $ß $ „È$ . The general solution is C œ -" /#> -# /$> -$ /Š$ È$‹> <% (<$ '<# $!< $' œ a< $ba< #bˆ<# '< '‰. -% /Š$ È$‹> Þ 28. The characteristic equation is <% '<$ "(<# ##< "% œ !Þ It can be shown that <% '<$ "(<# ##< "% œ a<# #< #ba<# %< (b. Hence the roots are < œ " „ 3 , # „ 3È$ . The general solution is C œ /> c-" -9= > -# =38 > d /#> ’-$ -9=È$ > -% =38È$ > “Þ È# 30. Ca>b œ " />Î # =38Š>ÎÈ#‹ " />Î # È# =38Š>ÎÈ#‹. 32. The characteristic equation is <$ <# < " œ ! , with roots < œ " , „ 3 . Hence the general solution is Ca>b œ -" /> -# -9= > -$ =38 > . Invoking the initial conditions, we obtain the system of equations -" - # œ # -" - $ œ " -" - # œ # with solution -" œ ! , -# œ # , -$ œ " . Therefore the solution of the initial value problem is Ca>b œ #-9= > =38 > . ________________________________________________________________________ page 153 —————————————————————————— CHAPTER 4. —— 33. The characteristic equation is #<% <$ *<# %< % œ ! , with roots < œ "Î# , " , „ # . Hence the general solution is Ca>b œ -" />Î# -# /> -$ /#> -% /#> . Applying the initial conditions, we obtain the system of equations -" -# - $ - % œ # " -" -# #-$ #-% œ ! # " -" -# %-$ %-% œ # % " -" -# )-$ )-% œ ! ) with solution -" œ "'Î"& , -# œ #Î$ , -$ œ "Î' , -% œ "Î"! . Therefore the " solution of the initial value problem is Ca>b œ "' />Î# # /> " /#> "! /#> . "& $ ' The solution decreases without bound. 34. Ca>b œ # > "$ / #% />Î# "$ -9= > $ ‘ "$ =38 > . ________________________________________________________________________ page 154 —————————————————————————— CHAPTER 4. —— The solution is an oscillation with increasing amplitude. 35. The characteristic equation is ' <$ &<# < œ ! , with roots < œ ! , "Î$ , "Î# . The general solution is Ca>b œ -" -# />Î$ -$ />Î# Þ Invoking the initial conditions, we require that -" - # - $ œ # " " -# - $ œ # $ # " " -# - $ œ ! * % with solution -" œ ) , -# œ ") , -$ œ ) . Therefore the solution of the initial value problem is Ca>b œ ) ")/>Î$ )/>Î# . 36. The general solution is derived in Prob.a#)b as Ca>b œ /> c-" -9= > -# =38 > d /#> ’-$ -9=È$ > -% =38È$ > “Þ Invoking the initial conditions, we obtain the system of equations ________________________________________________________________________ page 155 —————————————————————————— CHAPTER 4. —— with solution -" œ #"Î"$ , -# œ $)Î"$ , -$ œ )Î"$ , -% œ "(È$ Î$* . #-" #-# "!-$ *È$ -% œ $ -" - $ œ " -" -# #-$ È$ -% œ # #-# -$ %È$ -% œ ! The solution is a rapidly-decaying oscillation. 38. [ ˆ/> ß /> ß -9= >ß =38 >‰ œ ) [ a-9=2 >ß =382 >ß -9= >ß =38 >b œ % 40. Suppose that -" /<" > -# /<# > â -8 /<8 > œ ! , and each of the <5 are real and different. Multiplying this equation by /<" > , -" -# /a<# <" b> â -8 /a<8 <" b> œ ! . Differentiation results in Now multiplying the latter equation by /a<# <" b> , and differentiating, we obtain -# a<# <" b/a<# <" b> â -8 a<8 <" b/a<8 <" b> œ ! . -$ a<$ <# ba<$ <" b/a<$ <# b> â -8 a<8 <# ba<8 <" b/a<8 <# b> œ ! . -8 a<8 <8" bâa<8 <" b/a<8 <8" b> œ ! . -" /<" > -# /<# > â -8" /<8" > œ ! , _ > _ . Following the above steps in a similar manner, it follows that Since these equations hold for all >, and all the <5 are different, we have -8 œ ! . Hence The same procedure can now be repeated, successively, to show that -" œ - # œ â œ - 8 œ ! . ________________________________________________________________________ page 156 —————————————————————————— CHAPTER 4. —— Section 4.3 2. The general solution of the homogeneous equation is C- œ -" /> -# /> -$ -9= > -% =38 > . Let 1" a>b œ $> and 1# a>b œ -9= >. By inspection, we find that ]" a>b œ $>. Since 1# a>b is a solution of the homogeneous equation, set ]# a>b œ >aE-9= > F=38 >bÞ Substitution into the given ODE and comparing the coefficients of similar term results in E œ ! and F œ "Î% . Hence the general solution of the nonhomogeneous problem is > Ca>b œ C- a>b $> =38 > . % 3. The characteristic equation corresponding to the homogeneous problem can be written as a< "ba<# "b œ ! . The solution of the homogeneous equation is C- œ -" /> -# -9= > -$ =38 > . Let 1" a>b œ /> and 1# a>b œ %>. Since 11 a>b is a solution of the homogeneous equation, set ]1 a>b œ E>/> Þ Substitution into the ODE results in E œ "Î#. Now let ]# a>b œ F> G . We find that F œ G œ % . Hence the general solution of the nonhomogeneous problem is Ca>b œ C- a>b >/> Î# %a> "b . 4. The characteristic equation corresponding to the homogeneous problem can be written as <a< "ba< "b œ ! . The solution of the homogeneous equation is C- œ -" -# /> -$ /> . Since 1a>b œ # =38 > is not a solution of the homogeneous problem, we can set ] a>b œ E -9= > F =38 > . Substitution into the ODE results in E œ " and F œ ! . Thus the general solution is Ca>b œ -" -# /> -$ /> -9= > . 6. The characteristic equation corresponding to the homogeneous problem can be written as a<# "b# œ ! . It follows that C- œ -" -9= > -# =38 > >a-$ -9= > -% =38 >bÞ Since 1a>b is not a solution of the homogeneous problem, set ] a>b œ E F-9= #> G=38 #> . Substitution into the ODE results in E œ $, F œ "Î*, G œ ! . Thus the general solution is Ca>b œ C- a>b $ " -9= #> . * 7. The characteristic equation corresponding to the homogeneous problem can be written as <$ a<$ "b œ ! . Thus the homogeneous solution is Note the 1a>b œ > is a solution of the homogenous problem. Consider a particular solution of the form ] a>b œ >$ aE> F bÞ Substitution into the ODE results in E œ "Î#% and F œ !. Thus the general solution is Ca>b œ C- a>b >% Î#% . 8. The characteristic equation corresponding to the homogeneous problem can be written as <$ a< "b œ !. Hence the homogeneous solution is C- œ -" -# > -$ ># -% /> . Since 1a>b is not a solution of the homogeneous problem, set ] a>b œ E-9= #> F=38 #> . Substitution into the ODE results in E œ "Î%! and F œ "Î#! . Thus the general solution ________________________________________________________________________ page 157 C- œ -" -# > -$ ># -% /> />Î# ’-& -9=ŠÈ$ >Î#‹ -& =38ŠÈ$ >Î#‹“Þ —————————————————————————— CHAPTER 4. —— is Ca>b œ C- a>b a-9= #> #=38 #>bÎ%! . 10. From Prob. ## in Section %Þ# , the homogeneous solution is C- œ -" -9= > -# =38 > >c-$ -9= > -% =38 > dÞ Since 1a>b is not a solution of the homogeneous problem, substitute ] a>b œ E> F into the ODE to obtain E œ $ and F œ %. Thus the general solution is C a>b œ C- a>b $> %. Invoking the initial conditions, we find that -" œ % , -# œ % , -$ œ " , -% œ $Î# . Therefore the solution of the initial value problem is Ca>b œ a> %b-9= > a$>Î# %b=38 > $> % . 11. The characteristic equation can be written as < a<# $< #b œ !. Hence the homogeneous solution is C- œ -" -# /> -$ /#> . Let 1" a>b œ /> and 1# a>b œ >. Note that 1" is a solution of the homogeneous problem. Set ]" a>b œ E>/> . Substitution into the ODE results in E œ ". Now let ]# a>b œ F># G>. Substitution into the ODE results in F œ "Î% and G œ $Î% . Therefore the general solution is Ca>b œ -" -# /> -$ /#> >/> ˆ># $>‰Î% . Invoking the initial conditions, we find that -" œ " , -# œ -$ œ ! . The solution of the initial value problem is Ca>b œ " >/> a># $>bÎ% . ________________________________________________________________________ page 158 —————————————————————————— CHAPTER 4. —— 12. The characteristic equation can be written as a< "ba< $ba<# %b œ !. Hence the homogeneous solution is C- œ -" /> -# /$> -$ -9= #> -% =38 #>. None of the terms in 1a>b is a solution of the homogeneous problem. Therefore we can assume a form ] a>b œ E/> F-9= > G=38 > Þ Substitution into the ODE results in E œ "Î#! , F œ #Î& , G œ %Î& . Hence the general solution is Ca>b œ -" /> -# /$> -$ -9= #> -% =38 #> /> Î#! a#-9= > %=38 >bÎ& . Invoking the initial conditions, we find that -" œ )"Î%! , -# œ ($Î&#! , -$ œ ((Î'& , -% œ %*Î"$! . 14. From Prob. %, the homogeneous solution is C- œ -" -# /> -$ /> . Consider the terms 1" a>b œ >/> and 1# a>b œ #-9= >. Note that since < œ " is a simple root of the characteristic equation, Table %Þ$Þ" suggests that we set ]" a>b œ >aE> F b/> Þ The function #-9= > is not a solution of the homogeneous equation. We can simply choose ]# a>b œ G-9= > H=38 > . Hence the particular solution has the form ] a>b œ >aE> F b/> G-9= > H=38 > . # 15. The characteristic equation can be written as a<# "b œ !. The roots are given ________________________________________________________________________ page 159 —————————————————————————— CHAPTER 4. —— as < œ „ ", each with multiplicity two. Hence the solution of the homogeneous problem is C- œ -" /> -# >/> -$ /> -% >/> . Let 1" a>b œ /> and 1# a>b œ =38 >. The function /> is a solution of the homogeneous problem. Since < œ " has multiplicity >A9, we set ]" a>b œ E># /> . The function =38 > is not a solution of the homogeneous equation. We can set ]# a>b œ F-9= > G=38 > . Hence the particular solution has the form ] a>b œ E># /> F-9= > G=38 > . 16. The characteristic equation can be written as <# a<# %b œ ! , with roots < œ !, „#3Þ The root < œ ! has multiplicity >A9, hence the homogeneous solution is C- œ -" -# > -$ -9= #> -% =38 #> . The functions 1" a>b œ =38 #> and 1# a>b œ % are solutions of the homogenous equation. The complex roots have multiplicity one, therefore we need to set ]" a>b œ E> -9= #> F> =38 #> . Now 1# a>b œ % is associated with the double root < œ !Þ Based on Table %Þ$Þ", set ]# a>b œ G># . Finally, 1$ a>b œ >/> aand its derivativesb is independent of the homogeneous solution. Therefore set ]$ a>b œ aH> I b/> . Conclude that the particular solution has the form ] a>b œ E> -9= #> F> =38 #> G># aH> I b/> Þ 18. The characteristic equation can be written as <# a<# #< #b œ ! , with roots < œ !, with multiplicity two, and < œ " „ 3 . The homogeneous solution is C- œ -" -# > -$ /> -9= > -% /> =38 > . The function 1" a>b œ $/> #>/> , and all of its derivatives, is independent of the homogeneous solution. Therefore set ]" a>b œ E/> aF> G b/> Þ Now 1# a>b œ /> =38 > is a solution of the homogeneous equation, associated with the complex roots. We need to set ]# a>b œ >aH /> -9= > I /> =38 >bÞ It follows that the particular solution has the form ] a>b œ E/> aF> G b/> >ˆH /> -9= > I /> =38 >‰Þ 19. Differentiating C œ ?a>b@a>b, successively, we have Setting @a>b œ /!> , @a4b œ !4 /!> . So for any : œ "ß #ß âß 8 , C It follows that a:b C w œ ? w @ ?@ w C ww œ ? ww @ #? w @ w ?@ ww ã 8 8 a8b "Œ ?a84b @a4b Cœ 4 4œ ! : œ / "Œ !4 ?a: 4b . 4 4œ ! : !> ________________________________________________________________________ page 160 —————————————————————————— CHAPTER 4. —— 8 : P/!> ?‘ œ /!> " –+ 8: "Œ !4 ?a: 4b — 4 :œ! 4œ ! : a‡bÞ It is evident that the right hand side of Eq. a‡b is of the form Hence operator equation Pc/!> ?d œ /!> a,! >7 ," >7" â , 7" > ,7 b can be written as 5! ?a8b 5" ?a8"b â 5 8" ? w 58 ? œ œ ,! >7 ," >7" â , 7" > ,7 Þ The coefficients 5 3 , 3 œ !ß "ß âß 8 can be determined by collecting the like terms in the double summation in Eq. a‡b. For example, 5! is the coefficient of ?a8b . The only term that contains ?a8b is when : œ 8 and 4 œ ! . Hence 5! œ +! . On the other hand, 58 is the coefficient of ?a>b. The inner summation in a‡b contains terms with ?, given by !: ? awhen 4 œ :b, for each : œ !ß "ß âß 8 . Hence 58 œ "+ 8: !: . 8 :œ! /!> 5! ?a8b 5" ?a8"b â 5 8" ? w 58 ?‘ Þ 21a+b. Clearly, /#> is a solution of C w #C œ ! , and >/> is a solution of the differential equation C ww #C w C œ ! . The latter ODE has characteristic equation a< "b# œ ! . Hence aH #bc$/#> d œ $aH #bc/#> d œ ! and aH "b# c>/> d œ ! . Furthermore, we have aH #baH "b# c>/> d œ aH #bc!d œ !, and aH #baH "b# c$/#> d œ œ aH "b# aH #bc$/#> d œ aH "b# c!d œ ! . a,b. Based on Part a+b, aH #baH "b# aH #b$ aH "b] ‘ œ aH #baH "b# $/#> >/> ‘ œ !, since the operators are linear. The implied operations are associative and commutative. Hence a H #b % a H "b $ ] œ ! Þ The operator equation corresponds to the solution of a linear homogeneous ODE with characteristic equation a< #b% a< "b$ œ ! . The roots are < œ # , with multiplicity % and < œ " , with multiplicity $ . It follows that the given homogeneous solution is ] a>b œ -" /#> -# >/#> -$ ># /#> -% >$ /#> -& /> -' >/> -( ># /> , which is a linear combination of seven independent solutions. ________________________________________________________________________ page 161 —————————————————————————— CHAPTER 4. —— 22a"&b. Observe that aH "bc/> d œ ! and aH# "bc=38 >d œ ! . Hence the operator L aHb œ aH "baH# "b is an annihilator of /> =38 > . The operator corresponding # to the left hand side of the given ODE is aH# "b Þ It follows that aH "b# aH "b$ ˆH# "‰] œ ! . The resulting ODE is homogeneous, with solution a>b œ -" /> -# >/> -$ /> -% >/> -& >$ /> -' -9= > -( =38 > . ] a>b œ -& >$ /> -' -9= > -( =38 > . After examining the homogeneous solution of Prob. "& , and eliminating duplicate terms, we have 22a"'b. We find that Hc%d œ ! , aH "b# c>/> d œ ! , and aH# %bc=38 #>d œ ! . The operator L aHb œ HaH "b# aH# %bis an annihilator of ># >/> =38 #>. The operator corresponding to the left hand side of the ODE is H# aH# %bÞ It follows that H$ aH "b# ˆH# %‰ ] œ ! . # The resulting ODE is homogeneous, with solution a>b œ -" -# > -$ ># -% /> -& >/> -' -9= #> -( =38 #> -) >-9= #> -* >=38 #> . ] a>b œ -$ ># -% /> -& >/> -) >-9= #> -* >=38 #> . After examining the homogeneous solution of Prob. "' , and eliminating duplicate terms, we have 22a")b. Observe that aH "bc/> d œ ! , aH "b# c>/> d œ ! Þ The function /> =38 > is a solution of a second order ODE with characteristic roots < œ " „ 3 . It follows that aH# #H #bc/> =38 >d œ ! . Therefore the operator L aHb œ aH "baH "b# ˆH# #H #‰ is an annihilator of $/> #>/> /> =38 > . The operator corresponding to the left hand side of the given ODE is H# aH# #H #bÞ It follows that H# aH "baH "b# ˆH# #H #‰ ] œ ! . # The resulting ODE is homogeneous, with solution a>b œ -" -# > -$ /> -% /> -& >/> /> a-' -9= > -( =38 >b >/> a-) -9= > -* =38 > b Þ After examining the homogeneous solution of Prob. ") , and eliminating duplicate terms, ________________________________________________________________________ page 162 —————————————————————————— CHAPTER 4. —— we have a>b œ -$ /> -% /> -& >/> >/> a-) -9= > -* =38 > b. ________________________________________________________________________ page 163 —————————————————————————— CHAPTER 4. —— Section 4.4 2. The characteristic equation is <a<# "b œ ! . Hence the homogeneous solution is C- a>b œ -" -# /> -$ /> . The Wronskian is evaluated as [ a"ß /> ß /> b œ # . Now compute the three determinants â â /> â â ! /> â â [" a>b œ â ! /> /> â œ # â â /> â â " /> â â" â [# a>b œ â ! â â! ! ! " â /> â â /> â œ /> â /> â â !â â ! â œ /> â "â The solution of the system of equations a"!b is ?"w a>b œ â â" â [$ a>b œ â ! â â! /> /> /> ?#w a>b œ ?$w a>b œ > [ # a >b œ >/> Î# [ a>b > [ $ a >b œ >/> Î# [ a>b > [ " a >b œ > [ a>b Hence ?" a>b œ ># Î# , ?2 a>b œ /> a> "bÎ# , ?$ a>b œ /> a> "bÎ# Þ The particular solution becomes ] a>b œ ># Î# a> "bÎ# a> "bÎ# œ ># Î# " . The constant is a solution of the homogeneous equation, therefore the general solution is Ca>b œ -" -# /> -$ /> ># Î# . 3. From Prob. "$ in Section %Þ#, C- a>b œ -" /> -# /> -$ /#> Þ The Wronskian is evaluated as [ a/> ß /> ß /#> b œ ' /#> . Now compute the three determinants â â â ! /> /#> â â â [" a>b œ â ! /> #/#> â œ /$> â â â " /> %/#> â ________________________________________________________________________ page 164 —————————————————————————— CHAPTER 4. —— â > â/ â [# a>b œ â /> â > â/ â ! /#> â â ! #/#> â œ $/> â " %/#> â /> /> /> â !â â !â œ # â "â Hence ?"w a>b œ /&> Î' , ?#w a>b œ /$> Î# , ?$w a>b œ /#> Î$ . Therefore the particular solution can be expressed as ] a>b œ /> /&> Î$!‘ /> /$> Î'‘ /#> /#> Î'‘ œ /%> Î$! Þ â > â/ â [$ a>b œ â /> â > â/ 6. From Prob. ## in Section %Þ#, C- a>b œ -" -9= > -# =38 > >c-$ -9= > -% =38 > dÞ The Wronskian is evaluated as [ a-9= >ß =38 >ß > -9= >ß > =38 >b œ % . Now compute the four auxiliary determinants â â â! =38 > > -9= > > =38 > â â â -9= > -9= > > =38 > =38 > > -9= > â â! [" a>b œ â â œ #=38 > #> -9= > #-9= > > =38 > â â ! =38 > #=38 > > -9= > â â â " -9= > $-9= > > =38 > $=38 > > -9= > â â â â â [# a>b œ â â â â -9= > ! =38 > ! -9= > ! =38 > " > -9= > -9= > > =38 > #=38 > > -9= > $-9= > > =38 > =38 > ! -9= > ! =38 > ! -9= > " =38 > -9= > =38 > -9= > â > =38 > â â =38 > > -9= > â â œ #> =38 > #-9= > #-9= > > =38 > â â $=38 > > -9= > â â â â â [$ a>b œ â â â â It follows that ?"w a>b œ c =38# > > =38 > -9= >dÎ# , ?#w a>b œ c>=38# > =38 > -9= >dÎ# , ?$w a>b œ =38 > -9= >Î# , ?%w a>b œ =38# >Î# . Hence ?" a>b œ $=38 > -9= > #> -9=# > >‘Î) â â â â [% a>b œ â â â â -9= > =38 > -9= > =38 > -9= > =38 > -9= > =38 > â > -9= > !â â -9= > > =38 > !â â œ #=38 > #=38 > > -9= > ! â â $-9= > > =38 > " â â â > =38 > â =38 > > -9= > â â œ #-9= > #-9= > > =38 > â â $=38 > > -9= > â ________________________________________________________________________ page 165 —————————————————————————— CHAPTER 4. —— ?# a>b œ =38# > #-9=# > #> =38 > -9= > ># ‘Î) ?% a>b œ c-9= > =38 > >dÎ% Therefore the particular solution can be expressed as ] a>b œ -9= > c?" a>bd =38 > c?# a>bd > -9= > c?$ a>bd > =38 > c?% a>bd œ =38 > $> -9= > ># =38 >‘Î) Þ Ca>b œ -" -9= > -# =38 > >c-$ -9= > -% =38 > d ># =38 > Î) Þ ?$ a>b œ =38# >Î% Note that only the last term is not a solution of the homogeneous equation. Hence the general solution is 8. Based on the results in Prob. # , C- a>b œ -" -# /> -$ /> . It was also shown that [ a"ß /> ß /> b œ # , with [" a>b œ # , [# a>b œ /> , [$ a>b œ /> . Therefore we have ?"w a>b œ -=- > , ?#w a>b œ /> -=- > Î# , ?$w a>b œ /> -=- > Î# . The particular solution can be expressed as ] a>b œ c?" a>bd /> c?# a>bd /> c?$ a>bdÞ More specifically, /> > = /> > = ] a>b œ 68k-=- a>b -9>a>bk ( / -=- a=b.= ( / -=- a=b.= # >! # >! œ 68k-=- a>b -9>a>bk ( -9=2 a> =b-=- a=b.= . > >! 9. Based on Prob. % , ?"w a>b œ =/- > , ?#w a>b œ " , ?$w a>b œ >+8 > . The particular solution can be expressed as ] a>b œ c?" a>bd -9= > c?# a>bd =38 > c?$ a>bdÞ That is, ] a>b œ 68k=/- a>b >+8a>bk > -9= > =38 > 68k-9=a>bk. Hence the general solution of the initial value problem is Ca>b œ -" -# -9= > -$ =38 > 68k=/- a>b >+8a>bk > -9= > =38 > 68k-9=a>bkÞ Ca>b œ #-9= > =38 > 68k=/- a>b >+8a>bk > -9= > =38 > 68k-9=a>bk Invoking the initial conditions, we require that -" -# œ # , -$ œ " , -# œ # . Therefore ________________________________________________________________________ page 166 —————————————————————————— CHAPTER 4. —— 10. From Prob. ' , Ca>b œ -" -9= > -# =38 > -$ > -9= > -% > =38 > ># =38 > Î) Þ In order to satisfy the initial conditions, we require that -" œ # , -# -$ œ ! , -" #-% œ " , $Î% -# $-$ œ " Þ Therefore Ca>b œ #-9= > (=38 > (> -9= > %> =38 > ># =38 > ‘Î) Þ 12. From Prob. ) , the general solution of the initial value problem is Ca>b œ -" -# /> -$ /> 68k-=- a>b -9>a>bk /> > = /> > = ( / -=- a=b.= ( / -=- a=b.=. # >! # >! In this case, >! œ 1Î# . Observe that C a1Î#b œ C- a1Î#b , C w a1Î#b œ C-w a1Î#b , and C ww a1Î#b œ C-ww a1Î#b . Therefore we obtain the system of equations -" -# /1Î# -$ /1Î# œ # -# /1Î# -$ /1Î# œ " -# /1Î# -$ /1Î# œ " Hence the solution of the initial value problem is ________________________________________________________________________ page 167 —————————————————————————— CHAPTER 4. —— Ca>b œ $ />1Î# 68k-=- a>b -9>a>bk ( -9=2 a> =b-=- a=b.= Þ > >! 13. First write the equation as C www B" C ww #B# C w #B$ C œ #B Þ The Wronskian is evaluated as [ aBß B# ß "ÎBb œ 'ÎB . Now compute the three determinants â â â ! B# "ÎB â â â [" aBb œ â ! #B "ÎB# â œ $ â â #ÎB$ â â" # â âB ! â [# aBb œ â " ! â â! " â "ÎB â â "ÎB# â œ #ÎB â #ÎB$ â Hence ?"w aBb œ B# , ?#w aBb œ #BÎ$ , ?$w aBb œ B% Î$ . Therefore the particular solution can be expressed as ] aBb œ B B$ Î$‘ B# B# Î$‘ œ B% Î"& Þ 15. The homogeneous solution is C- a>b œ -" -9= > -# =38 > -$ -9=2 > -% =382 > Þ The Wronskian is evaluated as [ a-9= >ß =38 >ß -9=2 >ß =382 >b œ % . Now the four additional determinants are given by [" a>b œ # =38 > , [# a>b œ # -9= > , [$ a>b œ # =382 > , [% a>b œ # -9=2 > . If follows that ?"w a>b œ 1a>b =38a>bÎ# , ?#w a>b œ 1a>b -9=a>bÎ# , ?$w a>b œ 1a>b =382a>bÎ# , ?%w a>b œ 1a>b -9=2a>bÎ# . Therefore the particular solution ________________________________________________________________________ page 168 â â â B B# ! â â â [$ aBb œ â " #B ! â œ B# â â â! # "â "& B Î"&‘ B —————————————————————————— CHAPTER 4. —— can be expressed as ] a>b œ -9=a>b > =38a>b > ( 1a=b =38a=b .= ( 1a=b -9=a=b .= # # >! >! -9=2a>b > =382a>b > ( 1a=b =382a=b .= ( 1a=b -9=2a=b .= Þ # # >! >! "> "> 1a=b =382a> =b .= ( 1a=b =38a> =b .= Þ ( # >! # >! Using the appropriate identities, the integrals can be combined to obtain ] a>b œ 17. First write the equation as C www $ B" C ww ' B# C w ' B$ C œ 1aBb Î B$ Þ It can be shown that C- aBb œ -" B -# B# -$ B$ is a solution of the homogeneous equation. The Wronskian of this fundamental set of solutions is [ aBß B# ß B$ b œ #B$ Þ The three additional determinants are given by [" aBb œ B% , [# aBb œ #B$ , [$ aBb œ B# Þ Hence ?"w aBb œ 1aBbÎ#B# , ?#w aBb œ 1aBb ÎB$ , ?$w aBb œ 1aBbÎ#B% . Therefore the particular solution can be expressed as ] aBb œ B( B B 1a>b 1a>b 1a>b .> B# ( .> B$ ( .> # $ % B! #> B! > B! #> " BB # B# B$ œ ( ” # $ % •1a>b.> Þ # B! > > > B ________________________________________________________________________ page 169 —————————————————————————— CHAPTER 5. —— Chapter Five Section 5.1 1. Apply the ratio test : lim ¸aB $b8" ¸ k a B $b 8 k Hence the series converges absolutely for kB $k " . The radius of convergence is 3 œ " . The series diverges for B œ # and B œ % , since the n-th term does not approach zero. 3. Applying the ratio test, k8x B#8# k B# œ lim œ !Þ 8 Ä _ ka8 "bx B#8 k 8Ä_ 8 " lim 8Ä_ œ lim kB $k œ kB $kÞ 8Ä_ The series converges absolutely for all values of B . Thus the radius of convergence is 3 œ _. 4. Apply the ratio test : k#8" B8" k œ lim #kBk œ #kBkÞ 8 Ä _ k#8 B8 k 8Ä_ lim Hence the series converges absolutely for #kBk, or kBk "Î# . The radius of convergence is 3 œ "Î# . The series diverges for B œ „"Î# , since the n-th term does not approach zero. 6. Applying the ratio test, 8 Ä _ k a8 lim ¸8aB B! b8" ¸ Hence the series converges absolutely for kaB B! bk " . The radius of convergence is 3 œ " . At B œ B! " , we obtain the harmonic series, which is divergent. At the other endpoint, B œ B! " , we obtain " _ "baB B! b k 8 œ lim 8 kaB B! bk œ kaB B! bkÞ 8Ä_ 8 " a "b 8 , 8 8œ" which is conditionally convergent. 7. Apply the ratio test : ________________________________________________________________________ page 169 —————————————————————————— CHAPTER 5. —— ¸$ 8 a8 "b# aB #b8" ¸ k$ 8" 8# aB #b8 k a 8 "b # " œ lim k a B # b k œ k aB # b k Þ # 8Ä_ $8 $ Hence the series converges absolutely for " kB #k ", or kB #k $ . The radius of $ convergence is 3 œ $ . At B œ & and B œ " , the series diverges, since the n-th term does not approach zero. 8. Applying the ratio test, lim k88 a8 "bx B8" k 88 " œ lim 8 kBk œ kBk, 8" 8 Ä _ ¸ a 8 "b 8 Ä _ a 8 "b / 8x B8 ¸ 88 8œ 8 Ä _ a 8 "b lim lim Œ" 8Ä_ " 8 " œ/ . 8 8Ä_ lim since Hence the series converges absolutely for kBk /. The radius of convergence is 3 œ / . At B œ „ /, the series diverges, since the n-th term does not approach zero. This follows from the fact that 8Ä_ lim 8x /8 œ ". 88 È # 1 8 10. We have 0 aBb œ /B , with 0 a8b aBb œ /B , for 8 œ "ß #ß â . Therefore 0 a8b a!b œ ". Hence the Taylor expansion about B! œ ! is /B œ " _ B8 . 8x 8œ! Applying the ratio test, k8xB8" k " lim œ lim kBk œ !. 8 Ä _ ka8 "bx B8 k 8Ä_ 8 " The radius of convergence is 3 œ _ . 11. We have 0 aBb œ B , with 0 w aBb œ " and 0 a8b aBb œ ! , for 8 œ #ß â . Clearly, 0 a"b œ " and 0 w a"b œ " , with all other derivatives equal to zero. Hence the Taylor expansion about B! œ " is B œ " aB "bÞ Since the series has only a finite number of terms, the converges absolutely for all B . 14. We have 0 aBb œ "Îa" Bb, 0 w aBb œ "Îa" Bb# , 0 ww aBb œ #Îa" Bb$ , â with 0 a8b aBb œ a "b8 8xÎa" Bb8" , for 8 " . It follows that 0 a8b a!b œ a "b8 8x ________________________________________________________________________ page 170 —————————————————————————— CHAPTER 5. —— for 8 ! . Hence the Taylor expansion about B! œ ! is _ " œ " a "b8 B8 . " B 8œ! Applying the ratio test, The series converges absolutely for kBk " , but diverges at B œ „ " . kB8" k œ lim kBk œ kBk. lim 8 Ä _ k B8 k 8Ä_ 15. We have 0 aBb œ "Îa" Bb, 0 w aBb œ "Îa" Bb# , 0 ww aBb œ #Îa" Bb$ , â with 0 a8b aBb œ 8xÎa" Bb8" , for 8 " . It follows that 0 a8b a!b œ 8x, for 8 ! . Hence the Taylor expansion about B! œ ! is _ " œ " B8 . " B 8œ! Applying the ratio test, The series converges absolutely for kBk " , but diverges at B œ „ " . kB8" k œ lim kBk œ kBk. 8 Ä _ k B8 k 8Ä_ lim 16. We have 0 aBb œ "Îa" Bb, 0 w aBb œ "Îa" Bb# , 0 ww aBb œ #Îa" Bb$ , â with 0 a8b aBb œ 8xÎa" Bb8" , for 8 " . It follows that 0 a8b a#b œ a "b8" 8x for 8 ! . Hence the Taylor expansion about B! œ # is _ " œ " a " b 8 aB # b 8 . "B 8œ! Applying the ratio test, lim The series converges absolutely for kB #k " , but diverges at B œ " and B œ $ . 17. Applying the ratio test, lim ka8 "bB8" k 8" œ lim kBk œ kBk. 8k 8Ä_ 8Ä_ k8B 8 8Ä_ ¸aB #b8" ¸ k a B #b 8 k œ lim kB #k œ kB #k. 8Ä_ The series converges absolutely for kBk " . Term-by-term differentiation results in ________________________________________________________________________ page 171 —————————————————————————— CHAPTER 5. —— C œ " 8# B8" œ " %B *B# "' B$ â _ w 8œ" C œ " 8# a8 "b B8# œ % ")B %)B# "!!B$ â _ ww 8œ# Shifting the indices, we can also write C w œ " a8 "b# B8 and C ww œ " a8 #b# a8 "b B8 Þ _ _ 8œ! 8œ! 20. Shifting the index in the second series, that is, setting 8 œ 5 " , " +5 B5" œ " +8" B8 . _ _ 8œ" 5œ! Hence " +5" B " +5 B _ _ 5 5œ! 5" 5œ! œ " +5" B " +5" B5 _ _ 5 œ +" " a+5" +5" bB5" Þ _ 5œ" 5œ! 5œ" 21. Shifting the index by # , that is, setting 7 œ 8 # , " 8a8 "b+8 B8# œ " a7 #ba7 "b+7# B7 _ _ 8œ# œ " a8 #ba8 "b+8# B8 Þ 8œ! 7œ! _ 22. Shift the index down by # , that is, set 7 œ 8 # . It follows that " +8 B8# œ " +7# B7 _ _ 8œ! œ " +8# B8 Þ 8œ# 7œ# _ 24. Clearly, ________________________________________________________________________ page 172 —————————————————————————— CHAPTER 5. —— ˆ" B# ‰" 8a8 "b+8 B8# œ " 8a8 "b+8 B8# " 8a8 "b+8 B8 Þ _ _ _ 8œ# 8œ# 8œ# Shifting the index in the first series, that is, setting 5 œ 8 # , " 8a8 "b+8 B8# œ " a5 #ba5 "b+5# B5 _ _ 5œ! _ 8œ# œ " a8 #ba8 "b+8# B8 Þ 8œ! Hence ˆ" B #‰ 8œ# " 8a8 "b+8 B _ 8# œ " a8 #ba8 "b+8# B " 8a8 "b+8 B8 . _ _ 8 8œ! 8œ# Note that when 8 œ ! and 8 œ " , the coefficients in the second series are zero. So that ˆ" B# ‰" 8a8 "b+8 B8# _ 8œ# œ " ca8 #ba8 "b+8# 8a8 "b+8 dB8 Þ _ 8œ! 26. Clearly, " 8+8 B8" B " +8 B8 œ " 8+8 B8" " +8 B8" Þ _ _ _ _ 8œ! 8œ" 8œ! 8œ" Shifting the index in the first series, that is, setting 5 œ 8 " , " 8+8 B8" œ " a5 "b+5" B5 Þ _ _ 5œ! 8œ" Shifting the index in the second series, that is, setting 5 œ 8 " , " +8 B _ 8" 8œ! œ " +5" B5 Þ _ 5œ" Combining the series, and starting the summation at 8 œ " , " 8+8 B8" B " +8 B8 œ +" " ca8 "b+8" +8" dB8 Þ _ _ _ 8œ! 8œ" 8œ" 27. We note that ________________________________________________________________________ page 173 —————————————————————————— CHAPTER 5. —— B " 8a8 "b+8 B8# " +8 B8 œ " 8a8 "b+8 B8" " +8 B8 Þ _ _ _ _ 8œ# 8œ! 8œ# 8œ! Shifting the index in the first series, that is, setting 5 œ 8 " , " 8a8 "b+8 B8" œ " 5 a5 "b+5" B5 _ _ 5œ" _ 5œ! 8œ# œ " 5 a5 "b+5" B5 , since the coefficient of the term associated with 5 œ ! is zero. Combining the series, B " 8a8 "b+8 B _ 8œ# 8# " +8 B œ " c8a8 "b+8" +8 dB8 Þ _ _ 8 8œ! 8œ! ________________________________________________________________________ page 174 —————————————————————————— CHAPTER 5. —— Section 5.2 1. Let C œ +! +" B +# B# â +8 B8 â . Then C ww œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ _ _ 8œ# 8œ! Substitution into the ODE results in " a8 #ba8 "b+8# B8 " +8 B8 œ ! _ _ 8œ! 8œ! or " ca8 #ba8 "b+8# +8 dB8 œ ! Þ _ 8œ! Equating all the coefficients to zero, a8 #ba8 "b+8# +8 œ ! , +8# œ +8 , a8 "ba8 #b 8 œ !ß "ß #ß â Þ We obtain the recurrence relation 8 œ !ß "ß #ß â Þ The subscripts differ by two, so for 5 œ "ß #ß â +#5# +#5% +! +#5 œ œ œâœ a#5 "b#5 a#5 $ba#5 #ba#5 "b#5 a#5 bx and +#5" œ Hence C œ +! " _ _ B#5 B#5" +" " Þ a#5 bx a#5 "bx 5œ! 5œ! +#5" +#5$ +" œ œâœ Þ #5 a#5 "b a#5 #ba#5 "b#5 a#5 "b a#5 "bx The linearly independent solutions are C" œ +! Œ" C# œ +" ŒB B# B% B' â œ +! -9=2 B #x %x 'x B$ B& B( â œ +" =382 B . $x &x (x ________________________________________________________________________ page 175 —————————————————————————— CHAPTER 5. —— 4. Let C œ +! +" B +# B# â +8 B8 â . Then C ww œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ _ _ 8œ# 8œ! Substitution into the ODE results in " a8 #ba8 "b+8# B8 5 # B# " +8 B8 œ ! . _ _ 8œ! 8œ! Rewriting the second summation, " a8 #ba8 "b+8# B8 " 5 # +8# B8 œ ! , _ _ 8œ# 8œ! that is, #+# $ † # +$ B " a8 #ba8 "b+8# 5 # +8# ‘B8 œ ! Þ _ 8œ# Setting the coefficients equal to zero, we have +# œ ! , +$ œ ! , and a8 #ba8 "b+8# 5 # +8# œ ! , for 8 œ #ß $ß %ß â Þ The recurrence relation can be written as +8# œ 5 # +8# , 8 œ #ß $ß %ß â . a8 #ba8 "b The indices differ by four, so +% , +) , +"# , â are defined by 5 # +! 5 # +% 5 # +) +% œ , +) œ , +"# œ , â. %†$ )†( "# † "" Similarly, +& , +* , +"$ , â are defined by 5 # +" 5 # +& 5 # +* +& œ , +* œ , +"$ œ , â. &†% *†) "$ † "# The remaining coefficients are zero. Therefore the general solution is 5# % 5% 5' ) C œ + ! ”" B B B"# â• %†$ )†(†%†$ "# † "" † ) † ( † % † $ # % 5 5 5' + " ”B B& B* B"$ â•. &†% *†)†&†% "$ † "# † * † ) † % † % Note that for the even coefficients, ________________________________________________________________________ page 176 —————————————————————————— CHAPTER 5. —— +%7 œ and for the odd coefficients, +%7" œ 5 # +%7% , 7 œ "ß #ß $ß â a%7 "b%7 5 # +%7$ , 7 œ "ß #ß $ß â . %7a%7 "b Hence the linearly independent solutions are C" aBb œ " " _ _ 7" a "b7" ˆ5 # B% ‰ $ † % † ( † ) â a%7 $ba%7 %b 7œ! 7" a "b7" ˆ5 # B% ‰ C# aBb œ B–" " Þ % † & † ) † * â a%7 %ba%7 &b — 7œ! 6. Let C œ +! +" B +# B# â +8 B8 â . Then C w œ " 8+8 B8" œ " a8 "b+8" B8 _ _ 8œ" 8œ! and C ww œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ _ _ 8œ# 8œ! Substitution into the ODE results in ˆ# B# ‰" a8 #ba8 "b+8# B8 B" a8 "b+8" B8 %" +8 B8 œ ! . _ _ _ 8œ! 8œ! 8œ! Before proceeding, write B# " a8 #ba8 "b+8# B8 œ " 8a8 "b+8 B8 _ _ 8œ! 8œ# and B" a8 "b+8" B8 œ " 8 +8 B8 Þ _ _ 8œ! 8œ" It follows that %+! %+# a$+" "#+$ bB " c#a8 #ba8 "b+8# 8a8 "b+8 8 +8 %+8 dB _ 8œ# 8 œ !. ________________________________________________________________________ page 177 —————————————————————————— CHAPTER 5. —— Equating the coefficients to zero, we find that +# œ +! , +$ œ +" Î% , and +8# œ 8# #8 % +8 , 8 œ !ß "ß #ß â . #a8 #ba8 "b a#5 b# %5 % +#5 #a#5 #ba#5 "b The indices differ by two, so for 5 œ !ß "ß #ß â +#5# œ and +#5$ a#5 "b# %5 # œ +#5" Þ #a#5 $ba#5 #b Hence the linearly independent solutions are B% B' C" aBb œ " B â ' $! # C# aBb œ B B$ (B& "*B( âÞ % "'! "*#! 7. Let C œ +! +" B +# B# â +8 B8 â . Then C œ " 8+8 B8" œ " a8 "b+8" B8 _ _ w 8œ" 8œ! and C ww œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ _ _ 8œ# 8œ! Substitution into the ODE results in " a8 #ba8 "b+8# B B" a8 "b+8" B #" +8 B8 œ ! . _ _ _ 8 8 8œ! 8œ! 8œ! First write B" a8 "b+8" B8 œ " 8 +8 B8 Þ _ _ 8œ! 8œ" We then obtain #+# #+! " ca8 #ba8 "b+8# 8 +8 #+8 dB8 œ ! Þ _ 8œ" ________________________________________________________________________ page 178 —————————————————————————— CHAPTER 5. —— It follows that +# œ +! and +8# œ +8 Îa8 "b , 8 œ !ß "ß #ß â . Note that the indices differ by two, so for 5 œ "ß #ß â +#5 and +#5" +#5# +#5% a "b 5 + ! œ œ œâœ #5 " a#5 $ba#5 "b " † $ † & âa#5 "b +#5" +#5$ a "b 5 + " œ œ œâœ Þ #5 a#5 #b#5 # † % † ' âa#5 b Hence the linearly independent solutions are _ B# B% B' a "b8 B#8 " C" aBb œ " âœ" " "†$ "†$†& " † $ † & âa#8 "b 8œ" _ B$ B& B( a "b8 B#8" " C# aBb œ B âœB Þ # #†% #†%†' # † % † ' âa#8b 8œ" 9. Let C œ +! +" B +# B# â +8 B8 â . Then C œ " 8+8 B _ w 8œ" 8" œ " a8 "b+8" B8 _ 8œ! and C ww œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ _ _ 8œ# 8œ! Substitution into the ODE results in ˆ" B# ‰" a8 #ba8 "b+8# B8 %B" a8 "b+8" B8 '" +8 B8 œ ! . _ _ _ 8œ! 8œ! 8œ! Before proceeding, write B# " a8 #ba8 "b+8# B8 œ " 8a8 "b+8 B8 _ _ 8œ! 8œ# and B" a8 "b+8" B8 œ " 8 +8 B8 Þ _ _ 8œ! 8œ" It follows that ________________________________________________________________________ page 179 —————————————————————————— CHAPTER 5. — '+! #+# a#+" '+$ bB " ca8 #ba8 "b+8# 8a8 "b+8 %8 +8 '+8 dB _ 8œ# 8 œ !. Setting the coefficients equal to zero, we obtain +# œ $+! , +$ œ +" Î$ , and +8# œ a8 #ba8 $b +8 , 8 œ !ß "ß #ß â . a8 "ba8 #b Observe that for 8 œ # and 8 œ $ , we obtain +% œ +& œ ! . Since the indices differ by two, we also have +8 œ ! for 8 % . Therefore the general solution is a polynomial C œ +! +" B $+! B# +" B$ Î$ . Hence the linearly independent solutions are C" aBb œ " $B# and C# aBb œ B B$ Î$ . 10. Let C œ +! +" B +# B# â +8 B8 â . Then C ww œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ _ _ 8œ# 8œ! Substitution into the ODE results in ˆ% B# ‰" a8 #ba8 "b+8# B8 #" +8 B8 œ ! Þ _ _ 8œ! 8œ! First write B# " a8 #ba8 "b+8# B8 œ " 8a8 "b+8 B8 . _ _ 8œ! 8œ# It follows that #+! )+# a#+" #%+$ bB " c%a8 #ba8 "b+8# 8a8 "b+8 #+8 dB8 œ ! . _ 8œ# We obtain +# œ +! Î% , +$ œ +" Î"# and %a8 #b+8# œ a8 #b+8 , 8 œ !ß "ß #ß â . Note that for 8 œ # , +% œ ! . Since the indices differ by two, we also have +#5 œ ! for 5 œ #ß $ß â . On the other hand, for 5 œ "ß #ß â , +#5" œ +" a#5 $b+#5" a#5 &ba#5 $b+#5$ œ œâœ 5 Þ # a#5 "ba#5 "b %a#5 "b % % a#5 "ba#5 "b Therefore the general solution is ________________________________________________________________________ page 180 —————————————————————————— CHAPTER 5. —— _ B# B#8" "8 C œ +! + " B + ! + " . % % a#8 "ba#8 "b 8œ" Hence the linearly independent solutions are C" aBb œ " B# Î% and C# aBb œ B _ B#8" B$ B& B( . âœB" 8 % a#8 "ba#8 "b "# #%! ##%! 8œ" 11. Let C œ +! +" B +# B# â +8 B8 â . Then C w œ " 8+8 B8" œ " a8 "b+8" B8 _ _ 8œ" 8œ! and C œ " 8a8 "b+8 B _ ww 8œ# 8# œ " a8 #ba8 "b+8# B8 Þ _ 8œ! Substitution into the ODE results in ˆ$ B# ‰" a8 #ba8 "b+8# B8 $B" a8 "b+8" B8 " +8 B8 œ ! . _ _ _ 8œ! 8œ! 8œ! Before proceeding, write B and B" a8 "b+8" B8 œ " 8 +8 B8 Þ _ _ 8œ! 8œ" _ #" 8œ! a8 #ba8 "b+8# B œ " 8a8 "b+8 B8 _ 8 8œ# It follows that 8 '+# +! a %+" ")+$ bB " c$a8 #ba8 "b+8# 8a8 "b+8 $8 +8 +8 dB œ !. _ 8œ# We obtain +# œ +! Î' , #+$ œ +" Î* , and $a8 #b+8# œ a8 "b+8 , 8 œ !ß "ß #ß â . The indices differ by two, so for 5 œ "ß #ß â +#5 œ and $ † & âa#5 "b +! a#5 "b+#5# a#5 $ba#5 "b+#5% œ œâœ 5 # a#5 #ba#5 b $a#5 b $ $ † # † % âa#5 b ________________________________________________________________________ page 181 —————————————————————————— CHAPTER 5. —— a#5 b+#5" a#5 #ba#5 b+#5$ # † % † ' âa#5 b +" œ# œâœ 5 Þ $a#5 "b $ a#5 "ba#5 "b $ † $ † & âa#5 "b +#5" œ Hence the linearly independent solutions are _ B# B% &B' $ † & âa#8 "b B#8 C" aBb œ " âœ"" 8 ' #% %$# $ † # † % âa#8b 8œ" C# aBb œ B _ #B$ )B& "'B( # † % † ' âa#8b B#8" âœB" 8 Þ * "$& *%& $ † $ † & âa#8 "b 8œ" 12. Let C œ +! +" B +# B# â +8 B8 â . Then C w œ " 8+8 B8" œ " a8 "b+8" B8 _ _ 8œ" 8œ! and C œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ _ _ ww 8œ# 8œ! Substitution into the ODE results in a" Bb" a8 #ba8 "b+8# B8 B" a8 "b+8" B8 " +8 B8 œ ! . _ _ _ 8œ! 8œ! 8œ! Before proceeding, write B " a8 #ba8 "b+8# B8 œ " a8 "b8 +8" B8 _ _ 8œ! 8œ" and B" a8 "b+8" B8 œ " 8 +8 B8 Þ _ _ 8œ! 8œ" It follows that #+# +! " ca8 #ba8 "b+8# a8 "b8 +8" 8 +8 +8 dB8 œ !. _ 8œ" We obtain +# œ +! Î# and a8 #ba8 "b+8# a8 "b8 +8" a8 "b+8 œ ! for 8 œ !ß "ß #ß â . Writing out the individual equations, ________________________________________________________________________ page 182 —————————————————————————— CHAPTER 5. —— The coefficients can be calculated successively as +$ œ +! Îa# † $b, +% œ +$ Î# +# Î"# œ +! Î#% , +& œ $+% Î& +$ Î"! œ +! Î"#! , â . We can now see that for 8 # , +8 is proportional to +! . In fact, for 8 # , +8 œ +! Îa8xb . Therefore the general solution is C œ +! + " B Hence the linearly independent solutions are C# aBb œ B and C" aBb œ " " _ $ † # +$ # † " + # % † $ +% $ † # + $ + # & † % +& % † $ + % # + $ ' † & +' & † % + & $ + % ã œ! œ! œ! œ! +! B# +! B$ + ! B% â. #x $x %x B8 Þ 8x 8œ# 13. Let C œ +! +" B +# B# â +8 B8 â . Then C œ " 8+8 B _ w 8œ" 8" œ " a8 "b+8" B8 _ 8œ! and C ww œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ _ _ 8œ# 8œ! Substitution into the ODE results in # " a8 #ba8 "b+8# B8 B" a8 "b+8" B8 $" +8 B8 œ ! . _ _ _ 8œ! 8œ! 8œ! First write B" a8 "b+8" B8 œ " 8 +8 B8 Þ _ _ 8œ! 8œ" We then obtain %+# $+! " c#a8 #ba8 "b+8# 8 +8 $+8 dB8 œ ! Þ _ 8œ" It follows that +# œ $+! Î% and #a8 #ba8 "b+8# a8 $b+8 œ ! ________________________________________________________________________ page 183 —————————————————————————— CHAPTER 5. —— for 8 œ !ß "ß #ß â . The indices differ by two, so for 5 œ "ß #ß â +#5 œ a#5 "b+#5# a#5 "ba#5 "b+#5% œ# œâ #a#5 "ba#5 b # a#5 $ba#5 #ba#5 "ba#5 b a "b5 $ † & âa#5 "b œ +! Þ #5 a#5 bx a#5 #b+#5" a#5 ba#5 #b+#5$ œ# œâ #a#5 ba#5 "b # a#5 #ba#5 "ba#5 ba#5 "b a "b5 % † ' âa#5 ba#5 #b œ +" Þ #5 a#5 "bx and +#5" œ Hence the linearly independent solutions are _ " " "( a "b8 % † ' âa#8 #b #8" C# aBb œ B B$ B& B âœB" B Þ $ #! #"! #8 a#8 "bx 8œ" _ $ & (' a "b8 $ † & âa#8 "b #8 C" aBb œ " B# B% B ✠" B % $# $)% #8 a#8bx 8œ! 15a+b. From Prob. # , we have C" aBb œ " _ Since +! œ C a!b and +" œ C w a!b , we have C aBb œ # C" aBb C# aBb . That is, " " " " CaBb œ # B B# B$ B% B& B' â . $ % "& #% :% œ # B B# B$ Î$ :& œ # B B# B$ Î$ B% Î% Þ B#8 #8 8x 8œ! and C# aBb œ " _ #8 8x B#8" . a#8 "bx 8œ! The four- and five-term polynomial approximations are ________________________________________________________________________ page 184 —————————————————————————— CHAPTER 5. —— a, b . a- b. The four-term approximation :% appears to be reasonably accurate awithin "!%b on the interval kBk !Þ( . 17a+b. From Prob. (, the linearly independent solutions are C" aBb œ " " _ a "b8 B#8 " † $ † & âa#8 "b 8œ" a "b8 B#8" Þ # † % † ' âa#8b 8œ" _ Since +! œ C a!b and +" œ C w a!b , we have C aBb œ % C" aBb C# aBb . That is, " % " % CaBb œ % B %B# B$ B% B& B' â . # $ ) "& " :% œ % B %B# B$ # " % :& œ % B %B# B$ B% Þ # $ C# aBb œ B " The four- and five-term polynomial approximations are ________________________________________________________________________ page 185 —————————————————————————— CHAPTER 5. —— a, b . a- b. The four-term approximation :% appears to be reasonably accurate awithin "!%b on the interval kBk !Þ& . 18a+b. From Prob. "# , we have C" aBb œ " " _ Since +! œ C a!b and +" œ C w a!b , we have C aBb œ $ C" aBb # C# aBb . That is, $ " " " "' CaBb œ $ #B B# B$ B% B& B â. # # ) %! #%! $ :% œ $ #B B# # $ :& œ $ #B B# # "$ B # "$ "% B BÞ # ) B8 8x 8œ# and C# aBb œ B . The four- and five-term polynomial approximations are ________________________________________________________________________ page 186 —————————————————————————— CHAPTER 5. —— a, b . a- b. The four-term approximation :% appears to be reasonably accurate awithin "!%b on the interval kBk !Þ* . 20. Two linearly independent solutions of Airy's equation aabout B! œ !b are C" aBb œ " " _ _ B$8 # † $ âa$8 "ba$8b 8œ" Applying the ratio test to the terms of C" aBb , lim C# aBb œ B " B$8" Þ $ † % âa$8ba$8 "b 8œ" Similarly, applying the ratio test to the terms of C# aBb , 8 Ä _ k$ k# † $ âa$8 "ba$8b B$8$ k " œ lim kBk$ œ ! Þ $8 k 8 Ä _ k# † $ âa$8 #ba$8 $b B 8 Ä _ a$8 "ba$8 #ba$8 $b ¸$ † % âa$8ba$8 "b B$8% ¸ † % âa$8 $ba$8 %b B$8" k " kBk$ œ ! Þ 8 Ä _ a$8 #ba$8 $ba$8 %b lim œ lim Hence both series converge absolutely for all B . 21. Let C œ +! +" B +# B# â +8 B8 â . Then C œ " 8+8 B _ w 8œ" 8" œ " a8 "b+8" B8 _ 8œ! and ________________________________________________________________________ page 187 —————————————————————————— CHAPTER 5. —— C œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ _ _ ww 8œ# 8œ! Substitution into the ODE results in " a8 #ba8 "b+8# B8 #B" a8 "b+8" B8 - " +8 B8 œ ! . _ _ _ 8œ! 8œ! 8œ! First write B" a8 "b+8" B8 œ " 8 +8 B8 Þ _ _ 8œ! 8œ" We then obtain #+# - +! " ca8 #ba8 "b+8# #8 +8 - +8 dB8 œ ! Þ _ 8œ" Setting the coefficients equal to zero, it follows that +8# œ a#8 -b +8 a8 "ba8 #b for 8 œ !ß "ß #ß â . Note that the indices differ by two, so for 5 œ "ß #ß â +#5 œ a%5 % -b+#5# a%5 ) -ba%5 % -b+#5% œ œâ a#5 "b#5 a#5 $ba#5 #ba#5 "b#5 - âa- %5 )ba- %5 %b œ a "b 5 +! Þ a#5 bx a%5 # -b+#5" a%5 ' -ba%5 # -b+#5$ œ œâ #5 a#5 "b a#5 #ba#5 "b#5 a#5 "b a- #b âa- %5 'ba- %5 #b œ a "b 5 +" Þ a#5 "bx - # -a- %b % -a- %ba- )b ' B B B â #x %x 'x and +#5" œ Hence the linearly independent solutions of the Hermite equation aabout B! œ !b are C" aBb œ " C# aBb œ B - # $ a- #ba- 'b & a- #ba- 'ba- "!b ( B B B âÞ $x &x (x a,b. Based on the recurrence relation ________________________________________________________________________ page 188 —————————————————————————— CHAPTER 5. —— a#8 -b +8 , a8 "ba8 #b +8# œ the series solution will terminate as long as - is a nonnegative even integer. If - œ #7, then one or the other of the solutions in Part a,b will contain at most 7Î# " terms. In particular, we obtain the polynomial solutions corresponding to - œ !ß #ß %ß 'ß )ß "! À -œ! -œ# -œ% -œ' -œ) - œ "! C" aBb œ " C# aBb œ B C" aBb œ " #B# C# aBb œ B #B$ Î$ C" aBb œ " %B# %B% Î$ C# aBb œ B %B$ Î$ %B& Î"& a- b. Observe that if - œ #8 , and +! œ +" œ " , then +#5 œ a "b5 #8 âa#8 %5 )ba#8 %5 %b a#5 bx and +#5" œ a "b5 for 5 œ "ß #ß â c8Î#d. It follows that the coefficient of B8 , in C" and C# , is +8 œ Û 5 %5 5x Ü a "b a#5"bx for 8 œ #5 " 5 a#8 #b âa#8 %5 'ba#8 %5 #b Þ a#5 "bx % a "b5 a#55xx for 8 œ #5 b Ú Then by definition, L8 aBb œ a#5 bx a#5 bx a "b5 #8 %5 5x C" aBb œ a "b5 5x C" aBb for 8 œ #5 a#5"bx # a#5"bx a "b5 #8 %5 5x C# aBb œ a "b5 C# aBb for 8 œ 5x #5 " Therefore the first six Hermite polynomials are L! aBb œ " L" aBb œ #B L# aBb œ %B# # L$ aBb œ )B) "#B L% aBb œ "'B% %)B# "# L& aBb œ $#B& "'!B$ "#!B 23. The series solution is given by ________________________________________________________________________ page 189 —————————————————————————— CHAPTER 5. —— " " " " CaBb œ " B# # B% $ B' % B) â Þ # # #x # $x # %x 24. The series solution is given by C aBb œ " B# B% B' B) â. ' $! "#! 25. The series solution is given by C aBb œ B B$ B& B( B* â. # #†% #†%†' #†%†'†) ________________________________________________________________________ page 190 —————————————————————————— CHAPTER 5. —— 26. The series solution is given by C aBb œ B B$ B& B( B* âÞ "# #%! ##%! "'"#) 27. The series solution is given by CaBb œ " B% B) B"# â. "# '(# ))(!% ________________________________________________________________________ page 191 —————————————————————————— CHAPTER 5. —— 28. Let C œ +! +" B +# B# â +8 B8 â . Then C œ " 8+8 B8" œ " a8 "b+8" B8 _ _ w 8œ" 8œ! and C ww œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ _ _ 8œ# 8œ! Substitution into the ODE results in a" Bb" a8 #ba8 "b+8# B8 B" a8 "b+8" B8 # " +8 B8 œ ! . _ _ _ 8œ! 8œ! 8œ! After appropriately shifting the indices, it follows that #+# #+! " ca8 #ba8 "b+8# a8 "b8 +8" 8 +8 # +8 dB8 œ !. _ 8œ" We find that +# œ +! and a8 #ba8 "b+8# a8 "b8 +8" a8 #b+8 œ ! $ † # +$ # † " + # + " % † $ +% $ † # + $ & † % +& % † $ + % + $ ' † & +' & † % + & # + % ã œ! œ! œ! œ! for 8 œ "ß #ß â . Writing out the individual equations, Since +! œ ! and +" œ " , the remaining coefficients satisfy the equations ________________________________________________________________________ page 192 —————————————————————————— CHAPTER 5. —— $ † # +$ " œ ! % † $ +% $ † # + $ œ ! & † % +& % † $ + % + $ œ ! ' † & +' & † % + & # + % œ ! ã That is, +$ œ "Î' , +% œ "Î"# , +& œ "Î#% , +' œ "Î%& , â . Hence the series solution of the initial value problem is " " " " "$ ( CaBb œ B B$ B% B& B' B âÞ ' "# #% %& "!!) ________________________________________________________________________ page 193 —————————————————————————— CHAPTER 5. —— Section 5.3 2. Let C œ 9aBb be a solution of the initial value problem. First note that C ww œ a=38 BbC w a-9= BbC Þ Differentiating twice, Given that 9a!b œ ! and 9 w a!b œ " , the first equation gives 9 ww a!b œ ! and the last two equations give 9 www a!b œ # and 9 3@ a!b œ ! . 3. Let C œ 9aBb be a solution of the initial value problem. First write C ww œ Differentiating twice, C www œ " B w $ 68 B C CÞ B# B# C www œ a=38 BbC ww #a-9= BbC w a=38 BbC C 3@ œ a=38 BbC www $a-9= BbC ww $a=38 BbC w a-9= BbC Þ " ˆB B# ‰C ww a$B 68 B B #bC w a$ ' 68 BbC ‘Þ B$ " ’ˆB# B$ ‰C www ˆ$B# 68 B #B# %B‰C ww B% a' )B "#B 68 BbC w a") 68 B "&bC “Þ C 3@ œ Given that 9a"b œ # and 9 w a"b œ ! , the first equation gives 9 ww a"b œ ! and the last two equations give 9 www a!b œ ' and 9 3@ a!b œ %# . 4. Let C œ 9aBb be a solution of the initial value problem. First note that C ww œ B# C w a=38 BbC Þ Differentiating twice, Given that 9a!b œ +! and 9 w a!b œ +" , the first equation gives 9 ww a!b œ ! and the last two equations give 9 www a!b œ +! and 9 3@ a!b œ %+" . C www œ B# C ww a#B =38 BbC w a-9= BbC C 3@ œ B# C www a%B =38 BbC ww a# #-9= BbC w a=38 BbC Þ 5. Clearly, :aBb œ % and ; aBb œ 'B are analytic for all B . Hence the series solutions converge everywhere. 7. The zeroes of T aBb œ " B$ are the three cube roots of " . They all lie on the unit circle in the complex plane. So for B! œ ! , 3738 œ " . For B! œ # , the nearest ________________________________________________________________________ page 194 —————————————————————————— CHAPTER 5. —— root is /31Î$ œ Š" 3È$ ‹Î# , hence 3738 œ È$ . 9a,b. :aBb œ B and ; aBb œ " are analytic for all B . a- b. :aBb œ B and ; aBb œ " are analytic for all B . a. b. :aBb œ ! and ; aBb œ 5B# are analytic for all B . a/b. The only root of T aBb œ " B is " . Hence 3738 œ " . a1b. :aBb œ B and ; aBb œ # are analytic for all B . a3bÞ The zeroes of T aBb œ " B# are „ 3 . Hence 3738 œ " . a4bÞ The zeroes of T aBb œ % B# are „# . Hence 3738 œ # . a5 bÞ The zeroes of T aBb œ $ B# are „È$ . Hence 3738 œ È$ . a6bÞ The only root of T aBb œ " B is " . Hence 3738 œ " . a7b. :aBb œ BÎ# and ; aBb œ $Î# are analytic for all B . a8b. :aBb œ a" BbÎ# and ; aBb œ $Î# are analytic for all B . 12. The Taylor series expansion of /B , about B! œ ! , is / œ" _ B 8. The only root of T aBb œ B is zero. Hence 3738 œ " . B8 Þ 8x 8œ! Let C œ +! +" B +# B# â +8 B8 â . Substituting into the ODE, –" _ _ _ B8 " a8 #ba8 "b+8# B8 — B " +8 B8 œ ! Þ 8 x —–8 œ ! 8œ! 8œ! First note that B " +8 B8 œ " +8" B8 œ +! B +" B# +# B$ â +8" B8 â . _ _ 8œ! 8œ" The coefficient of B8 in the product of the two series is -8 œ #+# " " " '+$ "#+% â a8 "b8 +8" a8 #ba8 "b+8# Þ a 8 "b x a8 #b x 8x Expanding the individual series, it follows that #+# a#+# '+$ bB a+# '+$ "#+% bB# a+# '+$ "#+% #!+& bB$ â + ! B + " B# + # B $ â œ ! Þ Setting the coefficients equal to zero, we obtain the system #+# œ !, #+# '+$ +! œ !, +# '+$ "#+% +" œ ! , +# '+$ "#+% #!+& +# œ ! , â . Hence the general solution is ________________________________________________________________________ page 195 —————————————————————————— CHAPTER 5. —— CaBb œ +! +" B +! B$ B% B& B' % a+! +" b a#+" +! b Œ +! #+" â. $ ' "# %! "#! C" aBb œ " C# aBb œ B We find that two linearly independent solutions are B$ B% B& â ' "# %! B% B& B' â "# #! '! Since :aBb œ ! and ; aBb œ B/B converge everywhere, 3 œ _ . 13. The Taylor series expansion of -9= B , about B! œ ! , is -9= B œ " _ a "b8 B#8 8œ! a#8b x Þ Let C œ +! +" B +# B# â +8 B8 â . Substituting into the ODE, 8 #8 " a "b B —– " a8 #ba8 "b+8# B8 — " 8+8 B8 #" +8 B8 œ ! Þ – a#8b x _ _ _ _ 8œ! 8œ! 8œ" 8œ! The coefficient of B8 in the product of the two series is -8 œ #+# ,8 '+$ ,8" "#+% ,8# â a8 "b8 +8" ," a8 #ba8 "b+8# ,! , #+# #+! " -8 B8 " a8 #b+8 B8 œ ! Þ _ _ 8œ" 8œ" in which - 9= B œ ,! ," B ,# B# â ,8 B8 â . It follows that Expanding the product of the series, it follows that #+# #+! '+$ B a +# "#+% bB# a $+$ #!+& bB$ â +1 B +$ B$ #+% B% â œ ! Þ Setting the coefficients equal to zero, +# +! œ ! , '+$ +" œ ! , +# "#+% œ ! , $+$ #!+& +$ œ ! , â Þ Hence the general solution is CaBb œ +! +" B +! B# +" B$ B% B& B' B( +! +" +! +" â. ' "# '! "#! &'! B% B' â "# "#! We find that two linearly independent solutions are C" aBb œ " B# ________________________________________________________________________ page 196 —————————————————————————— CHAPTER 5. —— C# aBb œ B The nearest zero of T aBb œ -9= B is at B œ „1Î# Þ Hence 3738 œ 1Î# Þ 14. The Taylor series expansion of 68a" Bb , about B! œ ! , is 68a" Bb œ " _ B$ B& B( â ' '! &'! a "b8" B8 8œ" 8 Þ Let C œ +! +" B +# B# â +8 B8 â . Substituting into the ODE, –" _ a "b8 B8 _ –" 8œ! a "b 8x 8 — " a8 #ba8 "b+8# B _ 8œ! 8" B8 8œ" 8 8 8 — " a8 "b+8" B B " +8 B œ ! Þ _ _ 8œ! 8œ! The first product is the series #+# a #+# '+$ bB a+# '+$ "#+% bB# a +# '+$ "#+% #!+& bB$ â . The second product is the series +" B a#+# +" Î#bB# a$+$ +# +" Î$bB$ a%+% $+$ Î# #+# Î$ +" Î%bB$ â Þ Combining the series and equating the coefficients to zero, we obtain #+# œ ! #+# '+$ +" +! œ ! "#+% '+$ $+# $+" Î# œ ! #!+& "#+% *+$ $+# +" Î$ œ ! ã Hence the general solution is B$ B% (B& B' & CaBb œ +! +" B a+! +" b a#+! +" b +" Œ +" + ! â. $ ' #% "#! "#! We find that two linearly independent solutions are C" aBb œ " C# aBb œ B B$ B% B' â ' "# "#! B$ B% (B& â ' #% "#! The coefficient :aBb œ /B 68a" Bb is analytic at B! œ ! , but its power series has a radius of convergence 3 œ " . ________________________________________________________________________ page 197 —————————————————————————— CHAPTER 5. —— 15. If C" œ B and C# œ B# are solutions, then substituting C# into the ODE results in Setting B œ ! , we find that T a!b œ ! . Similarly, substituting C" into the ODE results in Ua!b œ ! . Therefore T aBbÎUaBb and V aBbÎT aBb may not be analytic. If they were, Theorem $Þ#Þ" would guarantee that C" and C# were the only two solutions. But note that an arbitrary value of Ca!b cannot be a linear combination of C" a!b and C# a!bÞ Hence B! œ ! must be a singular point. 16. Let C œ +! +" B +# B# â +8 B8 â . Substituting into the ODE, " a8 "b+8" B8 " +8 B8 œ ! Þ _ _ 8œ! # T aBb #B UaBb B# V aBb œ ! . 8œ! That is, " ca8 "b+8" +8 dB8 œ ! Þ _ 8œ! Setting the coefficients equal to zero, we obtain +8" œ for 8 œ !ß "ß #ß â . It is easy to see that +8 œ +! Îa8 xb . Therefore the general solution is CaBb œ +! ”" B œ +! / B Þ B# B$ â• #x $x +8 8" The coefficient +! œ C a!b, which can be arbitrary. 17. Let C œ +! +" B +# B# â +8 B8 â . Substituting into the ODE, " a8 "b+8" B8 B " +8 B8 œ ! Þ _ _ 8œ! 8œ! That is, " a8 "b+8" B8 " +8" B8 œ ! Þ _ _ 8œ" 8œ! Combining the series, we have ________________________________________________________________________ page 198 —————————————————————————— CHAPTER 5. —— +" " ca8 "b+8" +8" d B8 œ ! Þ _ Setting the coefficient equal to zero, +" œ ! and +8" œ +8" Îa8 "b for 8 œ "ß #ß â. Note that the indices differ by two, so for 5 œ "ß #ß â +#5# +#5% +! +#5 œ œ œâœ a#5 b a#5 #ba#5 b # † % âa#5 b and +#5" œ ! Þ Hence the general solution is CaBb œ +! ”" B# B% B' B#8 # $ â 8 â• # # #x # $x # 8x œ +! /B:ˆB# Î#‰. 8œ" The coefficient +! œ C a!b, which can be arbitrary. _ 19. Let C œ +! +" B +# B# â +8 B8 â . Substituting into the ODE, a" Bb" a8 "b+8" B8 " +8 B8 œ ! Þ _ 8œ! 8œ! That is, " a8 "b+8" B8 " 8 +8 B8 " +8 B8 œ ! Þ _ _ _ 8œ" 8œ! 8œ! Combining the series, we have +" +! " ca8 "b+8" 8 +8 +8 d B8 œ ! Þ _ 8œ" Setting the coefficients equal to zero, +" œ +! and +8" œ +8 for 8 œ !ß "ß #ß â . Hence the general solution is CaBb œ +! " B B# B$ â B8 â‘ " œ +! Þ "B The coefficient +! œ C a!b, which can be arbitrary. 21. Let C œ +! +" B +# B# â +8 B8 â . Substituting into the ODE, ________________________________________________________________________ page 199 —————————————————————————— CHAPTER 5. —— " a8 "b+8" B8 B " +8 B8 œ " B Þ _ _ 8œ! 8œ! That is, " a8 "b+8" B8 " +8" B8 œ " B Þ _ _ 8œ" 8œ! Combining the series, and the nonhomogeneous terms, we have a+" "b a#+# +! "bB " ca8 "b+8" +8" d B8 œ ! Þ _ 8œ# Setting the coefficients equal to zero, we obtain +" œ " , #+# +! " œ ! , and +8# +8 œ , 8 œ $ß %ß â . 8 The indices differ by two, so for 5 œ #ß $ß â +#5 œ +#5# +#5% a "b5" +# a " b 5 a+ ! " b œ œâœ œ , % † ' âa#5 b # † % † ' âa#5 b a#5 b a#5 #ba#5 b +#5" +#5$ a "b 5 œ œâœ Þ $ † & âa#5 "b a#5 "b a#5 "ba#5 "b " +! # B$ B% B& B' B +! # +! $ â # $ # #x $ † & # $x and for 5 œ "ß #ß â +#5" œ œ Hence the general solution is CaBb œ +! B CaBb œ +! ”" Collecting the terms containing +! , B# B% B' # $ â• # # #x # $x # $ B B B% B& B' B( ”B # $ â •Þ # $ # #x $ † & # $x $ † & † ( B# B$ B% B& B' B( # $ â •Þ # $ # #x $ † & # $x $ † & † ( # Upon inspection, we find that CaBb œ +! /B:ˆ B# Î#‰ ”B Note that the given ODE is first order linear, with integrating factor .a>b œ /B Î# . The general solution is given by ________________________________________________________________________ page 200 —————————————————————————— CHAPTER 5. —— CaBb œ /B Î# ( /? Î# .? aC a!b "b/B Î# " Þ # B # # ! 23. If ! œ ! , then C" aBb œ " . If ! œ #8 , then +#7 œ ! for 7 8 " . As a result, C" aBb œ " " a "b7 8 7œ" #7 8a8 "bâa8 7 "ba#8 "ba#8 $bâa#8 #7 "b #7 BÞ a#7bx !œ! !œ# !œ% C# aBb œ B " a "b7 8 7œ" " " $B# " "!B# $& % $B If ! œ #8 " , then +#7" œ ! for 7 8 " . As a result, #7 8a8 "bâa8 7 "ba#8 $ba#8 &bâa#8 #7 "b #7" Þ B a#7 "bx !œ" !œ$ !œ& 24a+b. Based on Prob. #$, !œ! !œ# !œ% B B & B$ $ B "% B$ $ #" & &B " " $B# " "!B# T! aBb œ " $& % $B C" a"b œ " C" a"b œ # C" a"b œ ) $ Normalizing the polynomials, we obtain T# aBb œ " $# B ## $ "& $& T% aBb œ B# B% ) % ) !œ" !œ$ !œ& Similarly, B B & B$ $ B "% B$ $ C# a " b œ " C# a"b œ ) C# a"b œ "& # $ #" & &B ________________________________________________________________________ page 201 —————————————————————————— CHAPTER 5. —— T" aBb œ B a, b . $ & T$ aBb œ B B$ # # "& $& $ '$ & T& aBb œ B B B ) % ) a- b. T! aBb has no roots. T" aBb has one root at B œ ! . The zeros of T# aBb are at B œ „ "ÎÈ$ Þ The zeros of T$ aBb are B œ !, „È$Î& . The roots of T% aBb are given by B# œ Š"& #È$!‹Î$& , Š"& #È$!‹Î$& . The roots of T& aBb are given by B œ ! and B# œ Š$& #È(!‹Î'$ , Š$& #È(!‹Î'$ . 25. Observe that T8 a "b œ But T8 a"b œ " for all nonnegative integers 8. 27. We have ˆB# "‰8 œ " 8 œ a " b 8 T 8 a "b . a "b 8 a "b5 a#8 #5 bx " #8 5 xa8 5 bxa8 #5 bx 5œ! Ò8Î#Ó a "b85 8 x #5 B, 5 x a 8 5 bx 5œ! which is a polynomial of degree #8. Differentiating n times, in which the lower index is . œ c8Î#d " . Note that if 8 œ #7 ", then . œ 7 " . ________________________________________________________________________ page 202 8 .8 # a "b85 8 x ˆB "‰8 œ " a#5 ba#5 "bâa#5 8 "bB#58 , .B8 5 xa8 5 bx 5œ. —————————————————————————— CHAPTER 5. —— Now shift the index, by setting 5 œ 8 4. Hence a "b 4 8 x .8 # ˆB "‰8 œ " a#8 #4ba#8 #4 "bâa8 #4 "bB8#4 a8 4bx4 x .B8 4œ! c8Î#d œ 8x " c8Î#d a "b4 a#8 #4bx 8#4 B Þ a8 4bx4 xa8 #4bx 4œ! Based on Prob. #&, .8 # ˆB "‰8 œ 8x #8 T8 aBbÞ 8 .B 29. Since the 8 " polynomials T! , T" , â, T8 are linearly independent, and the degree of T5 is 5 , any polynomial, 0 , of degree 8 can be expressed as a linear combination 0 aBb œ " +5 T5 aBb . 8 5œ! Multiplying both sides by T7 and integrating, ( 0 aBbT7 aBb.B œ " +5 ( T5 aBbT7 aBb.B . " 8 " " 5œ! " Based on Prob. #), ( T5 aBbT7 aBb.B œ " " # $57 . #7 " Hence ( 0 aBbT7 aBb.B œ " " # +7 Þ #7 " ________________________________________________________________________ page 203 —————————————————————————— CHAPTER 5. —— Section 5.4 2. We see that T aBb œ ! when B œ ! and " . Since the three coefficients have no factors in common, both of these points are singular points. Near B œ !, BÄ! lim B :aBb œ lim B BÄ! B# a" #B Bb# % Bb# œ #Þ BÄ! lim B# ; aBb œ lim B# BÄ! B# a" œ %Þ The singular point B œ ! is regular. Considering B œ ", BÄ" lim aB "b:aBb œ lim aB "b BÄ" B# a" Bb# #B Þ The latter limit does not exist. Hence B œ " is an irregular singular point. 3. T aBb œ ! when B œ ! and " . Since the three coefficients have no common factors, both of these points are singular points. Near B œ !, BÄ! lim B :aBb œ lim B BÄ! B# Þ B# a" Bb B# a" B# œ "Þ Bb The limit does not exist, and so B œ ! is an irregular singular point. Considering B œ ", BÄ" lim aB "b:aBb œ lim aB "b BÄ" BÄ" lim aB "b# ; aBb œ lim aB "b# BÄ" $B œ !Þ B# a" Bb Hence B œ " is a regular singular point. 4. T aBb œ ! when B œ ! and „ " . Since the three coefficients have no common factors, both of these points are singular points. Near B œ !, BÄ! lim B :aBb œ lim B BÄ! B$ a" # Þ B# b The limit does not exist, and so B œ ! is an irregular singular point. Near B œ ", BÄ" lim aB "b:aBb œ lim aB "b BÄ" # œ "Þ B$ a" B# b ________________________________________________________________________ page 204 —————————————————————————— CHAPTER 5. —— lim aB "b# ; aBb œ lim aB "b# BÄ" BÄ" B$ a" # œ !Þ B# b Hence B œ " is a regular singular point. At B œ " , BÄ" lim aB "b:aBb œ lim aB "b BÄ" B$ a" # œ "Þ B# b # œ !Þ B# b BÄ" lim aB "b# ; aBb œ lim aB "b# BÄ" B$ a" Hence B œ " is a regular singular point. 6. The only singular point is at B œ ! . We find that B lim B :aBb œ lim B # œ " Þ BÄ! BÄ! B lim B# ; aBb œ lim B# BÄ! BÄ! B# / # œ / #Þ B# Hence B œ ! is a regular singular point. 7. The only singular point is at B œ $ . We find that BÄ$ lim aB $b:aBb œ lim aB $b BÄ$ #B œ 'Þ B$ " B# œ !Þ B$ BÄ$ lim aB $b# ; aBb œ lim aB $b# BÄ$ Hence B œ $ is a regular singular point. 8. Dividing the ODE by Ba" B# b , we find that $ :aBb œ " # and ; aBb œ . Ba" B# b Ba" Bb# a" Bb$ lim B :aBb œ lim B BÄ! The singular points are at B œ ! and „ " . For B œ !, BÄ! " œ "Þ Ba" B# b # œ !Þ BÄ! lim B# ; aBb œ lim B# BÄ! Ba" Bb# a" Bb$ Hence B œ ! is a regular singular point. For B œ ", ________________________________________________________________________ page 205 —————————————————————————— CHAPTER 5. —— lim aB "b:aBb œ lim aB "b BÄ" BÄ" " " œ Þ #b Ba" B # # # $ BÄ" lim aB "b# ; aBb œ lim aB "b# BÄ" Ba" Bb a" Bb œ " Þ % Hence B œ " is a regular singular point. For B œ ", BÄ" lim aB "b:aBb œ lim aB "b BÄ" " " œ Þ #b Ba" B # Ba" Bb# a" Bb$ # Þ BÄ" lim aB "b# ; aBb œ lim aB "b# BÄ" The latter limit does not exist. Hence B œ " is an irregular singular point. 9. Dividing the ODE by aB #b# aB "b, we find that :aBb œ a B #b $ # and ; aBb œ # . aB #baB "b a B #b # $ $ The singular points are at B œ # and " . For B œ #, BÄ# lim aB #b:aBb œ lim aB #b BÄ# Þ The limit does not exist. Hence B œ # is an irregular singular point. For B œ ", BÄ" lim aB "b:aBb œ lim aB "b BÄ" a B #b # œ !Þ BÄ" lim aB "b# ; aBb œ lim aB "b# BÄ" # œ !Þ aB #baB "b Hence B œ " is a regular singular point. 10. T aBb œ ! when B œ ! and $ . Since the three coefficients have no common factors, both of these points are singular points. Near B œ !, BÄ! lim B :aBb œ lim B BÄ! B" " œÞ Ba$ Bb $ # œ !Þ Ba$ Bb BÄ! lim B# ; aBb œ lim B# BÄ! Hence B œ ! is a regular singular point. For B œ $, ________________________________________________________________________ page 206 —————————————————————————— CHAPTER 5. —— lim aB $b:aBb œ lim aB $b BÄ$ BÄ$ B" % œ Þ Ba$ Bb $ # œ !Þ Ba$ Bb BÄ$ lim aB $b# ; aBb œ lim aB $b# BÄ$ Hence B œ $ is a regular singular point. 11. Dividing the ODE by aB# B #b, we find that :aBb œ B" # and ; aBb œ . aB #baB "b aB #baB "b lim aB #b:aBb œ lim B" " œÞ BÄ# B " $ The singular points are at B œ # and " . For B œ #, BÄ# BÄ# lim aB #b# ; aBb œ lim #aB #b œ !Þ BÄ# B " Hence B œ # is a regular singular point. For B œ ", BÄ" lim aB "b:aBb œ lim B" # œÞ BÄ" B # $ #aB "b œ !Þ a B #b BÄ" lim aB "b# ; aBb œ lim BÄ" Hence B œ " is a regular singular point. 13. Note that :aBb œ 68kBk and ; aBb œ $B . Evidently, :aBb is not analytic at B! œ ! . Furthermore, the function B :aBb œ B 68kBk does not have a Taylor series about B! œ !. Hence B œ ! is an irregular singular point. 14. T aBb œ ! when B œ ! . Since the three coefficients have no common factors, B œ ! is a singular point. The Taylor series of /B ", about B œ !, is Hence the function B :aBb œ #a/B "bÎB is analytic at B œ ! . Similarly, the Taylor series of /B -9= B , about B œ !, is The function B# ; aBb œ /B -9= B is also analytic at B œ ! . Hence B œ ! is a regular singular point. /B -9= B œ " B B$ Î$ B% Î' â . /B " œ B B# Î# B$ Î' â . ________________________________________________________________________ page 207 —————————————————————————— CHAPTER 5. —— 15. T aBb œ ! when B œ ! . Since the three coefficients have no common factors, B œ ! is a singular point. The Taylor series of =38 B , about B œ !, is Hence the function B :aBb œ $=38 BÎB is analytic at B œ ! . On the other hand, ; aBb is a rational function, with BÄ! =38 B œ B B$ Î$x B& Î&x â . lim B# ; aBb œ lim B# BÄ! " B# œ "Þ B# Hence B œ ! is a regular singular point. 16. T aBb œ ! when B œ ! . Since the three coefficients have no common factors, B œ ! is a singular point. We find that BÄ! Although the function V aBb œ -9> B does not have a Taylor series about B œ ! , note that B# ; aBb œ B -9> B œ " B# Î$ B% Î%& #B' Î*%& â . Hence B œ ! is a regular singular point. Furthermore, ; aBb œ -9> BÎB# is undefined at B œ „ 81 . Therefore the points B œ „ 81 are also singular points. First note that BÄ„81 lim B :aBb œ lim B BÄ! " œ "Þ B lim aB…81b:aBb œ lim aB…81b BÄ„81 " œ !Þ B Furthermore, since -9> B has period 1 , ; aBb œ -9> BÎB œ -9>aB … 81bÎB " œ -9>aB … 81b Þ a B … 81 b „ 8 1 a B … 81 b •Þ a B … 81 b „ 8 1 Therefore aB … 81b# ; aBb œ aB … 81b-9>aB … 81b” From above, aB … 81b-9>aB … 81b œ " aB … 81b# Î$ aB … 81b% Î%& â . Note that the function in brackets is analytic near B œ „ 81. It follows that the function aB … 81b# ; aBb is also analytic near B œ „ 81 . Hence all the singular points are regular. 18. The singular points are located at B œ „ 81 , 8 œ !ß "ß â . Dividing the ODE by B =38 B , we find that B :aBb œ $ -=- B and B# ; aBb œ B# -=- B . Evidently, B :aBb is not even defined at B œ !. Hence B œ ! is an irregular singular point. On the other hand, the Taylor series of B -=- B, about B œ !, is ________________________________________________________________________ page 208 —————————————————————————— CHAPTER 5. —— Noting that -=- aB … 81b œ a "b8 -=- B , B -=- B œ " B# Î' (B% $'! â . aB … 81b:aBb œ $a "b8 aB … 81b-=- aB … 81bÎB œ $a "b8 aB … 81b-=- aB … 81b” It is apparent that aB … 81b:aBb is analytic at B œ „ 81 . Similarly, " •Þ a B … 81 b „ 8 1 aB … 81b# ; aBb œ aB … 81b# -=- B œ a "b8 aB … 81b# -=- aB … 81b, which is also analytic at B œ „ 81 . Hence all other singular points are regular. 20. B œ ! is the only singular point. Dividing the ODE by #B# , we have :aBb œ $Îa#Bb and ; aBb œ B# a" BbÎ#Þ It follows that BÄ! lim B :aBb œ lim B BÄ! $ $ œ, #B # BÄ! lim B# ; aBb œ lim B# BÄ! a" Bb " œ Þ # #B # Hence B œ ! is a regular singular point. Let C œ +! +" B +# B# â +8 B8 â . Substitution into the ODE results in #B# " a8 #ba8 "b+8# B8 $B" a8 "b+8" B8 a" Bb " +8 B8 œ ! . _ _ _ 8œ! 8œ! 8œ! That is, #" 8a8 "b+8 B $" 8 +8 B " +8 B " +8" B8 œ ! . _ _ _ _ 8 8 8 8œ# 8œ" 8œ! 8œ" It follows that +! a#+" +! bB " c#8a8 "b+8 $8 +8 +8 +8" dB8 œ ! Þ _ 8œ# Equating the coefficients to zero, we find that +! œ ! , #+" +! œ ! , and We conclude that all the +8 are equal to zero. Hence CaBb œ ! is the only solution that can be obtained. 22. Based on Prob. #", the change of variable, B œ "Î0 , transforms the ODE into the ________________________________________________________________________ page 209 a#8 "ba8 "b+8 œ +8" , 8 œ #ß $ß â . —————————————————————————— CHAPTER 5. —— form 0% .# C .C #0$ C œ !. # .0 .0 Evidently, 0 œ ! is a singular point. Now :a0b œ #Î0 and ; a0b œ "Î0% . Since the value of lim 0# ; a0b does not exist, 0 œ ! , that is, B œ _ , is an irregular singular point. 0 Ä! 24. Under the transformation B œ "Î0 , the ODE becomes 0 % Œ" that is, ˆ0% 0# ‰ : a0 b œ It follows that 0Ä! " .# C " " .C ”#0$ Œ" # #0# • !a! "bC œ ! , # . 0# 0 0 0 .0 .# C .C #0$ !a! "bC œ ! . # .0 .0 #0 ! a ! "b and ; a0b œ # # . 0 a0 "b " #0 œ !, " Therefore 0 œ ! is a singular point. Note that 0# lim0 :a0b œ lim 0 0Ä! 0# Hence 0 œ ! aB œ _b is a regular singular point. 0Ä! lim0# ; a0b œ lim 0# 0Ä! ! a ! "b œ ! a ! "b Þ 0 # a 0 # "b 26. Under the transformation B œ "Î0 , the ODE becomes 0% that is, 0% .# C .C #ˆ0$ 0‰ -C œ !. . 0# .0 #a0# "b and ; a0b œ % . $ 0 0 0Ä! .# C " .C ”#0$ #0# • -C œ !, # .0 0 .0 Therefore 0 œ ! is a singular point. Note that : a0 b œ It immediately follows that the limit lim0 :a0b does not exist. Hence 0 œ ! aB œ _b ________________________________________________________________________ page 210 —————————————————————————— CHAPTER 5. —— is an irregular singular point. 27. Under the transformation B œ "Î0 , the ODE becomes 0% .# C .C " #0$ C œ !. # .0 .0 0 # " and ; a0b œ & . 0 0 Therefore 0 œ ! is a singular point. Note that : a0 b œ We find that 0Ä! lim0 :a0b œ lim 0 0Ä! # œ #, 0 a "b . 0& but 0Ä! The latter limit does not exist. Hence 0 œ ! aB œ _b is an irregular singular point. lim0# ; a0b œ lim 0# 0Ä! ________________________________________________________________________ page 211 —————————————————————————— CHAPTER 5. —— Section 5.5 1. Substitution of C œ B< results in the quadratic equation J a<b œ ! , where J a<b œ <a< "b %< # œ <# $< # Þ The roots are < œ # , " . Hence the general solution, for B Á ! , is C œ -" B# -# B" Þ 3. Substitution of C œ B< results in the quadratic equation J a<b œ ! , where J a<b œ <a< "b $< % œ <# %< % Þ C œ a-" -# 68kBkb B# Þ The root is < œ # , with multiplicity two . Hence the general solution, for B Á ! , is 5. Substitution of C œ B< results in the quadratic equation J a<b œ ! , where J a < b œ < a < "b < " œ <# #< " Þ C œ a-" -# 68kBkb B Þ The root is < œ " , with multiplicity two . Hence the general solution, for B Á ! , is 6. Substitution of C œ aB "b< results in the quadratic equation J a<b œ ! , where J a<b œ <# (< "# Þ The roots are < œ $ , % . Hence the general solution, for B Á " , is C œ -" aB "b$ -# aB "b% Þ J a<b œ <# &< " Þ -# kBk 7. Substitution of C œ B< results in the quadratic equation J a<b œ ! , where The roots are < œ Š& „ È#*‹Î# . Hence the general solution, for B Á ! , is C œ -" kBk Š& È#*‹Î# Š& È#*‹Î# Þ 8. Substitution of C œ B< results in the quadratic equation J a<b œ ! , where ________________________________________________________________________ page 212 —————————————————————————— CHAPTER 5. —— J a<b œ <# $< $ Þ The roots are complex, with < œ Š$ „ 3È$ ‹Î# . Hence the general solution, for B Á !, is C œ -" kBk$Î# -9=Œ È$ # 68kBk -# kBk$Î# =38Œ È$ # 68kBkÞ 10. Substitution of C œ aB #b< results in the quadratic equation J a<b œ ! , where J a<b œ <# %< ) Þ The roots are complex, with < œ # „ #3 . Hence the general solution, for B Á #, is C œ -" aB #b# -9=a# 68kB #kb -# aB #b# =38a# 68kB #kbÞ J a< b œ < # < % Þ 11. Substitution of C œ B< results in the quadratic equation J a<b œ ! , where The roots are complex, with < œ Š" „ 3È"& ‹Î# . Hence the general solution, for B Á !, is C œ -" kBk"Î# -9=Œ È"& # 68kBk -# kBk"Î# =38Œ È"& # 68kBkÞ 12. Substitution of C œ B< results in the quadratic equation J a<b œ ! , where J a<b œ <# &< % Þ C œ - " B - # B% Þ 14. Substitution of C œ B< results in the quadratic equation J a<b œ ! , where J a<b œ %<# %< "( Þ The roots are complex, with < œ "Î# „ #3 . Hence the general solution, for B !, is C œ -" B"Î# -9=a# 68 Bb -# B"Î# =38a# 68 BbÞ The roots are < œ " , % . Hence the general solution, for B Á ! , is Invoking the initial conditions, we obtain the system of equations ________________________________________________________________________ page 213 —————————————————————————— CHAPTER 5. —— -" œ # " -" #-# œ $ # Hence the solution of the initial value problem is CaBb œ # B"Î# -9=a# 68 Bb B"Î# =38a# 68 BbÞ As B p ! , the solution decreases without bound. 15. Substitution of C œ B< results in the quadratic equation J a<b œ ! , where J a<b œ <# %< % Þ The root is < # , with multiplicity two . Hence the general solution, for B ! , is C œ a-" -# 68 kBkb B# Þ -" œ # #-" -# œ $ Hence the solution of the initial value problem is CaBb œ a# ( 68 kBkb B# Þ Invoking the initial conditions, we obtain the system of equations ________________________________________________________________________ page 214 —————————————————————————— CHAPTER 5. —— We find that CaBb p ! as B p ! . 18. Substitution of C œ B< results in the quadratic equation <# < " œ ! . The roots are <œ " „ È " %" . # If " "Î% , the roots are complex, with <"ß# œ ˆ" „ 3È%" " ‰Î# . Hence the general solution, for B Á ! , is " " C œ -" kBk"Î# -9=Œ È%" " 68kBk -# kBk"Î# =38Œ È%" " 68kBkÞ # # Since the trigonometric factors are bounded, CaBb p ! as B p ! . If " œ "Î% , the roots are equal, and Since lim ÈkBk 68 kBk œ ! , C aBb p ! as B p ! . If " "Î% , the roots are real, with <"ß# œ ˆ" „ È" %" ‰Î# . Hence the general solution, for B Á ! , is BÄ! C œ -" kBk"Î# -# kBk"Î# 68 kBk Þ Evidently, solutions approach zero as long as "Î# È" %" Î# ! Þ That is, ! " "Î% . Hence all solutions approach zero, for " ! . 19. Substitution of C œ B< results in the quadratic equation <# < # œ ! . The roots are < œ " , # . Hence the general solution, for B Á ! , is C œ -" B" -# B# Þ Invoking the initial conditions, we obtain the system of equations ________________________________________________________________________ page 215 C œ -" kBk"Î# È"%" Î# -# kBk"Î# È"%" Î# Þ —————————————————————————— CHAPTER 5. —— -" - # œ " -" #-# œ # Hence the solution of the initial value problem is CaBb œ # # " " # # B BÞ $ $ The solution is bounded, as B p ! , if # œ # . 20. Substitution of C œ B< results in the quadratic equation <# a! "b< &Î# œ ! . Formally, the roots are given by <œ " ! „ È ! # #! * # a3b The roots <"ß# will be complex, if k" !k È"! . For solutions to approach zero, as B p _ , we need È"! " ! ! Þ a33b The roots will be equal, if k" !k œ È"! Þ In this case, all solutions approach zero as long as " ! œ È"! . a333b The roots will be real and distinct, if k" !k È"! . It follows that For solutions to approach zero, we need " ! È!# #! * ! Þ That is, " ! È"! . Hence all solutions approach zero, as B p _ , as long as ! " Þ 23a+b. Given that B œ /D , C aBb œ C a/D b œ AaD b. By the chain rule, .C . .A .D " .A œ A aD b œ œ . .B .B .D .B B .D <7+B œ " ! È ! # #! * . # œ " ! „ ÊŠ! " È"!‹Š! " È"!‹ # Þ Similarly, .# C . " .A " .A " . # A .D œ ” •œ # .B# .B B .D B .D B .D # .B " .A " .#A œ # # Þ B .D B .D # a,b. Direct substitution results in ________________________________________________________________________ page 216 —————————————————————————— CHAPTER 5. —— B# ” that is, .# A .A a ! "b "A œ !. # .D .D " .# A " .A " .A # • ! B” • "A œ !, B# .D # B .D B .D The associated characteristic equation is <# a! "b< " œ !. Since D œ 68 B , it follows that CaBb œ Aa68 BbÞ a- b. If the roots <"ß# are real and distinct, then C œ - " / <" D - # / < # D œ -" B<" -# B<# Þ a. b. If the roots <"ß# are real and equal, then C œ - " / <" D - # D / < " D œ -" B<" -# B<" 68 B Þ a/b. If the roots are complex conjugates, then < œ - „ 3. , and C œ /-D a-" -9= .D -# =38 .D b œ B- c-" -9=a. 68 Bb -# =38a. 68 BbdÞ 24. Based on Prob. #$, the change of variable B œ /D transforms the ODE into . # A .A #A œ ! . .D # .D The associated characteristic equation is <# < # œ ! , with roots < œ " , # . Hence AaD b œ -" /D -# /#D , and C aBb œ -" B" -# B# Þ 26. The change of variable B œ /D transforms the ODE into .# A .A ' &A œ /D . # .D .D The associated characteristic equation is <# ' < & œ ! , with roots < œ & , " . Hence A- aD b œ -" /D -# /&D Þ Since the right hand side is not a solution of the homogeneous equation, we can use the method of undetermined coefficients to show that a particular solution is [ œ /D Î"# . Therefore the general solution is given by AaD b œ -" /D -# /&D /D Î"# , that is, C aBb œ -" B" -# B& BÎ"# Þ 27. The change of variable B œ /D transforms the given ODE into ________________________________________________________________________ page 217 —————————————————————————— CHAPTER 5. —— .# A .A $ #A œ $/#D #D . .D # .D The associated characteristic equation is <# $ < # œ ! , with roots < œ " , # . Hence A- aD b œ -" /D -# /#D Þ Using the method of undetermined coefficients, let [ œ E/#D FD/#D GD H . It follows that the general solution is given by AaD b œ -" /D -# /#D $D/#D D $Î# , that is, CaBb œ -" B -# B# $ B# 68 B 68 B $Î# Þ 28. The change of variable B œ /D transforms the given ODE into .# A %A œ =38 D . .D # The solution of the homogeneous equation is A- aD b œ -" -9= #D -# =38 #D Þ The right hand side is not a solution of the homogeneous equation. We can use the method of undetermined coefficients to show that a particular solution is [ œ " =38 D . Hence $ the general solution is given by AaD b œ -" -9= #D -# =38 #D " =38 D , that is, $ CaBb œ -" -9=a# 68 Bb -# =38a# 68 Bb " =38a68 Bb Þ $ 29. After dividing the equation by $ , the change of variable B œ /D transforms the ODE into .# A .A $ $A œ ! . # .D .D The associated characteristic equation is <# $ < $ œ ! , with complex roots < œ Š$ „ 3È$ ‹Î# . Hence the general solution is AaD b œ /$DÎ# ’-" -9=ŠÈ$ DÎ#‹ -# =38ŠÈ$ DÎ#‹“, È$ # and therefore CaBb œ B$Î# ”-" -9=Œ 68 B -# =38Œ È$ # 68 B•. 30. Let B !. Setting C œ a Bb< , successive differentiation gives C w œ <a Bb<" and C ww œ <a< "ba Bb<# . It follows that Since B# œ a Bb# , we find that Pca Bb< d œ <a< "bB# a Bb<# ! < Ba Bb<" " a Bb< Þ ________________________________________________________________________ page 218 —————————————————————————— CHAPTER 5. —— Pca Bb< d œ <a< "ba Bb< ! <a Bb< " a Bb< œ a Bb< c<a< "b ! < " dÞ Given that <" and <# are roots of J a<b œ <a< "b ! < " , we have Pca Bb<3 d œ !. Therefore C" œ a Bb<" and C# œ a Bb<# are linearly independent solutions of the differential equation, PcCd œ ! , for B ! , as long as <" Á <# . ________________________________________________________________________ page 219 —————————————————————————— CHAPTER 5. —— Section 5.6 1. T aBb œ ! when B œ ! . Since the three coefficients have no common factors, B œ ! is a singular point. Near B œ !, BÄ! lim B :aBb œ lim B BÄ! " " œÞ #B # " œ !Þ # _ BÄ! lim B# ; aBb œ lim B# BÄ! Hence B œ ! is a regular singular point. Let C œ B< ˆ+! +" B +# B# â +8 B8 ≠œ " +8 B<8 . 8œ! Then C w œ " a< 8b+8 B<8" _ 8œ! and C œ " a< 8ba< 8 "b+8 B<8# Þ _ ww 8œ! Substitution into the ODE results in # " a< 8ba< 8 "b+8 B<8" " a< 8b+8 B<8" _ _ 8œ! " +8 B<8" œ ! . 8œ! 8œ! _ That is, # " a< 8ba< 8 "b+8 B<8 " a< 8b+8 B<8 " +8# B<8 œ ! . _ _ _ 8œ! 8œ! 8œ# It follows that +! c#<a< "b <dB< +" c#a< "b< < "dB<" _ 8œ# " c#a< 8ba< 8 "b+8 a< 8b+8 +8# dB<8 œ ! Þ Assuming that +! Á ! , we obtain the indicial equation #<# < œ !, with roots <" œ "Î# ________________________________________________________________________ page 220 —————————————————————————— CHAPTER 5. —— and <# œ ! . It immediately follows that +" œ ! . Setting the remaining coefficients equal to zero, we have +8# , 8 œ #ß $ß â . +8 œ a< 8bc#a< 8b "d For < œ "Î# , the recurrence relation becomes +8# +8 œ , 8 œ #ß $ß â . 8a" #8b Since +" œ ! , the odd coefficients are zero. Furthermore, for 5 œ "ß #ß â , +#5 œ +#5# +#5% a "b5 + ! œ œ5 Þ #5 a" %5 b a#5 #ba#5 ba%5 $ba%5 "b # 5x & † * † "$ âa%5 "b For < œ ! , the recurrence relation becomes +8# +8 œ , 8 œ #ß $ß â . 8a#8 "b Since +" œ ! , the odd coefficients are zero, and for 5 œ "ß #ß â , +#5 œ +#5# +#5% a "b5 + ! œ œ5 Þ #5 a%5 "b # 5x $ † ( † ""âa%5 "b a#5 #ba#5 ba%5 &ba%5 "b _ a "b5 B#5 C" aBb œ ÈB –" " 5 # 5x & † * † "$ âa%5 "b — 5œ" The two linearly independent solutions are C# aBb œ " " _ a "b5 B#5 Þ #5 5x $ † ( † ""âa%5 "b 5œ" 3. Note that B :aBb œ ! and B# ; aBb œ B , which are both analytic at B œ ! . Set C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in 8œ! " a< 8ba< 8 "b+8 B<8" " +8 B<8 œ ! , _ _ 8œ! and after multiplying both sides of the equation by B , " a< 8ba< 8 "b+8 B<8 " +8" B<8 œ ! . _ _ 8œ" 8œ! It follows that ________________________________________________________________________ page 221 —————————————————————————— CHAPTER 5. —— +! c<a< "bdB " ca< 8ba< 8 "b+8 +8" dB<8 œ ! Þ _ < 8œ" Setting the coefficients equal to zero, the indicial equation is <a< "b œ ! . The roots are <" œ " and <# œ ! . Here <" <# œ " . The recurrence relation is +8" , 8 œ "ß #ß â . +8 œ a< 8ba< 8 "b For < œ " , +8 œ Hence for 8 " , +8 œ +8" , 8 œ "ß #ß â . 8a8 "b +8" +8# a "b8 + ! œ œâœ . a8 "b8# a8 "b 8a8 "b 8xa8 "bx C" aBb œ B " _ Therefore one solution is a "b8 B8 . 8xa8 "bx 8œ! 5. Here B :aBb œ #Î$ and B# ; aBb œ B# Î$ , which are both analytic at B œ ! . Set C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in $" a< 8ba< 8 "b+8 B _ 8œ! <8 #" a< 8b+8 B _ 8œ! <8 " +8 B<8# œ ! Þ _ 8œ! It follows that +! c$<a< "b #<dB< +" c$a< "b< #a< "bdB<" _ 8œ# " c$a< 8ba< 8 "b+8 #a< 8b+8 +8# dB<8 œ ! . Assuming +! Á ! , the indicial equation is $<# < œ ! , with roots <" œ "Î$ , <# œ ! Þ Setting the remaining coefficients equal to zero, we have +" œ ! , and +8# +8 œ , 8 œ #ß $ß â . a< 8bc$a< 8b "d It immediately follows that the odd coefficients are equal to zero. For < œ "Î$ , +8# , 8 œ #ß $ß â . +8 œ 8a" $8b So for 5 œ "ß #ß â , ________________________________________________________________________ page 222 —————————————————————————— CHAPTER 5. —— +#5# +#5% a "b5 + ! œ œ œ5 . #5 a'5 "b # 5x ( † "$ âa'5 "b a#5 #ba#5 ba'5 &ba'5"b +8 œ So for 5 œ "ß #ß â , +#5 œ +8# , 8 œ #ß $ß â . 8a$8 "b +#5 For < œ ! , a "b5 + ! +#5# +#5% . œ œ5 a#5 #ba#5 ba'5 (ba'5 "b #5 a'5 "b # 5x & † "" âa'5 "b C" aBb œ B –" " _ _ 5 a "b 5 B# Œ — 5x ( † "$ âa'5 "b # 5œ" The two linearly independent solutions are "Î$ C# aBb œ " " 5 a "b 5 B# Œ Þ 5x & † ""âa'5 "b # 5œ" 6. Note that B :aBb œ " and B# ; aBb œ B #, which are both analytic at B œ ! . Set C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in 8œ! " a< 8ba< 8 "b+8 B _ <8 " a< 8b+8 B<8 _ 8œ! _ " +8 B<8" #" +8 B<8 œ ! Þ _ 8œ! 8œ! After adjusting the indices in the second-to-last series, we obtain +! c<a< "b < #dB " ca< 8ba< 8 "b+8 a< 8b+8 # +8 +8" dB<8 œ !Þ _ < 8œ" Assuming +! Á ! , the indicial equation is <# # œ ! , with roots < œ „ È# . Setting the remaining coefficients equal to zero, the recurrence relation is +8" , 8 œ "ß #ß â . +8 œ a < 8b # # First note that a< 8b# # œ Š< 8 È# ‹Š< 8 È# ‹. So for < œ È# , +8 œ +8" , 8 œ "ß #ß â . 8Š8 #È# ‹ ________________________________________________________________________ page 223 —————————————————————————— CHAPTER 5. —— It follows that For < œ È# , a "b 8 + ! , 8 œ "ß #ß â . +8 œ 8xŠ" #È# ‹Š# #È# ‹âŠ8 #È# ‹ +8" , 8 œ "ß #ß â , 8Š8 #È# ‹ +8 œ and therefore a "b 8 + ! , 8 œ "ß #ß â . +8 œ 8xŠ" #È# ‹Š# #È# ‹âŠ8 #È# ‹ Ô × _ a "b8 B8 Ö" " Ù È# ‹Š# #È# ‹âŠ8 #È# ‹ Ø 8 œ " 8xŠ" # Õ The two linearly independent solutions are C" aBb œ B È# C# aBb œ B È# Ö Ô × _ a "b8 B8 ÙÞ "" 8xŠ" #È# ‹Š# #È# ‹âŠ8 #È# ‹ Ø 8œ" Õ 7. Here B :aBb œ " B and B# ; aBb œ B , which are both analytic at B œ ! . Set C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in 8œ! " a< 8ba< 8 "b+8 B<8" " a< 8b+8 B<8" _ _ 8œ! _ " a< 8b+8 B<8 " +8 B<8 œ ! Þ _ 8œ! 8œ! After multiplying both sides by B , " a< 8ba< 8 "b+8 B<8 " a< 8b+8 B<8 _ _ 8œ! 8œ! " a < 8 b+ 8 B _ 8œ! <8+1 " +8 B<8+1 œ !Þ _ 8œ! After adjusting the indices in the last two series, we obtain ________________________________________________________________________ page 224 —————————————————————————— CHAPTER 5. —— +! c<a< "b <dB " ca< 8ba< 8 "b+8 a< 8b+8 a< 8b+8" dB<8 œ ! Þ _ < 8œ" Assuming +! Á ! , the indicial equation is <# œ ! , with roots <" œ <# œ ! . Setting the remaining coefficients equal to zero, the recurrence relation is +8" +8 œ , 8 œ "ß #ß â . <8 With < œ ! , +8 œ Hence one solution is C" aBb œ " B B# B8 â â œ /B . "x #x 8x +8" , 8 œ "ß #ß â . 8 8. Note that B :aBb œ $Î# and B# ; aBb œ B# "Î#, which are both analytic at B œ ! . Set C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in #" a< 8ba< 8 "b+8 B<8 $" a< 8b+8 B<8 _ _ 8œ! 8œ! _ #" +8 B<8# " +8 B<8 œ ! Þ _ 8œ! 8œ! After adjusting the indices in the second-to-last series, we obtain _ +! c#<a< "b $< "dB< +" c#a< "b< $a< "b "d 8œ# " c#a< 8ba< 8 "b+8 $a< 8b+8 +8 # +8# dB<8 œ !Þ Assuming +! Á ! , the indicial equation is #<# < " œ ! , with roots <" œ "Î# and <# œ " . Setting the remaining coefficients equal to zero, the recurrence relation is +8 œ #+8# , 8 œ #ß $ß â . a< 8 "bc#a< 8b "d Setting the remaining coefficients equal to zero, we have +" œ ! , which implies that all of the odd coefficients are zero. With < œ "Î# , +8 œ So for 5 œ "ß #ß â , #+8# , 8 œ #ß $ß â . 8a#8 $b ________________________________________________________________________ page 225 —————————————————————————— CHAPTER 5. —— +#5# +#5% a "b5 + ! œ œ œ . 5 a%5 $b 5x ( † "" â a%5 $b a5 "b5 a%5 &ba%5 $b #+8# , 8 œ #ß $ß â . 8a#8 $b +#5 With < œ " , +8 œ So for 5 œ "ß #ß â , +#5 œ +#5# +#5% a "b5 + ! œ œ . 5 a%5 $b a5 "b5 a%5 ""ba%5 $b 5x & † * â a%5 $b C" aBb œ B"Î# –" " _ The two linearly independent solutions are a "b8 B#8 8x ( † "" â a%8 $b — 8œ" _ C# aBb œ B" –" " a "b8 B#8 Þ 8x & † * â a%8 $b — 8œ" 9. Note that B :aBb œ B $ and B# ; aBb œ B $, which are both analytic at B œ ! . Set C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in 8œ! " a< 8ba< 8 "b+8 B<8 " a< 8b+8 B<8" $" a< 8b+8 B<8 _ _ _ 8œ! " +8 B<8" $" +8 B<8 œ ! Þ _ _ 8œ! 8œ! 8œ! After adjusting the indices in the second-to-last series, we obtain +! c<a< "b $< $dB< _ 8œ" " ca< 8ba< 8 "b+8 a< 8 #b+8" $a< 8 "b+8 dB<8 œ !Þ Assuming +! Á ! , the indicial equation is <# %< $ œ ! , with roots <" œ $ and <# œ " . Setting the remaining coefficients equal to zero, the recurrence relation is +8 œ With < œ $ , a< 8 #b+8" , 8 œ "ß #ß â . a< 8 "ba< 8 $b ________________________________________________________________________ page 226 —————————————————————————— CHAPTER 5. —— a8 "b+8" , 8 œ "ß #ß â . 8a8 #b +8 œ It follows that for 8 " , +8 œ +8# # +! a8 "b+8" œ œâœ . 8a8 #b 8x a8 #b a8 "ba8 #b # B8 C" aBb œ B –" " . 8x a8 #b — _ $ 8œ" Therefore one solution is 10. Here B :aBb œ ! and B# ; aBb œ B# "Î% , which are both analytic at B œ ! . Set C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in 8œ! " a< 8ba< 8 "b+8 B<8 " +8 B<8# _ _ 8œ! "_ " +8 B<8 œ ! Þ % 8œ! After adjusting the indices in the second series, we obtain " " +! ”<a< "b •B< +" ”a< "b< •B<" % % _ " " ”a< 8ba< 8 "b+8 +8 +8# •B<8 œ ! Þ % 8œ# Assuming +! Á ! , the indicial equation is <# < " œ !, with roots <" œ <# œ "Î#Þ % Setting the remaining coefficients equal to zero, we find that +" œ ! . The recurrence relation is +8 œ With < œ "Î# , +8 œ +8# , 8 œ #ß $ß â . 8# a#< #8 "b# %+8# , 8 œ #ß $ß â . Since +" œ ! , the odd coefficients are zero. So for 5 " , +#5 œ Therefore one solution is +#5# +#5% a "b5 + ! œ œâœ . %5 # %# a5 "b# 5 # %5 a5xb# ________________________________________________________________________ page 227 —————————————————————————— CHAPTER 5. —— C" aBb œ ÈB –" " _ a "b8 B#8 ##8 a8xb# 8œ" —. 12a+b. Dividing through by the leading coefficient, the ODE can be written as C ww For B œ " , :! œ lim aB "b:aBb œ lim BÄ" B !# Cw C œ !. " B# " B# B " œÞ BÄ" B " # ;! œ lim aB "b# ; aBb œ lim BÄ" !# a" Bb œ !Þ BÄ" B" For B œ " , :! œ lim aB "b:aBb œ lim BÄ" # B " œÞ BÄ" B " # ! # a B "b ;! œ lim aB "b ; aBb œ lim œ !Þ BÄ" BÄ" a" Bb Hence both B œ " and B œ " are regular singular points. As shown in Example " , the indicial equation is given by < a < "b : ! < ; ! œ ! . a,b. Let > œ B " , and ?a>b œ C a> "bÞ Under this change of variable, the differential equation becomes Based on Part a+b, > œ ! is a regular singular point. Set ? œ ! +8 ><8 Þ Substitution _ In this case, both sets of roots are <" œ "Î# and <# œ ! . ˆ># #>‰? ww a> "b? w !# ? œ ! Þ into the ODE results in _ _ 8œ! 8œ! " a< 8ba< 8 "b+8 ><8 #" a< 8ba< 8 "b+8 ><8" " a< 8b+8 ><8 " a< 8b+8 ><8" !# " +8 ><8 œ ! Þ _ _ _ 8œ! 8œ! 8œ! 8œ! Upon inspection, we can also write ________________________________________________________________________ page 228 —————————————————————————— CHAPTER 5. —— " a< 8b# +8 ><8 #" a< 8bŒ< 8 " +8 ><8" !# " +8 ><8 œ !Þ _ _ _ 8œ! 8œ! # 8œ! After adjusting the indices in the second series, it follows that +! ”#<Œ< •><" " ”a< 8b# +8 #a< 8 "bŒ< 8 +8" !# +8 •><8 œ !Þ # # " _ " 8œ! Assuming that +! Á ! , the indicial equation is #<# < œ ! , with roots < œ ! , "Î# . The recurrence relation is " a< 8b# +8 #a< 8 "bŒ< 8 +8" !# +8 œ ! , 8 œ !ß "ß #ß â . # With <" œ "Î# , we find that for 8 " , %!# a#8 "b# +8 œ +8" %8a#8 "b # # # #‘ 8 c" %! dc* %! dâ a#8 "b %! œ a "b +! Þ #8 a#8 "bx With <# œ ! , we find that for 8 " , ! # a8 "b # +8 œ +8" 8a#8 "b !a !bc" !# dc% !# dâa8 "b# !# ‘ œ a "b 8 +! Þ 8x † $ † & âa#8 "b The two linearly independent solutions of the Chebyshev equation are C" aBb œ kB "k"Î# –" " a "b8 _ 8œ" c" %!# dc* %!# dâa#8 "b# %!# ‘ aB "b 8 — #8 a#8 "bx C# aBb œ " " a "b8 _ 8œ" !a !bc" !# dc% !# dâa8 "b# !# ‘ aB " b 8 Þ 8x † $ † & âa#8 "b 13. Here B :aBb œ " B and B# ; aBb œ - B , which are both analytic at B œ ! . In fact, Hence the indicial equation is <a< "b < œ ! , with roots <"ß# œ ! . Set ________________________________________________________________________ page 229 :! œ lim B :aBb œ " and ;! œ lim B# ; aBb œ ! . BÄ! BÄ! —————————————————————————— CHAPTER 5. —— C œ +! + " B + # B# â + 8 B 8 â . Substitution into the ODE results in " 8a8 "b+8 B8" " 8+8 B8" _ _ 8œ" _ 8œ# " 8+8 B8 -" +8 B8 œ ! Þ _ 8œ! 8œ! That is, " 8a8 "b+8" B " a8 "b+8" B8 _ _ 8 8œ! _ _ 8œ" " 8+8 B8 -" +8 B8 œ ! Þ 8œ" 8œ! It follows that +" - +! " a8 "b# +8" a8 -b+8 ‘B8 œ ! . _ 8œ" Setting the coefficients equal to zero, we find that +" œ -+! , and +8 œ That is, for 8 # , +8 œ a8 " - b +8" , 8 œ #ß $ß â . 8# a8 " - b a -ba" -bâa8 " -b +8" œ â œ +! . 8# a8xb# C" aBb œ " " _ Therefore one solution of the Laguerre equation is a -ba" -bâa8 " -b a8xb # B8 Þ 8œ" Note that if - œ 7 , a positive integer, then +8 œ ! for 8 7 " . In that case, the solution is a polynomial C" aBb œ " " 7 a -ba" -bâa8 " -b a8xb # B8 Þ 8œ" ________________________________________________________________________ page 230 —————————————————————————— CHAPTER 5. —— Section 5.7 2. T aBb œ ! only for B œ ! . Furthermore, B :aBb œ # B and B# ; aBb œ # B# Þ It follows that :! œ lim a # Bb œ # ;! œ lim ˆ# B# ‰ œ # BÄ! BÄ! and therefore B œ ! is a regular singular point. The indicial equation is given by <a< "b #< # œ ! , 4. The coefficients T aBb , UaBb , and V aBb are analytic for all B − ‘ . Hence there are no singular points. 5. T aBb œ ! only for B œ ! . Furthermore, B :aBb œ $ =38 B and B# ; aBb œ # Þ It B follows that =38 B œ$ BÄ! B ;! œ lim # œ # :! œ lim $ BÄ! that is, <# $< # œ ! , with roots <" œ # and <# œ " . and therefore B œ ! is a regular singular point. The indicial equation is given by that is, <# #< # œ ! , with roots <" œ " È$ and <# œ " È$ . <a< "b $< # œ ! , 6. T aBb œ ! for B œ ! and B œ # . We note that :aBb œ B" aB #b" Î# , and ; aBb œ aB #b" Î# Þ For the singularity at B œ ! , " " œ BÄ! #aB #b % # B ;! œ lim œ! BÄ! #aB #b :! œ lim and therefore B œ ! is a regular singular point. The indicial equation is given by " < a < "b < œ ! , % that is, <# $ < œ ! , with roots <" œ % $ % and <# œ ! . For the singularity at B œ # , ________________________________________________________________________ page 231 —————————————————————————— CHAPTER 5. —— :! œ lim aB #b:aBb œ lim " " œ BÄ# BÄ# #B % aB #b ;! œ lim aB #b# ; aBb œ lim œ! BÄ# BÄ# # and therefore B œ # is a regular singular point. The indicial equation is given by " < a < "b < œ ! , % that is, <# & < œ ! , with roots <" œ % 7. T aBb œ ! only for B œ ! . Furthermore, B :aBb œ follows that BÄ! & % and <# œ ! . " # =38 B #B and B# ; aBb œ " Þ It :! œ lim B:aBb œ " BÄ! ;! œ lim B# ; aBb œ " and therefore B œ ! is a regular singular point. The indicial equation is given by < a < "b < " œ ! , that is, <# " œ ! , with complex conjugate roots < œ „ 3 . 8. Note that T aBb œ ! only for B œ " . We find that :aBb œ $aB "bÎaB "b , and ; aBb œ $ÎaB "b# . It follows that :! œ lim aB "b:aBb œ lim $aB "b œ ' ;! œ lim aB "b ; aBb œ lim $ œ $ BÄ" BÄ" # BÄ" BÄ" and therefore B œ " is a regular singular point. The indicial equation is given by that is, <# (< $ œ ! , with roots <" œ Š( È$( ‹Î# and <# œ Š( È$( ‹Î# . <a< "b '< $ œ ! , 10. T aBb œ ! for B œ # and B œ # . We note that :aBb œ #BaB #b# aB #b" , and ; aBb œ $aB #b" aB #b" Þ For the singularity at B œ # , BÄ# lim aB #b:aBb œ lim BÄ# B# #B , % which is undefined. Therefore B œ ! is an irregular singular point. For the singularity at B œ # , ________________________________________________________________________ page 232 —————————————————————————— CHAPTER 5. —— :! œ lim aB #b:aBb œ lim BÄ# a B #b $aB #b ;! œ lim aB #b# ; aBb œ lim œ! BÄ# BÄ# B# BÄ# # #B œ " % and therefore B œ # is a regular singular point. The indicial equation is given by " < a < "b < œ ! , % that is, <# & < œ ! , with roots <" œ % 11. T aBb œ ! for B œ # and B œ # . We note that :aBb œ #BÎa% B# b , and ; aBb œ $Îa% B# bÞ For the singularity at B œ # , :! œ lim aB #b:aBb œ lim #B œ " BÄ# BÄ# B # $a# Bb ;! œ lim aB #b# ; aBb œ lim œ! BÄ# BÄ# B# and therefore B œ # is a regular singular point. The indicial equation is given by < a < "b < œ ! , & % and <# œ ! . that is, <# #< œ ! , with roots <" œ # and <# œ ! . For the singularity at B œ # , :! œ lim aB #b:aBb œ lim #B œ " BÄ# BÄ# # B $aB #b ;! œ lim aB #b# ; aBb œ lim œ! BÄ# BÄ# #B and therefore B œ # is a regular singular point. The indicial equation is given by < a < "b < œ ! , that is, <# #< œ ! , with roots <" œ # and <# œ ! . 12. T aBb œ ! for B œ ! and B œ $ . We note that :aBb œ #B" aB $b" , and ; aBb œ "ÎaB $b# Þ For the singularity at B œ ! , :! œ lim B :aBb œ lim # # œ BÄ! BÄ! B $ $ B# ;! œ lim B# ; aBb œ lim œ! BÄ! BÄ! aB $b# and therefore B œ ! is a regular singular point. The indicial equation is given by ________________________________________________________________________ page 233 —————————————————————————— CHAPTER 5. —— # < a < "b < œ ! , $ that is, <# & < œ ! , with roots <" œ $ :! œ lim aB $b:aBb œ lim BÄ$ & $ and <# œ ! . For the singularity at B œ $ , # # œ BÄ$ BÄ$ B $ # ;! œ lim aB $b ; aBb œ lim a "b œ " BÄ$ and therefore B œ $ is a regular singular point. The indicial equation is given by # < a < "b < " œ ! , $ 13a+b. Note the :aBb œ "ÎB and ; aBb œ "ÎB . Furthermore, B :aBb œ " and B# ; aBb œ B Þ It follows that :! œ lim a"b œ " BÄ! BÄ! that is, <# " < " œ ! , with roots <" œ Š" È$( ‹Î' and <# œ Š" È$( ‹Î' . $ ;! œ lim a Bb œ ! and therefore B œ ! is a regular singular point. a,bÞ The indicial equation is given by < a < "b < œ ! , that is, <# œ ! , with roots <" œ <# œ ! . a- b. Let C œ +! +" B +# B# â +8 B8 â . Substitution into the ODE results in 8œ! " a8 #ba8 "b+8# B8" " a8 "b+8" B8 " +8 B8 œ ! . _ _ _ 8œ! 8œ! After adjusting the indices in the first series, we obtain +" +! " c8a8 "b+8" a8 "b+8" +8 dB8 œ ! . _ 8œ" Setting the coefficients equal to zero, it follows that for 8 ! , +8 +8" œ . a 8 "b # So for 8 " , ________________________________________________________________________ page 234 —————————————————————————— CHAPTER 5. —— +8 œ +8" +8# " œ +! Þ # œâœ # 8 8# a 8 " b a8xb# With +! œ " , one solution is For a second solution, set C# aBb œ C" aBb 68 B ," B ,# B# â ,8 B8 â . Substituting into the ODE, we obtain PcC" aBbd † 68 B # C" aBb P– " ,8 B8 — œ ! . _ w 8œ" " " " 8 C" aBb œ " B B# B$ â # B â. % $' a8xb Since PcC" aBbd œ ! , it follows that P– " ,8 B8 — œ # C"w aBb . _ 8œ" More specifically, ," " c8a8 "b,8" a8 "b,8" ,8 dB8 œ _ 8œ" " " "% œ # B B# B $ B â. ' (# "%%! Equating the coefficients, we obtain the system of equations ," %,# ," *,$ ,# "',% ,$ ã Solving these equations for the coefficients, ," œ #, ,# œ $Î%, ,$ œ ""Î"!), ,% œ #&Î$%&' , â . Therefore a second solution is $ "" $ #& % C# aBb œ C" aBb 68 B ” #B B# B B â•Þ % "!) $%&' 14a+b. Here B :aBb œ #B and B# ; aBb œ ' B/B Þ Both of these functions are analytic at B œ ! , therefore B œ ! is a regular singular point. Note that :! œ ;! œ ! Þ a,bÞ The indicial equation is given by œ œ œ œ # " "Î' "Î(# ________________________________________________________________________ page 235 —————————————————————————— CHAPTER 5. —— < a < "b œ ! , that is, <# < œ ! , with roots <" œ " and <# œ ! . a- b. In order to find the solution corresponding to <" œ " , set C œ B ! +8 B8 Þ Upon _ substitution into the ODE, we have _ _ 8œ! 8œ! " a8 #ba8 "b+8" B8" #" a8 "b+8 B8" ' /B " +8 B8" œ ! . _ 8œ! 8œ! After adjusting the indices in the first two series, and expanding the exponential function, " 8a8 "b+8 B #" 8 +8" B8 ' +! B a'+! '+" bB# _ _ 8 8œ" $ 8œ" a'+# '+" $+! bB a'+$ '+# $+" +! bB% â œ ! . #+" #+! '+! '+# %+" '+! '+" "#+$ '+# '+# '+" $+! #!+% )+$ '+$ '+# $+" +! œ! œ! œ! œ! ã Equating the coefficients, we obtain the system of equations Setting +! œ " , solution of the system results in +" œ %, +# œ "(Î$, +$ œ %(Î"#, +% œ "*"Î"#! , â . Therefore one solution is C" aBb œ B %B# "( $ %( % B B â. $ "# The exponents differ by an integer. So for a second solution, set C# aBb œ + C" aBb 68 B " -" B -# B# â -8 B8 â . _ C" aBb P–" " -8 B8 — œ ! . B 8œ" Substituting into the ODE, we obtain + PcC" aBbd † 68 B #+ C"w aBb #+ C" aBb + Since PcC" aBbd œ ! , it follows that _ 8œ" P–" " -8 B8 — œ #+ C"w aBb #+ C" aBb + C" aBb . B More specifically, ________________________________________________________________________ page 236 —————————————————————————— CHAPTER 5. —— " 8a8 "b-8" B8 #" 8 -8 B8 ' a' '-" bB _ _ 8œ" a'-# '-" $bB# â œ + "!+B 8œ" '" # "*$ $ +B +B â . $ "# Equating the coefficients, we obtain the system of equations 'œ + #-# )-" ' œ "!+ '" '-$ "!-# '-" $ œ + $ "*$ "#-% "#-$ '-# $-" " œ + "# ã Solving these equations for the coefficients, + œ ' . In order to solve the remaining equations, set -" œ ! . Then -# œ $$, -$ œ %%*Î' , -% œ "&*&Î#% , â Þ Therefore a second solution is C# aBb œ ' C" aBb 68 B ”" $$B# %%* $ "&*& % B B â•Þ ' #% 15a+b. Note the :aBb œ 'BÎaB "b and ; aBb œ $B" aB "b" . Furthermore, B :aBb œ 'B# ÎaB "b and B# ; aBb œ $BÎaB "b Þ It follows that 'B# œ! BÄ! B " $B ;! œ lim œ! BÄ! B " :! œ lim and therefore B œ ! is a regular singular point. a,bÞ The indicial equation is given by < a < "b œ ! , _ that is, <# < œ ! , with roots <" œ " and <# œ ! . a- b. In order to find the solution corresponding to <" œ " , set C œ B ! +8 B8 Þ Upon substitution into the ODE, we have 8œ! ________________________________________________________________________ page 237 —————————————————————————— CHAPTER 5. —— " 8a8 "b+8 B8" " 8a8 "b+8 B8 _ _ 8œ" _ 8œ" '" a8 "b+8 B8# $" +8 B8" œ ! . _ 8œ! 8œ! After adjusting the indices, it follows that " 8a8 "b+8" B8 " 8a8 "b+8 B8 _ _ 8œ" _ 8œ# '" a8 "b+8# B8 $" +8" B8 œ ! . _ 8œ# 8œ" That is, #+" $+! " c 8a8 "b+8 ˆ8# 8 $‰+8" 'a8 "b+8# dB8 œ !Þ _ 8œ# Setting the coefficients equal to zero, we have +" œ $+! Î# , and for 8 # , 8a8 "b+8 œ ˆ8# 8 $‰+8" 'a8 "b+8# . If we assign +! œ " , then we obtain +" œ $Î# , +# œ *Î% , +$ œ &"Î"' , â . Hence one solution is $ * &" """ & C" aBb œ B B# B$ B% B â. # % "' %! The exponents differ by an integer. So for a second solution, set C# aBb œ + C" aBb 68 B " -" B -# B# â -8 B8 â . _ C" aBb P –" " - 8 B8 — œ ! , B 8œ" Substituting into the ODE, we obtain #+B C"w aBb #+ C"w aBb '+B C" aBb + C" aBb + since PcC" aBbd œ ! . It follows that _ 8œ" P–" " -8 B8 — œ #+ C"w aBb #+B C"w aBb + C" aBb '+B C" aBb + C" aBb . B Now P–" " -8 B8 — œ $ a #-# $-" bB a '-$ &-# '-" bB# 8œ" a "#-% *-$ "#-# bB$ a #!-& "&-% ")-$ bB% â Þ _ Substituting for C" aBb , the right hand side of the ODE is ________________________________________________________________________ page 238 —————————————————————————— CHAPTER 5. —— ( $ $$ )'( % %%" & + +B +B# +B$ +B +B â . # % "' )! "! Equating the coefficients, we obtain the system of equations $œ+ ( #-# $-" œ + # $ '-$ &-# '-" œ + % $$ "#-% *-$ "#-# œ + "' ã We find that + œ $. In order to solve the second equation, set -" œ !. Solution of the remaining equations results in -# œ #"Î% , -$ œ "*Î% , -% œ &*(Î'% , â. Hence a second solution is C# aBb œ $ C" aBb 68 B ”" #" # "* $ &*( % B B B â•Þ % % '% 16a+b. After multiplying both sides of the ODE by B , we find that B :aBb œ ! and B# ; aBb œ B . Both of these functions are analytic at B œ ! , hence B œ ! is a regular singular point. a,b. Furthermore, :! œ ;! œ ! . So the indicial equation is <a< "b œ ! , with roots <" œ " and <# œ ! . a- b. In order to find the solution corresponding to <" œ " , set C œ B ! +8 B8 Þ Upon _ substitution into the ODE, we have " 8a8 "b+8 B8 " +8 B8" œ ! . _ _ 8œ! 8œ! 8œ" That is, " c8a8 "b+8 +8" d B8 œ ! . _ 8œ" Setting the coefficients equal to zero, we find that for 8 " , +8" +8 œ . 8a8 "b It follows that ________________________________________________________________________ page 239 —————————————————————————— CHAPTER 5. —— a "b8 + ! +8" +8# . œ œâœ a8 "b8# a8 "b 8a8 "b a8xb# a8 "b +8 œ Hence one solution is " " "% "& C" aBb œ B B# B$ B B â. # "# "%% #))! The exponents differ by an integer. So for a second solution, set C# aBb œ + C" aBb 68 B " -" B -# B# â -8 B8 â . _ C" aBb P–" " -8 B8 — œ ! . B 8œ" Substituting into the ODE, we obtain + PcC" aBbd † 68 B #+ C"w aBb + Since PcC" aBbd œ ! , it follows that _ 8œ" P–" " -8 B8 — œ #+ C"w aBb + C" aBb . B Now P–" " -8 B8 — œ " a#-# -" bB a'-$ -# bB# a"#-% -$ bB$ 8œ" a#!-& -% bB% a$!-' -& bB& â Þ _ Substituting for C" aBb , the right hand side of the ODE is $ & ( " + +B +B# +B$ +B% â . # "# "%% $#! Equating the coefficients, we obtain the system of equations "œ + $ #-# -" œ + # & '-$ -# œ + "# ( "#-% -$ œ + "%% ã Evidently, + œ " . In order to solve the second equation, set -" œ ! . We then find that -# œ $Î% , -$ œ (Î$' , -% œ $&Î"(#) , â . Therefore a second solution is $ ( $& % C# aBb œ C" aBb 68 B ”" B# B$ B â•Þ % $' "(#) ________________________________________________________________________ page 240 —————————————————————————— CHAPTER 5. —— 19a+b. After dividing by the leading coefficient, we find that :! œ lim B :aBb œ lim BÄ! # a " ! " bB œ #Þ BÄ! "B !" B œ !Þ "B Hence B œ ! is a regular singular point. The indicial equation is <a< "b # < œ ! , with roots <" œ " # and <# œ ! . a,bÞ For B œ ", :! œ lim aB "b:aBb œ lim BÄ" ;! œ lim B# ; aBb œ lim BÄ! BÄ! BÄ" # a" ! " bB œ "#!"Þ B !" aB "b œ !Þ BÄ" B ;! œ lim aB "b# ; aBb œ lim BÄ" Hence B œ " is a regular singular point. The indicial equation is < # a# ! " b < œ ! , with roots <" œ # ! " and <# œ ! . a- b. Given that <" <# is not a positive integer, we can set C œ ! +8 B8 Þ Substitution _ into the ODE results in _ _ 8œ! Ba" Bb" 8a8 "b+8 B8# c# a" ! " bBd" 8 +8 B8" !" " +8 B8 œ !Þ _ 8œ# 8œ" 8œ! That is, " 8a8 "b+8" B8 " 8a8 "b+8 B8 # " a8 "b+8" B8 _ _ _ 8œ# 8œ! _ _ 8œ" a" ! " b" 8 +8 B8 !" " +8 B8 œ !Þ 8œ" 8œ! Combining the series, we obtain # +" !" +! ca# ## b+# a" ! " !" b+" dB " E8 B8 œ ! , _ 8œ# in which ________________________________________________________________________ page 241 —————————————————————————— CHAPTER 5. —— E8 œ a8 "ba8 # b+8" c8a8 "b a" ! " b8 !" d+8 . a8 !ba8 " b +8 a8 "ba8 # b Note that 8a8 "b a" ! " b8 !" œ a8 !ba8 " b . Setting the coefficients equal to zero, we have # +" !" +! œ ! , and +8" œ for 8 " . Hence one solution is C" aBb œ " Since the nearest other singularity is at B œ " , the radius of convergence of C" aBb will be at least 3 œ " . a. b. Given that <" <# is not a positive integer, we can set C œ B"# ! ,8 B8 Þ Then _ !" !a! "b" a" "b # B B # † "x # a# "b † #x !a! "ba! #b" a" "ba" #b $ B âÞ # a# "ba# #b † $x Substitution into the ODE results in Ba" Bb" a8 " # ba8 # b+8 B8# " _ 8œ! _ 8œ! c# a" ! " bBd" a8 " # b+8 B8# !" " +8 B8"# œ !Þ _ 8œ! 8œ! That is, 8œ! " a8 " # ba8 # b+8 B8# " a8 " # ba8 # b+8 B8"# _ _ 8œ! _ _ # " a8 " # b+8 B8# a" ! " b" a8 " # b+8 B8"# !" " +8 B8"# œ !Þ _ 8œ! 8œ! 8œ! After adjusting the indices, 8œ! " a8 " # ba8 # b+8 B8# " a8 # ba8 " # b+8" B8# _ _ 8œ" _ _ # " a8 " # b+8 B8# a" ! " b" a8 # b+8" B8# !" " +8" B8# œ !Þ _ 8œ! 8œ" 8œ" Combining the series, we obtain " F8 B8# œ ! , _ 8œ" in which ________________________________________________________________________ page 242 —————————————————————————— CHAPTER 5. —— F8 œ 8a8 " # b,8 ca8 # ba8 # ! " b !" d,8" . a8 ! # ba8 " # b ,8" Þ 8a8 " # b Note that a8 # ba8 # ! " b !" œ a8 ! # ba8 " # b . Setting F8 œ !, it follows that for 8 " , ,8 œ Therefore a second solution is C# aBb œ B"# ”" a" ! # ba" " # b B a# # b"x a" ! # ba# ! # ba" " # ba# " # b # B â •Þ a# # ba$ # b# x a/b. Under the transformation B œ "Î0 , the ODE becomes " " .# C " " " .C 0% Œ" # œ#0$ Œ" 0# ”# a" ! " b • !" C œ ! . .0 0 0 .0 0 0 0 ˆ0 $ 0 # ‰ .# C .C #0# # 0# a " ! " b0‘ !" C œ ! . . 0# .0 a# # b 0 a " ! " b !" and ; a0b œ $ . #0 0 0 0# a# # b 0 a " ! " b œ "!", 0Ä! 0" !" œ !" Þ 0" That is, Therefore 0 œ ! is a singular point. Note that : a0 b œ It follows that 0Ä! :! œ lim0 :a0b œ lim Hence 0 œ ! aB œ _b is a regular singular point. The indicial equation is or <# a! " b< !" œ ! Þ Evidently, the roots are < œ ! and < œ " . 21a+b. Note that :aBb œ ! " and ; a0b œ > . = B B <a< "b a" ! " b< !" œ ! , ;! œ lim0# ; a0b œ lim 0Ä! 0Ä! ________________________________________________________________________ page 243 —————————————————————————— CHAPTER 5. —— It follows that BÄ! lim B :aBb œ lim ! B"= , BÄ! 0Ä! lim0# ; a0b œ lim " B#= Þ 0Ä! Hence if = " or > # , one or both of the limits does not exist. Therefore B œ ! is an irregular singular point. a- b. Let C œ +! B< +" B<" â +8 B<8 â Þ Write the ODE as B$ C ww ! B# C w " C œ ! . Substitution of the assumed solution results in " a8 <ba8 < "b+8 B8<" !" a8 <b+8 B8<" " " +8 B8< œ !Þ _ _ _ 8œ! 8œ! 8œ! Adjusting the indices, we obtain " a8 " <ba8 < #b+8" B8< !" a8 " <b+8" B8< " " +8 B8< œ !Þ _ _ _ 8œ" 8œ! 8œ" Combining the series, " +! " E8 B8< œ ! , _ in which E8 œ " +8 a8 " <ba8 < ! #b+8" . Setting the coefficients equal to zero, we have +! œ ! . But for 8 " , +8 œ a8 " <ba8 < ! #b +8" Þ " 8œ" Therefore, regardless of the value of <, it follows that +8 œ ! , for 8 œ "ß #ß â . ________________________________________________________________________ page 244 —————————————————————————— CHAPTER 5. —— Section 5.8 3. Here B :aBb œ " and B# ; aBb œ #B , which are both analytic everywhere. We set C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in 8œ! " a< 8ba< 8 "b+8 B<8 " a< 8b+8 B<8 #" +8 B<8" œ ! Þ _ _ _ 8œ! 8œ! After adjusting the indices in the last series, we obtain +! c<a< "b <dB< " ca< 8ba< 8 "b+8 a< 8b+8 # +8" dB<8 œ !Þ _ 8œ" Assuming +! Á ! , the indicial equation is <# œ ! , with double root < œ ! . Setting the remaining coefficients equal to zero, we have for 8 " , + 8 a< b œ It follows that + 8 a< b œ a8 < b # # +8" a<b . ca8 <ba8 < "bâa" <bd# C" aBb œ " _ a "b 8 # 8 +! , 8 " . Since < œ ! , one solution is given by a "b 8 # 8 a8xb# B8 . 8œ! For a second linearly independent solution, we follow the discussion in Section &Þ( . First note that w + 8 a< b " " " œ #” â •. + 8 a< b 8< 8<" "< w + 8 a !b Setting < œ ! , œ # L8 +8 a!b œ # L8 a "b 8 # 8 a8xb# Þ Therefore, C# aBb œ C" aBb 68 B #" _ a "b 8 # 8 L 8 a8xb# B8 . 8œ! 4. Here B :aBb œ % and B# ; aBb œ # B , which are both analytic everywhere. We set C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in ________________________________________________________________________ page 245 —————————————————————————— CHAPTER 5. —— " a< 8ba< 8 "b+8 B<8 %" a< 8b+8 B<8 _ _ 8œ! _ 8œ! " +8 B<8" #" +8 B<8 œ ! Þ _ 8œ! 8œ! After adjusting the indices in the second-to-last series, we obtain +! c<a< "b %< #dB< " ca< 8ba< 8 "b+8 %a< 8b+8 # +8 +8" dB<8 œ !Þ _ 8œ" Assuming +! Á ! , the indicial equation is <# $< # œ ! , with roots <" œ " and <# œ # . Setting the remaining coefficients equal to zero, we have for 8 " , + 8 a< b œ It follows that + 8 a< b œ " +8" a<b . a8 < "ba8 < #b a "b 8 +! , 8 " . ca8 < "ba8 <bâa# <bdca8 < #ba8 <bâa$ <bd C" aBb œ B" " _ Since <" œ " , one solution is given by a "b 8 B8 . a8bxa8 "bx 8œ! For a second linearly independent solution, we follow the discussion in Section &Þ( . Since <" <# œ R œ " , we find that + " a< b œ " , a< #ba< $b with +! œ " . Hence the leading coefficient in the solution is + œ lim a< #b +" a<b œ " . <Ä# Further, a < #b + 8 a < b œ Let E8 a<b œ a< #b +8 a<b . It follows that a8 < #b ca8 < "ba8 <bâa$ <bd# a "b 8 Þ w E8 a<b " " " " œ #” â •. E8 a<b 8<# 8<" 8< $< Setting < œ <# œ # , ________________________________________________________________________ page 246 —————————————————————————— CHAPTER 5. —— w E8 a #b " " " œ #” â "• E8 a #b 8 8" 8# œ L8 L8" Þ Hence -8 a #b œ aL8 L8" b E8 a #b a "b 8 œ aL8 L8" b . 8xa8 "bx C# aBb œ C" aBb 68 B B –" " _ # Therefore, a "b8 aL8 L8" b 8 B —. 8xa8 "bx 8œ" 6. Let CaBb œ @aBbÎÈB . Then C w œ B"Î# @ w B$Î# @Î# and C ww œ B"Î# @ ww B$Î# @ w $ B&Î# @Î% . Substitution into the ODE results in " B$Î# @ ww B"Î# @ w $ B"Î# @Î%‘ B"Î# @ w B"Î# @Î#‘ ŒB# B"Î# @ œ ! . % Simplifying, we find that with general solution @aBb œ -" -9= B -# =38 B . Hence @ ww @ œ ! , CaBb œ -" B"Î# -9= B -# B"Î# =38 B . 8. The absolute value of the ratio of consecutive terms is +#7# B#7# kBk#7# ##7 a7 "bx 7x kBk# œ . º ºœ +#7 B#7 %a7 #ba7 "b kBk#7 ##7# a7 #bxa7 "bx lim º 7Ä_ +#7# B#7# kBk# œ lim œ !. º 7Ä_ +#7 B#7 %a7 #ba7 "b Applying the ratio test, Hence the series for N" aBb converges absolutely for all values of B . Furthermore, since the series for N! aBb also converges absolutely for all B, term-by-term differentiation results in ________________________________________________________________________ page 247 ————————————————————————— CHAPTER 5. —— N! aBb œ " _ w _ a "b7" B#7" œ " #7" # a7 "bx 7x 7œ! œ a "b7 B#7" ##7" 7xa7 "bx 7œ" Therefore, N!w aBb œ N" aBb . B_ a "b7 B#7 " #7 . # 7 œ ! # a7 "bx 7x 9a+b. Note that B :aBb œ " and B# ; aBb œ B# / # , which are both analytic at B œ ! . Thus B œ ! is a regular singular point. Furthermore, :! œ " and ;! œ / # . Hence the indicial equation is <# / # œ ! , with roots <" œ / and <# œ / . a,b. Set C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in " a< 8ba< 8 "b+8 B<8 " a< 8b+8 B<8 _ _ 8œ! _ 8œ! " +8 B<8# / # " +8 B<8 œ ! Þ _ 8œ! 8œ! After adjusting the indices in the second-to-last series, we obtain _ +! <a< "b < / # ‘B< +" a< "b< a< "b / # ‘ 8œ# " a< 8ba< 8 "b+8 a< 8b+8 / # +8 +8# ‘B<8 œ !Þ Setting the coefficients equal to zero, we find that +" œ ! , and +8 œ a < 8b # / # " +8# , for 8 # . It follows that +$ œ +& œ â œ +#7" œ â œ ! . Furthermore, with <œ/, +8 œ So for 7 œ "ß #ß â , +#7 œ " +#7# #7a#7 #/ b a "b 7 œ #7 +! . # 7xa" / ba# / bâa7 " / ba7 / b " +8# . 8a8 #/ b ________________________________________________________________________ page 248 —————————————————————————— CHAPTER 5. —— Hence one solution is / C" aBb œ B –" " _ a "b 7 B #7 Š ‹ —. 7xa" / ba# / bâa7 " / ba7 / b # 7œ" a- b. Assuming that <" <# œ #/ is not an integer, simply setting < œ / in the above results in a second linearly independent solution C# aBb œ B/ –" " _ a "b 7 B #7 Š ‹ —Þ 7xa" / ba# / bâa7 " / ba7 / b # 7œ" a. b. The absolute value of the ratio of consecutive terms in C" aBb is +#7# B#7# kBk#7# ##7 7xa" @bâa7 / b œ º º +#7 B#7 kBk#7 ##7# a7 "bxa" @bâa7 " / b kBk# œ . %a7 "ba7 " / b lim º 7Ä_ +#7# B#7# kBk# œ lim œ !. º 7Ä_ +#7 B#7 %a7 "ba7 " / b Applying the ratio test, Hence the series for C" aBb converges absolutely for all values of B . The same can be shown for C# aBb . Note also, that if / is a positive integer, then the coefficients in the series for C# aBb are undefined. 10a+b. It suffices to calculate PcN! aBb 68 Bd. Indeed, cN! aBb 68 Bd w œ N!w aBb 68 B N! aBb B and N!w aBb N! aBb cN! aBb 68 Bd œ N! aBb 68 B # . B B# ww ww Hence Since B# N!ww aBb B N!w aBb B# N! aBb œ ! , PcN! aBb 68 Bd œ B# N!ww aBb 68 B #B N!w aBb N! aBb B N!w aBb 68 B N! aBb B# N! aBb 68 B Þ PcN! aBb 68 Bd œ #B N!w aBb . ________________________________________________________________________ page 249 —————————————————————————— CHAPTER 5. —— a,b. Given that PcC# aBbd œ ! , after adjusting the indices in Part a+b, we have ," B ## ,# B# " ˆ8# ,8 ,8# ‰B8 œ #B N!w aBb . _ 8œ$ Using the series representation of N!w aBb in Problem ) , _ # # 8œ$ ," B # ,# B " ˆ8# ,8 ,8# ‰B8 œ # " _ a "b8 a#8bB#8 ##8 a8xb# . 8œ" a- b. Equating the coefficients on both sides of the equation, we find that Also, with 8 œ " , ## ,# œ "Îa"xb# , that is, ,# œ "Î## a"xb# ‘. Furthermore, for 7 #, a#7b# ,#7 ,#7# œ # a "b7 a#7b ##7 a7xb# . ," œ ,$ œ â œ ,#7" œ â œ ! Þ More explicitly, " " " #Œ % # " "" ,' œ # # # Œ" #%' #$ ã ,% œ ## It can be shown, in general, that ,#7 œ a "b7" 11. Bessel's equation of order one is ##7 a7xb# L7 . B# C ww B C w ˆB# "‰C œ ! . Based on Problem *, the roots of the indicial equation are <" œ " and <# œ " . Set C œ B< a+! +" B +# B# +8 B8 âb. Substitution into the ODE results in 8œ! " a< 8ba< 8 "b+8 B<8 " a< 8b+8 B<8 _ _ 8œ! _ " +8 B<8# " +8 B<8 œ ! Þ _ 8œ! 8œ! After adjusting the indices in the second-to-last series, we obtain ________________________________________________________________________ page 250 —————————————————————————— CHAPTER 5. —— +! c<a< "b < "dB< +" ca< "b< a< "b "d _ 8œ# " ca< 8ba< 8 "b+8 a< 8b+8 +8 +8# dB<8 œ !Þ + 8 a< b œ +8# a<b a < 8b # " " œ +8# a<b , a8 < "ba8 < "b " a "b 7 Setting the coefficients equal to zero, we find that +" œ ! , and for 8 # . It follows that +$ œ +& œ â œ +#7" œ â œ !. Solving the recurrence relation, +#7 a<b œ With < œ <" œ ", a#7 < "ba#7 < "b# âa< $b# a< "b a "b 7 +#7 a"b œ #7 +! Þ # a7 "bx 7x +! . For a second linearly independent solution, we follow the discussion in Section &Þ( . Since <" <# œ R œ # , we find that + # a< b œ " , a< $ba< "b " . # with +! œ " . Hence the leading coefficient in the solution is + œ lim a< "b +# a<b œ <Ä" Further, a< "b +#7 a<b œ Let E8 a<b œ a< "b +8 a<b . It follows that a#7 < "b ca#7 < "bâa$ <bd# a "b 7 Þ w E#7 a<b " " " œ #” â •. E#7 a<b #7 < " #7 < " $< Setting < œ <# œ " , we calculate ________________________________________________________________________ page 251 —————————————————————————— CHAPTER 5. —— " -#7 a "b œ aL7 L7" bE#7 a "b # " a "b 7 œ aL7 L7" b # #7 ca#7 #bâ#d# " a "b 7 œ aL7 L7" b #7" Þ # # 7xa7 "bx -#7" a "b œ ” . E#7" a<b• œ !. .< <œ<# Note that +#7" a<b œ ! implies that E#7" a<b œ ! , so Therefore, Based on the definition of N" aBb, _ _ " " B #7 a "b 7 a "b7 aL7 L7" b B #7 C# aBb œ –B " Š ‹ —68 B –" " Š ‹ —Þ # 7 œ ! a7 "bx 7x # 7xa7 "bx # B 7œ" _ " a "b7 aL7 L7" b B #7 C# aBb œ N" aBb 68 B –" " Š ‹ —Þ B 7xa7 "bx # 7œ" 12. Consider a solution of the form CaBb œ ÈB 0 ˆ!B" ‰. .0 !" B" 0 a0 b † . 0 ÈB #È B Then Cw œ in which 0 œ !B" . Hence C ww œ and B# C ww œ !# " # B#" ÈB Substitution into the ODE results in !# " # B#" . # 0 !# " # B#" .0 !" # B" 0 a0b † † , . 0# . 0 BÈB BÈB %BÈB .# 0 .0 " !" # B" ÈB ÈB 0 a0b Þ . 0# .0 % .# 0 .0 " " !" # B" 0 a0b Œ!# " # B#" / # " # 0 a0b œ ! Þ # .0 .0 % % Simplifying, and setting 0 œ !B" , we find that ________________________________________________________________________ page 252 —————————————————————————— CHAPTER 5. —— .# 0 .0 0 ˆ0# / # ‰0 a0b œ ! , a‡b . 0# .0 0# which is a Bessel equation of order / . Therefore, the general solution of the given ODE is in which 0" a0b and 0# a0b are the linearly independent solutions of a‡b. CaBb œ ÈB -" 0" ˆ!B" ‰ -# 0# ˆ!B" ‰‘, ________________________________________________________________________ page 253 —————————————————————————— CHAPTER 6. —— Chapter Six Section 6.1 3. The function 0 a>b is continuous. 4. The function 0 a>b has a jump discontinuity at > œ " . 7. Integration is a linear operation. It follows that ( E ! -9=2 ,> † /=> .> œ " E ,> => " E ,> => ( / † / .> ( / † / .> #! #! E " "E œ ( /a,=b> .> ( /a,=b> .> Þ #! #! Hence ( E -9=2 ,> † / ! => " " /a,=bE " " /a,=bE .> œ – —Þ # = , — #– =, Taking a limit, as E p _ , ________________________________________________________________________ page 254 —————————————————————————— CHAPTER 6. —— ( _ ! -9=2 ,> † /=> .> œ Note that the above is valid for = k,k . 8. Proceeding as in Prob. ( , ( E ! " " " " ” • ” • # =, # =, = œ# Þ = ,# =382 ,> † /=> .> œ " " /a,=bE " " /a,=bE – —Þ #– = , — # =, " " " " ” • ” • # =, # =, , œ# Þ = ,# Taking a limit, as E p _ , ( _ ! =382 ,> † /=> .> œ The limit exists as long as = k,k . ( 10. Observe that /+> =382 ,> œ ˆ/Ð+,Ñ> /Ð+,Ñ> ‰Î# Þ It follows that E ! /+> =382 ,> † /=> .> œ " " /a+,=bE " " /a,+=bE – Þ #– = + , — # =,+ — " " " " ” • ” • # =+, # =,+ , œ Þ a= + b # , # Taking a limit, as E p _ , ( _ ! /+> =382 ,> † /=> .> œ The limit exists as long as = + k,k . _c =38 ,>d œ Since ( we have _ ! 11. Using the linearity of the Laplace transform, " " _/3,> ‘ _/3,> ‘Þ #3 #3 " , = + 3, /a+3,b> /=> .> œ ________________________________________________________________________ page 255 —————————————————————————— CHAPTER 6. —— ( Therefore _c=38 ,>d œ " " " ” • #3 = 3, = 3, , . œ# = ,# _ ! /„ 3,> /=> .> œ " . = … 3, 12. Using the linearity of the Laplace transform, _c -9= ,>d œ From Prob. "" , we have ( Therefore _c-9= ,>d œ _ ! " " _/3,> ‘ _/3,> ‘Þ # # /„ 3,> /=> .> œ " . = … 3, " " " ” • # = 3, = 3, = œ# . = ,# 14. Using the linearity of the Laplace transform, _ /+> -9= ,>‘ œ ( Therefore _/+> -9= ,>‘ œ _ ! " " _/a+3,b> ‘ _/a+3,b> ‘Þ # # Based on the integration in Prob. "" , /a+ „ 3,b> /=> .> œ " Þ = + … 3, " " " ” • # = + 3, = + 3, =+ œ . a= + b # , # The above is valid for = + . 15. Integrating by parts, ________________________________________________________________________ page 256 —————————————————————————— CHAPTER 6. —— ( E > /a+=b> E " a+=b> .> œ / .> º ( =+ ! ! =+ " /Ea+=b Ea+ =b/Ea+=b œ Þ a= + b # E ! >/ † / +> => Taking a limit, as E p _ , ( _ ! >/+> † /=> .> œ a= + b # " Þ Note that the limit exists as long as = + . 17. Observe that > -9=2 +> œ a> /+> > /+> bÎ# Þ For any value of - , ( E ! E > /a-=b> E " a-=b> .> œ / .> º ( =- ! ! =" /Ea-=b Ea- =b/Ea-=b œ Þ a= - b # >/ † / -> => Taking a limit, as E p _ , ( _ ! >/-> † /=> .> œ Note that the limit exists as long as = k- k . Therefore, ( _ ! a= - b # " Þ > -9=2 +> † /=> .> œ œ " " " – — # # a= + b a= + b # a= + b # a= + b # =# + # . 18. Integrating by parts, ( E >8 /+> † /=> .> œ œ ! E >8 /a+=b> E 8 8" a+=b> ( >/ .> º =+ ! ! =+ E E8 /a=+bE 8 8" a+=b> ( >/ .> Þ =+ ! =+ Continuing to integrate by parts, it follows that ________________________________________________________________________ page 257 —————————————————————————— CHAPTER 6. —— ( E8 /a+=bE 8E8" /a+=bE =+ a= + b # 8xˆ/a+=bE "‰ 8xE/a+=bE â Þ a8 #bxa= +b$ a= +b8" a= +b8" 8x E ! >8 /+> † /=> .> œ That is, ( E ! in which :8 a0b is a polynomial of degree 8 . For any given polynomial, EÄ_ >8 /+> † /=> .> œ :8 aEb † /a+=bE , lim :8 aEb † /a=+bE œ ! , as long as = + . Therefore, ( _ ! >8 /+> † /=> .> œ a= +b8" 8x . 20. Observe that ># =382 +> œ a># /+> ># /+> bÎ# Þ Using the result in Prob. ") , ( _ ! ># =382 +> † /=> .> œ œ " #x #x – — # a= + b $ a= + b $ #+a$=# +# b a =# + # b $ . The above is valid for = k+k. 22. Integrating by parts, ( Taking a limit, as E p _ , ( Hence the integral converges . _ E ! > /> .> œ > /> º ( E ! E /> .> Þ œ"/ E E/ ! E > /> .> œ " /E Þ ! 23. Based on a series expansion, note that for > ! , /> " > ># Î# ># Î# . ________________________________________________________________________ page 258 —————————————————————————— CHAPTER 6. —— It follows that for > ! , ># /> Hence for any finite E " , ( E " " . # ># /> .> E" . # It is evident that the limit as E p _ does not exist. 24. Using the fact that k-9= >k Ÿ " , and the fact that ( _ /> .> œ " , ! it follows that the given integral converges. 25a+b. Let : ! . Integrating by parts, ( E ! /B B: .B œ /B B: º :( E ! E /B B:" .B œ E: /E :( _ ! ! E /B B:" .BÞ ! Taking a limit, as E p _ , ( _ ! That is, >a: "b œ : >a:b . a,b. Setting : œ ! , /B B: .B œ :( /B B:" .B . >a"b œ ( _ /B .B œ " . ! a- b. Let : œ 8 . Using the result in Part a,b, Since >a"b œ " , >a8 "b œ 8x . a. b. Using the result in Part a,b, >a8 "b œ 8 >a8b œ 8a8 "b>a8 "b ã œ 8a8 "ba8 #bâ# † " † >a"b . ________________________________________________________________________ page 259 —————————————————————————— CHAPTER 6. —— >a: 8b œ a: 8 "b >a: 8 "b œ a: 8 "ba: 8 #b>a: 8 #b ã œ a: 8 "ba: 8 #bâa: "b: >a:b . Hence >a: 8b œ :a: "ba: "bâa: 8 "b Þ >a : b È1 $ " " >Œ œ >Œ œ # # # # Given that >a"Î#b œ È1 , it follows that and >Œ *%&È1 "" *(&$ $ . œ † † † >Œ œ # #### # $# ________________________________________________________________________ page 260 —————————————————————————— CHAPTER 6. —— Section 6.2 1. Write the function as $ $# œ . =# % # =# % $ # Hence _" c] a=bd œ =38 #> . 3. Using partial fractions, =# # # " " œ” •. $= % & =" =% Hence _" c] a=bd œ # ˆ/> /%> ‰. & 5. Note that the denominator =# #= & is irreducible over the reals. Completing the square, =# #= & œ a= "b# % . Now convert the function to a rational function of the variable 0 œ = " . That is, =# We know that _" ” 0# #0 • œ # -9= #> . % #= # #a= "b œ . #= & a = "b # % Using the fact that _c/+> 0 a>bd œ _c0 a>bd=p=+ , _" ” =# #= # > • œ #/ -9= #> . #= & 6. Using partial fractions, #= $ " " ( œ” •. #% = % =# =# Hence _" c] a=bd œ " a/#> (/#> b. Note that we can also write % #= $ = $# œ# # . #% = = % # =# % 8. Using partial fractions, )=# %= "# " = # œ$ & # # # . # %b =a = = = % = % ________________________________________________________________________ page 261 —————————————————————————— CHAPTER 6. —— Hence _" c] a=bd œ $ & -9= #> # =38 #> . 9. The denominator =# %= & is irreducible over the reals. Completing the square, =# %= & œ a= #b# " . Now convert the function to a rational function of the variable 0 œ = # . That is, =# We find that _" ” 0# & #0 # • œ & =38 > # -9= > . " 0 " " #= & #a = # b œ . %= & a = #b # " Using the fact that _c/+> 0 a>bd œ _c0 a>bd=p=+ , _" ” =# " #= #> • œ / a& =38 > # -9= >b . %= & 10. Note that the denominator =# #= "! is irreducible over the reals. Completing the square, =# #= "! œ a= "b# * . Now convert the function to a rational function of the variable 0 œ = " . That is, =# We find that _" ” 0# #0 & & # • œ # -9= $> =38 $> . * 0 * $ #= $ #a= "b & œ . #= "! a = "b # * Using the fact that _c/+> 0 a>bd œ _c0 a>bd=p=+ , _" ” =# & #= $ > • œ / Œ# -9= $> =38 $> . $ #= "! 12. Taking the Laplace transform of the ODE, we obtain =# ] a=b = Ca!b C w a!b $c= ] a=b C a!bd # ] a=b œ ! . =# ] a=b $= ] a=b # ] a=b = $ œ ! . ] a =b œ =$ . $= # Applying the initial conditions, Solving for ] a=b, the transform of the solution is =# ________________________________________________________________________ page 262 —————————————————————————— CHAPTER 6. —— Using partial fractions, # " =$ œ . =" =# =# $= # Hence Ca>b œ _" c] a=bd œ # /> /#> . 13. Taking the Laplace transform of the ODE, we obtain =# ] a=b = Ca!b C w a!b #c= ] a=b C a!bd # ] a=b œ ! . =# ] a=b #= ] a=b # ] a=b " œ ! . ] a =b œ " . =# #= # Applying the initial conditions, Solving for ] a=b, the transform of the solution is Since the denominator is irreducible, write the transform as a function of 0 œ = " . That is, =# First note that _" ” 0# " • œ =38 > . " " " œ . #= # a = "b # " Using the fact that _c/+> 0 a>bd œ _c0 a>bd=p=+ , Hence Ca>b œ /> =38 > . _" ” =# " > • œ / =38 > . #= # 15. Taking the Laplace transform of the ODE, we obtain =# ] a=b = Ca!b C w a!b #c= ] a=b C a!bd # ] a=b œ ! . =# ] a=b #= ] a=b # ] a=b #= % œ ! . ] a =b œ #= % . #= # Applying the initial conditions, Solving for ] a=b, the transform of the solution is =# Since the denominator is irreducible, write the transform as a function of 0 œ = " . Completing the square, ________________________________________________________________________ page 263 —————————————————————————— CHAPTER 6. —— #= % #a= "b # œ . #= # a = "b # $ =# First note that _" ” Using the fact that _c/+> 0 a>bd œ _c0 a>bd=p=+ , the solution of the IVP is Ca>b œ _" ” =# 0# #0 # # # • œ # -9=2 È$ > È =382 È$ > . $ 0 $ $ #= % # > • œ / # -9=2 È$ > È =382 È$ > . #= # $ 16. Taking the Laplace transform of the ODE, we obtain =# ] a=b = Ca!b C w a!b #c= ] a=b C a!bd & ] a=b œ ! . =# ] a=b #= ] a=b & ] a=b #= $ œ ! . ] a =b œ #= $ . #= & Applying the initial conditions, Solving for ] a=b, the transform of the solution is =# Since the denominator is irreducible, write the transform as a function of 0 œ = " . That is, #= $ #a= "b " œ . =# #= & a = "b # % We know that _" ” 0# #0 " " # • œ # -9= #> =38 #> . % 0 % # Using the fact that _c/+> 0 a>bd œ _c0 a>bd=p=+ , the solution of the IVP is Ca>b œ _" ” =# " #= $ > • œ / Œ# -9= #> =38 #> . #= & # 17. Taking the Laplace transform of the ODE, we obtain =% ] a=b =$ Ca!b =# C w a!b = C ww a!b C www a!b %=$ ] a=b =# Ca!b = C w a!b C ww a!b‘ '=# ] a=b = Ca!b C w a!b‘ %c= ] a=b C a!bd ] a=b œ ! Applying the initial conditions, ________________________________________________________________________ page 264 —————————————————————————— CHAPTER 6. —— =% ] a=b %=$ ] a=b '=# ] a=b %= ] a=b ] a=b =# %= ( œ ! . Solving for the transform of the solution, ] a =b œ Using partial fractions, =# %= ( a= "b % =# %= ( =# %= ( œ . =% %=$ '=# %= " a = "b % a= "b % a= "b # a= "b# " œ % $ . Note that _c >8 d œ a8xbÎ=8" and _c/+> 0 a>bd œ _c0 a>bd=p=+ . Hence the solution of the IVP is Ca>b œ _" – =# %= ( a = "b % #$> #> > — œ $ > / > / >/ . 18. Taking the Laplace transform of the ODE, we obtain =% ] a=b =$ C a!b =# C w a!b = C ww a!b C www a!b ] a=b œ ! Þ =% ] a=b ] a=b = $ = œ ! . ] a =b œ = . " Applying the initial conditions, Solving for the transform of the solution, By inspection, it follows that Ca>b œ _" =#= ‘ œ -9=2 > Þ " 19. Taking the Laplace transform of the ODE, we obtain =% ] a=b =$ C a!b =# C w a!b = C ww a!b C www a!b % ] a=b œ ! Þ =% ] a=b %] a=b =$ #= œ ! . ] a =b œ = . # =# Applying the initial conditions, Solving for the transform of the solution, It follows that Ca>b œ _" =#= ‘ œ -9= È# > Þ # =# 20. Taking the Laplace transform of both sides of the ODE, we obtain ________________________________________________________________________ page 265 —————————————————————————— CHAPTER 6. —— =# ] a=b = Ca!b C w a!b =# ] a=b œ Applying the initial conditions, =# ] a=b =# ] a=b = œ = . % =# Solving for ] a=b, the transform of the solution is = = ] a =b œ # # . # ba=# %b a= = = =# Using partial fractions on the first term, a =# First note that _" ” =# = • œ -9= => =# and _" ” =# = • œ -9= #> . % =# ba=# = %b œ " = = # •. # ” =# =# %= = % =# = . % Hence the solution of the IVP is Ca>b œ " " -9= => -9= #> -9= => # %= % =# & =# " œ -9= => -9= #> . # %= % =# 21. Taking the Laplace transform of both sides of the ODE, we obtain =# ] a=b = Ca!b C w a!b #c= ] a=b C a!bd # ] a=b œ =# ] a=b #= ] a=b # ] a=b = # œ ] a =b œ =# = . " Applying the initial conditions, Solving for ] a=b, the transform of the solution is a =# =# = . " = =# # . # "b #= #ba= = #= # Using partial fractions on the first term, a =# Thus we can write = " =# =% œ ”# # •. # "b #= #ba= & = " = #= # ________________________________________________________________________ page 266 —————————————————————————— CHAPTER 6. —— ] a =b œ For the last term, we note that =# #= # œ a= "b# " . So that =# We know that _" ” 0# #0 " # • œ # -9= > =38 > . " 0 " #= $ #a= "b " œ . #= # a = "b # " "= #" # #= $ . #" #" &= &= & =# #= # Based on the translation property of the Laplace transform, _" ” C a >b œ #= $ œ /> a# -9= > =38 >b. # #= # • = Combining the above, the solution of the IVP is " # # -9= > =38 > /> a# -9= > =38 >b Þ & & & 23. Taking the Laplace transform of both sides of the ODE, we obtain =# ] a=b = Ca!b C w a!b #c= ] a=b C a!bd ] a=b œ Applying the initial conditions, =# ] a=b #= ] a=b ] a=b #= $ œ ] a =b œ % . =" % . =" Solving for ] a=b, the transform of the solution is a= "b % $ a= "b# #= $ . First write a= "b We note that _" ” #= $ # œ #a= " b " a= "b # œ # " Þ = " a= "b# % # " # • œ # ># # > . $ 0 0 0 So based on the translation property of the Laplace transform, the solution of the IVP is ________________________________________________________________________ page 267 —————————————————————————— CHAPTER 6. —— Ca>b œ # ># /> > /> # /> Þ 25. Let 0 a>b be the forcing function on the right-hand-side. Taking the Laplace transform of both sides of the ODE, we obtain =# ] a=b = Ca!b C w a!b ] a=b œ _c0 a>bd. =# ] a=b ] a=b œ _c0 a>bd. _ ! Applying the initial conditions, Based on the definition of the Laplace transform, _ c 0 a >b d œ ( 0 a>b /=> .> œ ( > /=> .> " ! " /= /= œ # #. = = = Solving for the transform, ] a =b œ Using partial fractions, =# a =# and =# a =# We find, by inspection, that _" ” =# a =# " • œ > =38 > . "b = " = œ # Þ "b = = " " " " œ # # "b = = " =# a =# " =" /= # # Þ "b = a = "b Referring to Line "$ , in Table 'Þ#Þ" , _c ?- a>b0 a> - bd œ /-= _c0 a>bd. =# a =# =" " " = " œ # # # . "b == = " = " Let _c 1a>b d œ ________________________________________________________________________ page 268 —————————————————————————— CHAPTER 6. —— Then 1a>b œ " > -9= > =38 > . It follows, therefore, that _" ”/= † =# a =# =" • œ ?" a>bc " a> "b -9=a> "b =38a> "b d . "b Combining the above, the solution of the IVP is Ca>b œ > =38 > ?" a>bc " a> "b -9=a> "b =38a> "b d . 26. Let 0 a>b be the forcing function on the right-hand-side. Taking the Laplace transform of both sides of the ODE, we obtain =# ] a=b = Ca!b C w a!b % ] a=b œ _c0 a>bd. =# ] a=b % ] a=b œ _c0 a>bd. _ Applying the initial conditions, Based on the definition of the Laplace transform, _ c 0 a >b d œ ( 0 a>b /=> .> => œ ( >/ " ! ! " /= œ # # . = = Solving for the transform, ] a =b œ Using partial fractions, =# a =# We find that _" ” =# a =# =# a =# .> ( _ " /=> .> " " /= # # Þ %b = a = %b " "" " œ ” # # •Þ %b %= = % " " " • œ > =38 > . %b % ) Referring to Line "$ , in Table 'Þ#Þ" , _c ?- a>b0 a> - bd œ /-= _c0 a>bd. It follows that ________________________________________________________________________ page 269 —————————————————————————— CHAPTER 6. —— _" ”/= † C a >b œ " " " • œ ?" a>b” a> "b =38a> "b•. %b % ) =# a =# Combining the above, the solution of the IVP is " " " " > =38 > ?" a>b” a> "b =38a> "b•. % ) % ) 28a+b. Assuming that the conditions of Theorem 'Þ#Þ" are satisfied, . _ => J a =b œ ( / 0 a>b.> .= ! _ ` => / 0 a>b‘.> œ( ! `= w œ( œ( _ ! _ ! /=> c >0 a>bd.> Þ > /=> 0 a>b‘.> a,b. Using mathematical induction, suppose that for some 5 " , J a5 b a=b œ ( _ ! /=> ’a >b5 0 a>b“.> . Differentiating both sides, J a5"b a=b œ . _ => 5 ( / ’a >b 0 a>b“.> .= ! _ ` => œ( ’/ a >b5 0 a>b“.> ! `= œ( œ( 29. We know that _ ! _ ! /=> ’a >b5" 0 a>b“.> . ’ > /=> a >b5 0 a>b“.> _/+> ‘ œ _ > /+> ‘ œ " . =+ . " ” •. .= = + Based on Prob. #) , ________________________________________________________________________ page 270 —————————————————————————— CHAPTER 6. —— Therefore, _> /+> ‘ œ a= + b # " . 31. Based on Prob. #) , .8 _ c a >b d œ 8 _ c " d .= .8 " œ 8 ” •Þ .= = 8 Therefore, _c >8 d œ a "b8 œ 8x =8" Þ a "b8 8x =8" 33. Using the translation property of the Laplace transform, _/+> =38 ,>‘ œ a= + b # , # , Þ Therefore, _> /+> =38 ,>‘ œ œ . , – .= a= +b# ,# — #,a= +b . a=# #+= +# ,# b# 34. Using the translation property of the Laplace transform, =+ _/+> -9= ,>‘ œ Þ a= + b # , # Therefore, _> /+> -9= ,>‘ œ œ . =+ – .= a= +b# ,# — a= + b # , # . a=# #+= +# ,# b# 35a+b. Taking the Laplace transform of the given Bessel equation, ________________________________________________________________________ page 271 —————————————————————————— CHAPTER 6. —— _c > C ww d _c C w d _c > C d œ ! . Using the differentiation property of the transform, That is, It follows that . . _c C ww d _c C w d _c C d œ ! . .= .= .# . = ] a=b = C a!b C w a!b‘ = ] a=b C a!b ] a=b œ ! . .= .= ˆ" =# ‰] w a=b = ] a=b œ ! . a,b. We obtain a first-order linear ODE in ] a=b: = ] w a =b # ] a=b œ ! , = " with integrating factor .a=b œ /B:Œ( The first-order ODE can be written as =# = .= œ È=# " . " . È# ’ = " † ] a=b“ œ ! , .= ] a =b œ È =# " - with solution . a- b. In order to obtain negative powers of = , first write " Expanding Š" " "Î# =# ‹ " " "Î# œ ”" # • Þ È =# " = = in a binomial series, valid for =# " . Hence, we can formally express ] a=b as " " " † $ % " † $ † & ' œ " =# = = â, È" a"Î=# b # #†% #†%†' " "" "†$ " "†$†& " ] a =b œ - ” â•Þ = # =$ # † % =& # † % † ' =( ________________________________________________________________________ page 272 —————————————————————————— CHAPTER 6. —— Assuming that term-by-term inversion is valid, C a >b œ - ” " " ># " † $ >% " † $ † & >' â• # #x # † % %x # † % † ' 'x #x ># %x >% 'x >' œ - ”" # # # # # # â•Þ # #x # † % %x # † % † ' 'x "# " " > # # >% # # # > ' â • # # # †% # †% †' 8 _ a "b #8 œ -" >Þ # #8 8 œ ! # a8xb It follows that Ca>b œ - ”" The series is evidently the expansion, about B œ ! , of N! a>bÞ 36a,b. Taking the Laplace transform of the given Legendre equation, _c C ww d _ ># C ww ‘ # _c > C w d !a! "b_c C d œ ! . Using the differentiation property of the transform, .# . _c C d # _c C ww d # _c C w d !a! "b_c C d œ ! . .= .= ww That is, =# ] a=b = Ca!b C w a!b‘ .# # = ] a=b = C a!b C w a!b‘ # .= . # c= ] a=b C a!bd !a! "b] a=b œ ! Þ .= Invoking the initial conditions, we have .# # . = ] a=b " # = ] a=b "‘ # c= ] a=bd !a! "b] a=b œ ! Þ .= .= # After carrying out the differentiation, the equation simplifies to .# # . = ] a=b‘ # c= ] a=bd =# !a! "b‘] a=b œ " Þ .=# .= That is, =# .# . ] a=b #= ] a=b =# !a! "b‘] a=b œ " . .=# .= 37. By definition of the Laplace transform, given the appropriate conditions, ________________________________________________________________________ page 273 —————————————————————————— CHAPTER 6. —— _c 1a>bd œ ( œ( /=> ”( 0 a7 b. 7 •.> > => ( / 0 a7 b. 7 .> Þ > ! ! _ ! _ ! Assuming that the order of integration can be exchanged, _c 1a>bd œ ( œ( _ ! _ ! 0 a7 b ”( _ 7 cNote the region of integration is the area between the lines 7 a>b œ > and 7 a>b œ ! Þd Hence _c 1a>bd œ "_ =7 ( 0 a7 b / . 7 =! " œ _ c 0 a> b d Þ = /=7 0 a7 b ” •. 7 Þ = /=> .>•. 7 ________________________________________________________________________ page 274 —————————————————————————— CHAPTER 6. —— Section 6.3 1. 3. 5. ________________________________________________________________________ page 275 —————————————————————————— CHAPTER 6. —— 6. 7. Using the Heaviside function, we can write 0 a >b œ a > # b # ? # a > b . The Laplace transform has the property that _c ?- a>b0 a> - bd œ /-= _c0 a>bd. _ a> #b# ?# a>b‘ œ # /2= . =# Hence 9. The function can be expressed as 0 a>b œ a> 1bc?1 a>b ?#1 a>bdÞ Before invoking the ><+8=6+>398 :<9:/<>C of the transform, write the function as 0 a>b œ a> 1b ?1 a>b a> #1b ?#1 a>b 1 ?#1 a>b Þ _ c 0 a >b d œ / 1 = /#1= 1/#1= # . =# = = It follows that 10. It follows directly from the ><+8=6+>398 :<9:/<>C of the transform that _ c 0 a >b d œ /= /$= /%= # ' . = = = 11. Before invoking the ><+8=6+>398 :<9:/<>C of the transform, write the function as 0 a>b œ a> #b ?# a>b ?# a>b a> $b ?$ a>b ?$ a>b Þ ________________________________________________________________________ page 276 —————————————————————————— CHAPTER 6. —— It follows that _ c 0 a >b d œ /#= /#= /$= /$= # . =# = = = 12. It follows directly from the ><+8=6+>398 :<9:/<>C of the transform that _ c 0 a >b d œ " /= #. =# = 13. Using the fact that _c/+> 0 a>bd œ _c0 a>bd=p=+ , _" – 15. First consider the function K a =b œ K a =b œ It follows that a = #b $x %— œ >$ /#> . =# #a= "b Þ #= # #a= "b Completing the square in the denominator, a = "b # " Þ _" cKa=bd œ # /> -9= > Þ _" /#= Ka=b‘ œ # /a>#b -9= a> #b ?# a>b Þ Hence 16. The inverse transform of the function #Îa=# %b is 0 a>b œ =382 #> Þ Using the ><+8=6+>398 :<9:/<>C of the transform, _" ” 17. First consider the function K a =b œ =# # /#= • œ =382 #a> #b † ?# a>b Þ =# % a = #b Þ %= $ Completing the square in the denominator, ________________________________________________________________________ page 277 —————————————————————————— CHAPTER 6. —— K a =b œ It follows that a = #b a = #b # " Þ _" cKa=bd œ /#> -9=2 > Þ _" ” a= #b/= œ /#a>"b -9=2 a> "b ?" a>b Þ # %= $ • = Hence 18. Write the function as J a =b œ _" ” /= /#= /$= /%= Þ = = = = It follows from the ><+8=6+>398 :<9:/<>C of the transform, that /= /#= /$= /%= • œ ?" a>b ?# a>b ?$ a>b ?% a>b Þ = 19a+b. By definition of the Laplace transform, _c 0 a->bd œ ( _ ! /=> 0 a->b.> Þ Making a change of variable, 7 œ -> , we have _c 0 a->bd œ Hence _c 0 a->bd œ " _ =a7 Î-b 0 a 7 b. 7 (/ -! " _ a=Î-b7 œ(/ 0 a 7 b. 7 Þ -! a,b. Using the result in Part a+b, " - = J ˆ - ‰ , where =Î- + . > _” 0 Œ • œ 5 J a5=bÞ 5 " > 0 Œ Þ 5 5 Hence _" cJ a5=bd œ ________________________________________________________________________ page 278 —————————————————————————— CHAPTER 6. —— a- b. From Part a,b, _" cJ a+=bd œ " > 0 Œ Þ + + Note that += , œ +a= ,Î+b. Using the fact that _c/-> 0 a>bd œ _c0 a>bd=p=- , _" cJ a+= ,bd œ /,>Î+ " > 0Œ . + + 20. First write J a =b œ 8x . = ˆ # ‰8" Let Ka=b œ 8xÎ=8" Þ Based on the results in Prob. "*, in which 1a>b œ >8 . Hence " " = _ ’KŠ ‹“ œ 1a#>b, # # _" cJ a=bd œ # a#>b8 œ #8" >8 Þ /%a="Î#b J a =b œ Þ #a= "Î#b K a =b œ /#= . = " > 1 Œ , ## 23. First write Now consider Using the result in Prob. "*a,b, in which 1a>b œ ?# a>b. Hence _" cKa#=bd œ _" cJ a=bd œ _" cKa#=bd œ " # " >Î# / ?% a >b . # ?# a>Î#b œ " # ?% a>b. It follows that 24. By definition of the Laplace transform, ________________________________________________________________________ page 279 —————————————————————————— CHAPTER 6. —— _ c 0 a >b d œ ( That is, _c 0 a>bd œ ( /=> .> " _ ! /=> ?" a>b.> Þ " /= œ Þ = ! 25. First write the function as 0 a>b œ ?! a>b ?" a>b ?# a>b ?$ a>b . It follows that _c 0 a>bd œ ( /=> .> ( /=> .> Þ " $ ! # That is, " /= /#= /$= _ c 0 a >b d œ = = " /= /#= /$= œ Þ = 26. The transform may be computed directly. On the other hand, using the ><+8=6+>398 :<9:/<>C of the transform, " #8" /5= " a "b 5 _ c 0 a >b d œ = 5œ" = œ " " a /= b#8# . œ = " /= That is, " a/#= b _ c 0 a >b d œ . =a" /= b 8" " #8" " a /= b5 — = –5 œ ! 29. The given function is periodic, with X œ # . Using the result of Prob. #), _ c 0 a >b d œ That is, # " " " => => ( / 0 a>b.> œ ( / .> . " /#= ! " /#= ! ________________________________________________________________________ page 280 —————————————————————————— CHAPTER 6. —— _ c 0 a >b d œ " /= =a" /#= b " . œ =a" /= b 31. The function is periodic, with X œ " . Using the result of Prob. #), _ c 0 a >b d œ It follows that _ c 0 a >b d œ " " => ( > / .> Þ " /= ! " /= a" =b Þ =# a" /= b 32. The function is periodic, with X œ 1 . Using the result of Prob. #), _ c 0 a >b d œ We first calculate => ( =38 > † / .> œ 1 ! 1 " => ( =38 > † / .> Þ " / 1 = ! " / 1 = Þ " =# Hence _ c 0 a >b d œ 33a+b. " / 1 = Þ a" /1= ba" =# b _c0 a>bd œ _c"d _c?" a>bd " /= . œ = = ________________________________________________________________________ page 281 —————————————————————————— CHAPTER 6. —— a, b . Let J a=b œ _c" ?" a>bd. Then > ! _”( c" ?" a7 bd. 7 • œ " " /= J a= b œ Þ = =# a- b . Let Ka=b œ _c1a>bd. Then _c2a>bd œ Ka=b /= Ka=b " /= " /= œ /= =# =# = # a" / b . œ =# ________________________________________________________________________ page 282 —————————————————————————— CHAPTER 6. —— 34a+b. a,b. The given function is periodic, with X œ # . Using the result of Prob. #), Based on the piecewise definition of :a>b, (/ # ! => " ! _ c 0 a >b d œ # " => ( / :a>b.> Þ " /#= ! :a>b.> œ ( > / => " œ # a" /= b# Þ = .> ( a# >b/=> .> # " Hence _ c : a >b d œ a" /= b Þ =# a" /= b a- b. Since :a>b satisfies the hypotheses of Theorem 'Þ#Þ" , Using the result of Prob. $!, _ c : w a >b d œ _ c : a >b d œ _c: w a>bd œ = _c:a>bd :a!b . a" /= b . =a" /= b We note the :a!b œ ! , hence " a" /= b ” •. = =a" /= b ________________________________________________________________________ page 283 —————————————————————————— CHAPTER 6. —— Section 6.4 2. Let 2a>b be the forcing function on the right-hand-side. Taking the Laplace transform of both sides of the ODE, we obtain =# ] a=b = Ca!b C w a!b #c= ] a=b C a!bd # ] a=b œ _c2 a>bd. =# ] a=b #= ] a=b # ] a=b " œ _c2 a>bd. _ c 2 a >b d œ /1= /#1= . = Applying the initial conditions, The forcing function can be written as 2a>b œ ?1 a>b ?#1 a>b Þ Its transform is Solving for ] a=b, the transform of the solution is ] a =b œ " /1= /#1= . =# #= # =a=# #= #b " " œ . #= # a = "b # " First note that =# Using partial fractions, =a =# _” Now let K a =b œ Then _" cKa=bd œ Using Theorem 'Þ$Þ" , _" c/-= Ka=bd œ " " ?- a>b /a>-b c-9=a> - b =38a> - bd?- a>b . # # " " > " / -9= > /> =38 > . ## # =a =# " . #= #b " " " " a = "b " œ Þ #= #b # = # a = "b # " Taking the inverse transform, term-by-term, " " > • œ _– — œ / =38 > . # # #= # = a = "b " Hence the solution of the IVP is ________________________________________________________________________ page 284 —————————————————————————— CHAPTER 6. —— " " Ca>b œ /> =38 > ?1 a>b /a>1b c-9=a> 1b =38a> 1bd?1 a>b # # " " a>#1b ?#1 a>b / c-9=a> #1b =38a> #1bd?#1 a>b . # # That is, " " Ca>b œ /> =38 > c?1 a>b ?#1 a>bd /a>1b c-9= > =38 >d ?1 a>b # # " a>#1b / c-9= > =38 >d?#1 a>b . # The solution starts out as free oscillation, due to the initial conditions. The amplitude increases, as long as the forcing is present. Thereafter, the solution rapidly decays. 4. Let 2a>b be the forcing function on the right-hand-side. Taking the Laplace transform of both sides of the ODE, we obtain =# ] a=b = Ca!b C w a!b % ] a=b œ _c2 a>bd. Applying the initial conditions, ________________________________________________________________________ page 285 —————————————————————————— CHAPTER 6. —— =# ] a=b % ] a=b œ _c2a>bd. The transform of the forcing function is _c2a>bd œ " / 1 = # . =# " = " Solving for ] a=b, the transform of the solution is ] a =b œ " / 1 = # . a=# %ba=# "b a= %ba=# "b Using partial fractions, a =# It follows that _" ” Based on Theorem 'Þ$Þ" , _" ” / 1 = " " œ ”=38a> 1b =38a#> #1b•?1 a>b . # %ba=# "b • a= $ # " " " " ”=38 > =38 #>• ”=38 > =38 #>•?1 a>b . $ # $ # a =# " " " œ ”=38 > =38 #>•. # "b • %ba= $ # " " " " œ ”# # •. # "b %ba= $ = " = % Hence the solution of the IVP is Ca>b œ ________________________________________________________________________ page 286 —————————————————————————— CHAPTER 6. —— Since there is no damping term, the solution follows the forcing function, after which the response is a steady oscillation about C œ ! . 5. Let 0 a>b be the forcing function on the right-hand-side. Taking the Laplace transform of both sides of the ODE, we obtain =# ] a=b = Ca!b C w a!b $c= ] a=b C a!bd # ] a=b œ _c0 a>bd. =# ] a=b $= ] a=b # ] a=b œ _c0 a>bd. _ c 0 a >b d œ Solving for the transform, ] a =b œ Using partial fractions, =a =# Hence _" ” Based on Theorem 'Þ$Þ" , _" ” " /"!= œ " /#a>"!b #/a>"!b ‘?"! a>b . # $= #b • =a = # " " /#> œ /> . • =a=# $= #b # # " "" " # œ” •Þ $= #b # = =# =" " /"!= . =a=# $= #b =a=# $= #b " /"!= . = = Applying the initial conditions, The transform of the forcing function is Hence the solution of the IVP is ________________________________________________________________________ page 287 —————————————————————————— CHAPTER 6. —— C a >b œ " /#> " c" ?"! a>bd /> /a#>#!b #/a>"!b ‘?"! a>b . # # # The solution increases to a temporary steady value of C œ "Î#. After the forcing ceases, the response decays exponentially to C œ ! . 6. Taking the Laplace transform of both sides of the ODE, we obtain =# ] a=b = Ca!b C w a!b $c= ] a=b C a!bd # ] a=b œ Applying the initial conditions, =# ] a=b $= ] a=b # ] a=b " œ Solving for the transform, ] a =b œ Using partial fractions, " /#= . =# $= # =a=# $= #b /#= . = /#= . = ________________________________________________________________________ page 288 —————————————————————————— CHAPTER 6. —— =# and =a =# " " " œ $= # =" =# " "" " # œ” •Þ $= #b # = =# =" Taking the inverse transform. term-by-term, the solution of the IVP is " " Ca>b œ /> /#> ” /a>#b /#a>#b •?# a>b Þ # # Due to the initial conditions, the response has a transient overshoot, followed by an exponential convergence to a steady value of C= œ "Î# . 7. Taking the Laplace transform of both sides of the ODE, we obtain = # ] a = b = C a!b C w a!b ] a= b œ Applying the initial conditions, /$1= . = ________________________________________________________________________ page 289 —————————————————————————— CHAPTER 6. —— =# ] a=b ] a=b = œ Solving for the transform, ] a =b œ Using partial fractions, " " = œ # . = a = # "b = = " Hence ] a =b œ =# = " = /$1= ” # •. " = = " = /$1= . =# " = a= # " b /$1= . = Taking the inverse transform, the solution of the IVP is Ca>b œ -9= > c" -9=a> $1bd?$1 a>b œ -9= > c" -9= > d?$1 a>b Þ Due to initial conditions, the solution temporarily oscillates about C œ ! . After the forcing is applied, the response is a steady oscillation about C7 œ " . ________________________________________________________________________ page 290 —————————————————————————— CHAPTER 6. —— 9. Let 1a>b be the forcing function on the right-hand-side. Taking the Laplace transform of both sides of the ODE, we obtain =# ] a=b = Ca!b C w a!b ] a=b œ _c1a>bd. =# ] a=b ] a=b " œ _c1a>bd. 1a>b œ > c" ?' a>bd $ ?' a>b # > " œ a> 'b?' a>b ## Applying the initial conditions, The forcing function can be written as with Laplace transform " /'= _c1a>bd œ . #=# #=# Solving for the transform, ] a =b œ Using partial fractions, #=# a=# " "b œ "" " ” # # •Þ #= = " " " /'= ## ## . =# " #= a= "b #= a= "b Taking the inverse transform, and using Theorem 'Þ$Þ" , the solution of the IVP is " " Ca>b œ =38 > c> =38 >d ca> 'b =38a> 'bd?' a>b # # " " œ c> =38 >d ca> 'b =38a> 'bd?' a>bÞ # # ________________________________________________________________________ page 291 —————————————————————————— CHAPTER 6. —— The solution increases, in response to the ramp input, and thereafter oscillates about a mean value of C7 œ $ . 11. Taking the Laplace transform of both sides of the ODE, we obtain = # ] a = b = C a!b C w a!b % ] a= b œ Applying the initial conditions, =# ] a=b % ] a=b œ Solving for the transform, ] a =b œ Using partial fractions, =a =# C a >b œ " "" = œ”# •Þ %b % = = % / 1 = /$1= . = a = # % b = a = # %b / 1 = /$1= . = = / 1 = /$1= . = = Taking the inverse transform, and applying Theorem 'Þ$Þ" , " " c" -9=a#> #1bd?1 a>b c" -9=a#> '1bd?$1 a>b % % " " œ c?1 a>b ?$1 a>bd -9= #> † c?1 a>b ?$1 a>bd . % % ________________________________________________________________________ page 292 —————————————————————————— CHAPTER 6. —— Since there is no damping term, the solution responds immediately to the forcing input. There is a temporary oscillation about C œ "Î% Þ 12. Taking the Laplace transform of the ODE, we obtain =% ] a=b =$ C a!b =# C w a!b = C ww a!b C www a!b ] a=b œ Applying the initial conditions, =% ] a=b ] a= b œ Solving for the transform of the solution, ] a =b œ Using partial fractions, " " % " " #= œ ” # •Þ = a = % "b % = =" =" = " It follows that /= /#= . = a = % " b = a = % "b /= /#= . = = /= /#= Þ = = ________________________________________________________________________ page 293 —————————————————————————— CHAPTER 6. —— _" ” " " > > • œ % / / # -9= >‘Þ "b % =a =% Based on Theorem 'Þ$Þ" , the solution of the IVP is " Ca>b œ c?" a>b ?# a>bd /a>"b /a>"b # -9=a> "b‘?" a>b % " a>#b a>#b / / # -9=a> #b‘?# a>b . % The solution increases without bound, exponentially. 13. Taking the Laplace transform of the ODE, we obtain =% ] a=b =$ C a!b =# C w a!b = C ww a!b C www a!b " / 1 = &=# ] a=b = C a!b C w a!b‘ % ] a=b œ Þ = = " / 1 = = ] a=b &= ] a=b %] a=b œ . = = % # Applying the initial conditions, Solving for the transform of the solution, ________________________________________________________________________ page 294 —————————————————————————— CHAPTER 6. —— ] a =b œ Using partial fractions, " "$ = %= œ # ”# •Þ =a=% &=# %b "# = = % = " It follows that _" ” =a =% " " œ c$ -9= #> % -9= >dÞ # %b • &= "# " / 1 = . =a=% &=# %b =a=% &=# %b Based on Theorem 'Þ$Þ" , the solution of the IVP is C a >b œ " " c" ?1 a>bd c-9= #> % -9= >d % "# " c-9= #a> 1b % -9=a> 1bd?1 a>b . "# " " c" ?1 a>bd c-9= #> % -9= >d % "# " c-9= #> % -9= >d?1 a>b . "# That is, C a >b œ ________________________________________________________________________ page 295 —————————————————————————— CHAPTER 6. —— After an initial transient, the solution oscillates about C7 œ ! . 14. The specified function is defined by 0 a >b œ Û > ! bß Ü 2ß 2 5 a> Ú !ß ! Ÿ > >! >! Ÿ > > ! 5 > >! 5 which can conveniently be expressed as 0 a >b œ 2 2 a> >! b ?>! a>b a> >! 5 b ?>! 5 a>b Þ 5 5 15. The function is defined by Ú !ß Ý Ý2 a > > ! bß 1a>b œ Û 5 2 Ý 5 a> >! #5 bß Ý Ü !ß ! Ÿ > >! >! Ÿ > > ! 5 >! 5 Ÿ > >! #5 > >! #5 which can also be written as 1a>b œ 2 #2 2 a> >! b ?>! a>b a> >! 5 b ?>! 5 a>b a> >! #5 b ?>! #5 a>bÞ 5 5 5 $ & ?a>b œ %5 ?$Î# a>b 2Œ> %5 ?&Î# a>b 2 Œ> , # # " È( >Î) $È( > " >Î) $È ( > / =38 -9= Þ %/ % )% ) ) 16a. b. From Part a- b, the solution is where 2a>b œ Due to the damping term, the solution will decay to zero. The maximum will occur ________________________________________________________________________ page 296 —————————————————————————— CHAPTER 6. —— shortly after the forcing ceases. By plotting the various solutions, it appears that the solution will reach a value of C œ # , as long as 5 #Þ&" . a/ bÞ Based on the graph, and numerical calculation, k?a>bk !Þ" for > #&Þ'(($ . 17. We consider the initial value problem C ww %C œ with Ca!b œ C w a!b œ ! . " ca> &b ?& a>b a> & 5 b ?&5 a>bd, 5 a+b. The specified function is defined by 0 a >b œ Û & bß Ü "ß Ú !ß " 5 a> !Ÿ>& &Ÿ>&5 > &5 a,bÞ Taking the Laplace transform of both sides of the ODE, we obtain ________________________________________________________________________ page 297 —————————————————————————— CHAPTER 6. —— /&= /a&5 b= = ] a = b = C a!b C a!b % ] a= b œ . 5=# 5=# # w Applying the initial conditions, =# ] a=b % ] a= b œ Solving for the transform, ] a =b œ Using partial fractions, =# a =# It follows that _" ” C a >b œ =# a =# /&= /a&5 b= . 5=# 5=# /&= /a&5 b= ## . 5=# a=# %b 5= a= %b " "" " œ ” # # •Þ %b %= = % " " " • œ > =38 #> . %b % ) Using Theorem 'Þ$Þ" , the solution of the IVP is " c2a> &b ?& a>b 2a> & 5 b ?&5 a>bd , 5 in which 2a>b œ " > " =38 #> . % ) Ca>b œ a- b. Note that for > & 5 , the solution is given by " " " =38a#> "!b =38a#> "! #5 b % )5 )5 " =38 5 œ -9=a#> "! 5 b Þ % %5 k=38a5 bk . %5 So for > & 5 , the solution oscillates about C7 œ "Î% , with an amplitude of Eœ ________________________________________________________________________ page 298 —————————————————————————— CHAPTER 6. —— ________________________________________________________________________ page 299 —————————————————————————— CHAPTER 6. —— 18a+b. a,b. The forcing function can be expressed as 0 5 a >b œ " c?%5 a>b ?%5 a>bdÞ #5 Taking the Laplace transform of both sides of the ODE, we obtain # w " /a%5 b= /a%5 b= = ] a=b = Ca!b C a!b c= ] a=b C a!bd % ] a=b œ . $ #5= #5= " /a%5 b= /a%5 b= = # ] a = b = ] a = b % ] a= b œ . $ #5= #5= ] a =b œ $ /a%5 b= $ /a%5 b= . #5=a$=# = "#b #5=a$=# = "#b Applying the initial conditions, Solving for the transform, Using partial fractions, =a$=# " "" " $= œ ” # • = "#b "# = $= = "# " " " " 'ˆ= " ‰ ' œ Þ "# – = ' ˆ= " ‰# "%$ — ' $' " ' Let " "" ' L a =b œ –= ˆ " ‰# )5 = ' "%$ $' ˆ= " ‰ # ' = "%$ $' —Þ It follows that ________________________________________________________________________ page 300 —————————————————————————— CHAPTER 6. —— 2a>b œ _" cL a=bd œ È"%$ > È"%$ > " />Î' " =38 -9= Þ )5 )5 – È"%$ ' ' — Based on Theorem 'Þ$Þ" , the solution of the IVP is Ca>b œ 2a> % 5 b ?%5 a>b 2a> % 5 b ?%5 a>b Þ a- b . As the parameter 5 decreases, the solution remains null for a longer period of time. ________________________________________________________________________ page 301 —————————————————————————— CHAPTER 6. —— Since the magnitude of the impulsive force increases, the initial overshoot of the response also increases. The duration of the impulse decreases. All solutions eventually decay to C œ ! Þ 19a+b. a- bÞ From Part a,b, ?a>b œ " -9= > #" a "b5 c" -9=a> 5 1bd?51 a>b Þ 8 5œ" ________________________________________________________________________ page 302 —————————————————————————— CHAPTER 6. —— 21a+b. a,b. Taking the Laplace transform of both sides of the ODE, we obtain =# Y a=b = ?a!b ? w a!b Y a=b œ 8 " a "b5 /5 1= " . = 5œ" = Applying the initial conditions, = # Y a = b Y a =b œ Solving for the transform, Y a =b œ Using partial fractions, =a =# Let 8 " a "b5 /5 1= " . = 5œ" = 8 " a "b5 /5 1= " . = a = # " b 5 œ " = a = # "b " " = . œ # "b = = " ________________________________________________________________________ page 303 —————————————————————————— CHAPTER 6. —— 2a>b œ _" ” " • œ " -9= > . "b =a =# 8 Applying Theorem 'Þ$Þ" , term-by-term, the solution of the IVP is ?a>b œ 2a>b " a "b5 2a> 5 1b ?51 a>b . 5œ" Note that 2a> 5 1b œ ?! a> 5 1b -9=a> 5 1b œ ?51 a>b a "b5 -9= > Þ 8 Hence ?a>b œ " -9= > " a "b5 ?51 a>b a-9= >b"?51 a>b Þ 8 5œ" 5œ" a- b . The ODE has no damping term. Each interval of forcing adds to the energy of the system. Hence the amplitude will increase. For 8 œ "& , 1a>b œ ! when > "&1 . Therefore the oscillation will eventually become steady, with an amplitude depending on the values of ?a"&1b and ? w a"&1b . a. b. As 8 increases, the interval of forcing also increases. Hence the amplitude of the transient will increase with 8 . Eventually, the forcing function will be constant. In fact, for large values of > , 1a>b œ œ Further, for > 81 , ________________________________________________________________________ page 304 "ß !ß 8 even 8 odd —————————————————————————— CHAPTER 6. —— ?a>b œ " -9= > 8 -9= > " a "b 8 Þ # Hence the steady state solution will oscillate about ! or " , depending on 8 , with an amplitude of E œ 8 " . In the limit, as 8 p _ , the forcing function will be a periodic function, with period #1 . From Prob. #(, in Section 'Þ$ , _c1a>bd œ " . =a" /= b As 8 increases, the duration and magnitude of the transient will increase without bound. 22a+b. Taking the initial conditions into consideration, the transform of the ODE is =# Y a=b !Þ" = Y a=b Y a=b œ 8 " a "b5 /5 1= " . = 5œ" = Solving for the transform, 8 " a "b5 /5 1= Y a =b œ " . =a=# !Þ"= "b 5 œ " =a=# !Þ"= "b Using partial fractions, =a =# " " = !Þ" œ # Þ !Þ"= "b = = !Þ"= " Since the denominator in the second term is irreducible, write =# Let = !Þ" a= !Þ!&b !Þ!& œ Þ !Þ"= " a= !Þ!&b# Ð$**Î%!!Ñ " a= !Þ!&b !Þ!& 2a>b œ _" – # = a= !Þ!&b Ð$**Î%!!Ñ a= !Þ!&b# Ð$**Î%!!Ñ — œ " />Î#! –-9= È$** #! > È$** " =38 È$** #! >—Þ Applying Theorem 'Þ$Þ" , term-by-term, the solution of the IVP is ?a>b œ 2a>b " a "b5 2a> 5 1b ?51 a>b . 8 5œ" ________________________________________________________________________ page 305 —————————————————————————— CHAPTER 6. —— For odd values of 8, the solution approaches C œ ! . For even values of 8, the solution approaches C œ " . a,b. The solution is a sum of damped sinusoids, each of frequency = œ È$** Î#! ¸ " . Each term has an 'initial' amplitude of approximately " Þ For any given 8 , the solution contains 8 " such terms. Although the amplitude will increase with 8, the amplitude will also be bounded by 8 " . a- b. Suppose that the forcing function is replaced by 1a>b œ =38 > . Based on the methods in Chapter $, the general solution of the differential equation is ?a>b œ />Î#! –-" -9= È$** #! > -# =38 È$** #! >— ?: a>b Þ Note that ?: a>b œ E -9= > F =38 > . Using the method of undetermined coefficients, E œ "! and F œ ! Þ Based on the initial conditions, the solution of the IVP is ________________________________________________________________________ page 306 —————————————————————————— CHAPTER 6. —— ?a>b œ "! />Î#! –-9= È$** #! È$** #! Observe that both solutions have the same frequency, = œ È$**Î#! ¸ " . > È$** " =38 >— "! -9= > Þ 23a+b. Taking the initial conditions into consideration, the transform of the ODE is = # Y a = b Y a =b œ 8 " a "b5 /a""5Î%b= #" . = = 5œ" Solving for the transform, Y a =b œ Using partial fractions, 8 " a "b5 /a""5Î%b= #" . = a = # "b = a = # "b 5œ" =a =# Let " " = œ # . "b = = " " • œ " -9= > . = a = # "b ""5 ?""5Î% a>b . % 2a>b œ _" ” Applying Theorem 'Þ$Þ" , term-by-term, the solution of the IVP is ?a>b œ 2a>b # " a "b5 2 Œ> 8 5œ" That is, ________________________________________________________________________ page 307 —————————————————————————— CHAPTER 6. —— ?a>b œ " -9= > # " a "b5 ”" -9=Œ> 8 5œ" ""5 • ?""5Î% a>b . % a, b . a- b. Based on the plot, the 'slow period' appears to be )) . The 'fast period' appears to be about ' . These values correspond to a 'slow frequency' of == œ !Þ!("% and a 'fast frequency' =0 œ "Þ!%(# . a. b. The natural frequency of the system is =! œ " Þ The forcing function is initially periodic, with period X œ ""Î# œ &Þ& . Hence the corresponding forcing frequency is A œ "Þ"%#% . Using the results in Section $Þ*, the 'slow frequency' is given by == œ and the 'fast frequency' is given by =0 œ k= =! k œ !Þ!("# # k= =! k œ "Þ!("# . # Based on theses values, the 'slow period' is predicted as ))Þ#%( and the 'fast period' is given as &Þ)'&' . ________________________________________________________________________ page 308 —————————————————————————— CHAPTER 6. —— Section 6.5 2. Taking the Laplace transform of both sides of the ODE, we obtain =# ] a=b = Ca!b C w a!b %] a=b œ /1= /#1= . =# ] a=b %] a=b œ /1= /#1= . /1= /#1= / 1 = /#1= œ# # . =# % = % = % Applying the initial conditions, Solving for the transform, ] a =b œ C a >b œ Applying Theorem 'Þ$Þ" , the solution of the IVP is " " =38a#> #1b?1 a>b =38a#> %1b?#1 a>b # # " œ =38a#>bc ?1 a>b ?#1 a>b d Þ # 4. Taking the Laplace transform of both sides of the ODE, we obtain =# ] a=b = Ca!b C w a!b ] a=b œ #! /$= . =# ] a=b ] a=b = œ #! /$= . = #! /$= ] a =b œ # # . = " = " Using a table of transforms, and Theorem 'Þ$Þ" , the solution of the IVP is Ca>b œ -9=2 > #! =382 a> $b?$ a>b . Applying the initial conditions, Solving for the transform, ________________________________________________________________________ page 309 —————————————————————————— CHAPTER 6. —— 6. Taking the initial conditions into consideration, the transform of the ODE is =# ] a=b %] a=b =Î# œ /%1= . ] a =b œ C a >b œ =Î# /%1= # . =# % = % " =38a#> )1b?%1 a>b # " =38a#>b ?%1 a>b Þ # Solving for the transform, Using a table of transforms, and Theorem 'Þ$Þ" , the solution of the IVP is " -9= #> # " œ -9= #> # 8. Taking the Laplace transform of both sides of the ODE, we obtain =# ] a=b = Ca!b C w a!b %] a=b œ # /a1Î%b= . =# ] a=b %] a=b œ # /a1Î%b= . Applying the initial conditions, Solving for the transform, ________________________________________________________________________ page 310 —————————————————————————— CHAPTER 6. —— # /a1Î%b= ] a =b œ # . = % Applying Theorem 'Þ$Þ" , the solution of the IVP is Ca>b œ =38Š#> 1 ‹?1Î% a>b œ -9=a#>b ?1Î% a>b Þ # 9. Taking the initial conditions into consideration, the transform of the ODE is =# ] a=b ] a= b œ Solving for the transform, ] a =b œ Using partial fractions, " " = œ # Þ = a = # "b = = " Hence ] a =b œ /a1Î#b= = /a1Î#b= $ /a$1Î#b= /#1= = /#1= # # Þ = = " =# " = = " /a1Î#b= /#1= $ /a$1Î#b= . = = /a1Î#b= $ /a$1Î#b= /#1= . = a = # "b =# " = a= # " b Based on Theorem 'Þ$Þ" , the solution of the IVP is Ca>b œ ?1Î# a>b -9=Š> That is, 1 $1 ‹?1Î# a>b $ =38Œ> ?$1Î# a>b # # ?#1 a>b -9=a> #1b?#1 a>b . Ca>b œ c" =38a>bd ?1Î# a>b $ -9=a>b ?$1Î# a>b c" -9=a>bd ?#1 a>b . page 311 ________________________________________________________________________ —————————————————————————— CHAPTER 6. —— 10. Taking the transform of both sides of the ODE, #=# ] a=b =] a=b %] a=b œ ( _ " œ /a1Î'b= Þ # ! /=> $ Š> 1 ‹=38 > .> ' Solving for the transform, ] a =b œ First write /a1Î'b= . #a#=# = %b " " % œ # = %b " ‰# #a#= ˆ= % $" "' . It follows that Ca>b œ _" c] a=bd œ È$" " a>1Î'bÎ% 1 / † =38 Š> ‹?1Î' a>b Þ È$" % ' ________________________________________________________________________ page 312 —————————————————————————— CHAPTER 6. —— 11. Taking the initial conditions into consideration, the transform of the ODE is = =# ] a=b #= ] a=b # ] a=b œ # /a1Î#b= . = " Solving for the transform, /a1Î#b= ] a =b œ # . a= "ba=# #= #b =# #= # = = #= #b " = # =% # # ”# •Þ & = " = " = #= # Using partial fractions, a =# "ba=# œ We can also write =% a = "b $ œ Þ =# #= # a = "b # " ]" a=b œ a =# "ba=# = #= #b Þ Let Then _" c]" a=bd œ Applying Theorem 'Þ$Þ" , _" – /a1Î#b= 1 ˆ 1‰ œ / > # =38Š> ‹?1Î# a>b Þ # #= # — = # " # " -9= > =38 > /> c-9= > $ =38 >dÞ & & & Hence the solution of the IVP is C a >b œ " # " -9= > =38 > /> c-9= > $ =38 >d & & & ˆ> 1 ‰ # -9=a>b ? / 1Î# a>b Þ ________________________________________________________________________ page 313 —————————————————————————— CHAPTER 6. —— 12. Taking the initial conditions into consideration, the transform of the ODE is =% ] a=b ] a=b œ /= . ] a =b œ Using partial fractions, =% It follows that _" ” C a >b œ =% " " " • œ =382 > =38 > Þ " # # " " " " œ ”# # •Þ " # = " = " /= . =% " Solving for the transform, Applying Theorem 'Þ$Þ" , the solution of the IVP is " c=382a> "b =38a> "bd?" a>b Þ # ________________________________________________________________________ page 314 —————————————————————————— CHAPTER 6. —— 14a+b. The Laplace transform of the ODE is " =# ] a=b = ] a=b ] a=b œ /= . # ] a =b œ /= . =# =Î# " Solving for the transform of the solution, First write =# " " œ =Î# " ˆ= " ‰# % "& "' Þ Taking the inverse transform and applying both shifting theorems, Ca>b œ È"& % a>"bÎ% / =38 a> "b ?" a>b . È"& % a,b. As shown on the graph, the maximum is attained at some >" #. Note that for > #, Ca>b œ È"& % a>"bÎ% / =38 a> "b Þ È"& % Setting C w a>b œ ! , we find that >" ¸ #Þ$'"$ . The maximum value is calculated as Ca#Þ$'"$b ¸ !Þ(""&$ Þ a- b. Setting # œ "Î% , the transform of the solution is ] a =b œ /= . =# =Î% " Following the same steps, it follows that ________________________________________________________________________ page 315 —————————————————————————— CHAPTER 6. —— Ca>b œ ) a>"bÎ) $È( / =38 a> "b ?" a>b . ) $È( Once again, the maximum is attained at some >" # . Setting C w a>b œ ! , we find that >" ¸ #Þ%&'* , with C a>" b ¸ !Þ)$$& Þ a. b. Now suppose that ! # " . Then the transform of the solution is ] a =b œ /= . =# # = " First write =# It follows that 2a>b œ _" ” Hence the solution is =# " " œ Þ # #= " a= # Î#b a" # # Î%b " # /# >Î# =38ŠÈ" # # Î% † >‹Þ •œ È #= " % ## Ca>b œ 2a> "b ?" a>b . The solution is nonzero only if > " , in which case C a>b œ 2a> "b Þ Setting C w a>b œ ! , we obtain " >+8’È" # # Î% † a> "b“ œ È% # # , # >+8È" # # Î% † a> "b‘ # œ. # Î% È" # # that is, ________________________________________________________________________ page 316 —————————————————————————— CHAPTER 6. —— As # p ! , we obtain the formal equation >+8a> "b œ _ . Hence >" p " 1 . Setting # > œ 1Î# in 2a>b, and letting # p ! , we find that C" p " Þ These conclusions agree with the case # œ ! , for which it is easy to show that the solution is Ca>b œ =38a> "b ?" a>b . 15a+b. See Prob. "% . It follows that the solution of the IVP is C a >b œ This function is a multiple of the answer in Prob. "%a+b. Hence the peak value occurs at >" ¸ #Þ$'"$ . The maximum value is calculated as C a#Þ$'"$b ¸ !Þ(""&$ 5 Þ We find that the appropriate value of 5 is 5" œ #Î!Þ(""&$ ¸ #Þ)"!) . a,b. Based on Prob. "%a- b, the solution is Ca>b œ ) 5 a>"bÎ) $È ( / =38 a> "b ?" a>b . ) $È( È"& %5 a>"bÎ% / =38 a> "b ?" a>b . È"& % Since this function is a multiple of the solution in Prob. "%a- b, we have >" ¸ #Þ%&'* , with Ca>" b ¸ !Þ)$$& 5 Þ The solution attains a value of C œ # , for 5" œ #Î!Þ)$$& , that is, 5" ¸ #Þ$**& . a- b. Similar to Prob. "%a. b, for ! # " , the solution is Ca>b œ 2a> "b ?" a>b , in which 2a>b œ It follows that >" " p 1Î# . Setting > œ 1Î# in 2a>b, and letting # p ! , we find that C" p 5 Þ Requiring that the peak value remains at C œ # , the limiting value of 5 is 5" œ # . These conclusions agree with the case # œ ! , for which it is easy to show that the solution is Ca>b œ 5 =38a> "b ?" a>b . #5 /# >Î# =38ŠÈ" # # Î% † >‹Þ È% # # 16a+b. Taking the initial conditions into consideration, the transformation of the ODE is =# ] a=b ] a= b œ " /a%5 b= /a%5 b= —. #5 – = = Solving for the transform of the solution, ________________________________________________________________________ page 317 —————————————————————————— CHAPTER 6. —— ] a =b œ Using partial fractions, =a =# Now let 2a>b œ _" ” 9 a> ß 5 b œ That is, 9 a> ß 5 b œ " c ?%5 a>b ?%5 a>bd #5 " c-9=a> % 5 b ?%5 a>b -9=a> % 5 b ?%5 a>bdÞ #5 =a =# " • œ " -9= > Þ "b " " = œ # Þ "b = = " " /a%5 b= /a%5 b= Þ #5 – =a=# "b =a=# "b — Applying Theorem 'Þ$Þ" , the solution is " c2a> % 5 b ?%5 a>b 2 a> % 5 b ?%5 a>bdÞ #5 a,b. Consider various values of > . For any fixed > % , 9a> ß 5 b œ ! , as long as % 5 >Þ If > % , then for % 5 > , 9 a> ß 5 b œ It follows that 5Ä! " c-9=a> % 5 b -9=a> % 5 b dÞ #5 lim 9a> ß 5 b œ lim -9=a> % 5 b -9=a> % 5 b 5Ä! #5 œ =38a> %b Þ lim 9a> ß 5 b œ =38a> %b ?% a>b . Hence 5Ä! a- b. The Laplace transform of the differential equation with Ca!b œ C w a!b œ ! , is C ww C œ $ a> %b, =# ] a=b ] a=b œ /%= . Solving for the transform of the solution, ________________________________________________________________________ page 318 —————————————————————————— CHAPTER 6. —— ] a =b œ It follows that the solution is /%= Þ =# " 9! a>b œ =38a> %b ?% a>b . a. b . 18a,b. The transform of the ODE agiven the specified initial conditionsb is = ] a=b ] a=b œ " a "b5" /5 1= . #! # 5œ" Solving for the transform of the solution, #! " " a "b5" /5 1= . ] a =b œ # = " 5œ" Applying Theorem 'Þ$Þ" , term-by-term, Ca>b œ " a "b5" =38a> 5 1b ?51 a>b #! œ =38a>b † " ?51 a>b Þ #! 5œ" 5œ" ________________________________________________________________________ page 319 —————————————————————————— CHAPTER 6. —— a- b . 19a,b. Taking the initial conditions into consideration, the transform of the ODE is = ] a=b ] a=b œ " /a5 1Î#b= . #! # 5œ" Solving for the transform of the solution, #! " " /a5 1Î#b= . ] a =b œ # = " 5œ" Applying Theorem 'Þ$Þ" , term-by-term, Ca>b œ " =38Œ> #! 5œ" 51 ?51Î# a>b Þ # a- b . ________________________________________________________________________ page 320 —————————————————————————— CHAPTER 6. —— 20a,b. The transform of the ODE agiven the specified initial conditionsb is = ] a=b ] a=b œ " a "b5 +1 /a5 1Î#b= . #! # 5œ" Solving for the transform of the solution, ] a=b œ " a "b5 +1 #! 5œ" /a5 1Î#b= . =# " Applying Theorem 'Þ$Þ" , term-by-term, Ca>b œ " a "b5" =38Œ> #! 5œ" 51 ?51Î# a>b Þ # a- b . 22a,b. Taking the initial conditions into consideration, the transform of the ODE is =# ] a=b ] a=b œ " a "b5" /a""5Î%b= . %! 5œ" Solving for the transform of the solution, ] a=b œ " a "b5" %! 5œ" /a""5Î%b= . =# " Applying Theorem 'Þ$Þ" , term-by-term, ________________________________________________________________________ page 321 —————————————————————————— CHAPTER 6. —— Ca>b œ " a "b5" =38Œ> %! 5œ" ""5 ?""5Î% a>b Þ % a- b . 23a,bÞ The transform of the ODE agiven the specified initial conditionsb is = ] a=b !Þ"= ] a=b ] a=b œ " a "b5" /5 1= . #! # 5œ" Solving for the transform of the solution, ] a =b œ " #! /5 1= . =# !Þ"= " 5œ" First write =# It follows that _" ” " œ !Þ"= " ˆ= " ‰# #! " $** %!! . È$** " #! >Î#! œ / =38 >Þ •È =# !Þ"= " #! $** Applying Theorem 'Þ$Þ" , term-by-term, ________________________________________________________________________ page 322 —————————————————————————— CHAPTER 6. —— Ca>b œ " a "b5" 2a> 5 1b ?51 a>b , #! 5œ" in which 2a>b œ a- b . È$** #! >Î#! / =38 >Þ È$** #! 24a,bÞ Taking the initial conditions into consideration, the transform of the ODE is =# ] a=b !Þ"= ] a=b ] a=b œ "/a#5"b1= . "& 5œ" Solving for the transform of the solution, ] a =b œ " "& /a#5"b1= . =# !Þ"= " 5œ" As shown in Prob. #$ , _" ” =# È$** " #! >Î#! / =38 >Þ •œ È !Þ"= " #! $** Applying Theorem 'Þ$Þ" , term-by-term, Ca>b œ " 2c> a#5 "b1d ?a#5"b1 a>b , "& 5œ" in which ________________________________________________________________________ page 323 —————————————————————————— CHAPTER 6. —— 2a>b œ a- b . È$** #! >Î#! / =38 >Þ È$** #! 25a+b. A fundamental set of solutions is C" a>b œ /> -9= > and C# a>b œ /> =38 > . Based on Prob. ## , in Section $Þ( , a particular solution is given by C: a>b œ ( In the given problem, C: a>b œ ( > => > ! C" a=bC# a>b C" a>bC# a=b 0 a=b.= . [ aC" ß C# ba=b / œ ( /a>=b =38a> =b0 a=b.= . > ! ! c-9=a=b=38a>b =38a=b-9=a>bd 0 a=b.= . /B:a #=b Given the specified initial conditions, Ca>b œ ( /a>=b =38a> =b0 a=b.= . > ! a,bÞ Let 0 a>b œ $ a> 1b . It is easy to see that if > 1 , C a>b œ ! . If > 1 , Setting > œ 1 & , and letting & p ! , we find that C a1b œ ! . Hence a>=b =38a> =b$ a= 1b.= œ /a>1b =38a> 1b . (/ > ! ________________________________________________________________________ page 324 —————————————————————————— CHAPTER 6. —— Ca>b œ /a>1b =38a> 1b ?1 a>b Þ a- b. The Laplace transform of the solution is ] a =b œ œ Hence the solutions agree. a = "b # " / 1 = =# #= # / 1 = Þ ________________________________________________________________________ page 325 —————————————————————————— CHAPTER 6. —— Section 6.6 1a+b. The convolution integral is defined as > ! 0 ‡ 1 a>b œ ( 0 a> 7 b1a7 b. 7 Þ Consider the change of variable ? œ > 7 . It follows that ( 0 a> 7 b1a7 b. 7 œ ( 0 a?b1a> ?ba .?b > ! ! œ ( 1a> ?b0 a?b.? > > a,b. Based on the distributive property of the real numbers, the convolution is also distributive. a- b. By definition, œ 1 ‡ 0 a>b Þ ! 0 ‡ a1 ‡ 2ba>b œ ( 0 a> 7 bc1 ‡ 2 a7 bd. 7 > ! > 7 ! ! œ ( 0 a> 7 b”( 1a7 (b2a(b. (•. 7 œ ( ( 0 a> 7 b1a7 (b2a(b . (. 7 Þ > 7 ! ! The region of integration, in the double integral is the area between the straight lines ( œ ! , ( œ 7 and 7 œ > Þ Interchanging the order of integration, ( ( 0 a> 7 b1a7 (b2a(b . (. 7 œ ( ( 0 a> 7 b1a7 (b2a(b . 7 . ( > 7 > > ! ! ! ( > > ! ( œ ( ”( 0 a> 7 b1a7 (b. 7 •2a(b . ( Þ 0 a> ( ?b1a?b.? Now let 7 ( œ ? . Then ( 0 a > 7 b1 a 7 ( b . 7 œ ( > ( >( ! œ 0 ‡ 1 a> (b . > Hence ( 0 a> 7 bc1 ‡ 2 a7 bd. 7 œ ( c0 ‡ 1 a> 7 bd2a7 b . 7 Þ > ! ! ________________________________________________________________________ page 326 —————————————————————————— CHAPTER 6. —— 2. Let 0 a>b œ /> . Then 0 ‡ " a>b œ ( />7 † " . 7 > œ / > ( / 7 . 7 > ! ! œ /> " . 3. It follows directly that 0 ‡ 0 a>b œ ( =38a> 7 b =38a7 b . 7 > ! œ "> ( c-9=a> #7 b -9=a>bd. 7 #! " œ c=38a>b > -9=a>bd. # The range of the resulting function is ‘ Þ 5. We have _c/> d œ "Îa= "b and _c=38 >d œ "Îa=# "bÞ Based on Theorem 'Þ'Þ", _”( /a>7 b =38a7 b . 7 • œ > ! " " †# =" = " " œ Þ a= "ba=# "b 6. Let 1a>b œ > and 2a>b œ /> . Then 0 a>b œ 1 ‡ 2 a>b . Applying Theorem 'Þ'Þ" , _ ” ( 1 a > 7 b2 a 7 b . 7 • œ > ! " " † # =" = " œ# Þ = a = "b 7. We have 0 a>b œ 1 ‡ 2 a>b , in which 1a>b œ =38 > and 2a>b œ -9= > . The transform of the convolution integral is _ ” ( 1 a > 7 b2 a 7 b . 7 • œ > ! " = †# " = " = œ Þ a = # "b # =# 9. It is easy to see that ________________________________________________________________________ page 327 —————————————————————————— CHAPTER 6. —— _" ” " > •œ/ =" and _" ” = • œ -9= #> . % =# Applying Theorem 'Þ'Þ" , _" ” 10. We first note that _" – a = "b " #— > = œ ( /a>7 b -9= #7 . 7 Þ • a= "ba=# %b ! œ > /> and _" ” =# " " • œ =38 #> . % # Based on the convolution theorem , _" – "> œ ( a> 7 b/a>7 b =38 #7 . 7 — #! a= "b# a=# %b "> œ ( 7 /7 =38a#> #7 b . 7 Þ #! " 11. Let 1a>b œ _" cKa=bd. Since _" c"Îa=# "bd œ =38 > , the inverse transform of the product is _" ” > K a =b œ ( 1a> 7 b =38 7 . 7 • =# " ! > œ ( =38a> 7 b 1a7 b . 7 Þ ! 12. Taking the initial conditions into consideration, the transform of the ODE is =# ] a=b " =# ] a=b œ Ka=b. ] a =b œ _" ” " K a =b # . # = = =# Solving for the transform of the solution, =# As shown in a related situation, Prob. "" , K a =b "> • œ ( =38 =a> 7 b 1a7 b . 7 Þ =! =# =# Hence the solution of the IVP is ________________________________________________________________________ page 328 —————————————————————————— CHAPTER 6. —— C a >b œ " "> =38 => ( =38 =a> 7 b 1a7 b . 7 Þ = =! 14. The transform of the ODE agiven the specified initial conditionsb is %=# ] a=b %= ] a=b "( ] a=b œ Ka=b. ] a =b œ First write " " % œ Þ " ‰# %=# %= "( ˆ= # % Solving for the transform of the solution, %=# K a =b . %= "( Based on the elementary properties of the Laplace transform, _" ” %=# " " >Î# =38 #> Þ •œ / %= "( ) Applying the convolution theorem, the solution of the IVP is " > a>7 bÎ# C a >b œ ( / =38 #a> 7 b 1a7 b . 7 Þ )! 16. Taking the initial conditions into consideration, the transform of the ODE is =# ] a=b #= $ %c= ] a=b #d % ] a=b œ Ka=b. ] a =b œ We can write a= #b It follows that _" ” # #> • œ #/ =# and _" – a = #b " #— Solving for the transform of the solution, a= #b #= & # a= #b# K a= b . #= & # œ # " Þ = # a= #b# œ > /#> . Based on the convolution theorem, the solution of the IVP is ________________________________________________________________________ page 329 —————————————————————————— CHAPTER 6. —— Ca>b œ #/#> > /#> ( a> 7 b/#a>7 b 1a7 b . 7 Þ > ! 18. The transform of the ODE agiven the specified initial conditionsb is =% ] a=b ] a= b œ K a =b . ] a =b œ First write =% It follows that _" ” C a >b œ =4 " " • œ c=382 > =38 >d Þ " # " " " " œ ”# # •Þ " # = " = " K a =b . =% " Solving for the transform of the solution, Based on the convolution theorem, the solution of the IVP is "> ( c=382a> 7 b =38a> 7 bd1a7 b . 7 Þ #! 19. Taking the initial conditions into consideration, the transform of the ODE is =% ] a=b =$ &=# ] a=b &= % ] a=b œ Ka=b. ] a =b œ =$ &= K a= b # . # "ba=# %b a= a= "ba=# %b Solving for the transform of the solution, Using partial fractions, we find that =$ &= " %= = œ ”# # •, # "ba=# %b a= $ = " = % and a =# It follows that " " " " œ ”# # •Þ # %b "ba= $ = " = % ________________________________________________________________________ page 330 —————————————————————————— CHAPTER 6. —— _" ” and _" ” a =# " " " œ =38 > =38 #> Þ # %b • "ba= $ ' = a = # &b % " • œ -9= > -9= #>, a=# "ba=# %b $ $ Based on the convolution theorem, the solution of the IVP is % " "> Ca>b œ -9= > -9= #> ( c# =38a> 7 b =38 #a> 7 bd1a7 b . 7 Þ $ $ '! 21a+b. Let 9a>b œ ? ww a>b . Substitution into the integral equation results in ? a>b ( a> 0b ? ww a0b . 0 œ =38 #> Þ ww > ! Integrating by parts, ( a> 0 b ? a 0 b . 0 œ a> 0 b ? a 0 b º > ww w ! w 0œ> 0œ! œ > ? a!b ?a>b ?a!b Þ ! ( ? w a0 b . 0 > Hence ? ww a>b ?a>b > ? w a!b ?a!b œ =38 #> Þ ? ww a>b ?a>b œ =38 #> Þ a,b. Substituting the given initial conditions for the function ?a>b, Hence the solution of the IVP is equivalent to solving the integral equation in Part a+b. a- b. Taking the Laplace transform of the integral equation, with Fa=b œ _c9a>bd , Note that the convolution theorem was applied. Solving for the transform Fa=b , Fa=b œ #=# Þ a=# "ba=# %b Fa=b " # † Fa=b œ # Þ # = = % Using partial fractions, we can write ________________________________________________________________________ page 331 —————————————————————————— CHAPTER 6. —— #=# # % " œ ”# # •Þ a=# "ba=# %b $ = % = " Therefore the solution of the integral equation is 9a>b œ % # =38 #> =38 > . $ $ a. b. Taking the Laplace transform of the ODE, with Y a=b œ _c?a>bd , = # Y a = b Y a =b œ Y a =b œ Using partial fractions, we can write a =# # " # # œ ”# # •Þ # %b "ba= $ = " = % ?a>b œ # " =38 > =38 #> Þ $ $ =# # Þ % Solving for the transform of the solution, a =# # Þ "ba=# %b It follows that the solution of the IVP is We find that ? ww a>b œ # =38 > % =38 #> Þ $ $ 22a+b. First note that , 0 aC b " .C œ ‡ 0 a,bÞ (È ÈC ,C ! Take the Laplace transformation of both sides of the equation. Using the convolution theorem, with J a=b œ _c0 aC bd, X! " " œ J a= b † _ – Þ È#1 ÈC — = _– Hence " 1 — œ Ê= Þ ÈC It was shown in Prob. #(a- b, Section 'Þ" , that ________________________________________________________________________ page 332 —————————————————————————— CHAPTER 6. —— X! " 1 œ J a= b † Ê , È#1 = = with J a =b œ Ê Taking the inverse transform, we obtain 0 aC b œ X! #1 Þ 1ËC #1 X! † Þ 1 È= a,b. Combining equations a3b and a3@b, #1 X!# .B œ"Œ Þ #C 1 .C # Solving for the derivative .BÎ.C , .B #! C œË , .C C a- b. Consider the change of variable C œ #! =38# a)Î#bÞ Using the chain rule, .C .) œ #! =38a)Î#b-9=a)Î#b † .B .B in which ! œ 1X!# Î1# Þ and .B " .B œ † . .C #! =38a)Î#b-9=a)Î#b . ) It follows that .B -9=# a)Î#b œ #! =38a)Î#b-9=a)Î#bË .) =38# a)Î#b œ #! -9=# a)Î#b œ ! ! -9= ) . Direct integration results in Since the curve passes through the origin, we require Ca!b œ Ba!b œ ! Þ Hence G œ !, and Ba) b œ ! ) ! =38 ) . We also have ________________________________________________________________________ page 333 Ba) b œ ! ) ! =38 ) G . —————————————————————————— CHAPTER 6. —— Ca) b œ #! =38# a)Î#b œ ! ! -9= ) Þ ________________________________________________________________________ page 334 —————————————————————————— CHAPTER 7. —— Chapter Seven Section 7.1 1. Introduce the variables B" œ ? and B# œ ? w . It follows that B"w œ B# and B#w œ ? ww œ #? !Þ& ? w . In terms of the new variables, we obtain the system of two first order ODEs B"w œ B# B#w œ #B" !Þ& B# Þ 3. First divide both sides of the equation by ># , and write ? ww œ "w " ? Œ" # ? . > %> Set B" œ ? and B# œ ? w . It follows that B"w œ B# and B#w œ ? ww œ We obtain the system of equations B"w œ B# B#w œ Œ" " " B B# Þ # " %> > "w " ? Œ" # ? . > %> 6. One of the ways to transform the system is to assign the variables C" œ B" , C# œ B"w , C$ œ B# , C% œ B#w Þ Before proceeding, note that " c a5" 5# bB" 5# B# J" a>bd 7" " B#ww œ c5# B" a5# 5$ bB# J# a>bd Þ 7# B"ww œ Differentiating the new variables, we obtain the system of four first order equations ________________________________________________________________________ page 335 —————————————————————————— CHAPTER 7. —— C"w œ C# " C#w œ c a5" 5# bC" 5# C$ J" a>bd 7" C$w œ C% " C%w œ c5# C" a5# 5$ bC$ J# a>bd Þ 7# 7a+b. Solving the first equation for B# , we have B# œ B"w #B" . Substitution into the second equation results in aB"w #B" b œ B" #aB"w #B" b Þ w That is, B"ww % B"w $ B" œ ! . The resulting equation is a second order differential equation with constant coefficients. The general solution is B" a>b œ -" /> -# /$> . With B# given in terms of B" , it follows that B# a>b œ -" /> -# /$> . a,b. Imposing the specified initial conditions, we obtain -" - # œ # -" - # œ $ , with solution -" œ &Î# and -# œ "Î# . Hence B" a>b œ a- b . & > " $> & " // and B# a>b œ /> /$> . # # # # 10. Solving the first equation for B# , we obtain B# œ aB" B"w bÎ# . Substitution into ________________________________________________________________________ page 336 —————————————————————————— CHAPTER 7. —— the second equation results in aB" B"w b Î# œ $ B" #aB" B"w b Þ w Rearranging the terms, the single differential equation for B" is B"ww $ B"w # B" œ ! . The general solution is B" a>b œ -" /> -# /#> . $ B# a>b œ -" /> -# /$> . # With B# given in terms of B" , it follows that Invoking the specified initial conditions, -" œ ( and -# œ ' . Hence B" a>b œ (/> ' /#> and B# a>b œ (/> * /$> . 11. Solving the first equation for B# , we have B# œ B"w Î# . Substitution into the second equation results in B"ww Î# œ # B" Þ The resulting equation is B"ww % B" œ ! , with general solution B" a>b œ -" -9= #> -# =38 #> . With B# given in terms of B" , it follows that B# a>b œ -" =38 #> -# -9= #> . Imposing the specified initial conditions, we obtain -" œ $ and -# œ % . Hence B" a>b œ $ -9= #> %=38 #> and B# a>b œ $=38 #> %-9= #> . ________________________________________________________________________ page 337 —————————————————————————— CHAPTER 7. —— 12. Solving the first equation for B# , we obtain B# œ B"w Î# B" Î% . Substitution into the second equation results in B"ww Î# B"w Î% œ # B" aB"w Î# B" Î%bÎ# Þ "( B" œ ! . % Rearranging the terms, the single differential equation for B" is B"ww B"w The general solution is B" a>b œ />Î# c-" -9= #> -# =38 #>d. B# a>b œ />Î# c -" -9= #> -# =38 #>d . With B# given in terms of B" , it follows that Imposing the specified initial conditions, we obtain -" œ # and -# œ # . Hence B" a>b œ />Î# c #-9= #> #=38 #>d and B# a>b œ />Î# c#-9= #> #=38 #>d . 13. Solving the first equation for Z , we obtain Z œ P † M w . Substitution into the second equation results in ________________________________________________________________________ page 338 —————————————————————————— CHAPTER 7. —— P † M ww œ M Pw MÞ G VG Rearranging the terms, the single differential equation for M is PVG † M ww P † M w V † M œ ! . 15. Direct substitution results in a-" B" a>b -# B# a>bbw œ :"" a>bc-" B" a>b -# B# a>bd :"# a>bc-" C" a>b -# C# a>bd a-" C" a>b -# C# a>bbw œ :#" a>bc-" B" a>b -# B# a>bd :## a>bc-" C" a>b -# C# a>bd Þ -" B"w a>b -# B#w a>b œ -" c:"" a>bB" a>b :"# a>bC" a>bd -# c:"" a>bB# a>b :"# a>bC# a>bd Þ Expanding the left-hand-side of the first equation, Repeat with the second equation to show that the system of ODEs is identically satisfied. 16. Based on the hypothesis, B"w a>b œ :"" a>bB" a>b :"# a>bC" a>b 1" a>b B#w a>b œ :"" a>bB# a>b :"# a>bC# a>b 1" a>b Þ Subtracting the two equations, B"w a>b B#w a>b œ :"" a>bcB"w a>b B#w a>bd :"# a>bcC"w a>b C#w a>bd Þ C"w a>b C#w a>b œ :#" a>bcB"w a>b B#w a>bd :## a>bcC"w a>b C#w a>bd Þ Similarly, Hence the difference of the two solutions satisfies the homogeneous ODE. 17. For rectilinear motion in one dimension, Newton's second law can be stated as "J œ 7 B ww Þ The </=3=>381 force exerted by a linear spring is given by J= œ 5 $ , in which $ is the displacement of the end of a spring from its equilibrium configuration. Hence, with ! B" B# , the first two springs are in tension, and the last spring is in compression. The sum of the spring forces on 7" is J=" œ 5" B" 5# aB# B" b Þ The total force on 7" is ________________________________________________________________________ page 339 —————————————————————————— CHAPTER 7. —— "J " œ 5" B" 5# aB# B" b J" a>b . Similarly, the total force on 7# is "J # œ 5# aB# B" b 5$ B# J# a>b . a,b. Consider the right node. The current in is given by M" M# . The current leaving the node is M$ . Hence the current passing through the node is aM" M# b a M$ bÞ Based on Kirchhoff's first law, M" M# M$ œ ! . a- b. In the capacitor, In the resistor, Z# œ V M # . In the inductor, P M$w œ Z$ Þ a. b. Based on part a+b, Z$ œ Z# œ Z" . Based on part a,b, G Z"w It follows that G Z"w œ " Z" M$ and P M$w œ Z" Þ V " Z# M $ œ ! Þ V 18a+b. Taking a clockwise loop around each of the paths, it is easy to see that voltage drops are given by Z" Z# œ ! , and Z# Z$ œ ! . G Z"w œ M" Þ 20. Let M" ß M# ß M$ ß and M% be the current through the resistors, inductor, and capacitor, respectively. Assign Z" ß Z# ß Z$ ß and Z% as the respective voltage drops. Based on Kirchhoff's second law, the net voltage drops, around each loop, satisfy Z" Z$ Z% œ ! , Z" Z$ Z# œ ! and Z% Z# œ ! . Applying Kirchhoff's first law to the upper-right node, M $ aM # M % b œ ! Þ Likewise, in the remaining nodes, ________________________________________________________________________ page 340 —————————————————————————— CHAPTER 7. —— M" M$ œ ! and M# M% M" œ ! . That is, Z% Z# œ ! , Z" Z$ Z% œ ! and M# M% M$ œ ! . Using the current-voltage relations, Z" œ V" M" , Z# œ V# M# , P M$w œ Z$ , G Z%w œ M% . Combining these equations, V" M$ P M$w Z% œ ! and G Z%w œ M$ Now set M$ œ M and Z% œ Z , to obtain the system of equations P M w œ V" M Z and G Z w œ M 22a+b. Z . V# Z% . V# Let U" a>b and U# a>b be the amount of salt in the respective tanks at time > . Note that the volume of each tank remains constant. Based on conservation of mass, the rate of increase of salt, in any given tank, is given by <+>/ 90 38-</+=/ œ <+>/ 38 <+>/ 9?> . For Tank ", the rate of salt flowing into Tank " is ________________________________________________________________________ page 341 —————————————————————————— CHAPTER 7. —— <38 œ ’;" U# 9D 9D 1+6 1+6 “’$ “” •’" 738 “ 1+6 738 "!! 1+6 U# 9D Þ "!! 738 œ $ ;" The rate at which salt flow out of Tank " is <9?> œ ” Hence .U" U# U" œ $ ;" . .> "!! "& Similarly, for Tank #, .U# U" $U# . œ ;# .> $! "!! The process is modeled by the system of equations U" U# $ ;" "& "!! U" $U# w U# œ ;# Þ $! "!! w U" œ U" 9D U" 9D 1+6 Þ •’% 738 “ œ '! 1+6 "& 738 The initial conditions are U" a!b œ U! and U# a!b œ U! Þ " # U" U# $ ;" œ ! "& "!! U" $U# ;# œ ! Þ $! "!! a,b. The equilibrium values are obtain by solving the system Its solution leads to UI œ &% ;" ' ;# and UI œ '! ;" %! ;# Þ " # a- b. The question refers to possible solution of the system &% ;" ' ;# œ '! '! ;" %! ;# œ &! Þ It is possible for formally solve the system of equations, but the unique solution gives ;" œ which is not physically possible. ( 9D " 9D and ;# œ , ' 1+6 # 1+6 ________________________________________________________________________ page 342 —————————————————————————— CHAPTER 7. —— a. b. We can write UI " ' $ UI ;# œ ;" # , # %! ;# œ * ;" which are the equations of two lines in the ;" ;# -plane: The intercepts of the first line are UI Î&% and UI Î' . The intercepts of the second " " line are UI Î'! and UI Î%! . Therefore the system will have a unique solution, in the # # first quadrant, as long as UI Î&% Ÿ UI Î'! or UI Î%! Ÿ UI Î' . That is, " # # " "! UI #! . Ÿ #Ÿ I * $ U" ________________________________________________________________________ page 343 —————————————————————————— CHAPTER 7. —— Section 7.2 2a+b. a, b . a- b . A #B œ Œ " 3 #3 $ #3 % " #3 ' "3 œ Œ " #3 # 3 %3 $ '3 $ $ %3 œ ' $3 #3 Œ "" '3 ( #3 Þ # $3 '3 Þ ' &3 $A B œ Œ AB œ Œ $ $3 3 * '3 # a. b . a" 3b3 #a " #3b a$ #3b3 #a# 3b $ &3 ( &3 œŒ Þ #3 ( #3 $a" 3b a " #3ba #3b $a$ #3b a# 3ba #3b a" 3b3 $a$ #3b BA œ Œ #a" 3b a #3ba$ #3b ) (3 % %3 œŒ Þ ' %3 % a " #3b3 $a# 3b #a " #3b a #3ba# 3b 3. AX BX œ Î # " Ï# Î " $ œ Ï& œ aA BbX Þ 4a,b. " # Ñ Î" ! " # $ " Ò Ï$ % !Ñ " ! % "Ò $ " " #Ñ " !Ò $ #3 AœŒ #3 "3 Þ # $3 X a- b. By definition, A‡ œ ˆAX ‰ œ a Ab . 5. ________________________________________________________________________ page 344 —————————————————————————— CHAPTER 7. —— Î& #aA Bb œ # ! Ï" # Ñ Î "! & ! œ Ò Ï# $ %Ñ "! Þ 'Ò $ # # ' % % 7. Let A œ a+34 b and B œ a,34 b . The given operations in a+b a. b are performed elementwise. That is, a+ b . a, b . a- b . a. b . +34 ,34 œ ,34 +34 . +34 a,34 -34 b œ a+34 ,34 b -34 . !a+34 ,34 b œ ! +34 ! ,34 Þ a! " b +34 œ ! +34 ! +34 Þ In the following, let A œ a+34 b , B œ a,34 band C œ a-34 b . a/b. Calculating the generic element, aBCb34 œ " ,35 -54 Þ 8 5œ" Therefore cAaBCbd34 œ " +3< " ,<5 -54 8 8 œ " "+3< ,<5 -54 8 <œ" 5œ" 8 8 <œ" 8 5œ" œ " –" +3< ,<5 -54 —Þ 5œ" <œ" The last summation is recognized as " +3< ,<5 œ aABb35 , 8 <œ" which is the 35 -th element of the matrix AB Þ a0 b. Likewise, ________________________________________________________________________ page 345 —————————————————————————— CHAPTER 7. —— cAaB Cbd34 œ " +35 a ,54 -54 b 8 5œ" 8 8 œ " +35 ,54 " +35 -54 œ aABb34 aACb34 Þ 5œ" 5œ" 8a+b. a, b . a- b . a. b . xX y œ #a " 3b #a$3b a" 3ba$ 3b œ %3 . yX y œ a " 3b# ## a$ 3b# œ "# )3 Þ ax , yb œ #a " 3b #a$3b a" 3ba$ 3b œ # #3 . ay , yb œ a " 3ba " 3b ## a$ 3ba$ 3b œ "' . xX y œ " B4 C4 œ yX x , 8 4œ" 9. Indeed, and ax , yb œ " B4 C4 œ " C4 B4 œ " C4 B4 œ ay , xb Þ 8 8 8 4œ" 4œ" 4œ" 11. First augment the given matrix by the identity matrix: cA l I d œ Œ Divide the first row by $ , to obtain Œ " Œ ! Divide the s/-98. row by % , to obtain " ! " $ " " $ $ ' " # "! Þ ! " ! Þ " ! Þ " " ' " $ # " $ ! Adding ' times the first row to the second row results in " $ % " $ # ! " # " % Þ Finally, adding "Î$ times the second row to the first row results in ________________________________________________________________________ page 346 —————————————————————————— CHAPTER 7. —— " ! Hence $ Œ' 13. The augmented matrix is ! " " ' " # " "# " % Þ " Þ $ " # " œ " # Œ ' "# Î" # Ï" " " " " " # " ! !Ñ !"!Þ ! ! "Ò Combining the elements of the first row with the elements of the second and third rows results in Î" ! Ï! " $ ! " $ $ " ! !Ñ # " ! Þ " ! "Ò Divide the elements of the second row by $ , and the elements of the third row by $ Þ Now subtracting the new second row from the first row yields Î" Ð! Ï! Î" Ð! Ï! Hence Î" # Ï" " " " "Ñ " #Ò " ! " ! ! " " " $ # $ " $ " $ ! " $ !Ñ ! ÓÞ "Ò $ Finally, combine the third row with the second row to obtain ! " ! ! ! " " $ " $ " $ " $ ! " $ !Ñ "Ó Þ $ "Ò $ " "Î " œ $Ï " " " ! !Ñ "Þ "Ò 15. Elementary row operations yield ________________________________________________________________________ page 347 —————————————————————————— CHAPTER 7. —— Î# ! Ï! Î" Ð! Ï! !Ñ Î" ! pÐ! "Ò Ï! " # " % " # ! ! " ! ! " # ! " " ! ! " % ! " ! " # " # ! " # " # ! " # ! ! ! Î" Ð! Ï! ! Ñ Î" ! " Óp Ð ! " % "ÒÏ !! # " # " % " ! " ! ! ! !Ñ ! Óp "Ò # " % ! " Ñ ÓÞ "Ò " ) " % # " # " # " % ! ! ! !Ñ " ÓÞ % "Ò # Finally, combining the 0 3<=> and >23<. rows results in ! " ! ! ! " " # ! ! ! 16. Elementary row operations yield Î" # Ï$ Î" Ð! Ï! " " # ! " ! # $ "! $ " ! " " $ " $ " ! !Ñ Î" !"!p! ! ! "Ò Ï! # $ ( $ " $ " $ " $ !Ñ Î" ! ! Óp Ð ! " "Ò Ï! ! " "! $ "! # & " $ " " # % ! ! "! $ " ! !Ñ # " ! p $ ! "Ò $ "& ( $ " "! $ "! # & " $ Ñ ÓÞ "Ò " "! " & Finally, normalizing the last row results in Î" Ð! Ï! ! " ! ! ! " $ "& ( "! " "! Ñ ÓÞ $Ò " "! " & "! 17. Elementary row operations on the augmented matrix yield the row-reduced form of the augmented matrix Î" Ð! Ï! ! " ! $ ( " ( ! ! ! " " ( % ( # Ñ ÓÞ "Ò # ( " ( The left submatrix cannot be converted to the identity matrix. Hence the given matrix is singular. 18. Elementary row operations on the augmented matrix yield ________________________________________________________________________ page 348 —————————————————————————— CHAPTER 7. —— Î" Ð! Ð " Ï! Î" Ð! Ð ! Ï! ! " ! ! " " ! ! !Ñ Î" ! ! " ! !Ó Ð! Óp Ð ! !!"! ! " ! ! ! "Ò Ï! " ! " ! ! " ! " ! !Ñ Î" ! !Ó Ð! Óp Ð ! "! ! "Ò Ï! ! " ! ! " " ! ! !Ñ ! ! " ! !Ó Óp " " ! " ! " ! ! ! "Ò " ! " " ! " " ! "Ñ "Ó ÓÞ " "Ò ! " ! " ! " " ! ! " " " " ! " " ! " ! " ! ! " ! ! ! ! " ! " " " " " " ! 19. Elementary row operations on the augmented matrix yield Î" Ð " Ð " Ï # Î" Ð! Ð ! Ï! ! " ! ! " # ! # ! # " % # % " ! # # " " ! # $ " # " # # " ! ! !Ñ Î" ! " ! !Ó Ð! Óp Ð !!"! ! ! ! ! "Ò Ï! " " " ! " " " ! # # " % # % " & ! # $ " # $ # "! " ! ! !Ñ " " ! !Ó Óp " ! " ! # ! ! "Ò " " " % ! # " % !Ñ !Ó ÓÞ ! "Ò Ñ Ó ÓÞ Ó ! !Ñ Î" ! ! ! !Ó Ð! " ! Óp Ð !!" "! ! "Ò Ï! ! ! Normalizing the last row and combining it with the others results in Î" Ð! Ð! Ï! ! " ! ! ! ! " ! # % " " # $ # # " " " % & ! # " % & ! Ñ Î" ! Ó Ð! Ð ! Óp Ð ! " Ò Ï! & ! " ! ! ! ! " ! ! ! ! " ' & ! # "$ & "" & " & % & " & % & ) & ' & "Ò & # & % & " & 20. Suppose that A is nonsingular, and that there exist matrices B and C, such that AB œ I and AC œ I Þ Based on the properties of matrices, it follows that AaB Cb œ AY œ 08‚8 Þ Y œ y" l y# lâ l y8 ‘. Write the difference of the two matrices, Y , in terms of its columns as The 4-th column of the product matrix, AY , can be expressed as A yj . Now since all columns of the product matrix consist only of zeros, we end up with 8 homogeneous systems of linear equations A yj œ 08‚" , 4 œ "ß #ß âß 8 Þ Since A is nonsingular, each system must have a trivial solution. That is, yj œ 08‚" , for 4 œ "ß #ß âß 8 . Hence Y œ 08‚8 and B œ C Þ 21a+b. ________________________________________________________________________ page 349 —————————————————————————— CHAPTER 7. —— > Î/ #/> Ï /> > Î (/ /> Ï )/> A $B œ œ #/> /> $/> &/> (/> ! /#> Ñ Î '/> /#> $/> #/#> Ò Ï */> "!/#> Ñ #/#> Þ /#> Ò $/> '/> $/> */#> Ñ $/#> $/#> Ò a,b. Based on the standard definition of matrix multiplication, #> $> Î #/ # $/ AB œ %/#> " $/$> Ï #/#> $ '/$> " %/#> /> # #/#> /> " '/#> #/> $/$> #/> /%> Ñ Þ '/$> /> /%> $> > %> Ò $/ $/ #/ a- b . /> .A Î > œ #/ .> Ï /> #/> /> $/> #/#> Ñ #/#> Þ %/#> Ò a. b. Note that > Î/ #/> ( Aa>b.> œ Ï /> #/> /> $/> /#> Î# Ñ /#> Î# C . /#> Ò "Î# Ñ "Î# "Ò Therefore #/" /# Î# Ñ Î " Î/ # #/ /" /# Î# ( Aa>b.> œ ! Ï / $/" Ò Ï " # / " /# Î# "Î# Ñ Î / " # #/ œ #/ # " /" "Î# /# Î# . Ï " / $ $/" /# " Ò " # " $ The result can also be written as Î" a / "b Ð # Ï " # / " / $ / " # a / "b Ñ " # a/ "b ÓÞ /" Ò 23. First note that ________________________________________________________________________ page 350 —————————————————————————— CHAPTER 7. —— " " $/> #> /> x w œ Œ /> #Œ ˆ/> > /> ‰ œ Œ > . ! " #/ #> /> We also have # Œ$ " # " "> # x œ Œ $ # Œ ! / Œ $ # # # œ Œ /> Œ ˆ> /> ‰ $ # > > #/ #> / œŒ > . $/ #> /> " # > Œ # ˆ> / ‰ # It follows that # Œ$ 24. It is easy to see that " " $/> #> /> xŒ /> œ Œ > . # " #/ #> /> > Î ' Ñ > Î ! Ñ #> Î '/ Ñ ) % xw œ / /œ )/> %/#> . Ï%Ò Ï %Ò Ï %/> %/#> Ò On the other hand, Î" # Ï! " " " "Ñ " x "Ò œ œ Î" # Ï! Î ' Ñ > Î ! Ñ #> ) % / /. Ï%Ò Ï %Ò " " " " ÑÎ ' Ñ Î" " ) /> # Ï! " ÒÏ % Ò " " " " ÑÎ ! Ñ " # /#> ÒÏ # Ò " 26. Differentiation, elementwise, results in /> Gw œ %/> Ï /> On the other hand, Î #/#> #/#> #/#> $/$> Ñ '/$> Þ $/$> Ò ________________________________________________________________________ page 351 —————————————————————————— CHAPTER 7. —— Î" $ Ï# %Ñ Î" " G œ $ Ï# "Ò Î /#> % ÑÎ /> " %/> /#> " ÒÏ /> /#> #/#> $/$> Ñ #/#> '/$> Þ #/#> $/$> Ò /$> Ñ #/$> /$> Ò " # " " # " /> œ %/> Ï /> ________________________________________________________________________ page 352 ————————————————————————— CHAPTER 7. —— Section 7.3 4. The augmented matrix is Î" # Ï" !Ñ !Þ !Ò # " " " " # l l l Adding # times the first row to the second row and subtracting the first row from the third row results in Î" ! Ï! # $ $ " $ $ l l l !Ñ !Þ !Ò !Ñ !Þ !Ò Adding the negative of the second row to the third row results in Î" ! Ï! # $ ! " $ ! l l l We evidently end up with an equivalent system of equations B" #B# B$ œ ! B# B$ œ ! Þ Since there is no unique solution, let B$ œ ! , where ! is arbitrary. It follows that B# œ ! , and B" œ ! . Hence all solutions have the form xœ! Î "Ñ " Þ Ï"Ò 5. The augmented matrix is Î" $ Ï " ! " " " " # l l l !Ñ !Þ !Ò Adding $ times the first row to the second row and adding the first row to the last row yields Î" ! Ï! ! " " " l $ l " l !Ñ !Þ !Ò Now add the negative of the second row to the third row to obtain ________________________________________________________________________ page 353 —————————————————————————— CHAPTER 7. —— Î" ! Ï! !Ñ !Þ !Ò ! " ! " l $ l # l We end up with an equivalent linear system B" B$ œ ! B# $ B$ œ ! B$ œ ! Þ Hence the unique solution of the given system of equations is B" œ B# œ B$ œ ! Þ 7. Write the given vectors as columns of the matrix Xœ Î# " Ï! ! " ! "Ñ # Þ Ò ! It is evident that ./>aXb œ !. Hence the vectors are linearly dependent. In order to find a linear relationship between them, write -" xa"b -# xa#b -$ xa$b œ 0 Þ The latter equation is equivalent to Î# " Ï! ! " ! " ÑÎ -" Ñ Î ! Ñ # -# œ ! . ! ÒÏ -$ Ò Ï ! Ò !Ñ Î" !p! !Ò Ï! Performing elementary row operations, Î# " Ï! ! " ! " l # l ! l ! " ! "Î# &Î# ! l l l !Ñ !Þ !Ò We obtain the system of equations -" -$ Î# œ ! -# &-$ Î# œ ! Þ Setting -$ œ # , it follows that -" œ " and -$ œ & Þ Hence xa"b &xa#b #xa$b œ 0 Þ 9. The matrix containing the given vectors as columns is Î" Ð# XœÐ " Ï! # $ " " " ! # # $Ñ "Ó ÓÞ " $Ò ________________________________________________________________________ page 354 —————————————————————————— CHAPTER 7. —— We find that ./>aXb œ (!. Hence the given vectors are linearly independent. 10. Write the given vectors as columns of the matrix Î" # Xœ Ï # $ " ! # " " %Ñ $ Þ #Ò The four vectors are necessarily linearly dependent. Hence there are nonzero scalars such that -" xa"b -# xa#b -$ xa$b -% xa%b œ 0 Þ The latter equation is equivalent to Î" # Ï # Î" # Ï # $ " ! # " " ! % ÑÎ " Ñ Î Ñ -# Ó Ð ! Ó Ð $ Ð Ó œ Ð Ó. ! Ò -$ # Ï Ò Ï Ò -% ! Performing elementary row operations, $ " ! # " " % $ # l !Ñ Î" ! ! " l !Ñ l!p!"!"l!Þ l !Ò Ï! ! " ! l !Ò We end up with an equivalent linear system -" - % œ ! -# - % œ ! -$ œ ! Þ Let -% œ " . Then -" œ " and -# œ " Þ Therefore we find that xa"b xa#b xa%b œ 0 Þ 11. The matrix containing the given vectors as columns, X , is of size 8 ‚ 7 . Since ~ 8 7, we can augment the matrix with 7 8 rows of zeros. The resulting matrix, X , ~ is of size 7 ‚ 7 . Since X is square matrix, with at least one row of zeros, it follows ~ ~ that ./>ˆX‰ œ ! Þ Hence the column vectors of X are linearly dependent. That is, there ~ is a nonzero vector, c , such that X c œ 07‚" . If we write only the first 8 rows of the latter equation, we have X c œ 08‚" . Therefore the column vectors of X are linearly dependent. 12. By inspection, we find that xa"b a>b # xa#b a>b œ Œ /> Þ ! Hence $ xa"b a>b ' xa#b a>b xa$b a>b œ 0 , and the vectors are linearly dependent. 13. Two vectors are linearly dependent if and only if one is a nonzero scalar multiple ________________________________________________________________________ page 355 —————————————————————————— CHAPTER 7. —— of the other. However, there is no nonzero scalar, - , such that # =38 > œ - =38 > and =38 > œ #- =38 > for all > − a _ ß _b. Therefore the vectors are linearly independent. 16. The eigenvalues - and eigenvectors x satisfy the equation Œ $% B" ! # Œ œ Œ Þ "B# ! For a nonzero solution, we must have a$ -ba " -b ) œ ! , that is, - # #- & œ ! Þ The eigenvalues are -" œ " # 3 and -# œ " # 3 Þ The components of the eigenvector xa"b are solutions of the system Œ # #3 % B" ! # Œ œ Œ Þ # #3 B# ! The two equations reduce to a" 3bB" œ B# . Hence xa"b œ a" ß " 3bX Þ Now setting - œ -# œ " # 3 , we have Œ # #3 % B" ! # Œ œ Œ , # #3 B# ! with solution given by xa#b œ a" ß " 3bX Þ Œ #" 17. The eigenvalues - and eigenvectors x satisfy the equation B" ! " Œ œ Œ Þ #B# ! For a nonzero solution, we must have a # -ba # -b " œ ! , that is, - # %- $ œ ! Þ The eigenvalues are -" œ $ and -# œ " Þ For -" œ $ , the system of equations becomes " Œ" B" ! " Œ œ Œ , " B# ! which reduces to B" B# œ ! . A solution vector is given by xa"b œ a" ß "bX Þ Substituting - œ -# œ " , we have " Œ" B" ! " œ . " Œ B# Œ ! The equations reduce to B" œ B# . Hence a solution vector is given by xa#b œ a" ß "bX Þ 19. The eigensystem is obtained from analysis of the equation ________________________________________________________________________ page 356 —————————————————————————— CHAPTER 7. —— È$ B" ! Œ B œ Œ ! Þ # "È $ "- For a nonzero solution, the determinant of the coefficient matrix must be zero. That is, -# % œ ! Þ Hence the eigenvalues are -" œ # and -# œ # Þ Substituting the first eigenvalue, - œ # , yields $ È $ È$ " B" ! Œ B œ Œ ! Þ # The system is equivalent to the equation È$ B" B# œ ! . A solution vector is given X by xa"b œ Š" ß È$ ‹ Þ Substitution of - œ # results in " È $ È$ $ B" ! Œ B œ Œ ! , # which reduces to B" œ È$ B# . A corresponding solution vector is xa#b œ ŠÈ$ ß "‹ Þ X 20. The eigenvalues - and eigenvectors x satisfy the equation Œ $& B" ! $Î% Œ œ Œ Þ "B# ! For a nonzero solution, we must have a $ -ba" -b "&Î% œ ! , that is, -# #- $Î% œ ! Þ Hence the eigenvalues are -" œ $Î# and -# œ "Î# Þ In order to determine the eigenvector corresponding to -" , set - œ $Î# . The system of equations becomes $Î# Œ & B" ! $Î% œ , &Î# Œ B# Œ ! which reduces to # B" B# œ ! . A solution vector is given by xa"b œ a" ß #bX Þ Substitution of - œ -# œ "Î# results in &Î# Œ & B" ! $Î% Œ œ Œ , $Î# B# ! which reduces to "! B" œ $ B# . A corresponding solution vector is xa#b œ a$ ß "!bX Þ 22. The eigensystem is obtained from analysis of the equation ________________________________________________________________________ page 357 —————————————————————————— CHAPTER 7. —— Î$ " Ï # # ÑÎ B" Ñ Î ! Ñ " B# œ ! . ÒÏ B$ Ò Ï ! Ò "- # %% The characteristic equation of the coefficient matrix is -$ '-# ""- ' œ ! , with roots -" œ " , -# œ # and -$ œ $ . Setting - œ -" œ " , we have Î# " Ï # # $ % # ÑÎ B" Ñ Î ! Ñ " B# œ ! . # ÒÏ B$ Ò Ï ! Ò This system is reduces to the equations B" B$ œ ! B# œ ! Þ A corresponding solution vector is given by xa"b œ a" ß ! ß "bX Þ Setting - œ -# œ # , the reduced system of equations is B" # B# œ ! B$ œ ! Þ A corresponding solution vector is given by xa#b œ a # ß " ß !bX Þ Finally, setting - œ -$ œ $ , the reduced system of equations is B" œ ! B# B$ œ ! Þ A corresponding solution vector is given by xa$b œ a! ß " ß "bX Þ 23. For computational purposes, note that if - is an eigenvalue of B , then - - is an eigenvalue of the matrix A œ - B Þ Eigenvectors are unaffected, since they are only determined up to a scalar multiple. So with Bœ Î "" # Ï) # # "! )Ñ "! , &Ò the associated characteristic equation is .$ ").# )". "%&) œ ! , with roots ." œ * , .# œ * and .$ œ ") . Hence the eigenvalues of the given matrix, A , are -" œ " , -# œ " and -$ œ # . Setting - œ -" œ " , Ðwhich corresponds to using ." œ * in the modified problemÑ the reduced system of equations is # B" B$ œ ! B# B$ œ ! Þ A corresponding solution vector is given by xa"b œ a" ß # ß #bX Þ Setting - œ -# œ " , the reduced system of equations is ________________________________________________________________________ page 358 —————————————————————————— CHAPTER 7. —— A corresponding solution vector is given by xa#b œ a# ß # ß "bX Þ Finally, setting - œ -# œ " , the reduced system of equations is B" B$ œ ! # B# B$ œ ! Þ B" # B$ œ ! B# # B$ œ ! Þ A corresponding solution vector is given by xa$b œ a# ß " ß #bX Þ 25. Suppose that Ax œ 0 , but that x Á 0 . Let A œ a+34 bÞ Using elementary row operations, it is possible to transform the matrix into one that is not upper triangular. If it were upper triangular, backsubstitution would imply that x œ 0 . Hence a linear combination of all the rows results in a row containing only zeros. That is, there are 8 scalars, "3 , one for each row and not all zero, such that for each for column 4 , " "3 +34 œ ! . 8 Now consider A‡ œ a,34 b. By definition, ,34 œ +43 , or +34 œ ,43 Þ It follows that for each 4 , " "3 ,43 œ " ,45 "5 œ " ,45 "5 œ ! . 8 8 8 3œ" Let y œ ˆ"" ß "# ß âß "8 ‰ . We therefore have nonzero vector, y , such that A‡ y œ 0 . X 3œ" 5œ" 5œ" 26. By definition, aAx ß yb œ "aAxb3 C3 8 3œ! 8 8 œ " "+34 B4 C3 Þ 3œ! 4œ! Let ,34 œ +43 , so that +34 œ ,43 Þ Now interchanging the order or summation, aAx ß yb œ " B4 " +34 C3 8 8 œ " B4 " ,43 C3 Þ 4œ! 3œ! 4œ! 8 3œ! 8 Now note that ________________________________________________________________________ page 359 —————————————————————————— CHAPTER 7. —— " ,43 C3 œ " ,43 C3 œ aA‡ yb4 . 8 8 3œ! 3œ! Therefore aAx ß yb œ " B4 aA‡ yb4 œ ax ß A‡ yb . 8 4œ! 28. By linearity, Aˆxa!b ! 0 ‰ œ Axa!b ! A0 œb0 œ bÞ 29. Let -34 œ +43 . By the hypothesis, there is a nonzero vector, y , such that " -34 C4 œ " +43 C4 œ ! , 3 œ "ß #ß âß 8 . 8 8 4œ" 4œ" Taking the conjugate of both sides, and interchanging the indices, we have " +34 C3 œ ! Þ 8 3œ" This implies that a linear combination of each row of A is equal to zero. Now consider the augmented matrix cA lbd. Replace the last row by " C3 c+3" ß +3# ß âß +38 ß ,3 d œ –! ß ! ß âß ! ß " C3 ,3 —Þ 8 8 3œ" 3œ" We find that if ab ß yb œ !, then the last row of the augmented matrix contains only zeros. Hence there are 8 " remaining equations. We can now set B8 œ ! , some parameter, and solve for the other variables in terms of ! Þ Therefore the system of equations Ax œ b has a solution. 30. If - œ ! is an eigenvalue of A , then there is a nonzero vector, x , such that Ax œ - x œ 0 . That is, Ax œ 0 has a nonzero solution. This implies that the mapping defined by A is not 1-to-1, and hence not invertible. On the other hand, if A is singular, then ./>aAb œ !Þ Thus, Ax œ 0 has a nonzero solution. The latter equation can be written as Ax œ ! x Þ 31. As shown in Prob. #' , aAx ß yb œ ax ß A‡ yb . By definition of a Hermitian matrix, ________________________________________________________________________ page 360 —————————————————————————— CHAPTER 7. —— A œ A‡ Þ 32a+b. Based on Prob. $" , aAx ß xb œ ax ß Axb . a,b. Let x be an eigenvector corresponding to an eigenvalue - Þ It then follows that aAx ß xb œ a-x ß xb and ax ß Axb œ ax ß -xb Þ Based on the properties of the inner product, a-x ß xb œ -ax ß xb and ax ß -xb œ -ax ß xb . Then from Part a+b, -ax ß xb œ -ax ß xb Þ a- b. From Part a,b, Based on the definition of an eigenvector, ax ß xb œ lxl# ! Þ Hence we must have - - œ ! , which implies that - is real. 33. From Prob. $" , ˆAxa"b ß xa#b ‰ œ ˆxa"b ß Axa#b ‰ . -" ˆxa"b ß xa#b ‰ œ -# ˆxa"b ß xa#b ‰ œ -# ˆxa"b ß xa#b ‰ , a-" -# bˆxa"b ß xa#b ‰ œ ! . ˆ- -‰ax ß xb œ ! Þ Hence since the eigenvalues are real. Therefore Given that -" Á -# , we must have ˆxa"b ß xa#b ‰ œ ! . ________________________________________________________________________ page 361 —————————————————————————— CHAPTER 7. —— Section 7.4 3. Eq. a"%b states that the Wronskian satisfies the first order linear ODE .[ œ a:"" :## â :88 b[ Þ .> The general solution is [ a>b œ G /B:”( a:"" :## â :88 b.> •, in which G is an arbitrary constant. Let X" and X# be matrices representing two sets of fundamental solutions. It follows that ./>aX" b œ [" a>b œ G" /B:”( a:"" :## â :88 b.> • Hence . />aX" bÎ./>aX# b œ G" ÎG# . Note that G# Á !Þ [ xa"b ß xa#b ‘ œ - / ./>aX# b œ [# a>b œ G# /B:”( a:"" :## â :88 b.> • Þ 4. First note that :"" :## œ :a>bÞ As shown in Prob. a$b, For second order linear ODE, the Wronskian aas defined in Chap. $b satisfies the first order differential equation [ w :a>b[ œ ! . It follows that [ Ca"b ß Ca#b ‘ œ -" / ' :a>b.> ' :a>b.> Þ Þ Alternatively, based on the hypothesis, Ca"b œ !"" B"" !"# B"# Ca#b œ !#" B"" !## B"# Þ Direct calculation shows that ! B !"# B"# !#" B"" !## B"# [ Ca"b ß Ca#b ‘ œ º "" "" w w w w !"" B"" !"# B"# !#" B"" !## B"# º w w œ a!"" !## !"# !#" bB"" B"# a!"" !## !"# !#" bB"# B"" œ a!"" !## !"# !#" bB"" B## a!"" !## !"# !#" bB"# B#" Þ Here we used the fact that B"w œ B# . Hence [ Ca"b ß Ca#b ‘ œ a!"" !## !"# !#" b[ xa"b ß xa#b ‘ Þ w 5. The particular solution satisfies the ODE xa:b ‘ œ Pa>bxa:b ga>b Þ Now let ________________________________________________________________________ page 362 —————————————————————————— CHAPTER 7. —— x œ 9a>b be any solution of the homogeneous equation. That is, x w œ Pa>bx . We know that x œ x- , in which x- is a linear combination of some fundamental solution. By linearity of the differential equation, it follows that x œ xa:b x- is a solution of the ODE. Based on the uniqueness theorem, all solutions must have this form. 7a+b. By definition, [ xa"b ß xa#b ‘ œ º ># /> ˆ# ‰> > º œ > #> / Þ #> / a,b. The Wronskian vanishes at >! œ ! and >! œ # . Hence the vectors are linearly independent on W œ a _ ß !b a! ß #b a# ß _bÞ a- b. It follows from Theorem (Þ%Þ$ that one or more of the coefficients of the ODE must be discontinuous at >! œ ! and >! œ # . If not, the Wronskian would not vanish. a. b. Let ># /> x œ -" Œ - # Œ > Þ #> / x w œ -" Œ #> /> - # Œ > Þ # / Then On the other hand, :"" Œ: #" ># /> :"# :"" :"# :"" :"# x œ -" Œ : Œ -# Œ : Œ > :## :## :## #> / #" #" # > -" c:"" > #:"# >d -# c:"" :"# d/ œŒ Þ -" c:#" ># #:## >d -# c:#" :## d/> :"" ># #:"# > œ #> :"" :"# œ " # :#" > #:## > œ # :#" :## œ " Þ Comparing coefficients, we find that Solution of this system of equations results in # #> ># # :"" a>b œ ! ß :"# a>b œ " ß :#" a>b œ # ß :## a>b œ # Þ > #> > #> Hence the vectors are solutions of the ODE ________________________________________________________________________ page 363 —————————————————————————— CHAPTER 7. —— " ! # #> Œ # #> > ># #> x Þ ># # xw œ 8. Suppose that the solutions xa"b , xa#b ,â, xa7b are linearly dependent at > œ >! . Then there are constants -" ß -# ß âß -7 anot all zerob such that Now let za>b œ -" xa"b a>b -# xa#b a> b â -7 xa7b a> b . Then clearly, za>b is a solution of x w œ Pa>bx , with za>! b œ ! . Furthermore, ya>b ´ 0 is also a solution, with ya>! b œ ! . By the uniqueness theorem, za>b œ ya>b œ 0 Þ Hence -" xa"b a> b -# xa#b a> b â -7 xa7b a> b œ 0 on the entire interval ! > " . Going in the other direction is trivial. 9a+b. Let ya>b be any solution of x w œ Pa>bx . It follows that za>b ya>b œ -" xa"b a> b -# xa#b a> b â -8 xa8b a> b ya>b xa"b a>! bß xa#b a>! bß âß xa8b a>! bß ya>! b -" xa"b a>! b -# xa#b a>! b â -7 xa7b a>! b œ 0 . is also a solution. Now let >! − a! ß " b . Then the collection of vectors constitutes 8 " vectors, each with 8 components. Based on the assertion in Prob. "", Section (Þ$ , these vectors are necessarily linearly dependent. That is, there are 8 " constants ," ß ,# ß âß ,8 ß ,8" anot all zerob such that ," xa"b a>! b ,# xa#b a>! b â ,8 xa8b a>! b ,8" ya>! b œ 0 . ," xa"b a> b ,# xa#b a> b â ,8 xa8b a> b ,8" ya> b œ 0 From Prob. ), we have for all > − a! ß " bÞ Now ,8" Á ! , otherwise that would contradict the fact that the first 8 vectors are linearly independent. Hence ya> b œ and the assertion is true. a,b. Consider za>b œ -" xa"b a> b -# xa#b a> b â -8 xa8b a> b, and suppose that we also have za>b œ 5" xa"b a> b 5# xa#b a> b â 58 xa8b a> b Þ " ,8" ˆ," xa"b a> b ,# xa#b a> b â ,8 xa8b a> b‰, Based on the assumption, ________________________________________________________________________ page 364 —————————————————————————— CHAPTER 7. —— a5" -" bxa"b a> b a5# -# bxa#b a> b â a58 -8 bxa8b a> b œ 0 . The collection of vectors xa"b a> bß xa#b a> bß âß xa8b a> b is linearly independent on ! > " . It follows that 53 -3 œ ! , for 3 œ "ß #ß âß 8 . ________________________________________________________________________ page 365 —————————————————————————— CHAPTER 7. —— Section 7.5 2. Setting x œ 0 /<> , and substituting into the ODE, we obtain the algebraic equations Œ "< $ 0" ! # Œ œ Œ . %< 0# ! For a nonzero solution, we must have ./>aA < Ib œ <# $< # œ ! . The roots of the characteristic equation are <" œ " and <# œ # . For < œ " , the two equations reduce to 0" œ 0# . The corresponding eigenvector is 0 a"b œ a" ß "bX Þ Substitution of < œ # results in the single equation $ 0" œ # 0# . A corresponding eigenvector is 0 a#b œ a# ß $bX Þ Since the eigenvalues are distinct, the general solution is " # x œ -" Œ /> -# Œ /#> Þ " $ 3. Setting x œ 0 /<> results in the algebraic equations Œ #< $ 0" ! " Œ œ Œ . #< 0# ! For a nonzero solution, we must have ./>aA < Ib œ <# " œ ! . The roots of the characteristic equation are <" œ " and <# œ " . For < œ " , the system of equations reduces to 0" œ 0# . The corresponding eigenvector is 0 a"b œ a" ß "bX Þ Substitution of < œ " results in the single equation $ 0" œ 0# . A corresponding eigenvector is 0 a#b œ a" ß $bX Þ Since the eigenvalues are distinct, the general solution is " " x œ -" Œ /> -# Œ /> Þ " $ ________________________________________________________________________ page 366 —————————————————————————— CHAPTER 7. —— The system has an unstable eigendirection along 0 a"b œ a" ß "bX Þ Unless -" œ ! , all solutions will diverge. 4. Solution of the ODE requires analysis of the algebraic equations Œ "< % 0" ! " œ . # < Œ 0# Œ ! For a nonzero solution, we must have ./>aA < Ib œ <# < ' œ ! . The roots of the characteristic equation are <" œ # and <# œ $ . For < œ # , the system of equations reduces to 0" œ 0# . The corresponding eigenvector is 0 a"b œ a" ß "bX Þ Substitution of < œ $ results in the single equation % 0" 0# œ ! . A corresponding eigenvector is 0 a#b œ a" ß %bX Þ Since the eigenvalues are distinct, the general solution is " " $> x œ -" Œ /#> -# Œ / Þ " % The system has an unstable eigendirection along 0 a"b œ a" ß "bX Þ Unless -" œ ! , all solutions will diverge. 8. Setting x œ 0 /<> results in the algebraic equations ________________________________________________________________________ page 367 —————————————————————————— CHAPTER 7. —— $< Œ " 0" ! ' Œ œ Œ . #< 0# ! For a nonzero solution, we must have ./>aA < Ib œ <# < œ ! . The roots of the characteristic equation are <" œ " and <# œ ! . With < œ " , the system of equations reduces to 0" $0# œ !. The corresponding eigenvector is 0 a"b œ a$ ß "bX Þ For the case < œ !, the system is equivalent to the equation 0" # 0# œ ! . An eigenvector is 0 a#b œ a# ß "bX Þ Since the eigenvalues are distinct, the general solution is x œ -" Œ $ # > / - # Œ Þ " " The entire line along the eigendirection 0 a#b œ a# ß "bX consists of equilibrium points. All other solutions diverge. The direction field changes across the line B" #B# œ ! Þ Eliminating the exponential terms in the solution, the trajectories are given by B" $ B# œ -# . 10. The characteristic equation is given by #< º " #3 œ <# a" 3b< 3 œ ! Þ "3<º The equation has complex roots <" œ " and <# œ 3. For < œ ", the components of the solution vector must satisfy 0" a# 3b0# œ ! . Thus the corresponding eigenvector is 0 a"b œ a# 3 ß "bX Þ Substitution of < œ 3 results in the single equation 0" 0# œ !. A corresponding eigenvector is 0 a#b œ a" ß "bX Þ Since the eigenvalues are distinct, the general solution is x œ -" Œ #3 > " 3> / - # Œ / Þ " " 11. Setting x œ 0 /<> results in the algebraic equations ________________________________________________________________________ page 368 —————————————————————————— CHAPTER 7. —— Î" < " Ï# # ÑÎ 0" Ñ Î ! Ñ " 0# œ ! . ÒÏ 0$ Ò Ï ! Ò "< For a nonzero solution, we must have ./>aA < Ib œ <$ %<# < % œ ! . The roots of the characteristic equation are <" œ % , <# œ " and <$ œ " . Setting < œ % , we have Î $ " Ï# " # " # ÑÎ 0" Ñ Î ! Ñ " 0# œ ! . ÒÏ 0$ Ò Ï ! Ò $ " #< " This system is reduces to the equations 0" 0 $ œ ! 0# 0 $ œ ! Þ A corresponding solution vector is given by 0 a"b œ a" ß " ß "bX Þ Setting - œ " , the reduced system of equations is 0" 0 $ œ ! 0# # 0 $ œ ! Þ A corresponding solution vector is given by 0 a#b œ a" ß # ß "bX Þ Finally, setting - œ " , the reduced system of equations is 0" 0 $ œ ! 0# œ ! Þ A corresponding solution vector is given by 0 a$b œ a" ß ! ß "bX Þ Since the eigenvalues are distinct, the general solution is x œ -" Î " Ñ %> Î"Ñ Î " Ñ > " / -# # / > - $ ! /Þ Ï"Ò Ï"Ò Ï "Ò 12. The eigensystem is obtained from analysis of the equation Î$ < # Ï% # < # % ÑÎ 0" Ñ Î ! Ñ # 0# œ ! . $ < ÒÏ 0$ Ò Ï ! Ò The characteristic equation of the coefficient matrix is <$ '<# "&< ) œ ! , with roots <" œ ) , <# œ " and <$ œ " . Setting < œ <" œ ) , we have ________________________________________________________________________ page 369 —————————————————————————— CHAPTER 7. —— Î & # Ï% % ÑÎ 0" Ñ Î ! Ñ # 0# œ ! . ÒÏ 0$ Ò Ï ! Ò & # ) # This system is reduced to the equations 0" 0 $ œ ! # 0# 0 $ œ ! Þ A corresponding solution vector is given by 0 a"b œ a# ß " ß #bX Þ Setting < œ " , the system of equations is reduced to the single equation # 0" 0 # # 0 $ œ ! Þ Two independent solutions are obtained as 0 a2b œ a" ß # ß !bX and 0 a$b œ a! ß # ß "bX Þ Hence the general solution is Î # Ñ )> Î " Ñ > Î ! Ñ > x œ -" " / - # # / - $ # / Þ Ï#Ò Ï!Ò Ï"Ò 13. Setting x œ 0 /<> results in the algebraic equations Î" < # Ï ) " "< & For a nonzero solution, we must have ./>aA < Ib œ <$ <# %< % œ ! . The roots of the characteristic equation are <" œ # , <# œ # and <$ œ " . Setting < œ # , we have Î " # Ï ) " " & " ÑÎ 0" Ñ Î ! Ñ " 0# œ ! . & ÒÏ 0$ Ò Ï ! Ò " ÑÎ 0" Ñ Î ! Ñ " 0# œ ! . ÒÏ 0$ Ò Ï ! Ò $< This system is reduces to the equations 0" œ ! 0# 0 $ œ ! Þ A corresponding solution vector is given by 0 a"b œ a! ß " ß "bX Þ Setting - œ " , the reduced system of equations is # 0" $ 0 $ œ ! 0# # 0 $ œ ! Þ ________________________________________________________________________ page 370 —————————————————————————— CHAPTER 7. —— A corresponding solution vector is given by 0 a#b œ a$ ß % ß #bX Þ Finally, setting - œ # , the reduced system of equations is ( 0" % 0 $ œ ! (0# & 0$ œ ! Þ A corresponding solution vector is given by 0 a$b œ a% ß & ß (bX Þ Since the eigenvalues are distinct, the general solution is x œ -" Î ! Ñ #> Î$Ñ Î%Ñ " / -# % /> -$ & /#> Þ Ï "Ò Ï #Ò Ï (Ò 15. Setting x œ 0 /<> results in the algebraic equations Œ &< $ 0" ! " Œ œ Œ . "< 0# ! For a nonzero solution, we must have ./>aA < Ib œ <# '< ) œ ! . The roots of the characteristic equation are <" œ % and <# œ # . With < œ % , the system of equations reduces to 0" 0# œ !. The corresponding eigenvector is 0 a"b œ a" ß "bX Þ For the case < œ # , the system is equivalent to the equation $ 0" 0# œ ! . An eigenvector is 0 a#b œ a" ß $bX Þ Since the eigenvalues are distinct, the general solution is " " x œ -" Œ /%> -# Œ /#> Þ " $ Invoking the initial conditions, we obtain the system of equations -" - # œ # -" $ - # œ " Þ Hence -" œ (Î# and -# œ $Î# , and the solution of the IVP is xœ ( " %> $ " #> Œ / Œ / Þ #" #$ 17. Setting x œ 0 /<> results in the algebraic equations Î" < ! Ï " " #< " For a nonzero solution, we must have ./>aA < Ib œ <$ '<# ""< ' œ ! . The roots of the characteristic equation are <" œ " , <# œ # and <$ œ $ . Setting < œ " , we have # ÑÎ 0" Ñ Î ! Ñ # 0# œ ! . $ < ÒÏ 0$ Ò Ï ! Ò ________________________________________________________________________ page 371 —————————————————————————— CHAPTER 7. —— Î! ! Ï " " # ÑÎ 0" Ñ Î ! Ñ "# 0# œ ! . ÒÏ 0$ Ò Ï ! Ò "# 0" œ ! 0# # 0 $ œ ! Þ This system is reduces to the equations A corresponding solution vector is given by 0 a"b œ a! ß # ß "bX Þ Setting - œ # , the reduced system of equations is 0" 0 # œ ! 0$ œ ! Þ A corresponding solution vector is given by 0 a#b œ a" ß " ß !bX Þ Finally, upon setting - œ $ , the reduced system of equations is 0" # 0 $ œ ! 0# # 0 $ œ ! Þ A corresponding solution vector is given by 0 a$b œ a# ß # ß "bX Þ Since the eigenvalues are distinct, the general solution is Î ! Ñ> Î " Ñ #> Î # Ñ $> x œ -" # / - # " / - $ # / Þ Ï"Ò Ï!Ò Ï"Ò -# # - $ œ # # -" - # # - $ œ ! -" - $ œ " Þ It follows that -" œ " , -# œ # and -$ œ ! . Hence the solution of the IVP is xœ Î ! Ñ> Î"Ñ # / # " /#> Þ Ï"Ò Ï!Ò Invoking the initial conditions, the coefficients must satisfy the equations 18. The eigensystem is obtained from analysis of the equation Î < # Ï " ! < # " ÑÎ 0" Ñ Î ! Ñ ! 0# œ ! . % < ÒÏ 0$ Ò Ï ! Ò The characteristic equation of the coefficient matrix is <$ %<# < % œ ! , with roots <" œ " , <# œ " and <$ œ % . Setting < œ <" œ " , we have ________________________________________________________________________ page 372 —————————————————————————— CHAPTER 7. —— Î " # Ï " " ÑÎ 0" Ñ Î ! Ñ ! 0# œ ! . ÒÏ 0$ Ò Ï ! Ò $ ! " # This system is reduced to the equations 0" 0 $ œ ! 0# # 0 $ œ ! Þ A corresponding solution vector is given by 0 a"b œ a" ß # ß "bX Þ Setting < œ " , the system reduces to the equations 0" 0 $ œ ! 0# # 0 $ œ ! Þ The corresponding eigenvector is 0 a2b œ a" ß # ß "bX Þ Finally, upon setting < œ % , the system is equivalent to the equations % 0" 0 $ œ ! ) 0# 0 $ œ ! Þ The corresponding eigenvector is 0 a$b œ a# ß " ß )bX Þ Hence the general solution is x œ -" Î " Ñ > Î " Ñ> Î # Ñ %> # / -# # " / -$ /Þ Ï"Ò Ï "Ò Ï )Ò -" - # # - $ œ ( # -" # - # - $ œ & -" - # ) - $ œ & Þ It follows that -" œ $ , -# œ ' and -$ œ " . Hence the solution of the IVP is Î " Ñ > Î " Ñ > Î # Ñ %> # " xœ$ # / ' / /Þ Ï"Ò Ï "Ò Ï )Ò Invoking the initial conditions, 19. Set x œ 0 >< . Substitution into the system of differential equations results in > † <><" 0 œ A 0 >< , which upon simplification yields is, A 0 <0 œ 0 Þ Hence the vector 0 and constant < must satisfy aA <Ib0 œ 0 Þ 21. Setting x œ 0 >< results in the algebraic equations ________________________________________________________________________ page 373 —————————————————————————— CHAPTER 7. —— Œ 0" ! " Œ œ Œ . "< 0# ! For a nonzero solution, we must have ./>aA < Ib œ <# '< ) œ ! . The roots of the characteristic equation are <" œ % and <# œ # . With < œ % , the system of equations reduces to 0" 0# œ !. The corresponding eigenvector is 0 a"b œ a" ß "bX Þ For the case < œ # , the system is equivalent to the equation $ 0" 0# œ ! . An eigenvector is 0 a#b œ a" ß $bX Þ It follows that The Wronskian of this solution set is [ cxa"b ß xa#b d œ #>' . Thus the solutions are linearly independent for > ! . Hence the general solution is " " x œ - " Œ >% - # Œ > # Þ " $ 22. As shown in Prob. "* , solution of the ODE requires analysis of the equations Œ %< ) 0" ! $ Œ œ Œ . '< 0# ! " " xa"b œ Œ >% and xa#b œ Œ ># Þ " $ &< $ For a nonzero solution, we must have ./>aA < Ib œ <# #< œ ! . The roots of the characteristic equation are <" œ ! and <# œ # . For < œ ! , the system of equations reduces to % 0" œ $ 0# . The corresponding eigenvector is 0 a"b œ a$ ß %bX Þ Setting < œ # results in the single equation # 0" 0# œ ! . A corresponding eigenvector is 0 a#b œ a" ß #bX Þ It follows that The Wronskian of this solution set is [ cxa"b ß xa#b d œ # ># . These solutions are linearly independent for > ! . Hence the general solution is $ " x œ -" Œ -# Œ ># Þ % # 23. Setting x œ 0 >< results in the algebraic equations Œ $< # 0" ! # œ . # < Œ 0# Œ ! $ " xa"b œ Œ and xa#b œ Œ ># Þ % # For a nonzero solution, we must have ./>aA < Ib œ <# < # œ ! . The roots of the characteristic equation are <" œ # and <# œ " . Setting < œ # , the system of equations reduces to 0" # 0# œ !. The corresponding eigenvector is 0 a"b œ a# ß "bX Þ ________________________________________________________________________ page 374 —————————————————————————— CHAPTER 7. —— With < œ " , the system is equivalent to the equation # 0" 0# œ ! . An eigenvector is 0 a#b œ a" ß #bX Þ It follows that The Wronskian of this solution set is [ cxa"b ß xa#b d œ $ >. Thus the solutions are linearly independent for > ! . Hence the general solution is # " x œ -" Œ ># -# Œ >" Þ " # 24a+b. The general solution is # " xa"b œ Œ ># and xa#b œ Œ >" Þ " # x œ -" Œ " > " #> / - # Œ / Þ # # a, b . a- b . ________________________________________________________________________ page 375 —————————————————————————— CHAPTER 7. —— 26a+b. The general solution is a, b . x œ -" Œ " > " #> / - # Œ / Þ # # ________________________________________________________________________ page 376 —————————————————————————— CHAPTER 7. —— a, b . a- b . a,b. It follows that aA <# Ib0 a"b œ A 0 a"b <# 0 a"b œ <" 0 a"b <# 0 a"b . 28a+b. We note that aA <3 Ib0 a3b œ 0 , for 3 œ "ß # . a- b. Suppose that 0 a"b and 0 a#b are linearly dependent. Then there exist constants -" and -# , not both zero, such that -" 0 a"b -# 0 a#b œ 0 Þ Assume that -" Á ! Þ It is clear that aA <# Ibˆ-" 0 a"b -# 0 a#b ‰ œ 0 Þ On the other hand, aA <# Ibˆ-" 0 a"b -# 0 a#b ‰ œ -" a<" <# b0 a"b 0 œ -" a<" <# b0 a"b . Since <" Á <# , we must have -" œ ! , which leads to a contradiction. a. b. Note that aA <" Ib0 a#b œ a<# <" b0 a#b . ________________________________________________________________________ page 377 —————————————————————————— CHAPTER 7. —— a/b. Let 8 œ $, with <" Á <# Á <$ . Suppose that 0 a"b , 0 a#b and 0 a$b are indeed linearly dependent. Then there exist constants -" , -# and -$ , not all zero, such that Assume that -" Á ! Þ It is clear that aA <# Ibˆ-" 0 a"b -# 0 a#b -$ 0 a$b ‰ œ 0 Þ On the other hand, It follows that -" a<" <# b0 a"b -$ a<$ <# b0 a$b œ 0 Þ Based on the result of Part a+b, which is actually not dependent on the value of 8 , the vectors 0 a"b and 0 a$b are linearly independent. Hence we must have -" a<" <# b œ -$ a<$ <# b œ ! , which leads to a contradiction. 29a+b. Let B" œ C and B# œ C w . It follows that B"w œ B# and B#w œ C ww " œ a- C , C w b . + aA <# Ibˆ-" 0 a"b -# 0 a#b -$ 0 a$b ‰ œ -" a<" <# b0 a"b -$ a<$ <# b0 a$b Þ -" 0 a"b -# 0 a#b -$ 0 a$b œ 0 Þ In terms of the new variables, we obtain the system of two first order ODEs B"w œ B# " B#w œ a- B" , B# b Þ + a,b. The coefficient matrix is given by AœŒ ! + " Þ , + Setting x œ 0 /<> results in the algebraic equations < Œ+ " , + 0" ! . œ < Œ 0# Œ ! For a nonzero solution, we must have , ./>aA < Ib œ <# < œ ! . + + Multiplying both sides of the equation by + , we obtain + <# , < - œ ! Þ 30. Solution of the ODE requires analysis of the algebraic equations Œ "< % 0" ! " . œ # < Œ 0# Œ ! For a nonzero solution, we must have ./>aA < Ib œ ! . The characteristic equation is ________________________________________________________________________ page 378 —————————————————————————— CHAPTER 7. —— )! <# #% < " œ ! , with roots <" œ "Î% and <# œ "Î#! . With < œ "Î% , the system of equations reduces to # 0" 0# œ ! . The corresponding eigenvector is 0 a"b œ a" ß #bX Þ Substitution of < œ "Î#! results in the equation # 0" $ 0# œ ! . A corresponding eigenvector is 0 a#b œ a$ ß #bX Þ Since the eigenvalues are distinct, the general solution is x œ -" Œ " $ >Î% -# Œ />Î#! Þ / # # Invoking the initial conditions, we obtain the system of equations -" $ -# œ "( # -" # -# œ #" Þ Hence -" œ #*Î) and -# œ &&Î) , and the solution of the IVP is xœ a, b . #* " && $ >Î% Œ />Î#! Þ Œ / ) # )# a- b. Both functions are monotone increasing. It is easy to show that !Þ& Ÿ B" a>b ! and !Þ& Ÿ B# a>b ! provided that > X ¸ (%Þ$* Þ 31a+b. For ! œ "Î# , solution of the ODE requires that "< Œ "Î# 0" ! " Œ œ Œ . "< 0# ! The characteristic equation is # <# % < " œ ! , with roots <" œ " "ÎÈ# and <# œ " "ÎÈ# . With < œ " "ÎÈ# , the system of equations reduces to X È# 0" # 0# œ ! . The corresponding eigenvector is 0 a"b œ Š È# ß "‹ Þ Substitution ________________________________________________________________________ page 379 —————————————————————————— CHAPTER 7. —— of < œ " "ÎÈ# results in the equation È# 0" # 0# œ ! . An eigenvector is X 0 a#b œ ŠÈ# ß "‹ Þ The general solution is x œ -" È# È# Š#È#‹>Î# Š#È#‹>Î# / -# Þ / " " The eigenvalues are distinct and both 8/1+>3@/Þ The equilibrium point is a stable node. a,b. For ! œ # , the characteristic equation is given by <# # < " œ ! , with roots <" œ " È# and <# œ " È# . With < œ " È# , the system of equations X reduces to È# 0" 0# œ ! . The corresponding eigenvector is 0 a"b œ Š" ß È#‹ Þ Substitution of < œ " È# results in the equation È# 0" 0# œ ! . An eigenvector X is 0 a#b œ Š" ß È#‹ Þ The general solution is x œ -" Œ " " Š"È#‹> Š"È#‹> - # Œ È / Þ È # / # The eigenvalues are of opposite sign, hence the equilibrium point is a saddle point. 32. The system of differential equations is " .M œ $# Œ .> Z # " M # Þ & Œ Z # Solution of the system requires analysis of the eigenvalue problem " # $ # < & # ! 0" Œ 0 œ Œ ! . < " # # The characteristic equation is <# $ < #, with roots <" œ " and <# œ #. With < œ " , the equations reduce to 0" 0# œ ! . A corresponding eigenvector is given by 0 a"b œ a" ß "bX Þ Setting < œ # , the system reduces to the equation $ 0" 0# œ ! . An eigenvector is 0 a#b œ a" ß $bX Þ Hence the general solution is Œ M " > " #> œ -" Œ / - # Œ / Þ Z " $ a,bÞ The eigenvalues are distinct and both 8/1+>3@/Þ We find that the equilibrium point a! ß !b is a stable node. Hence all solutions converge to a! ß !bÞ 33a+b. Solution of the ODE requires analysis of the algebraic equations ________________________________________________________________________ page 380 —————————————————————————— CHAPTER 7. —— ! 0" Œ 0 œ Œ ! . < # V" P " G < " GV# " P The characteristic equation is <# Œ P GV" V# V" V # œ !Þ < PGV# PGV# The eigenvectors are real and distinct, provided that the discriminant is positive. That is, P GV" V# V" V # Œ %Œ !, PGV# PGV# # which simplifies to the condition " V" % !. Œ GV# P PG # a,b. The parameters in the ODE are all positive. Observe that the sum of the roots is Also, the product of the roots is V" V # !Þ PGV# It follows that both roots are negative. Hence the equilibrium solutio8 M œ ! ß Z œ ! represents a stable node, which attracts all solutions. a- b. If the condition in Part a+b is not satisfied, that is, # P GV" V# !Þ PGV# " V" % Ÿ !, Œ GV# P PG V/a<"ß# b œ then the real part of the eigenvalues is P GV" V# Þ # PGV# As long as the parameters are all positive, then the solutions will still converge to the equilibrium point a! ß !b. ________________________________________________________________________ page 381 —————————————————————————— CHAPTER 7. —— Section 7.6 2. Setting x œ 0 /<> results in the algebraic equations Œ "< " 0" ! % Œ œ Œ . "< 0# ! For a nonzero solution, we require that ./>aA < Ib œ <# #< & œ !. The roots of the characteristic equation are < œ " „ #3 . Substituting < œ " #3 , the two equations reduce to 0" #3 0# œ ! . The two eigenvectors are 0 a"b œ a #3 ß "bX and 0 a#b œ a#3 ß "bX Þ Hence one of the complex-valued solutions is given by xa"b œ Œ #3 a"#3b> / " #3 > œŒ / a-9= #> 3 =38 #>b " # =38 #> # -9= #> > œ /> Œ 3/ Œ Þ -9= #> =38 #> # =38 #> > # -9= #> -# / Œ Þ -9= #> =38 #> Based on the real and imaginary parts of this solution, the general solution is x œ -" /> Œ 3. Solution of the ODEs is based on the analysis of the algebraic equations Œ #< " 0" ! & Œ œ Œ . #< 0# ! For a nonzero solution, we require that ./>aA < Ib œ <# " œ !. The roots of the characteristic equation are < œ „3 . Setting < œ 3 , the equations are equivalent to 0" a# 3b0# œ ! . The eigenvectors are 0 a"b œ a# 3 ß "bX and 0 a#b œ a# 3 ß "bX Þ Hence one of the complex-valued solutions is given by ________________________________________________________________________ page 382 —————————————————————————— CHAPTER 7. —— xa"b œ Œ # 3 3> / " #3 œŒ a-9= > 3 =38 >b " # -9= > =38 > -9= > # =38 > œŒ 3Œ Þ -9= > =38 > # -9= > =38 > -9= > # =38 > -# Œ Þ -9= > =38 > Therefore the general solution is x œ -" Œ x œ -" Œ The solution may also be written as & -9= > & =38 > -# Œ Þ # -9= > =38 > -9= > # =38 > 4. Setting x œ 0 /<> results in the algebraic equations #< Œ *Î& 0" ! &Î# Œ œ Œ . "< 0# ! For a nonzero solution, we require that ./>aA < Ib œ <# < & œ !. The roots of # the characteristic equation are < œ a" „ $3bÎ# . With < œ a" $ 3bÎ# , the equations reduce to the single equation a$ $3b0" & 0# œ ! . The corresponding eigenvector is given by 0 a"b œ a& ß $ $ 3bX Þ Hence one of the complex-valued solutions is ________________________________________________________________________ page 383 —————————————————————————— CHAPTER 7. —— xa"b œ Œ & a"$3b>Î# / $ $3 # 3 >Î# $ $ œŒ / Œ-9= > 3 =38 > " # # œ />Î# The general solution is x œ -" / >Î# $ $ # -9= $ > =38 $ > >Î# -9= # > # =38 # > # # -# / Þ -9= $ > =38 $ > # # # -9= $ > =38 $ > -9= $ > # =38 $ > # # # # 3/>Î# Þ $ $ -9= # > =38 # > The solution may also be written as x œ -" / >Î# & -9= $ > & =38 $ > >Î# # # $ -9= $ > $=38 $ > -# / $ -9= $ > $=38 $ > Þ # # # # 5. Setting x œ 0 >< results in the algebraic equations Œ "< & 0" ! " Œ œ Œ . $< 0# ! The characteristic equation is <# # < # œ ! , with roots < œ " „ 3 . Substituting < œ " 3 reduces the system of equations to a# 3b0" 0# œ ! . The eigenvectors are 0 a"b œ a" ß # 3bX and 0 a#b œ a" ß # 3bX Þ Hence one of the complex-valued solutions is given by ________________________________________________________________________ page 384 —————————————————————————— CHAPTER 7. —— xa"b œ Œ " a"3b> / #3 " > œŒ / a-9= > 3 =38 >b #3 -9= > =38 > > œ /> Œ 3/ Œ Þ # -9= > =38 > -9= > #=38 > -9= > =38 > > -# / Œ Þ # -9= > =38 > -9= > #=38 > The general solution is x œ -" /> Œ 6. Solution of the ODEs is based on the analysis of the algebraic equations "< Œ & 0" ! # Œ œ Œ . "< 0# ! For a nonzero solution, we require that ./>aA < Ib œ <# * œ !. The roots of the characteristic equation are < œ „ $ 3 . Setting < œ $ 3 , the two equations reduce to a" $ 3b0" #0# œ ! . The corresponding eigenvector is 0 a"b œ a # ß " $3bX Þ Hence one of the complex-valued solutions is given by xa"b œ Œ # $3> / " $3 # œŒ a-9= $> 3 =38 $>b " $3 # -9= $> # =38 $> œŒ 3Œ Þ -9= $> $ =38 $> $-9= $> =38 $> The general solution is ________________________________________________________________________ page 385 —————————————————————————— CHAPTER 7. —— x œ -" Œ # -9= $> # =38 $> -# Œ Þ -9= $> $ =38 $> $-9= $> =38 $> 8. The eigensystem is obtained from analysis of the equation Î $< " Ï # ! "< " # ÑÎ 0" Ñ Î ! Ñ ! 0# œ ! . ÒÏ 0$ Ò Ï ! Ò < The characteristic equation of the coefficient matrix is <$ %<# (< ' œ ! , with roots <" œ # , <# œ " È# 3 and <$ œ " È# 3 . Setting < œ # , the equations reduce to 0" # 0 $ œ ! 0" 0 # œ ! Þ The corresponding eigenvector is 0 a"b œ a# ß # ß "bX Þ With < œ " È# 3 , the system of equations is equivalent to Š# 3È# ‹0" # 0$ œ ! 0" 3È # 0 # œ ! Þ complex-valued solutions is given by An eigenvector is given by 0 a#b œ Š 3È# ß " ß " 3È# ‹ Þ Hence one of the X ________________________________________________________________________ page 386 —————————————————————————— CHAPTER 7. —— 3È # Ñ È Ó/Š"3 #‹3> " Ï " 3È # Ò Î Ð Î x a#b œ œÐ 3È # Ñ Ó/> Š-9= È# > 3 =38 È# >‹ " Ï " 3È # Ò > Ð œ/ The other complex-valued solution is xa$b œ 0 a#b /<$ > . The general solution is Î # Ñ #> x œ -" # / Ï"Ò -# / > Ð È# =38 È# > È# -9= È# > Ñ Î Ñ Ó 3/> Ð ÓÞ È# > =38 È# > -9= Ï -9= È# > È# =38 È# > Ò Ï È# -9= È# > =38 È# > Ò Î It is easy to see that all solutions converge to the equilibrium point a! ß ! ß !b Þ 10. Solution of the system of ODEs requires that Œ $< " 0" ! # Œ œ Œ . "< 0# ! È# =38 È# > È# -9= È# > Ñ Î Ñ > Ð Ó -$ / ÓÞ È# > È# > -9= =38 Ï -9= È# > È# =38 È# > Ò Ï È# -9= È# > =38 È# > Ò Î The characteristic equation is <# % < & œ ! , with roots < œ # „ 3 . Substituting < œ # 3 , the equations are equivalent to 0" a" 3b0# œ ! . The corresponding eigenvector is 0 a"b œ a" 3 ß "bX Þ One of the complex-valued solutions is given by xa"b œ Œ " 3 a#3b> / " " 3 #> œŒ / a-9= > 3 =38 >b " -9= > =38 > -9= > =38 > #> œ /#> Œ 3/ Œ Þ -9= > =38 > Hence the general solution is x œ -" /#> Œ -9= > =38 > -9= > =38 > #> -# / Œ Þ -9= > =38 > Invoking the initial conditions, we obtain the system of equations -" - # œ " -" œ # Þ ________________________________________________________________________ page 387 —————————————————————————— CHAPTER 7. —— Solving for the coefficients, the solution of the initial value problem is x œ # /#> Œ -9= > =38 > -9= > =38 > #> $/ Œ -9= > =38 > -9= > & =38 > œ /#> Œ Þ # -9= > $ =38 > 11a+b. With xa!b œ a# ß #bX , the solution is 11a,b. x œ />Î% Œ # -9= > # =38 > Þ # -9= > ________________________________________________________________________ page 388 —————————————————————————— CHAPTER 7. —— 11a- b. 12. Solution of the ODEs is based on the analysis of the algebraic equations < " % & ' & # 0" ! Œ 0 œ Œ ! . < # The characteristic equation is #& <# "! < #' œ ! , with roots < œ " „ 3 . Setting & < œ "Î& 3 , the two equations reduce to 0" a" 3b0# œ ! . The corresponding eigenvector is 0 a"b œ a" 3 ß "bX Þ One of the complex-valued solutions is given by xa"b œ Œ " 3 ˆ " 3‰> / & " " 3 >Î& œŒ / a-9= > 3 =38 >b " -9= > =38 > -9= > =38 > >Î& œ />Î& Œ 3/ Œ Þ -9= > =38 > Hence the general solution is x œ -" />Î& Œ -9= > =38 > -9= > =38 > >Î& -# / Œ Þ -9= > =38 > a,b. Let xa!b œ aB! ß B! bX Þ The solution of the initial value problem is " # x œ B! />Î& Œ # With xa!b œ a" ß #bX , the solution is -9= > =38 > -9= > =38 > >Î& ! ! aB# B" b/ Œ Þ -9= > =38 > B! -9= > a# B! B! b=38 > # " œ />Î& Œ " ! Þ B# -9= > aB! B! b=38 > # " ________________________________________________________________________ page 389 —————————————————————————— CHAPTER 7. —— x œ />Î& Œ -9= > $ =38 > Þ # -9= > =38 > a- b . ________________________________________________________________________ page 390 —————————————————————————— CHAPTER 7. —— a. b . 13a+b. The characteristic equation of the coefficient matrix is <# #!< " !# , with roots < œ ! „ 3 . a,b. When ! ! and ! ! , the equilibrium point a! ß !b is a stable spiral and an unstable spiral, respectively. The equilibrium point is a center when ! œ ! Þ a- b . ________________________________________________________________________ page 391 —————————————————————————— CHAPTER 7. —— 14a+b. The roots of the characteristic equation, <# ! < & œ ! , are a,b. Note that the roots are complex when È#! ! È#! . For the case when ! − Š È#! ß !‹, the equilibrium point a! ß !b is a stable spiral. On the other hand, <"ß# œ ! "È # ! #! Þ „ ## when ! − Š! ß È#! ‹, the equilibrium point is an unstable spiral. For the case ! œ !, the roots are purely imaginary, so the equilibrium point is a center. When !# #! , the roots are real and distinct. The equilibrium point becomes a node, with its stability dependent on the sign of ! Þ Finally, the case !# œ #! marks the transition from spirals to nodes. a- b . ________________________________________________________________________ page 392 —————————————————————————— CHAPTER 7. —— 17. The characteristic equation of the coefficient matrix is <# #< " ! œ ! , with roots given formally as <"ß# œ " „ È ! . The roots are real provided that ! Ÿ ! Þ First note that the sum of the roots is # and the product of the roots is " ! . For negative values of ! , the roots are distinct, with one always negative. When ! ", the roots have opposite signs. Hence the equilibrium point is a saddle. For the case " ! ! , the roots are both negative, and the equilibrium point is a stable node. ! œ " represents a transition from saddle to node. When ! œ ! , both roots are equal. For the case ! ! , the roots are complex conjugates, with negative real part. Hence the equilibrium point is a stable spiral. ________________________________________________________________________ page 393 —————————————————————————— CHAPTER 7. —— 19. The characteristic equation for the system is given by <# a% !b< "! %! œ !Þ ! È# „ ! )! #% Þ # The roots are <"ß# œ # First note that the roots are complex when % #È"! ! % #È"! . We also find that when % #È"! ! # , the equilibrium point is a stable spiral. For the case ! œ # , the equilibrium point is a center. When # ! % #È"! , the equilibrium point is an unstable spiral. For all other cases, the roots are real. When ! #Þ& , the roots have opposite signs, with the equilibrium point being a saddle. For the case % #È"! ! #Þ& , the roots are both positive, and the equilibrium point is an unstable node. Finally, when ! % #È"! , both roots are negative, with the equilibrium point being a stable node. ________________________________________________________________________ page 394 —————————————————————————— CHAPTER 7. —— 20. The characteristic equation is <# # < a#% )!b œ ! , with roots <"ß# œ " „ È#& )! Þ The roots are complex when ! #&Î) . Since the real part is negative, the origin is a stable spiral. Otherwise the roots are real. When #& ! $ , both roots are negative, and hence the equilibrium point is a stable node. For ! $ , the roots are of opposite sign and the origin is a saddle. 22. Based on the method in Prob. "* of Section (Þ& , setting x œ 0 >< results in the ________________________________________________________________________ page 395 —————————————————————————— CHAPTER 7. —— algebraic equations Œ #< " 0" ! & Œ œ Œ . #< 0# ! The characteristic equation for the system is <# " œ ! , with roots <"ß# œ „ 3 . With < œ 3 , the equations reduce to the single equation 0" a# 3b0# œ !. A corresponding eigenvector is 0 a"b œ a# 3 ß "bX Þ One complex-valued solution is xa"b œ Œ #3 3 > Þ " We can write >3 œ /3 68 > Þ Hence xa"b œ Œ # 3 3 68 > / " #3 œŒ c-9=a68 >b 3 =38a68 >bd " # -9=a68 >b =38a68 >b -9=a68 >b # =38a68 >b œŒ 3Œ Þ -9=a68 >b =38a68 >b # -9=a68 >b =38a68 >b -9=a68 >b # =38a68 >b -# Œ Þ -9=a68 >b =38a68 >b Therefore the general solution is x œ -" Œ Other combinations are also possible. 24a+b. The characteristic equation of the system is # )" "( <$ <# < œ !, & )! "'! with eigenvalues <" œ "Î"! , and <#ß$ œ "Î% „ 3 . For < œ "Î"!, simple calculations reveal that a corresponding eigenvector is 0 a"b œ a! ß ! ß "bX Þ Setting < œ "Î% 3 , we obtain the system of equations 0" 3 0 # œ ! 0$ œ ! Þ A corresponding eigenvector is 0 a#b œ a3 ß " ß !bX Þ Hence one solution is xa"b œ Î ! Ñ >Î"! !/ Þ Ï"Ò Another solution, which is complex-valued, is given by ________________________________________________________________________ page 396 —————————————————————————— CHAPTER 7. —— Î 3 Ñ ˆ " 3‰> œ "/% Ï!Ò Î 3 Ñ >Î% œ"/ a-9= > 3 =38>b Ï!Ò œ />Î% x a#b Using the real and imaginary parts of xa#b , the general solution is constructed as Î ! Ñ >Î"! Î =38 > Ñ Î -9= > Ñ >Î% >Î% -9= > -$ / =38 > Þ x œ -" ! / -# / Ï"Ò Ï!Ò Ï!Ò Î =38 > Ñ Î -9= > Ñ -9= > 3/>Î% =38 > Þ Ï!Ò Ï!Ò a,b. Let xa!b œ aB! ß B! ß B! b . The solution can be written as " # $ Î With xa!b œ a" ß " ß "b, the solution of the initial value problem is Î!Ñ Î =38 > -9= > Ñ >Î% ! -9= > =38 > Þ xœ / Ï />Î"! Ò Ï Ò ! ! ! !Ñ Î B# =38 > B" -9= > Ñ >Î% ! ! xœ / B# -9= > B! =38 > Þ " Ï B! />Î"! Ò Ï Ò ! $ ________________________________________________________________________ page 397 —————————————————————————— CHAPTER 7. —— 25a+b. Based on Probs. ") #! of Section (Þ" , the system of differential equations is V" .M P Œ œ " .> Z G " # " P M Þ " Œ GV# Z With V" œ V# œ % ohms , G œ " # farads and P œ ) henrys , the eigenvalue problem is " # < # 0" ! Œ 0 œ Œ ! . < " ) # a,b. The characteristic equation of the system is <# < <"ß# œ "" „ 3Þ ## " # œ ! , with eigenvalues Setting < œ "Î# 3Î# , the algebraic equations reduce to %30" 0# œ ! . It follows that 0 a"b œ a" ß %3bX Þ Hence one complex-valued solution is ________________________________________________________________________ page 398 —————————————————————————— CHAPTER 7. —— M a"b " a"3b>Î# Œ œŒ / Z %3 " >Î# œŒ c-9=a>Î#b 3 =38a>Î#bd / %3 -9=a>Î#b =38a>Î#b >Î# œ />Î# Œ 3/ Œ Þ % =38a>Î#b % -9=a>Î#b Therefore the general solution is Œ M -9=a>Î#b =38a>Î#b >Î# >Î# œ -" / Œ -# / Œ Þ Z % =38a>Î#b % -9=a>Î#b a- b. Imposing the initial conditions, we arrive at the equations -" œ # and -# œ $ , % and $ M >Î# # -9=a>Î#b % =38a>Î#b Œ œ/ Œ Þ Z ) =38a>Î#b $ -9=a>Î#b a. b. Since the eigenvalues have negative real parts, all solutions converge to the origin. 26a+b. The characteristic equation of the system is <# with eigenvalues <"ß# œ " " %V # G Ê" „ Þ #VG #VG P %V # G !. P " " < œ !, VG GP The eigenvalues are real and different provided that " The eigenvalues are complex conjugates as long as %V # G " !Þ P a,b. With the specified values, the eigenvalues are <"ß# œ " „ 3 Þ The eigenvector corresponding to < œ " 3 is 0 a"b œ a" ß %3bX Þ Hence one complex-valued solution is ________________________________________________________________________ page 399 —————————————————————————— CHAPTER 7. —— M a"b " a"3b> Œ œŒ / Z "3 " > œŒ / a-9= > 3 =38 >b "3 -9= > =38 > > œ /> Œ 3/ Œ Þ -9= > =38 > -9= > =38 > Therefore the general solution is Œ M -9= > =38 > > > œ -" / Œ -# / Œ Þ Z -9= > =38 > -9= > =38 > a- b. Imposing the initial conditions, we arrive at the equations -" œ # -" - # œ " , with -" œ # and -# œ $ . Therefore the solution of the IVP is Œ M > # -9= > $ =38 > œ/ Œ Þ Z -9= > &=38 > a. b. Since V/a<"ß# b œ " , all solutions converge to the origin. 27a+b. Suppose that -" a -# b œ 0 . Since a and b are the real and imaginary parts of the vector 0 a"b , respectively, a œ ˆ0 a"b 0 a"b ‰Î# and b œ ˆ0 a"b 0 a"b ‰Î#3 . Hence -" ˆ0 a"b 0 a"b ‰ 3-# ˆ0 a"b 0 a"b ‰ œ 0 , a-" 3-# b0 a"b a-" 3-# b0 a"b œ 0 Þ -" 3-# œ ! -" 3-# œ ! Þ It follows that -" œ -# œ ! . a- b. Recall that ua>b œ /-> aa -9= .> b =38 .>b va>b œ /-> aa -9= .> b =38 .>b . which leads to Now since 0 a"b and 0 a"b are linearly independent, we must have Consider the equation -" ua>! b -# va>! b œ 0 , for some >! . We can then write ________________________________________________________________________ page 400 —————————————————————————— CHAPTER 7. —— -" /->! aa -9= .>! b =38 .>! b -# /->! aa -9= .>! b =38 .>! b œ 0 . a‡b Rearranging the terms, and dividing by the exponential, a-" -# b-9= .>! a a-# -" b=38 .>! b œ 0 . a-" -# b-9= .>! œ a-# -" b=38 .>! œ ! Þ From Part a,b, since a and b are linearly independent, it follows that Without loss of generality, assume that the trigonometric factors are nonzero. Otherwise proceed again from Equation a‡b, above. We then conclude that which leads to -" œ -# œ !. Thus ua>! b and va>! b are linearly independent for some >! , and hence the functions are linearly independent at every point. 28a+b. Let B" œ ? and B# œ ? w . It follows that B"w œ B# and B#w œ ? ww œ 5 ?. 7 -" -# œ ! and -# -" œ ! , In terms of the new variables, we obtain the system of two first order ODEs B"w œ B# B#w œ a,b. The associated eigenvalue problem is < Œ 5Î7 5 B" Þ 7 The characteristic equation is <# 5Î7 œ ! , with roots <"ß# œ „ 3È5Î7 Þ 0" ! " Œ œ Œ . < 0# ! a- b. Since the eigenvalues are purely imaginary, the origin is a center. Hence the phase curves are ellipses, with a clockwise flow. For computational purposes, let 5 œ " and 7 œ #. ________________________________________________________________________ page 401 —————————————————————————— CHAPTER 7. —— a. b. The general solution of the second order equation is ?a>b œ -" -9= Ê 5 5 > -# =38 Ê > . 7 7 The general solution of the system of ODEs is given by ________________________________________________________________________ page 402 —————————————————————————— CHAPTER 7. —— 5 5 Î È 7 =38É 7 > Ñ Î È 7 -9=É 7 > Ñ Ð5 Ó -# Ð 5 ÓÞ x œ -" 5 5 Ï -9=É 7 > Ò Ï =38É 7 > Ò It is evident that the natural frequency of the system is equal to M 7a<"ß# b . ________________________________________________________________________ page 403 —————————————————————————— CHAPTER 7. —— Section 7.7 1. The eigenvalues and eigenvectors were found in Prob. " , Section (Þ& . " # <" œ " , 0 a"b œ Œ à <# œ # , 0 a#b œ Œ Þ # " The general solution is x œ -" Œ /> # /#> -# Œ #> Þ # /> / # /#> Þ /#> # , " Hence a fundamental matrix is given by Ga>b œ Œ We now have Ga!b œ Œ So that Fa>b œ Ga>bG" a!b œ " /> %/#> Œ $ #/> #/#> #/> #/#> Þ %/> /#> " # " " # " and G a!b œ Œ # " $ /> # /> 3. The eigenvalues and eigenvectors were found in Prob. $ , Section (Þ& . The general solution of the system is /> /> x œ -" Œ > -# Œ > Þ / $/ Given the initial conditions xa!b œ ea"b , we solve the equations -" - # œ " -" $-# œ ! , to obtain -" œ $Î# , -# œ "Î# . The corresponding solution is xœ# $ / " /> # Þ > $ /> / # # $> Given the initial conditions xa!b œ ea#b , we solve the equations -" - # œ ! -" $-# œ " , to obtain -" œ "Î# , -# œ "Î# . The corresponding solution is ________________________________________________________________________ page 404 —————————————————————————— CHAPTER 7. —— xœ Therefore the fundamental matrix is Fa>b œ " /> " /> # # Þ " /> $ /> # # /> /> Þ /> $/> " $/> /> Œ # $/> $/> 5. The general solution, found in Prob. $ , Section (Þ' , is given by x œ -" Œ & -9= > & =38 > -# Œ Þ # -9= > =38 > -9= > # =38 > Given the initial conditions xa!b œ ea"b , we solve the equations &-" œ " #-" -# œ ! , resulting in -" œ "Î& , -# œ #Î& . The corresponding solution is xœŒ -9= > # =38 > Þ =38 > Given the initial conditions xa!b œ ea#b , we solve the equations &-" œ ! #-" -# œ " , resulting in -" œ ! , -# œ " . The corresponding solution is xœŒ Therefore the fundamental matrix is Fa>b œ Œ -9= > # =38 > =38 > & =38 > Þ -9= > # =38 > & =38 > Þ -9= > # =38 > 7. The general solution, found in Prob. "& , Section (Þ& , is given by x œ -" Œ /#> /%> -# Œ %> Þ $/#> / Given the initial conditions xa!b œ ea"b , we solve the equations -" - # œ " $-" -# œ ! , ________________________________________________________________________ page 405 —————————————————————————— CHAPTER 7. —— resulting in -" œ "Î# , -# œ $Î# . The corresponding solution is xœ " /#> $/%> Œ Þ # $/#> $/%> The initial conditions xa!b œ ea#b require that -" - # œ ! $-" -# œ " , resulting in -" œ "Î# , -# œ "Î# . The corresponding solution is xœ Therefore the fundamental matrix is Fa>b œ " /#> $/%> Œ # $/#> $/%> /#> /%> Þ $/#> /%> " /#> /%> Œ Þ # $/#> /%> 8. The general solution, found in Prob. & , Section (Þ' , is given by x œ -" /> Œ -9= > =38 > > -# / Œ Þ # -9= > =38 > -9= > # =38 > x œ /> Œ -9= > # =38 > Þ &=38 > =38 > Þ -9= > #=38 > =38 > Þ -9= > #=38 > The specific solution corresponding to the initial conditions xa!b œ ea"b is For the initial conditions xa!b œ ea#b , the solution is x œ /> Œ Therefore the fundamental matrix is Fa>b œ /> Œ -9= > # =38 > &=38 > 9. The general solution, found in Prob. "$ , Section (Þ& , is given by Given the initial conditions xa!b œ ea"b , we solve the equations #> > Î %/ Ñ Î $/ Ñ Î!Ñ > #> /#> x œ -" &/ -# %/ -$ Þ Ï (/#> Ò Ï #/> Ò Ï /#> Ò ________________________________________________________________________ page 406 —————————————————————————— CHAPTER 7. —— %-" $-# œ " &-" %-# -$ œ ! (-" #-# -$ œ ! , resulting in -" œ "Î# , -# œ " , -$ œ $Î# . The corresponding solution is Î #/ $/ Ñ x œ Ð & /#> %/> $ /#> ÓÞ #> > # The initial conditions xa!b œ ea#b , we solve the equations %-" $-# œ ! &-" %-# -$ œ " (-" #-# -$ œ ! , Ï ( /#> #/> $ /#> Ò # # # resulting in -" œ "Î% , -# œ "Î$ , -$ œ "$Î"# . The corresponding solution is /#> /> Î Ñ x œ Ð & /#> % /> "$ /#> ÓÞ Ï ( /#> # /> % $ % $ "# "$ #> "# / The initial conditions xa!b œ ea$b , we solve the equations %-" $-# œ ! &-" %-# -$ œ ! (-" #-# -$ œ " , Ò resulting in -" œ "Î% , -# œ "Î$ , -$ œ "Î"# . The corresponding solution is /#> /> Î Ñ Ð & /#> % /> " /#> ÓÞ xœ Ï ( /#> # /> % $ % $ "# " #> "# / Ò Therefore the fundamental matrix is Î #/ $/ Fa>b œ Ð & /#> %/> $ /#> #> > # Ï ( /#> #/> $ /#> # # # & #> %/ ( #> %/ /#> /> % /> "$ /#> $ "# "$ # /> "# /#> $ & #> %/ ( #> %/ /#> /> Ñ " % /> "# /#> ÓÞ $ " # /> "# /#> Ò $ 12. The solution of the initial value problem is given by ________________________________________________________________________ page 407 —————————————————————————— CHAPTER 7. —— x œ Fa>bxa!b /> -9= #> #/> =38 #> $ œ Œ " > Œ > / -9= #> " # / =38 #> $-9= #> #=38 #> œ /> Œ $ Þ =38 #> -9= #> # 13. Let a"b Î B" a>b Ga>b œ ã Ï Ba"b a>b 8 a"b Î B" a>! b Ga>! b œ ã Ï Ba"b a> b ! 8 â â Ba8b a>b Ñ " Þ ã Ò a8b B8 a>b Ba8b a>! b Ñ " ã a8b B8 a>! b Ò -"8 Ñ ãÞ -88 Ò It follows that â â is a scalar matrix, which is invertible, since the solutions are linearly independent. Let G" a>! b œ a-34 b. Then a"b Î B" a>b Ga>bG a>! b œ ã Ï Ba"b a>b 8 " â â a4b Ba8b a>b ÑÎ -"" " ã ã ÒÏ -8" a8b B8 a>b 8 â â The 4-th column of the product matrix is cGa>bG" a>! bd œ " -54 xa5 b , 5œ" which is a solution vector, since it is a linear combination of solutions. Furthermore, the columns are all linearly independent, since the vectors xa5 b areÞ Hence the product is a fundamental matrix. Finally, setting > œ >! , Ga>! bG" a>! b œ I . This is precisely the definition of Fa>b. 14. The fundamental matrix Fa>b for the system xw œ Œ " % " x " /$> /> Þ #/$> #/> is given by Fa>b œ Direct multiplication results in ________________________________________________________________________ page 408 " #/$> #/> Œ % %/$> %/> ————————————————————————— CHAPTER 7. —— Fa>bFa=b œ " #/$> #/> /$> /> #/$= #/= Œ $> Œ $= %/ %/= "' %/ %/> #/$> #/> " )a/$>$= />= b %a/$>$= />= b œ Œ Þ "' "'a/$>$= />= b )a/$>$= />= b " #/$a>=b #/a>=b Œ % %/$a>=b %/a>=b /$= /= #/$= #/= Hence Fa>bFa=b œ /$a>=b /a>=b Þ #/$a>=b #/a>=b 15a+b. Let = be arbitrary, but fixed, and > variable. Similar to the argument in Prob. "$ , the columns of the matrix Fa>bFa=b are linear combinations of fundamental solutions. Hence the columns of Fa>bFa=b are also solution of the system of equations. Further, setting > œ ! , Fa!bFa=b œ I Fa=b œ Fa=b Þ That is, Fa>bFa=b is a solution of the initial value problem Z w œ AZ , with Za!b œ Fa=b Þ Now consider the change of variable 7 œ > = . Let Wa7 b œ Za7 =b. The given initial value problem can be reformulated as . W œ AW , with Wa=b œ Fa=b Þ .7 Wa7 b œ Fa7 bF" a=b‘Fa=b œ Fa 7 b Þ Since Fa>b is a fundamental matrix satisfying F w œ AF , with Fa!b œ I , it follows that That is, Fa> =b œ Fa7 b œ Wa7 b œ Za>b œ Fa>bFa=b . Fa >b œ F" a>bÞ a,b. Based on Part a+b, Fa>bFa >b œ Fa> a >bb œ Fa!b œ I . Hence a- b. It also follows that Fa> =b œ Fa> a =bb œ Fa>bFa =b œ Fa>bF" a=bÞ 5 5 5 A5 œ +" ea"b ß +# ea#b ß âß +8 ea8b ‘. 16. Let A be a diagonal matrix, with A œ c+" ea"b ß +# ea#b ß âß +8 ea8b d. Note that for any positive integer, 5 , It follows, from basic matrix algebra, that ________________________________________________________________________ page 409 —————————————————————————— CHAPTER 7. —— 5 >5 Î ! +" 5x Ð 5œ ! Ð Ð 7 5 > ! " A5 œ Ð I Ð 5x Ð 5œ" Ð ã Ð ! Ï 7 ! 5 ! +# >5 7 5x â â 7 ! ! ã 5œ ! ã ! â 5œ ! 5 ! +8 >5 Ò 5x Ñ Ó Ó Ó Ó ÓÞ Ó Ó Ó It can be shown that the partial sums on the left hand side converge for all > . Taking the limit aas 7 p _b on both sides of the equation, we obtain Î/ Ð! /B:aA>b œ Ð ã Ï! +" > ! / +# > ã ! â â â !Ñ !Ó ÓÞ ã / +8 > Ò Alternatively, consider the system x w œ Ax . Since ODEs are uncoupled, the vectors xa4b œ /B:a+4 >b ea4b ß 4 œ "ß #ß â8 , are a set of linearly independent solutions. Hence the matrix is a fundamental matrix. Finally, since Xa!b œ I , it follows that X œ /B:a+" >b ea"b ß /B:a+# >b ea#b ß âß /B:a+8 >b ea8b ‘ /B:a+" >b ea"b ß /B:a+# >b ea#b ß âß /B:a+8 >b ea8b ‘ œ Fa>b œ /B:aA>b . 17a+b. Assuming that x œ 9a>b is a solution, then 9 w œ A9, with 9a!b œ x! Þ Integrate both sides of the equation to obtain 9a>b 9a!b œ ( A9a=b.= . > ! Hence 9a>b œ x! ( A9a=b.= . > ! a,b. Proceed with the iteration With 9a!b a>b œ x! , and noting that A is a constant matrix, 9a3"b a>b œ x! ( A9a3b a=b.= . > ! ________________________________________________________________________ page 410 —————————————————————————— CHAPTER 7. —— 9a"b a>b œ x! ( Ax! .= > ! That is, 9a"b a>b œ aI A>bx! . a- b. We then have œ x Ax! > Þ ! 9a#b a>b œ x! ( AaI A>bx! .= > ! œ x Ax > A x ! ! # !> # # ># ! œ ŒI A> A# x . # ># >8 â A8 x! . # 8x Now suppose that 9a8b a>b œ ŒI A> A# It follows that # ( AŒI A> A > ! ># >8 â A8 x! .= œ # 8x $ 8" ># #> 8> ! œ AŒI> A A âA x # $x a8 "bx # $ > > >8 œ ŒA> A# A$ â A8" x! Þ # $x 8x Therefore 9a8"b a>b œ ŒI A> A# ># >8" ! â A8" x . # a 8 "b x By induction, the asserted form of 9a8b a>b is valid for all 8 ! Þ 8 Ä_ a. b. Define 9a_b a>b œ lim 9a8b a>b. It can be shown that the limit does exist. In fact, 9a_b a>b œ /B:aA>bx! Þ Term-by-term differentiation results in ________________________________________________________________________ page 411 —————————————————————————— CHAPTER 7. —— . a_b . ># >8 9 a>b œ ŒI A> A# â A8 x! .> .> # 8x >8" œ ŒA A# > â A8 x! a 8 "b x # > >8" œ AŒI A> A# â A8" x! Þ a 8 "b x # That is, . a_b 9 a>b œ A9a_b a>b. .> Furthermore, 9a_b a!b œ x! . Based on uniqueness of solutions, 9a>b œ 9a_b a>b . ________________________________________________________________________ page 412 —————————————————————————— CHAPTER 7. —— Section 7.8 2. Setting x œ 0 >< results in the algebraic equations Œ %< ) 0" ! # Œ œ Œ . %< 0# ! The characteristic equation is <# œ ! , with the single root < œ ! . Substituting < œ ! reduces the system of equations to #0" 0# œ ! . Therefore the only eigenvector is 0 œ a" ß #bX Þ One solution is " xa"b œ Œ , # which is a constant vector. In order to generate a second linearly independent solution, we must search for a generalized eigenvector. This leads to the system of equations % Œ) (" " # Œ œ Œ . % (# # This system also reduces to a single equation, #(" (# œ "Î# . Setting (" œ 5 , some arbitrary constant, we obtain (# œ #5 "Î# . A second solution is " 5 xa#b œ Œ > Œ # #5 "Î# " ! " œ Œ > Œ 5 Œ Þ # "Î# # Note that the last term is a multiple of xa"b and may be dropped. Hence " ! xa#b œ Œ > Œ . # "Î# The general solution is " " ! x œ - " Œ - # ” Œ > Œ •Þ # # "Î# ________________________________________________________________________ page 413 —————————————————————————— CHAPTER 7. —— All of the points on the line B# œ #B" are equilibrium points. Solutions starting at all other points become unbounded. 3. Solution of the ODEs is based on the analysis of the algebraic equations $ # < " % " " # ! 0" Œ 0 œ Œ ! . < # The characteristic equation is <# # < " œ ! , with a single root < œ " . Setting < œ " , the two equations reduce to 0" #0# œ ! . The corresponding eigenvector is 0 œ a# ß "bX Þ One solution is # xa"b œ Œ /> . " A second linearly independent solution is obtained by finding a generalized eigenvector. We therefore analyze the system " # " % " " # " (" Œ ( œ Œ # . # The equations reduce to the single equation (" #(# œ #Þ Let (" œ #5 . We obtain (# œ " 5 , and a second linearly independent solution is # #5 > xa#b œ Œ >/> Œ / " "5 # ! # œ Œ >/> Œ /> 5 Œ /> Þ " " " Dropping the last term, the general solution is # # ! x œ -" Œ /> -# ”Œ >/> Œ /> •Þ " " " ________________________________________________________________________ page 414 —————————————————————————— CHAPTER 7. —— 4. Solution of the ODE requires analysis of the algebraic equations $< & # 0" ! Œ 0 œ Œ ! . #< & # # For a nonzero solution, we must have ./>aA < Ib œ <# < " œ ! . The only root % is < œ "Î# , which is an eigenvalue of multiplicity two. Setting < œ "Î# is the coefficient matrix reduces the system to the single equation 0" 0# œ ! . Hence the corresponding eigenvector is 0 œ a" ß "bX Þ One solution is " xa"b œ Œ />Î# . " In order to obtain a second linearly independent solution, we find a solution of the system & # & # & # & # (" " Œ ( œ Œ " . # There equations reduce to &(" &(# œ # . Set (" œ 5 , some arbitrary constant. Then (# œ 5 #Î& . A second solution is " 5 >Î# xa#b œ Œ >/>Î# Œ / " 5 #Î& " ! " >Î# œ Œ >/>Î# Œ 5 Œ />Î# Þ / " #Î& " Dropping the last term, the general solution is " " ! >Î# x œ -" Œ />Î# -# ”Œ >/>Î# Œ / •Þ " " #Î& ________________________________________________________________________ page 415 —————————————————————————— CHAPTER 7. —— 6. The eigensystem is obtained from analysis of the equation Î < " Ï" " < " " ÑÎ 0" Ñ Î ! Ñ " 0# œ ! . < ÒÏ 0$ Ò Ï ! Ò The characteristic equation of the coefficient matrix is <$ $< # œ ! , with roots <" œ # and <#ß$ œ " . Setting < œ # , we have Î # " Ï" " # " " ÑÎ 0" Ñ Î ! Ñ " 0# œ ! . ÒÏ 0$ Ò Ï ! Ò # This system is reduced to the equations 0" 0 $ œ ! 0# 0 $ œ ! Þ A corresponding eigenvector vector is given by 0 a"b œ a" ß " ß "bX Þ Setting < œ " , the system of equations is reduced to the single equation An eigenvector vector is given by 0 a#b œ a" ß ! ß "bX Þ Since the last equation has two free variables, a third linearly independent eigenvector aassociated with < œ "b is 0 a$b œ a! ß " ß "bX Þ Therefore the general solution may be written as Î " Ñ #> Î " Ñ > Î ! Ñ > ! " x œ -" " / - # / -$ /Þ Ï"Ò Ï "Ò Ï "Ò 0" 0 # 0 $ œ ! Þ 7. Solution of the ODE requires analysis of the algebraic equations Œ "< % 0" ! % . œ ( < Œ 0# Œ ! For a nonzero solution, we must have ./>aA < Ib œ <# '< * œ ! . The only root is < œ $ , which is an eigenvalue of multiplicity two. Substituting < œ $ into the coefficient matrix, the system reduces to the single equation 0" 0# œ ! . Hence the corresponding eigenvector is 0 œ a" ß "bX Þ One solution is " xa"b œ Œ /$> . " For a second linearly independent solution, we search for a generalized eigenvector. Its components satisfy ________________________________________________________________________ page 416 —————————————————————————— CHAPTER 7. —— % Œ% (" " % Œ œ Œ , % (# " that is, %(" %(# œ " . Let (# œ 5 , some arbitrary constant. Then (" œ 5 "Î% Þ It follows that a second solution is given by " 5 "Î% $> xa#b œ Œ >/$> Œ / " 5 " "Î% $> " $> œ Œ >/$> Œ / 5 Œ / Þ " ! " Dropping the last term, the general solution is " " "Î% $> x œ -" Œ /$> -# ”Œ >/$> Œ / •Þ " " ! Imposing the initial conditions, we require that " -" - # œ $ % -" œ # , which results in -" œ # and -# œ % Þ Therefore the solution of the IVP is $ % x œ Œ /$> Œ >/$> Þ # % ________________________________________________________________________ page 417 —————————————————————————— CHAPTER 7. —— 8. Solution of the ODEs is based on the analysis of the algebraic equations & # < $ # " # 0" ! Œ 0 œ Œ ! . < $ # # The characteristic equation is <# # < " œ ! , with a single root < œ " . Setting < œ ", the two equations reduce to 0" 0# œ !. The corresponding eigenvector is 0 œ a" ß "bX Þ One solution is " xa"b œ Œ /> . " A second linearly independent solution is obtained by solving the system $ # $ # $ # $ # (" " Œ ( œ Œ " . # The equations reduce to the single equation $(" $(# œ #Þ Let (" œ 5 . We obtain (# œ #Î$ 5 , and a second linearly independent solution is " 5 > xa#b œ Œ >/> Œ / " #Î$ 5 " ! " > > œ Œ >/> Œ / 5 Œ / Þ " #Î$ " Dropping the last term, the general solution is " " ! > x œ -" Œ /> -# ”Œ >/> Œ / •Þ " " #Î$ Imposing the initial conditions, find that ________________________________________________________________________ page 418 ————————————————————————— CHAPTER 7. —— -" œ $ # -" - # œ " , $ so that -" œ $ and -# œ ' Þ Therefore the solution of the IVP is xœŒ $ ' > > / Œ >/ Þ " ' 10. The eigensystem is obtained from analysis of the equation $< Œ " 0" ! * Œ œ Œ . $< 0# ! The characteristic equation is <# œ ! , with a single root < œ ! . Setting < œ ! , the two equations reduce to 0" $0# œ !. The corresponding eigenvector is 0 œ a $ ß "bX Þ Hence one solution is ________________________________________________________________________ page 419 —————————————————————————— CHAPTER 7. —— xa"b œ Œ $ , " which is a constant vector. A second linearly independent solution is obtained from the system $ Œ " (" $ * Œ œ Œ . $ (# " The equations reduce to the single equation (" $(# œ " Þ Let (# œ 5 . We obtain (" œ " $5 , and a second linearly independent solution is xa#b œ Œ $ " $5 > Œ " 5 $ " $ œŒ > Œ 5Œ Þ " ! " $ $ " -# ”Œ > Œ •Þ " " ! Dropping the last term, the general solution is x œ -" Œ Imposing the initial conditions, we require that $-" -# œ # -" œ % , which results in -" œ % and -# œ "% Þ Therefore the solution of the IVP is # $ x œ Œ "%Œ >Þ % " ________________________________________________________________________ page 420 —————————————————————————— CHAPTER 7. —— 12. The characteristic equation of the system is ) <$ '! <# "#' < %* œ ! . The eigenvalues are <" œ "Î# and <#ß$ œ (Î# . The eigenvector associated with <" is 0 a"b œ a" ß " ß "bX Þ Setting < œ (Î# , the components of the eigenvectors must satisfy the relation An eigenvector vector is given by 0 a#b œ a" ß ! ß "bX Þ Since the last equation has two free variables, a third linearly independent eigenvector aassociated with < œ (Î#b is 0 a$b œ a! ß " ß "bX Þ Therefore the general solution may be written as x œ -" Î " Ñ >Î# Î " Ñ (>Î# Î ! Ñ (>Î# "/ ! " -# / -$ / Þ Ï"Ò Ï "Ò Ï "Ò -" - # œ # -" - $ œ $ -" - # - $ œ " Þ Hence the solution of the IVP is " "Ñ !Ñ % Î Ñ >Î# # Î &Î (>Î# "/ ! " xœ / /(>Î# Þ $Ï Ò $Ï $Ï " "Ò "Ò 0" 0 # 0 $ œ ! Þ Invoking the initial conditions, we require that 13. Setting x œ 0 >< results in the algebraic equations Œ $< " 0" ! % Œ œ Œ . "< 0# ! The characteristic equation is <# #< " œ ! , with a single root of <"ß# œ " . With ________________________________________________________________________ page 421 —————————————————————————— CHAPTER 7. —— < œ " , the system reduces to a single equation 0" # 0# œ !. An eigenvector is given by 0 œ a# ß "bX Þ Hence one solution is # xa"b œ Œ > . " In order to find a second linearly independent solution, we search for a generalized eigenvector whose components satisfy # Œ" (" # % Œ œ Œ . # (# " These equations reduce to (" # (# œ " . Let (# œ 5 , some arbitrary constant. Then (" œ " #5 Þ Ò Before proceeding, note that if we set ? œ 68 > , the original equation is transformed into a constant coefficient equation with independent variable ? . Recall that a second solution is obtained by multiplication of the first solution by the factor ? . This implies that we must multiply first solution by a factor of 68 > . Ó Hence a second linearly independent solution is # " #5 xa#b œ Œ > 68 > Œ > " 5 # " # œ Œ > 68 > Œ > 5 Œ >Þ " ! " Dropping the last term, the general solution is # # " x œ -" Œ > -# ”Œ > 68 > Œ >•Þ " " ! 15. The characteristic equation is <# a+ . b< +. ,- œ ! Þ Hence the eigenvalues are <"ß# œ +. " „ Éa+ . b# %a+. ,- b Þ # # ÈP# %V # GP " „ Þ #VG #VGP 16a+b. Using the result in Prob. "& , the eigenvalues are <"ß# œ The discriminant vanishes when P œ %V # GP . ________________________________________________________________________ page 422 —————————————————————————— CHAPTER 7. —— a,b. The system of differential equations is .M ! Œ œŒ .> Z " Œ " % " Œ M Þ Z The associated eigenvalue problem is < Œ " " % "< 0" ! œ Œ . 0# ! The characteristic equation is <# < "Î% œ ! , with a single root of <"ß# œ "Î# . Setting < œ "Î# , the algebraic equations reduce to #0" 0# œ !. An eigenvector is given by 0 œ a" ß #bX Þ Hence one solution is M a"b " >Î# . Œ œŒ / Z # A second solution is obtained from a generalized eigenvector whose components satisfy " " # (" " Œ ( œ Œ # . # " % " # It follows that (" œ 5 and (# œ % #5 Þ A second linearly independent solution is M Œ Z a#b œŒ " 5 >Î# >Î# Œ > / / # % #5 " ! " >Î# >Î# œŒ Œ />Î# 5 Œ Þ > / / # % # Dropping the last term, the general solution is Œ M " " ! >Î# >Î# -# ” Œ Œ />Î# •Þ œ -" Œ / > / Z # # % Imposing the initial conditions, we require that -" œ " #-" %-# œ # , which results in -" œ " and -# œ " Þ Therefore the solution of the IVP is Œ M " >Î# " >Î# Œ Þ œ Œ / > / Z # # 18a+b. The eigensystem is obtained from analysis of the equation ________________________________________________________________________ page 423 —————————————————————————— CHAPTER 7. —— Î& < ) Ï % # ÑÎ 0" Ñ Î ! Ñ % 0# œ ! . ÒÏ 0$ Ò Ï ! Ò $< $ &< $ The characteristic equation of the coefficient matrix is <$ $<# $< " œ ! , with a single root of multiplicity three, < œ " . Setting < œ " , we have Î% ) Ï % $ ' $ # ÑÎ 0" Ñ Î ! Ñ % 0# œ ! . # ÒÏ 0$ Ò Ï ! Ò The system of algebraic equations reduce to a single equation An eigenvector vector is given by 0 a"b œ a" ß ! ß #bX Þ Since the last equation has two free variables, a second linearly independent eigenvector aassociated with < œ "b is 0 a#b œ a! ß # ß $bX Þ Therefore two solutions are obtained as xa"b œ Î"Ñ > Î ! Ñ> ! / and xa#b œ # /Þ Ï#Ò Ï $Ò %0" $0# #0$ œ ! Þ a,b. It follows directly that x w œ 0>/> 0/> ( /> Þ Hence the coefficient vectors must satisfy 0 >/> 0/> ( /> œ A0>/> A(/> . Rearranging the terms , we have Given an eigenvector 0 , it follows that aA Ib( œ 0 . 0 /> œ aA Ib0>/> aA Ib( /> . a- b. Note that a linear combination of two eigenvectors, associated with the same eigenvalue, is also an eigenvector. Consider the equation aA Ib( œ -" 0 a"b -# 0 a#b Þ The augmented matrix is Î% ) Ï % $ ' $ # % # l l l -" Ñ #-# Þ Ò #-" $-# Using elementary row operations, we obtain Î% ! Ï! $ ! ! # ! ! l l l -" Ñ #-" #-# Þ $-" $-# Ò It is evident that a solution exists provided -" œ -# . a. b. Let -" œ -# œ # . The components of the generalized eigenvector must satisfy ________________________________________________________________________ page 424 —————————————————————————— CHAPTER 7. —— Î% ) Ï % # ÑÎ (" Ñ Î # Ñ % (# œ % . ÒÏ ($ Ò Ï # Ò # Based on Part a- b, the equations reduce to the single equation %(" $(# #($ œ # Þ Let (" œ ! and (# œ #" , where ! and " are arbitrary constants. We then have ($ œ " #! $" , so that ! Ñ Î!Ñ Î"Ñ Î!Ñ #" ! # (œ œ ! ! " Þ Ï " #! $ " Ò Ï " Ò Ï#Ò Ï $Ò Î # Ñ > Î ! Ñ> % ! œ >/ /Þ Ï #Ò Ï "Ò Î $ ' $ Observe that ( œ ! 0 a"b " 0 a#b Þ Hence a third linearly independent solution is x a$b a/b. Given the three linearly independent solutions, a fundamental matrix is given by > Î/ Ga>b œ ! Ï #/> ! #/> $/> #> /> Ñ Þ %> /> > >Ò #> / / a0 b. We construct the transformation matrix Tœ with inverse T " Î" ! Ï# # % # !Ñ ! , Ò " !Ñ ! Þ "Ò !Ñ "Þ "Ò Î" œ! Ï# "Î# "Î% $Î# The Jordan form of the matrix A is Î" J œ T AT œ ! Ï! " ! " ! 20a+b. Direct multiplication results in ________________________________________________________________________ page 425 —————————————————————————— CHAPTER 7. —— Î# Jœ ! Ï! $ !Ñ Î#- ß J$ œ ! Ï! #Ò % !Ñ Î$-# ß J% œ ! Ï! $Ò - # ! -# ! ! -$ ! ! -% ! !Ñ %-$ Þ -% Ò a,b. Suppose that Î8 Jœ ! Ï! 8 ! -8 ! !Ñ 8-8" Þ -8 Ò Then J8+1 8 ! ! ÑÎ - ! ! Ñ Î8 !-" œ ! - 8-8" 8 ÒÏ ! Ï! ! -Ò ! 8 ! ! Î- † Ñ 8 8 œ ! - † - - 8- † -8" Þ Ï! Ò ! - † -8 Hence the result follows by mathematical induction. a- b. Note that J is block diagonal. Hence each block may be exponentiated. Using the result in Prob. a"*b, Î/ /B:aJ>b œ ! Ï! -> ! /- > ! !Ñ >/-> Þ /- > Ò a. b. Setting - œ " , and using the transformation matrix T in Prob. a")b, # ! ÑÎ /> ! ! Ñ Î" % ! T/B:aJ>b œ ! ! /> >/> Ï # # " ÒÏ ! ! /> Ò > #/> #> /> Î/ Ñ > œ Þ ! %/ %> /> Ï #/> #/> #> /> /> Ò Based on the form of J , /B:aJ>b is the fundamental matrix associated with the solutions ya"b œ 0 a"b /> , ya#b œ ˆ#0a"b #0a#b ‰/> and ya$b œ ˆ#0a"b #0a#b ‰>/> ( /> Þ Hence the resulting matrix is the fundamental matrix associated with the solution set ________________________________________________________________________ page 426 —————————————————————————— CHAPTER 7. —— ˜0 a"b /> ß ˆ#0 a"b #0 a#b ‰/> ß ˆ#0 a"b #0 a#b ‰>/> ( /> ™, ˜0 a"b /> ß 0 a#b /> ß ˆ#0 a"b #0 a#b ‰>/> ( /> ™. as opposed to the solution set in Prob. a")b, given by 21a+b. Direct multiplication results in ÎJœ ! Ï! # # #-# ! $ "Ñ Î$ #- ß J œ ! Ï! #Ò - $-# -$ ! % $- Ñ Î% $-# ß J œ ! Ï! $Ò - %-$ -% ! '-# Ñ %-$ Þ -% Ò a,b. Suppose that Î -8 8 J œÐ ! Ï! 8-8" -8 ! 8a8"b 8# # 8" 8-8 Ñ ÓÞ Ò Then J 8+1 Î -8 œÐ ! Ï! Ï Î - † -8 œÐ ! ! 8-8" -8 ! 8a8"b 8# # 8" 8-8 ÑÎ Ó! ÒÏ ! -8 8- † -8" - † -8 ! 8-8" 8a8"b - † -8# Ñ # ÓÞ -8 8- † -8" Ò - † -8 " ! !Ñ " -Ò The result follows by noting that 8-8" 8a8 "b 8a8 "b 8" - † -8# œ ”8 •# # 8# 8 8" œ -Þ # a- b. We first observe that ________________________________________________________________________ page 427 —————————————————————————— CHAPTER 7. —— " -8 _ 8a8 "b 8# >8 ># _ 8# >8# ># " œ "œ /-> Þ # 8x # 8œ# # a8 #bx 8œ! 8œ! _ " 8-8" _ 8œ! >8 œ /-> 8x _ >8 >8" " -8" œ> œ > /-> 8x a8 "bx 8œ" Therefore Î /- > /B:aJ>b œ Ð ! Ï! >/-> /- > ! ># - > #/ -> >/ /- > Ñ ÓÞ Ò a. b. Setting - œ # , and using the transformation matrix T in Prob. a"(b, Î! " T/B:aJ>b œ Ï " " # ÑÎ /#> " ! Ð! ! $ ÒÏ ! /#> >/#> /#> >/#> >/#> /#> ! ># #> #/ #> Î! œ Ð /#> Ï /#> >/#> #/#> Ñ ># #> #> Ó Þ # / >/ # > #> #> Ò / $/ # >/ /#> Ò Ñ Ó ________________________________________________________________________ page 428 —————————————————————————— CHAPTER 7. —— Section 7.9 5. As shown in Prob. # , Section (Þ) , the general solution of the homogeneous equation is " > x- œ -" Œ -# Œ Þ # #> " # An associated fundamental matrix is Ga>b œ Œ G" a>b œ Œ We can now compute G" a>bga>b œ and " ( G a>bga>b .> œ Œ " # > Þ #> " # #> # Þ %> & The inverse of the fundamental matrix is easily determined as %> $ )> ) " #># %> " Œ , >$ #> % " ># %>" #68 > # Þ #># #>" Finally, va>b œ Ga>b( G" a>bga>b .>, where " @" a>b œ ># #>" # 68 > # # " @# a>b œ &> % 68 > % Þ Note that the vector a# ß %bX is a multiple of one of the fundamental solutions. Hence we can write the general solution as " > " "Î# "# " x œ -" Œ - # Œ Œ #68 >Œ Þ " #Œ # #> # > ! >& # 6. The eigenvalues of the coefficient matrix are <" œ ! and <# œ & . It follows that the solution of the homogeneous equation is ________________________________________________________________________ page 429 —————————————————————————— CHAPTER 7. —— " #/&> x- œ -" Œ -# Œ &> Þ # / The coefficient matrix is symmetric. Hence the system is diagonalizable. Using the normalized eigenvectors as columns, the transformation matrix, and its inverse, are Tœ " " È& Œ # " # " " , T œ È Œ # " & # Þ " Setting x œ Ty , and ha>b œ T" ga>b , the transformed system is given, in scalar form, as C"w œ & )> È& > C#w œ &C# The solutions are readily obtained as C" a>b œ È& 68 > % Þ È& % % > -" and C# a>b œ -# /&> Þ È& &È& C" a>b # Œ " C# a>b Transforming back to the original variables, we have x œ Ty , with xœ œ " " È& Œ # " " " # Œ C" a>b È Œ C# a>bÞ È& # " & Hence the general solution is, " #/&> " %" % # x œ 5" Œ 5 # Œ &> Œ 68 > Œ > Œ Þ # / # &# #& " 7. The solution of the homogeneous equation is x- œ -" Œ /> /$> -# Œ $> Þ #/> #/ Based on the simple form of the right hand side, we use the method of undetermined coefficients. Set v œ a /> . Substitution into the ODE yields Œ +" > " / œ Œ % +# +" > # " > Œ / Œ / Þ " +# " In scalar form, after canceling the exponential, we have ________________________________________________________________________ page 430 —————————————————————————— CHAPTER 7. —— +" œ + " + # # +# œ %+" +# " , with +" œ "Î% and +# œ # . Hence the particular solution is vœŒ so that the general solution is x œ -" Œ /> /$> " /> -# Œ $> Œ Þ #/> #/ % )/> "Î% > / , # 8. The eigenvalues of the coefficient matrix are <" œ " and <# œ " . It follows that the solution of the homogeneous equation is " " x- œ -" Œ /> -# Œ /> Þ " $ Use the method of undetermined coefficients. Since the right hand side is related to one of the fundamental solutions, set v œ a >/> b /> . Substitution into the ODE yields Œ +" ," > +" # " > > > ˆ/ >/ ‰ Œ / œ Œ $ # Œ >/ +# ,# +# ," > " # " > Œ Œ / Œ / Þ $ # ,# " In scalar form, we have a+" ," b/> +" >/> œ a#+" +# b>/> a#," ,# b/> /> a+# ,# b/> +# >/> œ a$+" #+# b>/> a$," #,# b/> /> Þ +" + " ," +# +# ,# œ #+" +# œ #," ,# " œ $+" #+# œ $," #,# " Þ Equating the coefficients in these two equations, we find that It follows that +" œ +# . Setting +" œ +# œ + , the equations reduce to ," ,# œ + " $," $,# œ " + Þ Combining these equations, it is necessary that + œ # . As a result, ," œ ,# " . Choosing +" œ +# œ # , and ,# œ 5 , some arbitrary constant, a particular solution is ________________________________________________________________________ page 431 —————————————————————————— CHAPTER 7. —— # 5" > # "> "> > v œ Œ >/> Œ / œ Œ >/ 5 Œ / Œ / Þ # 5 # " ! Since the second vector is a fundamental solution, the general solution can be written as " " # " x œ -" Œ /> -# Œ /> Œ >/> Œ /> Þ " $ # ! 9. Note that the coefficient matrix is symmetric. Hence the system is diagonalizable. The eigenvalues and eigenvectors are given by <" œ " " " , 0 a"b œ Œ and <# œ # , 0 a#b œ Œ Þ # " " Using the normalized eigenvectors as columns, the transformation matrix, and its inverse, are Tœ " " Œ" È# " " " " , T œ È Œ" " # " Þ " Setting x œ Ty , and ha>b œ T" ga>b , the transformed system is given, in scalar form, as " "> C"w œ C" È# > / È# # "> C#w œ #C# È# > /Þ È# È# Using any elementary method for first order linear equations, the solutions are / > %È # #È # > $ "> " " C# a>b œ 5 # /#> / >Þ È# $È# #È# C" a>b œ 5" />Î# Transforming back to the original variables, x œ Ty , the general solution is " " " "( "& "" > #> x œ -" Œ />Î# -# Œ / Œ Œ > Œ / Þ " " % "& #$ '$ 10. Since the coefficient matrix is symmetric, the differential equations can be decoupled. The eigenvalues and eigenvectors are given by ________________________________________________________________________ page 432 —————————————————————————— CHAPTER 7. —— È# " a#b <" œ % , 0 a"b œ and <# œ " , 0 œ Œ È# Þ " Using the normalized eigenvectors as columns, the transformation matrix, and its inverse, are Tœ È# " È$ " È# " " , T" œ È# È$ " " Þ È# Setting x œ Ty , and ha>b œ T" ga>b , the transformed system is given, in scalar form, as " Š" È#‹/> È$ " C#w œ C# Š" È#‹ /> Þ È$ C"w œ %C" The solutions are easily obtained as " Š" È#‹/> È$ $ " C# a>b œ 5 # /> Š" È# ‹>/> Þ È$ C" a>b œ 5" /%> Transforming back to the original variables, the general solution is x œ -" Note that È# / " 4> " " # È# $È$ > " " È# -# Œ È /> È È >/> Þ / È * $ ' # " $ # # # # È # $È $ # È# " È œ È $È' È# " $ $ Œ È # Þ #" È# " # È# " " " È# /4> -# Œ È /> È /> È >/> Þ * # " $ # # " # The second vector is an eigenvector, hence the solution may be written as x œ -" 11. Based on the solution of Prob. $ of Section (Þ' , a fundamental matrix is given by Ga>b œ Œ & -9= > # -9= > =38 > & =38 > Þ -9= > # =38 > The inverse of the fundamental matrix is easily determined as ________________________________________________________________________ page 433 —————————————————————————— CHAPTER 7. —— G" a>b œ It follows that G" a>bga>b œ Œ and " ( G a>bga>b .> œ " -9= > #=38 > Œ & #-9= > =38 > &=38 > Þ &-9= > -9= > =38 > , -9=# > Þ " -9= > =38 > " > # # " # # =38 > A particular solution is constructed as va>b œ Ga>b( G" a>bga>b .>, where @" a>b œ & & -9= > =38 > -9=# > > " # # " " @# a>b œ -9= > =38 > -9=# > > Þ # # Hence the general solution is x œ -" Œ & -9= > & =38 > -# Œ # -9= > =38 > -9= > # =38 > &Î# ! > =38 >Œ a> -9= > =38 >bŒ Þ " "Î# 13a+b. As shown in Prob. #& of Section (Þ' , the solution of the homogeneous system is Ba-b -9=a>Î#b =38a>Î#b " >Î# >Î# Œ -# / Œ Þ Ba-b œ -" / % =38a>Î#b % -9=a>Î#b # Ga>b œ />Î# Œ -9=a>Î#b % =38a>Î#b =38a>Î#b Þ % -9=a>Î#b =38a>Î#b Þ -9=a>Î#b Therefore the associated fundamental matrix is given by a,bÞ The inverse of the fundamental matrix is " />Î# % -9=a>Î#b G a>b œ Œ % =38a>Î#b % ________________________________________________________________________ page 434 —————————————————————————— CHAPTER 7. —— It follows that G" a>bga>b œ and " ( G a>bga>b .> œ Œ " -9=a>Î#b Œ , # =38a>Î#b =38a>Î#b Þ -9=a>Î#b A particular solution is constructed as va>b œ Ga>b( G" a>bga>b .>, where @" a>b œ ! @# a>b œ % />Î# Þ x œ -" />Î# Œ -9=a>Î#b =38a>Î#b >Î# >Î# ! -# / Œ %/ Œ Þ % =38a>Î#b % -9=a>Î#b " -" œ ! %-# % œ ! , which results in -" œ ! and -# œ " Þ Therefore the solution of the IVP is x œ />Î# Œ =38a>Î#b Þ % % -9=a>Î#b Hence the general solution is Imposing the initial conditions, we require that 15. The general solution of the homogeneous problem is Ba-b " " ## " Ba-b œ -" Œ # > -# Œ " > , # which can be verified by substitution into the system of ODEs. Since the vectors are linearly independent, a fundamental matrix is given by Ga>b œ Œ The inverse of the fundamental matrix is >" #>" #># Þ ># ________________________________________________________________________ page 435 —————————————————————————— CHAPTER 7. —— G" a>b œ " > Œ $ #># #> Þ ># Dividing both equations by > , we obtain ga>b œ Œ G" a>bga>b œ and " ( G a>bga>b .> œ #& "# # "& > $ > $ > Þ "# > % >" " ># ' $ ' >$ # . >" Proceeding with the method of variation of parameters, #% # # $> $> $ , " % # " >$ $> $> $ Hence a particular solution is obtained as " >% $> " v œ " &% Þ $ "! > #> # The general solution is " # " # % $ " x œ œ -" Œ >" -# Œ ># Œ > Œ > Œ Þ # " "! " # $Î# 16. Based on the hypotheses, 9 w a>b œ Pa>b9a>b ga>b and v w a>b œ Pa>bva>b ga>b . 9 w a>b v w a>b œ Pa>b9a>b Pa>bva>b , c9a>b va>bd w œ Pa>bc9a>b va>bdÞ Subtracting the two equations results in that is, It follows that 9a>b va>b is a solution of the homogeneous equation. According to Theorem (Þ%Þ# , 9a>b va>b œ -" xa"b a>b -# xa#b a>b â -8 xa8b a>bÞ 9a>b œ ua>b va>b, Hence ________________________________________________________________________ page 436 —————————————————————————— CHAPTER 7. —— in which ua>b is the general solution of the homogeneous problem. x œ Fa>bx! Fa>b( F" a=bga=b.= > ! > ! 17a+b. Setting >! œ ! in Eq. a$%b, It was shown in Prob. "&a- b in Section (Þ( that Fa>bF" a=b œ Fa> =bÞ Therefore x œ Fa>bx! ( Fa> =bga=b.= Þ > ! œ Fa>bx! ( Fa>bF" a=bga=b.= Þ a,b. The principal fundamental matrix is identified as Fa>b œ /B:aA>b. Hence x œ /B:aA>bx! ( /B:cAa> =bdga=b.= Þ > ! In Prob. #' of Section $Þ(, the particular solution is given as ] a>b œ ( O a> =b1a=b.= , > >! in which the kernel O a>b depends on the nature of the fundamental solutions. ________________________________________________________________________ page 437 —————————————————————————— CHAPTER 8. —— Chapter Eight Section 8.1 2. The Euler formula for this problem is C8" œ C8 2ˆ& >8 $ÈC8 ‰, C8" œ C8 &82# $2 ÈC8 , in which >8 œ >! 82 Þ Since >! œ ! , we can also write a+b. Euler method with 2 œ !Þ!& À >8 C8 8œ# !Þ" "Þ&**)! 8œ% !Þ# "Þ#*#)) 8œ' !Þ$ "Þ!(#%# 8œ) !Þ% !Þ*$!"(& with C! œ # . a,b. Euler method with 2 œ !Þ!#& À >8 C8 8œ% !Þ" "Þ'""#% 8œ) !Þ# "Þ$"$'" 8 œ "# !Þ$ "Þ"!!"# 8 œ "' !Þ% !Þ*'#&&# The backward Euler formula is C8" œ C8 2ˆ& >8+1 $ÈC8+1 ‰, in which >8 œ >! 82 Þ Since >! œ ! , we can also write C8" œ C8 &a8 "b2# $2 ÈC8" , # $ "È # %C œ” 2 a#!8 #*b2 8• Þ # # with C! œ # . Solving for C8" , and choosing the positive root, we find that C8" ________________________________________________________________________ page 438 —————————————————————————— CHAPTER 8. —— a- b. Backward Euler method with 2 œ !Þ!& À >8 C8 8œ# !Þ" "Þ'%$$( 8œ% !Þ# "Þ$("'% 8œ' !Þ$ "Þ"(('$ 8œ) !Þ% "Þ!&$$% a. b. Backward Euler method with 2 œ !Þ!#& À >8 C8 8œ% !Þ" "Þ'$$!" 8œ) !Þ# "Þ$&#*& 8 œ "# !Þ$ "Þ"&#'( 8 œ "' !Þ% "Þ!#%!( 3. The Euler formula for this problem is C8" œ C8 2a# C8 $ >8 b, C8" œ C8 #2C8 $82 # , in which >8 œ >! 82 Þ Since >! œ ! , with C! œ " . a+b. Euler method with 2 œ !Þ!& À >8 C8 8œ# !Þ" "Þ#!#& 8œ% !Þ# "Þ%"'!$ 8œ' !Þ$ "Þ'%#)* 8œ) !Þ% "Þ))&*! a,b. Euler method with 2 œ !Þ!#& À >8 C8 8œ% !Þ" "Þ#!$)) 8œ) !Þ# "Þ%"*$' 8 œ "# !Þ$ "Þ'%)*' 8 œ "' !Þ% "Þ)*&(# The backward Euler formula is C8" œ C8 2a# C8+1 $ >8" b, in which >8 œ >! 82 Þ Since >! œ ! , we can also write C8" œ C8 #2 C8" $a8 "b2 # , C8 $a8 "b2# Þ " #2 with C! œ " . Solving for C8" , we find that C8" œ ________________________________________________________________________ page 439 —————————————————————————— CHAPTER 8. —— a- b. Backward Euler method with 2 œ !Þ!& À >8 C8 8œ# !Þ" "Þ#!)'% 8œ% !Þ# "Þ%$"!% 8œ' !Þ$ "Þ'(!%# 8œ) !Þ% "Þ*$!(' a. b. Backward Euler method with 2 œ !Þ!#& À >8 C8 4. The Euler formula is 8œ% !Þ" "Þ#!'*$ 8œ) !Þ# "Þ%#')$ 8 œ "# !Þ$ "Þ''#'& 8 œ "' !Þ% "Þ*")!# C8" œ C8 2c# >8 /B:a >8 C8 bdÞ C8" œ C8 #82# 2 /B:a 82 C8 b, Since >8 œ >! 82 and >! œ ! , we can also write with C! œ " Þ a+b. Euler method with 2 œ !Þ!& À >8 C8 8œ# !Þ" "Þ"!#%% 8œ% !Þ# "Þ#"%#' 8œ' !Þ$ "Þ$$%)% 8œ) !Þ% "Þ%'$** a,b. Euler method with 2 œ !Þ!#& À >8 C8 8œ% !Þ" "Þ"!$'& 8œ) !Þ# "Þ#"'&' 8 œ "# !Þ$ "Þ$$)"( 8 œ "' !Þ% "Þ%')$# The backward Euler formula is Since >! œ ! and >8" œ a8 "b2 , we can also write C8" œ C8 2c# >8" /B:a >8" C8" bdÞ C8" œ C8 #2# a8 "b 2 /B:c a8 "b2 C8" d, with C! œ " Þ This equation cannot be solved explicitly for C8" . At each step, given the current value of C8 , the equation must be solved numerically for C8" Þ ________________________________________________________________________ page 440 —————————————————————————— CHAPTER 8. —— a- b. Backward Euler method with 2 œ !Þ!& À >8 C8 8œ# !Þ" "Þ"!(#! 8œ% !Þ# "Þ##$$$ 8œ' !Þ$ "Þ$%(*( 8œ) !Þ% "Þ%)""! a. b. Backward Euler method with 2 œ !Þ!#& À >8 C8 8œ% !Þ" "Þ"!'!$ 8œ) !Þ# "Þ##""! 8 œ "# !Þ$ "Þ$%%($ 8 œ "' !Þ% "Þ%(')) 6. The Euler formula for this problem is # C8" œ C8 2ˆ># C8 ‰=38 C8 . 8 Here >! œ ! and >8 œ 82 . So that # C8" œ C8 2ˆ8# 2# C8 ‰=38 C8 , with C! œ " . a+b. Euler method with 2 œ !Þ!& À >8 C8 8œ# !Þ" !Þ*#!%*) 8œ% !Þ# !Þ)&(&$) 8œ' !Þ$ !Þ)!)!$! 8œ) !Þ% !Þ((!!$) a,b. Euler method with 2 œ !Þ!#& À >8 C8 8œ% !Þ" !Þ*##&(& 8œ) !Þ# !Þ)'!*#$ 8 œ "# !Þ$ !Þ)#$!! 8 œ "' !Þ% !Þ((%*'& ________________________________________________________________________ page 441 —————————————————————————— CHAPTER 8. —— The backward Euler formula is # C8" œ C8 2ˆ># C8" ‰=38 C8" Þ 8" Since >! œ ! and >8" œ a8 "b2 , we can also write # C8" œ C8 2a8 "b# 2 # C8" ‘=38 C8" , with C! œ " . Note that this equation cannot be solved explicitly for C8" . Given C8 , the transcendental equation # C8" 2 C8" =38 C8" œ C8 2 a8 "b# 2 # must be solved numerically for C8" Þ a- b. Backward Euler method with 2 œ !Þ!& À >8 C8 8œ# !Þ" !Þ*#)!&* 8œ% !Þ# !Þ)(!!&% 8œ' !Þ$ !Þ)#%!#" 8œ) !Þ% !Þ())')' a. b. Backward Euler method with 2 œ !Þ!#& À >8 C8 8œ% !Þ" !Þ*#'$%" 8œ) !Þ# !Þ)'("'$ 8 œ "# !Þ$ !Þ)#!#(* 8 œ "' !Þ% !Þ()%#(& 8. The Euler formula C8" œ C8 2ˆ& >8 $ÈC8 ‰, C8" œ C8 &82# $2 ÈC8 , in which >8 œ >! 82 Þ Since >! œ ! , we can also write a+b. Euler method with 2 œ !Þ!#& À >8 C8 8 œ #! !Þ& !Þ)*")$! 8 œ %! "Þ! "Þ#&##& 8 œ '! "Þ& #Þ$()") 8 œ )! #Þ! %Þ!(#&( with C! œ # . ________________________________________________________________________ page 442 —————————————————————————— CHAPTER 8. —— a,b. Euler method with 2 œ !Þ!"#& À >8 C8 8 œ %! !Þ& !Þ*!)*!# 8 œ )! "Þ! "Þ#')(# 8 œ "#! "Þ& #Þ$*$$' 8 œ "'! #Þ! %Þ!)(** The backward Euler formula is C8" œ C8 2ˆ& >8+1 $ÈC8+1 ‰, in which >8 œ >! 82 Þ Since >! œ ! , we can also write C8" œ C8 &a8 "b2# $2 ÈC8" , # $ "È # %C œ” 2 a#!8 #*b2 8• Þ # # with C! œ # . Solving for C8" , and choosing the positive root, we find that C8" a- b. Backward Euler method with 2 œ !Þ!#& À >8 C8 8 œ #! !Þ& !Þ*&)&'& 8 œ %! "Þ! "Þ$"()' 8 œ '! "Þ& #Þ%$*#% 8 œ )! #Þ! %Þ"$%(% a. b. Backward Euler method with 2 œ !Þ!"#& À >8 C8 8 œ %! !Þ& !Þ*%##'" 8 œ )! "Þ! "Þ$!"&$ 8 œ "#! "Þ& #Þ#%$)* 8 œ "'! #Þ! %Þ""*!) 9. The Euler formula for this problem is C8" œ C8 2È>8 C8 . Here >! œ ! and >8 œ 82 . So that C8" œ C8 2È82 C8 , with C! œ $ . ________________________________________________________________________ page 443 —————————————————————————— CHAPTER 8. —— 10. The Euler formula is C8" œ C8 2c# >8 /B:a >8 C8 bdÞ C8" œ C8 #82# 2 /B:a 82 C8 b, Since >8 œ >! 82 and >! œ ! , we can also write with C! œ " Þ a+b. Euler method with 2 œ !Þ!#& À >8 C8 8 œ #! !Þ& "Þ'!(#* 8 œ %! "Þ! #Þ%')$! 8 œ '! "Þ& $Þ(#"'( 8 œ )! #Þ! &Þ%&*'$ a,b. Euler method with 2 œ !Þ!"#& À >8 C8 8 œ %! !Þ& "Þ'!**' 8 œ )! "Þ! #Þ%(%'! 8 œ "#! "Þ& $Þ($$&' 8 œ "'! #Þ! &Þ%(((% The backward Euler formula is Since >! œ ! and >8" œ a8 "b2 , we can also write C8" œ C8 2c# >8" /B:a >8" C8" bdÞ C8" œ C8 #2# a8 "b 2 /B:c a8 "b2 C8" d, with C! œ " Þ This equation cannot be solved explicitly for C8" . At each step, given the current value of C8 , the equation must be solved numerically for C8" Þ a- b. Backward Euler method with 2 œ !Þ!#& À >8 C8 8 œ #! !Þ& "Þ'"(*# 8 œ %! "Þ! #Þ%*$&' 8 œ '! "Þ& $Þ('*%! 8 œ )! #Þ! &Þ&$##$ ________________________________________________________________________ page 444 —————————————————————————— CHAPTER 8. —— a. b. Backward Euler method with 2 œ !Þ!"#& À >8 C8 11. The Euler formula is 8 œ %! !Þ& "Þ'"&#) 8 œ )! "Þ! #Þ%)(#$ 8 œ "#! "Þ& $Þ(&(%# 8 œ "'! #Þ! &Þ&"%!% # C8" œ C8 2a% >8 C8 bΈ" C8 ‰Þ # C8" œ C8 ˆ%2 82 # C8 ‰Îˆ" C8 ‰, Since >8 œ >! 82 and >! œ ! , we can also write a+b. Euler method with 2 œ !Þ!#& À >8 C8 8 œ #! !Þ& "Þ%&)'& 8 œ %! "Þ! !Þ#"(&%& 8 œ '! "Þ& "Þ!&("& 8 œ )! #Þ! "Þ%"%)( with C! œ # Þ a,b. Euler method with 2 œ !Þ!"#& À >8 C8 8 œ %! !Þ& "Þ%&$## 8 œ )! "Þ! !Þ")!)"$ 8 œ "#! "Þ& "Þ!&*!$ 8 œ "'! #Þ! "Þ%"#%% The backward Euler formula is Since >! œ ! and >8" œ a8 "b2 , we can also write # C8" œ C8 2a% >8" C8" bΈ" C8" ‰Þ # # C8" ˆ" C8" ‰ œ C8 ˆ" C8" ‰ %2 a8 "b2 # C8" ‘, with C! œ # Þ This equation cannot be solved explicitly for C8" . At each step, given the current value of C8 , the equation must be solved numerically for C8" Þ ________________________________________________________________________ page 445 —————————————————————————— CHAPTER 8. —— a- b. Backward Euler method with 2 œ !Þ!#& À >8 C8 8 œ #! !Þ& "Þ%$'!! 8 œ %! "Þ! !Þ!')"'&( 8 œ '! "Þ& "Þ!'%)* 8 œ )! #Þ! "Þ%!&(& a. b. Backward Euler method with 2 œ !Þ!"#& À >8 C8 8 œ %! !Þ& "Þ%%"*! 8 œ )! "Þ! !Þ"!&($( 8 œ "#! "Þ& "Þ!'#*! 8 œ "'! #Þ! "Þ%!()* 12. The Euler formula is # # C8" œ C8 2ˆC8 # >8 C8 ‰Îˆ$ >8 ‰Þ Since >8 œ >! 82 and >! œ ! , we can also write # C8" œ C8 ˆ2 C8 # 82 # C8 ‰Îˆ$ 8# 2 # ‰, a+b. Euler method with 2 œ !Þ!#& À >8 C8 8 œ #! !Þ& !Þ&)(*)( 8 œ %! "Þ! !Þ(*"&)* 8 œ '! "Þ& "Þ"%(%$ 8 œ )! #Þ! "Þ(!*($ with C! œ !Þ& Þ a,b. Euler method with 2 œ !Þ!"#& À >8 C8 8 œ %! !Þ& !Þ&)*%%! 8 œ )! "Þ! !Þ(*&(&) 8 œ "#! "Þ& "Þ"&'*$ 8 œ "'! #Þ! "Þ(#*&& The backward Euler formula is Since >! œ ! and >8" œ a8 "b2 , we can also write # # C8" œ C8 2ˆC8" # >8" C8" ‰Îˆ$ >8" ‰Þ # C8" $ a8 "b# 2# ‘ 2 C8" œ C8 $ a8 "b# 2 # ‘ #a8 "b2 # C8" , with C! œ !Þ& Þ Note that although this equation can be solved explicitly for C8" , it is also possible to use a numerical equation solver. At each time step, given the current ________________________________________________________________________ page 446 —————————————————————————— CHAPTER 8. —— value of C8 , the equation may be solved numerically for C8" Þ a- b. Backward Euler method with 2 œ !Þ!#& À >8 C8 8 œ #! !Þ& !Þ&*$*!" 8 œ %! "Þ! !Þ)!)("' 8 œ '! "Þ& "Þ")')( 8 œ )! #Þ! "Þ(*#*" a. b. Backward Euler method with 2 œ !Þ!"#& À >8 C8 8 œ %! !Þ& !Þ&*#$*' 8 œ )! "Þ! !Þ)!%$"* 8 œ "#! "Þ& "Þ"(''% 8 œ "'! #Þ! "Þ((""" 13. The Euler formula for this problem is C8" œ C8 2a" >8 % C8 b, in which >8 œ >! 82 Þ Since >! œ ! , we can also write C8" œ C8 2 82 # %2 C8 , with C! œ " . With 2 œ !Þ!" , a total number of #!! iterations is needed to reach > œ # Þ With 2 œ !Þ!!" , a total of #!!! iterations are necessary. 14. The backward Euler formula is C8" œ C8 2a" >8" % C8" bÞ C8 2 2 >8" . " %2 Since the equation is linear, we can solve for C8" in terms of C8 À C8" œ Here >! œ ! and C! œ " Þ With 2 œ !Þ!" , a total number of #!! iterations is needed to reach > œ # Þ With 2 œ !Þ!!" , a total of #!!! iterations are necessary. 18. Let 9a>b be a solution of the initial value problem. The local truncation error for the Euler method, on the interval >8 Ÿ > Ÿ >8" , is given by /8" œ where >8 >8 >8" Þ Since 9 w a>b œ ># c9a>bd# , it follows that 9 ww a>b œ #> # 9a>b9 w a>b œ #> #># 9a>b #c9a>bd$ Þ " ww 9 a > 8 b2 # , # ________________________________________________________________________ page 447 —————————————————————————— CHAPTER 8. —— Hence $ k/8" k Ÿ >8" ># Q8" Q8" ‘2 # , 8" 20. Given that 9a>b is a solution of the initial value problem, the local truncation error for the Euler method, on the interval >8 Ÿ > Ÿ >8" , is where >8 >8 >8" Þ Based on the ODE, 9 w a>b œ È> 9a>b , and hence 9 ww a>b œ " 9 w a >b #È> 9a>b " " œ Þ # #È> 9a>b /8" œ " ww 9 a > 8 b2 # , # in which Q8" œ 7+Be9a>b l >8 Ÿ > Ÿ >8" fÞ Therefore k/8" k Ÿ " " # –" È —2 Þ % >8 9a >8 b 21. Let 9a>b be a solution of the initial value problem. The local truncation error for the Euler method, on the interval >8 Ÿ > Ÿ >8" , is given by /8" œ where >8 >8 >8" Þ Since 9 w a>b œ #> /B:c > 9a>bd, it follows that " ww 9 a > 8 b2 # , # 9 ww a>b œ # #c9a>b > 9 w a>bd † /B:c > 9a>bd œ # ˜9a>b #># > /B:c > 9a>bd™ † /B:c > 9a>bdÞ 2# # š9a >8 b # >8 >8 /B:c >8 9a >8 bd› † /B:c >8 9a >8 bdÞ # " &1 =38 &1> Hence /8" œ 2# 22a+b. Direct integration yields 9a>b œ "Þ ________________________________________________________________________ page 448 —————————————————————————— CHAPTER 8. —— a, bÞ >8 C8 8œ! !Þ! "Þ! 8œ" !Þ# "Þ# 8œ# !Þ% "Þ! 8œ$ !Þ' "Þ# a- bÞ >8 C8 8œ! !Þ! "Þ! 8œ# !Þ# "Þ" 8œ% !Þ% "Þ! 8œ' !Þ' "Þ" ________________________________________________________________________ page 449 —————————————————————————— CHAPTER 8. —— a. b. Since 9 ww a>b œ &1 =38 &1> , the local truncation error for the Euler method, on the interval >8 Ÿ > Ÿ >8" , is given by /8" œ In order to satisfy k/8" k Ÿ it is necessary that 2 25a+b. The Euler formula is È&!1 " ¸ !Þ!) Þ &12# !Þ!& , # &12# =38 &1 >8 Þ # C8" œ C8 2a" >8 % C8 b. 8œ# !Þ" "Þ&& 8œ% !Þ# #Þ$% 8œ' !Þ$ $Þ%' 8œ) !Þ% &Þ!( >8 C8 a,b. The Euler formula for this problem is C8" œ C8 2a$ >8 C8 b. 8œ# !Þ" "Þ#! 8œ% !Þ# "Þ$* 8œ' !Þ$ "Þ&( 8œ) !Þ% "Þ(% >8 C8 a- b. The Euler formula is ________________________________________________________________________ page 450 —————————————————————————— CHAPTER 8. —— C8" œ C8 2a# C8 $ >8 b. >8 C8 26a+b. a, b . a- b . 8œ# !Þ" "Þ#! 8œ% !Þ# "Þ%# 8œ' !Þ$ "Þ'& 8œ) !Þ% "Þ*! "!!! † º "!!! † º "!!! † º 'Þ! #Þ! ") œ "!!! † a!b œ ! Þ 'Þ! º ")Þ! œ "!!!a!Þ!'b œ '! Þ 'Þ!! º 'Þ!" #Þ!! 'Þ!"! #Þ!!% ")Þ!% œ "!!!a !Þ!*#"'b œ *# Þ"' Þ 'Þ!!! º 27. Rounding to three digits, +a, - b ¸ !Þ##% . Likewise, to three digits, +, ¸ !Þ(!# and +- ¸ !Þ%(( Þ It follows that +, +- ¸ !Þ##& Þ ________________________________________________________________________ page 451 —————————————————————————— CHAPTER 8. —— Section 8.2 1. The improved Euler formula for this problem is " " 2# C8" œ C8 2Œ$ >8 >8" C8 a$ >8 C8 bÞ # # # Since >8 œ >! 82 and >! œ ! , we can also write C8" œ C8 2a$ C8 b with C! œ " . a+b. 2 œ !Þ!& À >8 C8 a,b. 2 œ !Þ!#& À >8 C8 a- b. 2 œ !Þ!"#& À >8 C8 8œ) !Þ" "Þ"*&"' 8 œ "' !Þ# "Þ$)"#' 8 œ #% !Þ$ "Þ&&*") 8 œ $# !Þ% "Þ(#*'( 8œ% !Þ" "Þ"*&"& 8œ) !Þ# "Þ$)"#& 8 œ "# !Þ$ "Þ&&*"' 8 œ "' !Þ% "Þ(#*'& 8œ# !Þ" "Þ"*&"# 8œ% !Þ# "Þ$)"#! 8œ' !Þ$ "Þ&&*!* 8œ) !Þ% "Þ(#*&' 2# 82$ aC8 # #8b , # # 2. The improved Euler formula is C8" œ C8 in which O8 œ C8 2ˆ& >8 $ÈC8 ‰. Since >8 œ >! 82 and >! œ ! , we can also write C8" œ C8 with C! œ # . a+b. 2 œ !Þ!& À ________________________________________________________________________ page 452 2 2 ˆ& >8 $ÈC8 ‰ Š& >8" $ÈO8 ‹, # # 2 2 ˆ& 82 $ÈC8 ‰ ’& a8 "b2 $ÈO8 “, # # —————————————————————————— CHAPTER 8. —— >8 C8 a,b. 2 œ !Þ!#& À >8 C8 a- b. 2 œ !Þ!"#& À >8 C8 8œ# !Þ" "Þ'##)$ 8œ% !Þ# "Þ$$%'! 8œ' !Þ$ "Þ"#)#! 8œ) !Þ% !Þ**&%%& 8œ% !Þ" "Þ'##%$ 8œ) !Þ# "Þ$$$)' 8 œ "# !Þ$ "Þ"#(") 8 œ "' !Þ% !Þ**%#"& 8œ) !Þ" "Þ'##$% 8 œ "' !Þ# "Þ$$$') 8 œ #% !Þ$ "Þ"#'*$ 8 œ $# !Þ% !Þ**$*#" 3. The improved Euler formula for this problem is C8" œ C8 2 a% C8 $ >8 $ >8" b 2 # a# C8 $ >8 bÞ # 2# a% C8 $ '8b $82 $ , # Since >8 œ >! 82 and >! œ ! , we can also write C8" œ C8 #2 C8 with C! œ " . a+b. 2 œ !Þ!& À >8 C8 a,b. 2 œ !Þ!#& À >8 C8 a- b. 2 œ !Þ!"#& À 8œ% !Þ" "Þ#!&$$ 8œ) !Þ# "Þ%##*! 8 œ "# !$ "Þ'&&%# 8 œ "' !Þ% "Þ*!'#" 8œ# !Þ" "Þ#!&#' 8œ% !Þ# "Þ%##($ 8œ' !Þ$ "Þ'&&"" 8œ) !Þ% "Þ*!&(! ________________________________________________________________________ page 453 —————————————————————————— CHAPTER 8. —— >8 C8 8œ) !Þ" "Þ#!&$% 8 œ "' !Þ# "Þ%##*% 8 œ #% !Þ$ "Þ'&&&! 8 œ $# !Þ% "Þ*!'$% 5. The improved Euler formula is C8" in which # C8 # >8 C8 O8 œ C8 2 . $ ># 8 # # C8 # >8 C8 O8 # >8" O8 , œ C8 2 2 #a$ ># b #ˆ$ ># ‰ 8 8" Since >8 œ >! 82 and >! œ ! , we can also write C8" œ C8 2 with C! œ !Þ& . a+b. 2 œ !Þ!& À >8 C8 a,b. 2 œ !Þ!#& À >8 C8 a- b. 2 œ !Þ!"#& À >8 C8 8œ) !Þ" !Þ&"!'* 8 œ "' !Þ# !Þ&#%"$( 8œ% !Þ" !Þ&"!"') 8œ) !Þ# !Þ&#%"$& 8œ# !Þ" !Þ&"!"'% 8œ% !Þ# !Þ&#%"#' # C8 #82 C8 O # #a8 "b2O8 2 8 , #a$ 8# 2# b #$ a8 "b# 2# ‘ 8œ' !Þ$ !Þ&%!)$ 8œ) !Þ% !Þ&'%#&" 8 œ "# !Þ$ !Þ&%#"!! 8 œ "' !Þ% !Þ&'%#(( 8 œ #% !Þ$ !Þ&%#"!% 8 œ $# !Þ% !Þ&'%#)% 6. The improved Euler formula for this problem is C8" œ C8 2# 2# # # ˆ>8 C8 ‰=38 C8 ˆ>8" O8 ‰=38 O8 , # # ________________________________________________________________________ page 454 —————————————————————————— CHAPTER 8. —— in which # O8 œ C8 2 ˆ># C8 ‰=38 C8 . 8 Since >8 œ >! 82 and >! œ ! , we can also write C8" œ C8 with C! œ " . a+b. 2 œ !Þ!& À >8 C8 a,b. 2 œ !Þ!#& À >8 C8 8œ% !Þ" !Þ*#%&&! 8œ) !Þ# !Þ)'%"(( 8œ# !Þ" !Þ*#%'&! 8œ% !Þ# !Þ)'%$$) 2 ## 2 # # ˆ8 2 C8 ‰=38 C8 a8 "b# 2 # O8 ‘=38 O8 , # # 8œ' !Þ$ !Þ)"''%# 8œ) !Þ% !Þ()!!!) 8 œ "# !Þ$ !Þ)"'%%# 8 œ "' !Þ% !Þ((*()" a- b. 2 œ !Þ!"#& À >8 C8 8œ) !Þ" !Þ*#%&#& 8 œ "' !Þ# !Þ)'%"$) 8 œ #% !Þ$ !Þ)"'$*$ 8 œ $# !Þ% !Þ((*(#& 7. The improved Euler formula for this problem is C8" œ C8 2 a% C8 >8 >8" "b 2 # a# C8 >8 !Þ&bÞ # Since >8 œ >! 82 and >! œ ! , we can also write C8" œ C8 2a# C8 !Þ&b 2 # a# C8 8b 82 $ , with C! œ " . a+b. 2 œ !Þ!#& À >8 C8 8 œ #! !Þ& #Þ*'("* 8 œ %! "Þ! (Þ))$"$ 8 œ '! "Þ& #!Þ)""% 8 œ )! #Þ! &&Þ&"!' ________________________________________________________________________ page 455 —————————————————————————— CHAPTER 8. —— a,b. 2 œ !Þ!"#& À >8 C8 8 œ %! !Þ& #Þ*')!! 8 œ )! "Þ! (Þ))(&& 8 œ "#! "Þ& #!Þ)#*% 8 œ "'! #Þ! &&Þ&(&) 8. The improved Euler formula is C8" œ C8 in which O8 œ C8 2ˆ& >8 $ÈC8 ‰. Since >8 œ >! 82 and >! œ ! , we can also write C8" œ C8 with C! œ # . a+b. 2 œ !Þ!#& À >8 C8 a,b. 2 œ !Þ!"#& À >8 C8 8 œ %! !Þ& !Þ*#&)"& 8 œ )! "Þ! "Þ#)&#& 8 œ "#! "Þ& #Þ%!)'* 8 œ "'! #Þ! %Þ"!$&* 8 œ #! !Þ& !Þ*#'"$* 8 œ %! "Þ! "Þ#)&&) 8 œ '! "Þ& #Þ%!)*) 8 œ )! #Þ! %Þ"!$)' 2 2 ˆ& 82 $ÈC8 ‰ ’& a8 "b2 $ÈO8 “, # # 2 2 ˆ& >8 $ÈC8 ‰ Š& >8" $ÈO8 ‹, # # 9. The improved Euler formula for this problem is C8" œ C8 in which O8 œ C8 2È>8 C8 . Since >8 œ >! 82 and >! œ ! , we can also write C8" œ C8 with C! œ $ . a+b. 2 œ !Þ!#& À 2È 2 82 C8 Èa8 "b2 O8 , # # 2 2 È>8 C8 È>8" O8 , # # ________________________________________________________________________ page 456 —————————————————————————— CHAPTER 8. —— >8 C8 a,b. 2 œ !Þ!"#& À >8 C8 8 œ #! !Þ& $Þ*'#"( 8 œ %! "Þ! &Þ"!))( 8 œ '! "Þ& 'Þ%$"$% 8 œ )! #Þ! (Þ*#$$# 8 œ %! !Þ& $Þ*'#") 8 œ )! "Þ! &Þ"!))* 8 œ "#! "Þ& 'Þ%$"$) 8 œ "'! #Þ! (Þ*#$$( 10. The improved Euler formula is C8" œ C8 2 2 c# >8 /B:a >8 C8 bd c# >8" /B:a >8" O8 bd, # # in which O8 œ C8 2c# >8 /B:a >8 C8 bd. Since >8 œ >! 82 and >! œ ! , we can also write C8" œ C8 with C! œ " . a+b. 2 œ !Þ!#& À >8 C8 a,b. 2 œ !Þ!"#& À >8 C8 8 œ %! !Þ& "Þ'"#'$ 8 œ )! "Þ! #Þ%)!*# 8 œ "#! "Þ& $Þ(%&&! 8 œ "'! #Þ! &Þ%*&)* 8 œ #! !Þ& "Þ'"#'$ 8 œ %! "Þ! #Þ%)!*( 8 œ '! "Þ& $Þ(%&&' 8 œ )! #Þ! &Þ%*&*& 2 2 c# 82 /B:a 82 C8 bd e#a8 "b2 /B:c a8 "b2O8 df, # # 12. The improved Euler formula is C8" œ C8 2 in which O8 œ C8 2 # C8 # >8 C8 . $ ># 8 # C8 # >8 C8 O # # >8" O8 2 8 , #a$ ># b #ˆ$ ># ‰ 8 8" ________________________________________________________________________ page 457 —————————————————————————— CHAPTER 8. —— Since >8 œ >! 82 and >! œ ! , we can also write C8" œ C8 2 with C! œ !Þ& . a+b. 2 œ !Þ!#& À >8 C8 a,b. 2 œ !Þ!"#& À >8 C8 8 œ %! !Þ& !Þ&*!*!' 8 œ )! "Þ! !Þ(***)) 8 œ "#! "Þ& "Þ"'''$ 8 œ "'! #Þ! "Þ(%**# 8 œ #! !Þ& !Þ&*!)*( 8 œ %! "Þ! !Þ(***&! 8 œ '! "Þ& "Þ"''&$ 8 œ )! #Þ! "Þ(%*'* # C8 #82 C8 O # #a8 "b2O8 , 2 8 #a$ 8# 2# b #$ a8 "b# 2# ‘ 16. The exact solution of the initial value problem is 9a>b œ " " /#> Þ Based on the # # result in Prob. "%a- b, the local truncation error for a linear differential equation is /8" œ where >8 >8 >8" Þ Since 9 www a>b œ % /#> , the local truncation error is /8" œ Furthermore, with ! Ÿ >8 Ÿ " , # /B:a# >8 b2 $ . $ ##$ /2. $ " www 9 a > 8 b2 $ , ' k/8" k Ÿ It also follows that for 2 œ !Þ" , k/" k Ÿ # !Þ# " / a!Þ"b$ œ /!Þ# . $ "&!! Using the improved Euler method, with 2 œ !Þ" , we have C" ¸ "Þ""!!! . The exact value is given by 9a!Þ"b œ "Þ""!(!"% Þ 17. The exact solution of the initial value problem is given by 9a>b œ " > /#> Þ Using # the modified Euler method, the local truncation error for a linear differential equation is /8" œ " www 9 a > 8 b2 $ , ' ________________________________________________________________________ page 458 —————————————————————————— CHAPTER 8. —— where >8 >8 >8" Þ Since 9 www a>b œ ) /#> , the local truncation error is /8" œ % /B:a# >8 b2 $ . $ %#$ /2. $ Furthermore, with ! Ÿ >8 Ÿ " , the local error is bounded by k/8" k Ÿ It also follows that for 2 œ !Þ" , k/" k Ÿ % !Þ# " !Þ# / a!Þ"b$ œ /. $ (&! Using the improved Euler method, with 2 œ !Þ" , we have C" ¸ "Þ#(!!! . The exact value is given by 9a!Þ"b œ "Þ#("%!$ Þ 18. Using the Euler method, C" œ " !Þ"a!Þ& ! # † "b œ "Þ#& Þ Using the improved Euler method, C" œ " !Þ!&a!Þ& ! # † "b !Þ!&a!Þ& !Þ" # † "Þ#&b œ "Þ#( Þ The estimated error is /" ¸ "Þ#( "Þ#& œ !Þ!# Þ The step size should be adjusted by a factor of È!Þ!!#&Î!Þ!# ¸ !Þ$&% . Hence the required step size is estimated as 2 ¸ a!Þ"ba!Þ$'b œ !Þ!$' Þ ________________________________________________________________________ page 459 —————————————————————————— CHAPTER 8. —— 20. Using the Euler method, C" œ $ !Þ"È! $ œ $Þ"($#!& Þ Using the improved Euler method, C" œ $ !Þ!&È! $ !Þ!&È!Þ" $Þ"($#!& œ $Þ"((!'$ Þ The estimated error is /" ¸ $Þ"((!'$ $Þ"($#!& œ !Þ!!$)&) Þ The step size should be adjusted by a factor of È!Þ!!#&Î!Þ!!$)&) ¸ !Þ)!& . Hence the required step size is estimated as 2 ¸ a!Þ"ba!Þ)!&b œ !Þ!)!& Þ a!Þ&b# ! $! œ !Þ&!)$$% 21. Using the Euler method, C" œ !Þ& !Þ" Using the improved Euler method, a!Þ&b# ! a!Þ&!)$$%b# #a!Þ"ba!Þ&!)$$%b C" œ !Þ& !Þ!& !Þ!& $! $ a!Þ"b# œ !Þ&"!"%) Þ The estimated error is /" ¸ !Þ&"!"%) !Þ&!)$$% œ !Þ!!") Þ The local truncation error is less than the given tolerance. The step size can be adjusted by a factor of È!Þ!!#&Î!Þ!!") ¸ "Þ"()& . Hence it is possible to use a step size of 2 ¸ a!Þ"ba"Þ"()&b ¸ !Þ""( Þ 22. Assuming that the solution has continuous derivatives at least to the third order, 9a>8" b œ 9a>8 b 9 w a>8 b2 9 ww a>8 b # 9 www a>8 b $ 2 2, #x $x a+b. The local truncation error is given by The modified Euler formula is defined as where >8 >8 >8" Þ Suppose that C8 œ 9a>8 bÞ /8" œ 9a>8" b C8" Þ ________________________________________________________________________ page 460 —————————————————————————— CHAPTER 8. —— " " C8" œ C8 2 0 ”>8 2 ß C8 2 0 a>8 ß C8 b•Þ # # Observe that 9 w a>8 b œ 0 a>8 ß 9a>8 bb œ 0 a>8 ß C8 b . It follows that a,b. As shown in Prob. "%a,b, Furthermore, /8" œ 9a>8" b C8" 9 ww a>8 b # 9 www a>8 b $ œ 2 0 a>8 ß C8 b 2 2 #x $x " " 2 0 ”>8 2 ß C8 2 0 a>8 ß C8 b•Þ # # 9 ww a>8 b œ 0> a>8 ß C8 b 0C a>8 ß C8 b0 a>8 ß C8 b . " # 0 ”>8 2 ß C8 2 0 a>8 ß C8 b• œ 0 a>8 ß C8 b 0> a>8 ß C8 b " # in which 5 œ " 2 0 a>8 ß C8 b and >8 0 >8 2Î# , C8 ( C8 5 . Therefore # /8" œ 9 www a>8 b $ 2 2# 2 ” 0>> 25 0>C 5 # 0CC • Þ $x #x % >œ0 ß Cœ( " 2# , ” 0>> 25 0>C 5 # 0CC • #x % >œ0 ß Cœ( 2 0C a>8 ß C8 b 5 # Note that each term in the brackets has a factor of 2# . Hence the local truncation error is proportional to 2$ . a- b. If 0 a>ß Cb is linear, then 0>> œ 0>C œ 0CC œ ! , and /8" œ 9 www a>8 b $ 2Þ $x 23. The modified Euler formula for this problem is " " C8" œ C8 2œ$ >8 2 ”C8 2 a$ >8 C8 b• # # # 2 œ C8 2a$ >8 C8 b aC8 >8 #bÞ # Since >8 œ >! 82 and >! œ ! , we can also write C8" 2# œ C8 2a$ 82 C8 b aC8 82 #b , # with C! œ " . Setting 2 œ !Þ" , we obtain the following values À ________________________________________________________________________ page 461 —————————————————————————— CHAPTER 8. —— >8 C8 8œ" !Þ" "Þ"*&!! 8œ# !Þ# "Þ$)!*) 8œ$ !Þ$ "Þ&&)() 8œ% !Þ% "Þ(#*#! 25. The modified Euler formula is $ C8" œ C8 2”#C8 $>8 2 2 a#C8 $>8 b• # 2# œ C8 2a#C8 $>8 b a%C8 '>8 $bÞ # Since >8 œ >! 82 and >! œ ! , we can also write C8" 2# œ C8 2a#C8 $82 b a%C8 ' 82 $b , # with C! œ " . Setting 2 œ !Þ" , we obtain À >8 C8 8œ" !Þ" "Þ#!&!! 8œ# !Þ# "Þ%##"! 8œ$ !Þ$ "Þ'&$*' 8œ% !Þ% "Þ*!$)$ 26. The modified Euler formula for this problem is C8" œ C8 2œ#>8 2 /B:” Œ>8 2 O8 •, # in which O8 œ C8 2 c#>8 /B:a >8 C8 bdÞ Now >8 œ >! 82 , with >! œ ! and # C! œ " . Setting 2 œ !Þ" , we obtain the following values À >8 C8 8œ" !Þ" "Þ"!%))& 8œ# !Þ# "Þ#")*# 8œ$ !Þ$ "Þ$%"&( 8œ% !Þ% "Þ%(#(#% ________________________________________________________________________ page 462 —————————————————————————— CHAPTER 8. —— 27. Let 0 a>ß C b be linear in both variables. The improved Euler formula is " C8" œ C8 2c0 a>8 ß C8 b 0 a>8 2ß C8 20 a>8 ß C8 bbd # " " " œ C8 20 a>8 ß C8 b 20 a>8 ß C8 b 20 c2ß 20 a>8 ß C8 bd # # # " œ 20 a2ß C8 b 20 c2ß 20 a>8 ß C8 bdÞ # " " C8" œ C8 20 ”>8 2ß C8 20 a>8 ß C8 b• # # "" œ C8 20 a>8 ß C8 b 20 ” 2ß 20 a>8 ß C8 b•Þ ## "" " 0 ” 2ß 20 a>8 ß C8 b• œ 0 c2ß 20 a>8 ß C8 bdÞ ## # The modified Euler formula is Since 0 a>ß C b is linear in both variables, ________________________________________________________________________ page 463 —————————————————————————— CHAPTER 8. —— Section 8.3 1. The ODE is linear, with 0 a>ß C b œ $ > C . The Runge-Kutta algorithm requires the evaluations 5 8" œ 0 a>8 ß C8 b " " 5 8# œ 0 Œ>8 2 ß C8 25 8" # # " " 5 8$ œ 0 Œ>8 2 ß C8 25 8# # # 5 8% œ 0 a>8 2 ß C8 25 8$ b. The next estimate is given as the weighted average C8" œ C8 a+bÞ For 2 œ !Þ" À >8 C8 8œ" !Þ" "Þ"*&"' 2 a5 8" # 5 8# # 5 8$ 5 8% bÞ ' a,bÞ For 2 œ !Þ!& À 8œ# !Þ# "Þ$)"#( 8œ$ !Þ$ "Þ&&*") 8œ% !Þ% "Þ(#*') 2. In this problem, 0 a>ß C b œ &> $ÈC . At each time step, the Runge-Kutta algorithm requires the evaluations 5 8" œ 0 a>8 ß C8 b " " 5 8# œ 0 Œ>8 2 ß C8 25 8" # # " " 5 8$ œ 0 Œ>8 2 ß C8 25 8# # # 5 8% œ 0 a>8 2 ß C8 25 8$ b. The exact solution of the IVP is Ca>b œ # > /> Þ >8 C8 8œ# !Þ" "Þ"*&"' 8œ% !Þ# "Þ$)"#( 8œ' !Þ$ "Þ&&*") 8œ) !Þ% "Þ(#*') The next estimate is given as the weighted average C8" œ C8 2 a5 8" # 5 8# # 5 8$ 5 8% bÞ ' ________________________________________________________________________ page 464 —————————————————————————— CHAPTER 8. —— a+bÞ For 2 œ !Þ" À >8 C8 8œ" !Þ" "Þ'##$" 8œ# !Þ# "Þ$$$'# 8œ$ !Þ$ "Þ"#')' 8œ% !Þ% !Þ**$)$* a,bÞ For 2 œ !Þ!& À >8 C8 8œ# !Þ" "Þ'##$! 8œ% !Þ# "Þ$$$'# 8œ' !Þ$ "Þ"#')& 8œ) !Þ% !Þ**$)#' The exact solution of the IVP is given implicitly as È# " œ Þ &"# ˆ#ÈC &>‰& ˆ> ÈC ‰# 3. The ODE is linear, with 0 a>ß C b œ #C $> . The Runge-Kutta algorithm requires the evaluations 5 8" œ 0 a>8 ß C8 b " " 5 8# œ 0 Œ>8 2 ß C8 25 8" # # " " 5 8$ œ 0 Œ>8 2 ß C8 25 8# # # 5 8% œ 0 a>8 2 ß C8 25 8$ b. The next estimate is given as the weighted average C8" œ C8 a+bÞ For 2 œ !Þ" À >8 C8 8œ" !Þ" "Þ#!&$& 2 a5 8" # 5 8# # 5 8$ 5 8% bÞ ' a,bÞ For 2 œ !Þ!& À 8œ# !Þ# "Þ%##*& 8œ$ !Þ$ "Þ'&&&$ 8œ% !Þ% "Þ*!'$) The exact solution of the IVP is Ca>b œ /#> Î% $>Î# $Î% Þ >8 C8 8œ# !Þ" "Þ#!&$& 8œ% !Þ# "Þ%##*' 8œ' !Þ$ "Þ'&&&$ 8œ) !Þ% "Þ*!'$) 5. In this problem, 0 a>ß C b œ aC # #>C bÎa$ ># b . The Runge-Kutta algorithm ________________________________________________________________________ page 465 —————————————————————————— CHAPTER 8. —— requires the evaluations 5 8" œ 0 a>8 ß C8 b " " 5 8# œ 0 Œ>8 2 ß C8 25 8" # # " " 5 8$ œ 0 Œ>8 2 ß C8 25 8# # # 5 8% œ 0 a>8 2 ß C8 25 8$ b. 2 a5 8" # 5 8# # 5 8$ 5 8% bÞ ' The next estimate is given as the weighted average C8" œ C8 a+bÞ For 2 œ !Þ" À >8 C8 8œ" !Þ" !Þ&"!"(! a,bÞ For 2 œ !Þ!& À 8œ# !Þ# !Þ&#%"$) 8œ$ !Þ$ !Þ&%#"!& 8œ% !Þ% !Þ&'%#)' The exact solution of the IVP is Ca>b œ a$ ># bÎa' >b Þ >8 C8 8œ# !Þ" !Þ&#!"'* 8œ% !Þ# !Þ&#%"$) 8œ' !Þ$ !Þ&%#"!& 8œ) !Þ% !Þ&'%#)' 6. In this problem, 0 a>ß C b œ a># C# b=38 C . At each time step, the Runge-Kutta algorithm requires the evaluations 5 8" œ 0 a>8 ß C8 b " " 5 8# œ 0 Œ>8 2 ß C8 25 8" # # " " 5 8$ œ 0 Œ>8 2 ß C8 25 8# # # 5 8% œ 0 a>8 2 ß C8 25 8$ b. The next estimate is given as the weighted average C8" œ C8 2 a5 8" # 5 8# # 5 8$ 5 8% bÞ ' ________________________________________________________________________ page 466 —————————————————————————— CHAPTER 8. —— a+bÞ For 2 œ !Þ" À >8 C8 8œ" !Þ" !Þ*#%&"( 8œ# !Þ# !Þ)'%"#& 8œ$ !Þ$ !Þ)"'$(( 8œ% !Þ% !Þ((*(!' a,bÞ For 2 œ !Þ!& À >8 C8 8œ# !Þ" !Þ*#%&"( 8œ% !Þ# !Þ)'%"#& 8œ' !Þ$ !Þ)"'$(( 8œ) !Þ% !Þ((*(!' 7. a+bÞ For 2 œ !Þ" À >8 C8 8œ& !Þ& #Þ*')#& 8 œ "! "Þ! (Þ))))* 8 œ "& "Þ& #!Þ)$%* 8 œ #! #Þ! &&Þ&*&( a,bÞ For 2 œ !Þ!& À The exact solution of the IVP is Ca>b œ /#> >Î# Þ 8. See Prob. # . for the exact solution. a+bÞ For 2 !Þ" À >8 C8 8œ& !Þ& !Þ*#&(#& 8 œ "! "Þ! "Þ#)&"' >8 C8 8 œ "! !Þ& #Þ*')#) 8 œ #! "Þ! (Þ))*!% 8 œ $! "Þ& #!Þ)$&& 8 œ %! #Þ! &&Þ&*)! a,bÞ For 2 œ !Þ!& À 8 œ "& "Þ& #Þ%!)'! 8 œ #! #Þ! %Þ"!$&! >8 C8 8 œ "! !Þ& !Þ*#&("" 8 œ #! "Þ! "Þ#)&"& 8 œ $! "Þ& #Þ%!)'! 8 œ %! #Þ! %Þ"!$&! ________________________________________________________________________ page 467 —————————————————————————— CHAPTER 8. —— 9a+bÞ For 2 œ !Þ" À >8 C8 8œ& !Þ& $Þ*'#"* 8 œ "! "Þ! &Þ"!)*! 8 œ "& "Þ& 'Þ%$"$* 8 œ #! #Þ! (Þ*#$$) a,bÞ For 2 œ !Þ!& À >8 C8 68” 8 œ "! !Þ& $Þ*'#"* 8 œ #! "Þ! &Þ"!)*! 8 œ $! "Þ& 'Þ%$"$* 8 œ %! #Þ! (Þ*#$$) The exact solution is given implicitly as # • #È> C # +<->+82 È> C œ > #È$ # +<->+82 È$ Þ C>" 10. See Prob. % . a+bÞ For 2 œ !Þ" À >8 C8 8œ& !Þ& "Þ'"#'# 8 œ "! "Þ! #Þ%)!*" 8 œ "& "Þ& $Þ(%&%) 8 œ #! #Þ! &Þ%*&)( a,bÞ For 2 œ !Þ!& À >8 C8 8 œ "! !Þ& "Þ'"#'# 8 œ #! "Þ! #Þ%)!*" 8 œ $! "Þ& $Þ(%&%) 8 œ %! #Þ! &Þ%*&)( 12. See Prob. & . for the exact solution. a+bÞ For 2 œ !Þ" À >8 C8 8œ& !Þ& !Þ&*!*!* 8 œ "! "Þ! !Þ)!!!!! 8 œ "& "Þ& "Þ"''''( 8 œ #! #Þ! "Þ(&!!! a,bÞ For 2 œ !Þ!& À >8 C8 8 œ "! !Þ& !Þ&*!*!* 8 œ #! "Þ! !Þ)!!!!! 8 œ $! "Þ& "Þ"''''( 8 œ %! #Þ! "Þ(&!!! 13. The ODE is linear, with 0 a>ß C b œ " > %C . The Runge-Kutta algorithm requires ________________________________________________________________________ page 468 —————————————————————————— CHAPTER 8. —— the evaluations 5 8" œ 0 a>8 ß C8 b " " 5 8# œ 0 Œ>8 2 ß C8 25 8" # # " " 5 8$ œ 0 Œ>8 2 ß C8 25 8# # # 5 8% œ 0 a>8 2 ß C8 25 8$ b. 2 a5 8" # 5 8# # 5 8$ 5 8% bÞ ' "* %> "' / The next estimate is given as the weighted average C8" œ C8 The exact solution of the IVP is Ca>b œ a+bÞ For 2 œ !Þ" À "> % $ "' Þ a,bÞ For 2 œ !Þ!& À >8 C8 15a+b. >8 C8 8œ& !Þ& )Þ(!*$"(& 8 œ "! "Þ! '%Þ)&)"!( 8 œ "& "Þ& %()Þ)"*#) 8 œ #! #Þ! $&$&Þ)''( 8 œ "! !Þ& )Þ("")!'! 8 œ #! "Þ! '%Þ)*%)(& 8 œ $! "Þ& %(*Þ##'(% 8 œ %! #Þ! $&$*Þ))!% ________________________________________________________________________ page 469 —————————————————————————— CHAPTER 8. —— a,b. For the integral curve starting at a!ß !b, the slope becomes infinite near >Q ¸ "Þ& Þ Note that the exact solution of the IVP is defined implicitly as C$ %C œ >$ Þ Using the classic Runge-Kutta algorithm, with 2 œ !Þ!" , we obtain the values >8 C8 8 œ (! !Þ( !Þ!)&*" 8 œ )! !Þ) !Þ"#)&$ 8 œ *! !Þ* !Þ")$)! 8 œ *& !Þ*& !Þ#"')* a- b. Based on the direction field, the solution should decrease monotonically to the limiting value C œ #ÎÈ$ Þ In the following table, the value of >Q corresponds to the approximate time in the iteration process that the calculated values begin to increase. 2 !Þ" !Þ!& !Þ!#& !Þ!" >Q "Þ* "Þ'& "Þ&& "Þ%&& a. b. Numerical values will continue to be generated, although they will not be associated with the integral curve starting at a!ß !b. These values are approximations to nearby integral curves. a/b. We consider the solution associated with the initial condition Ca!b œ " . The exact solution is given by C$ %C œ >$ $ Þ ________________________________________________________________________ page 470 —————————————————————————— CHAPTER 8. —— For the integral curve starting at a! ß "b, the slope becomes infinite near >Q ¸ #Þ! Þ In the following table, the values of >Q corresponds to the approximate time in the iteration process that the calculated values begin to increase. 2 !Þ" !Þ!& !Þ!#& !Þ!" >Q "Þ)& "Þ)& "Þ)' "Þ)$& ________________________________________________________________________ page 471 —————————————————————————— CHAPTER 8. —— Section 8.4 1a+b. Using the notation 08 œ 0 a>8 ß C8 b , the predictor formula is C8" œ C8 With 08" œ 0 a>8" ß C8" b, the corrector formula is C8" œ C8 2 a&& 08 &* 08" $( 08# * 08$ bÞ #% 2 a* 08" "* 08 & 08" 08# b Þ #% We use the starting values generated by the Runge-Kutta method À >8 C8 8œ! !Þ! "Þ! 8œ" !Þ" "Þ"*&"' 8œ# !Þ# "Þ$)"#( 8œ$ !Þ$ "Þ&&*") 8 œ &a-9<b !Þ& "Þ)*$%'*($ >8 C8 8 œ %a:</b !Þ% "Þ(#*'('*! 8 œ %a-9<b !Þ% "Þ(#*)')!" 8 œ &a:</b !Þ& "Þ)*$%'%$' a,b. With 08" œ 0 a>8" ß C8" b, the fourth order Adams-Moulton formula is C8" œ C8 2 a* 08" "* 08 & 08" 08# b Þ #% In this problem, 08" œ $ >8" C8" Þ Since the ODE is linear, we can solve for C8" œ " c#% C8 #( 2 *2 >8" 2 a"* 08 & 08" 08# bd Þ #% * 2 8œ% !Þ% "Þ(#*')!! 8œ& !Þ& "Þ)*$%'*& >8 C8 a- b. The fourth order backward differentiation formula is C8" œ " c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ #& In this problem, 08" œ $ >8" C8" Þ Since the ODE is linear, we can solve for C8" œ " c$' 2 "#2 >8" %) C8 $' C8" "' C8# $ C8$ d Þ #& "# 2 ________________________________________________________________________ page 472 —————————————————————————— CHAPTER 8. —— The exact solution of the IVP is Ca>b œ # > /> Þ >8 C8 8œ% !Þ% "Þ(#*')!& 8œ& !Þ& "Þ)*$%("" 2a+b. Using the notation 08 œ 0 a>8 ß C8 b , the predictor formula is C8" œ C8 With 08" œ 0 a>8" ß C8" b, the corrector formula is C8" œ C8 2 a&& 08 &* 08" $( 08# * 08$ bÞ #% 2 a* 08" "* 08 & 08" 08# b Þ #% We use the starting values generated by the Runge-Kutta method À >8 C8 8œ! !Þ! #Þ! 8œ" !Þ" "Þ'##$" 8 œ %a-9<b !Þ% !Þ**$)&# 8œ# !Þ# "Þ$$$'# 8œ$ !Þ$ "Þ"#')' 8 œ &a-9<b !Þ& !Þ*#&('% >8 C8 8 œ %a:</b !Þ% !Þ**$(&" 8 œ &a:</b !Þ& !Þ*#&%'* a,b. With 08" œ 0 a>8" ß C8" b, the fourth order Adams-Moulton formula is In this problem, 08" œ &>8" $ÈC8" Þ Since the ODE is 898linear, an equation solver is needed to approximate the solution of C8" œ C8 2 %&>8" #(ÈC8" "* 08 & 08" 08# ‘ #% C8" œ C8 2 a* 08" "* 08 & 08" 08# b Þ #% at each time step. We obtain the approximate values: >8 C8 8œ% !Þ% !Þ**$)%( 8œ& !Þ& !Þ*#&(%' a- b. The fourth order backward differentiation formula is ________________________________________________________________________ page 473 —————————————————————————— CHAPTER 8. —— " c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ #& C8" œ Since the ODE is 898linear, an equation solver is used to approximate the solution of C8" œ " %) C8 $' C8" "' C8# $ C8$ "#2 ˆ&>8" $ÈC8" ‰‘ #& at each time step. >8 C8 8œ% !Þ% !Þ**$)'* 8œ& !Þ& !Þ*#&)$( The exact solution of the IVP is given implicitly by È# " & # œ &"# Þ ˆ#ÈC &>‰ ˆ> ÈC ‰ 3a+b. The predictor formula is C8" œ C8 With 08" œ 0 a>8" ß C8" b, the corrector formula is C8" œ C8 2 a&& 08 &* 08" $( 08# * 08$ bÞ #% 2 a* 08" "* 08 & 08" 08# b Þ #% Using the starting values generated by the Runge-Kutta method À >8 C8 8œ! !Þ! "Þ! 8œ" !Þ" "Þ#!&$&! 8 œ %a-9<b !Þ% "Þ*!'$)# 8œ# !Þ# "Þ%##*&% 8œ$ !Þ$ "Þ'&&&#( 8 œ &a-9<b !Þ& #Þ"(*&'( >8 C8 8 œ %a:</b !Þ% "Þ*!'$%! 8 œ &a:</b !Þ& #Þ"(*%&& a,b. With 08" œ 0 a>8" ß C8" b, the fourth order Adams-Moulton formula is C8" œ C8 2 a* 08" "* 08 & 08" 08# b Þ #% In this problem, 08" œ # C8" $ >8" Þ Since the ODE is linear, we can solve for ________________________________________________________________________ page 474 —————————————————————————— CHAPTER 8. —— " c#% C8 #(2 >8" 2 a"* 08 & 08" 08# bd Þ #% ") 2 8œ% !Þ% "Þ*!'$)& 8œ& !Þ& #Þ"(*&(' C8" œ >8 C8 a- b. The fourth order backward differentiation formula is C8" œ " c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ #& In this problem, 08" œ # C8" $ >8" Þ Since the ODE is linear, we can solve for C8" œ " c%) C8 $' C8" "' C8# $ C8$ $'2 >8" d Þ #& #% 2 8œ% !Þ% "Þ*!'$*& 8œ& !Þ& #Þ"(*'"" The exact solution of the IVP is Ca>b œ /#> Î% $>Î# $Î% Þ 5a+b. The predictor formula is C8" œ C8 2 a&& 08 &* 08" $( 08# * 08$ bÞ #% 2 a* 08" "* 08 & 08" 08# b Þ #% >8 C8 With 08" œ 0 a>8" ß C8" b, the corrector formula is C8" œ C8 ________________________________________________________________________ page 475 —————————————————————————— CHAPTER 8. —— Using the starting values generated by the Runge-Kutta method À >8 C8 8œ! !Þ! !Þ& 8œ" !Þ" !Þ&"!"'*&! 8œ# !Þ# !Þ&#%"$(*& 8œ$ !Þ$ !Þ&%#"!&#* 8 œ &a-9<b !Þ& !Þ&*!*!*") >8 C8 8 œ %a:</b !Þ% !Þ&'%#)&$# 8 œ %a-9<b !Þ% !Þ&'%#)&(( 8 œ &a:</b !Þ& !Þ&*!*!)"' a,b. With 08" œ 0 a>8" ß C8" b, the fourth order Adams-Moulton formula is C8" œ C8 In this problem, 08" œ # C8" # >8" C8" Þ $ ># " 8 2 a* 08" "* 08 & 08" 08# b Þ #% Since the ODE is 898linear, an equation solver is needed to approximate the solution of C8" œ C8 at each time step. >8 C8 8œ% !Þ% !Þ&'%#)&() 8œ& !Þ& !Þ&*!*!*#! C# # >8" C8" 2 * 8" "* 08 & 08" 08# • ” #% $ ># " 8 a- b. The fourth order backward differentiation formula is C8" œ " c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ #& Since the ODE is 898linear, an equation solver is needed to approximate the solution of C8" œ C# # >8" C8" " %) C8 $' C8" "' C8# $ C8$ "#2 8" ” • #& $ ># " 8 at each time step. We obtain the approximate values: >8 C8 8œ% !Þ% !Þ&'%#)&)) 8œ& !Þ& !Þ&*!*!*&# ________________________________________________________________________ page 476 —————————————————————————— CHAPTER 8. —— The exact solution of the IVP is Ca>b œ a$ ># bÎa' >b Þ 6a+b. The predictor formula is C8" œ C8 2 a&& 08 &* 08" $( 08# * 08$ bÞ #% 2 a* 08" "* 08 & 08" 08# b Þ #% With 08" œ 0 a>8" ß C8" b, the corrector formula is C8" œ C8 We use the starting values generated by the Runge-Kutta method À >8 C8 8œ! !Þ! "Þ! 8œ" !Þ" !Þ*#%&"( 8œ# !Þ# !Þ)'%"#& 8œ$ !Þ$ !Þ)"'$(( 8 œ &a-9<b !Þ& !Þ(&$"$& >8 C8 8 œ %a:</b !Þ% !Þ((*)$# 8 œ %a-9<b !Þ% !Þ((*'*$ 8 œ &a:</b !Þ& !Þ(&$$"" a,b. With 08" œ 0 a>8" ß C8" b, the fourth order Adams-Moulton formula is C8" œ C8 # In this problem, 08" œ a># C8" b=38 C8" Þ Since the ODE is 898linear, we obtain 8" the implicit equation 2 a* 08" "* 08 & 08" 08# b Þ #% C8" œ C8 2 # *ˆ># C8" ‰=38 C8" "* 08 & 08" 08# ‘Þ 8" #% 8œ% !Þ% !Þ((*(!! 8œ& !Þ& !Þ(&$"%% >8 C8 ________________________________________________________________________ page 477 —————————————————————————— CHAPTER 8. —— a- b. The fourth order backward differentiation formula is C8" œ " c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ #& Since the ODE is 898linear, we obtain the implicit equation C8" œ " # %) C8 $' C8" "' C8# $ C8$ "#2 ˆ># C8" ‰=38 C8" ‘Þ 8" #& 8œ% !Þ% !Þ((*')! 8œ& !Þ& !Þ(&$!)* >8 C8 8a+b. The predictor formula is C8" œ C8 With 08" œ 0 a>8" ß C8" b, the corrector formula is C8" œ C8 2 a&& 08 &* 08" $( 08# * 08$ bÞ #% 2 a* 08" "* 08 & 08" 08# b Þ #% We use the starting values generated by the Runge-Kutta method À >8 C8 8œ! !Þ! #Þ! 8œ" !Þ!& "Þ(**'#*' 8œ# !Þ" "Þ'##$!%# 8œ$ !Þ"& "Þ%'(#&!$ >8 C8 8 œ "! !Þ& !Þ*#&("$$ 8 œ #! "Þ! "Þ#)&"%) 8 œ $! "Þ& #Þ%!)&*& 8 œ %! #Þ! %Þ"!$%*& a,b. Since the ODE is 898linear, an equation solver is needed to approximate the solution of C8" œ C8 2 %&>8" #(ÈC8" "* 08 & 08" 08# ‘ #% at each time step. We obtain the approximate values: >8 C8 8 œ "! !Þ& !Þ*#&("#& 8 œ #! "Þ! "Þ#)&"%) 8 œ $! "Þ& #Þ%!)&*& 8 œ %! #Þ! %Þ"!$%*& ________________________________________________________________________ page 478 —————————————————————————— CHAPTER 8. —— a- b. The fourth order backward differentiation formula is C8" œ " c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ #& Since the ODE is 898linear, an equation solver is needed to approximate the solution of C8" œ " %) C8 $' C8" "' C8# $ C8$ "#2 ˆ&>8" $ÈC8" ‰‘ #& at each time step. >8 C8 8 œ "! !Þ& !Þ*#&(#%) 8 œ #! "Þ! "Þ#)&"&) 8 œ $! "Þ& #Þ%!)&*% 8 œ %! #Þ! %Þ"!$%*$ The exact solution of the IVP is given implicitly by È# " œ Þ &"# ˆ#ÈC &>‰& ˆ> ÈC ‰# 9a+b. The predictor formula is C8" œ C8 With 08" œ 0 a>8" ß C8" b, the corrector formula is C8" œ C8 2 a&& 08 &* 08" $( 08# * 08$ bÞ #% 2 a* 08" "* 08 & 08" 08# b Þ #% Using the starting values generated by the Runge-Kutta method À >8 C8 8œ! !Þ! $Þ! 8œ" !Þ!& $Þ!)(&)' 8œ# !Þ" $Þ"(("#( 8œ$ !Þ"& $Þ#')'!* >8 C8 8 œ "! !Þ& $Þ*'#")' 8 œ #! "Þ! &Þ"!)*!$ 8 œ $! "Þ& 'Þ%$"$*! 8 œ %! #Þ! (Þ*#$$)& a,b. With 08" œ 0 a>8" ß C8" b, the fourth order Adams-Moulton formula is C8" œ C8 2 a* 08" "* 08 & 08" 08# b Þ #% ________________________________________________________________________ page 479 —————————————————————————— CHAPTER 8. —— In this problem, 08" œ È>8" C8" Þ Since the ODE is 898linear, an equation solver must be implemented in order to approximate the solution of C8" œ C8 at each time step. >8 C8 8 œ "! !Þ& $Þ*'#")' 8 œ #! "Þ! &Þ"!)*!$ 8 œ $! "Þ& 'Þ%$"$*! 8 œ %! #Þ! (Þ*#$$)& 2 *È>8" C8" "* 08 & 08" 08# ‘ #% a- b. The fourth order backward differentiation formula is C8" œ " c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ #& Since the ODE is 898linear, an equation solver is needed to approximate the solution of C8" œ at each time step. >8 C8 68” 8 œ "! !Þ& $Þ*'#")' 8 œ #! "Þ! &Þ"!)*!$ 8 œ $! "Þ& 'Þ%$"$*! 8 œ %! #Þ! (Þ*#$$)& " %) C8 $' C8" "' C8# $ C8$ "#2 È>8" C8" ‘ #& The exact solution is given implicitly by # • #È> C # +<->+82 È> C œ > #È$ # +<->+82 È$ Þ C>" ________________________________________________________________________ page 480 —————————————————————————— CHAPTER 8. —— 10a+b. The predictor formula is C8" œ C8 With 08" œ 0 a>8" ß C8" b, the corrector formula is C8" œ C8 2 a&& 08 &* 08" $( 08# * 08$ bÞ #% 2 a* 08" "* 08 & 08" 08# b Þ #% We use the starting values generated by the Runge-Kutta method À >8 C8 8œ! !Þ! "Þ! 8œ" !Þ!& "Þ!&"#$! 8œ# !Þ" "Þ"!%)%$ 8œ$ !Þ"& "Þ"'!(%! >8 C8 8 œ "! !Þ& "Þ'"#'## 8 œ #! "Þ! #Þ%)!*!* 8 œ $! "Þ& $Þ(%&"%(* 8 œ %! #Þ! &Þ%*&)(# a,b. With 08" œ 0 a>8" ß C8" b, the fourth order Adams-Moulton formula is C8" œ C8 In this problem, 08" œ # >8" /B:a >8" C8" b Þ Since the ODE is 898linear, an equation solver must be implemented in order to approximate the solution of C8" œ C8 at each time step. >8 C8 8 œ "! !Þ& "Þ'"#'## 8 œ #! "Þ! #Þ%)!*!* 8 œ $! "Þ& $Þ(%&"%(* 8 œ %! #Þ! &Þ%*&)(# 2 e*c# >8" /B:a >8" C8" bd "* 08 & 08" 08# f #% 2 a* 08" "* 08 & 08" 08# b Þ #% a- b. The fourth order backward differentiation formula is C8" œ " c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ #& Since the ODE is 898linear, we obtain the implicit equation C8" œ " e%) C8 $' C8" "' C8# $ C8$ "#2 c# >8" /B:a >8" C8" bdfÞ #& ________________________________________________________________________ page 481 —————————————————————————— CHAPTER 8. —— >8 C8 8 œ "! !Þ& "Þ'"#'#$ 8 œ #! "Þ! #Þ%)!*!& 8 œ $! "Þ& $Þ(%&"%($ 8 œ %! #Þ! &Þ%*&)'* 11a+b. The predictor formula is C8" œ C8 With 08" œ 0 a>8" ß C8" b, the corrector formula is C8" œ C8 2 a&& 08 &* 08" $( 08# * 08$ bÞ #% 2 a* 08" "* 08 & 08" 08# b Þ #% Using the starting values generated by the Runge-Kutta method À >8 C8 8œ! !Þ! #Þ! 8œ" !Þ!& "Þ*&))$$ 8œ# !Þ" "Þ*"&##" 8œ$ !Þ"& "Þ)')*(& >8 C8 8 œ "! !Þ& "Þ%%('$* 8 œ #! "Þ! !Þ"%$'#)" 8 œ $! "Þ& "Þ!'!*%' 8 œ %! #Þ! "Þ%"!"## a,b. With 08" œ 0 a>8" ß C8" b, the fourth order Adams-Moulton formula is C8" œ C8 In this problem, 08" œ % >8" C8" Þ # " C8" 2 a* 08" "* 08 & 08" 08# b Þ #% Since the differential equation is 898linear, an equation solver is used to approximate the solution of C8" œ C8 at each time step. >8 C8 8 œ "! !Þ& "Þ%%('$) 8 œ #! "Þ! !Þ"%$'('( 8 œ $! "Þ& "Þ!'!*"$ 8 œ %! #Þ! "Þ%"!"!$ 2 % >8" C8" "* 08 & 08" 08# • ”* # #% " C8" ________________________________________________________________________ page 482 —————————————————————————— CHAPTER 8. —— a- b. The fourth order backward differentiation formula is C8" œ " c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ #& Since the ODE is 898linear, an equation solver must be implemented in order to approximate the solution of C8" œ at each time step. >8 C8 8 œ "! !Þ& "Þ%%('#" 8 œ #! "Þ! !Þ"%%('"* 8 œ $! "Þ& "Þ!'!("( 8 œ %! #Þ! "Þ%"!!#( " % >8" C8" ”%) C8 $' C8" "' C8# $ C8$ "#2 • # #& " C8" 12a+b. The predictor formula is C8" œ C8 With 08" œ 0 a>8" ß C8" b, the corrector formula is C8" œ C8 2 a&& 08 &* 08" $( 08# * 08$ bÞ #% 2 a* 08" "* 08 & 08" 08# b Þ #% We use the starting values generated by the Runge-Kutta method À >8 C8 8œ! !Þ! !Þ& 8œ" !Þ!& !Þ&!%'#") 8œ# !Þ" !Þ&"!"'*& 8œ$ !Þ"& !Þ&"''''' >8 C8 8 œ "! !Þ& !Þ&*!*!*" 8 œ #! "Þ! !Þ)!!!!!! 8 œ $! "Þ& "Þ"''''( 8 œ %! #Þ! "Þ(&!!!! a,b. With 08" œ 0 a>8" ß C8" b, the fourth order Adams-Moulton formula is C8" œ C8 In this problem, 2 a* 08" "* 08 & 08" 08# b Þ #% ________________________________________________________________________ page 483 —————————————————————————— CHAPTER 8. —— 08" œ # C8" # >8" C8" Þ $ ># " 8 Since the ODE is 898linear, an equation solver is needed to approximate the solution of C8" œ C8 at each time step. >8 C8 8 œ "! !Þ& !Þ&*!*!*" 8 œ #! "Þ! !Þ)!!!!!! 8 œ $! "Þ& "Þ"''''( 8 œ %! #Þ! "Þ(&!!!! C# # >8" C8" 2 * 8" "* 08 & 08" 08# • ” #% $ ># " 8 a- b. The fourth order backward differentiation formula is C8" œ " c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ #& Since the ODE is 898linear, we obtain the implicit equation C8" # C8" # >8" C8" " œ ”%) C8 $' C8" "' C8# $ C8$ "#2 •Þ #& $ ># " 8 The exact solution of the IVP is Ca>b œ a$ ># bÎa' >b Þ >8 C8 8 œ "! !Þ& !Þ&*!*!*# 8 œ #! "Þ! !Þ)!!!!!# 8 œ $! "Þ& "Þ"''''( 8 œ %! #Þ! "Þ(&!!!" 13. Both Adams methods entail the approximation of 0 a> ß C b, on the interval c>8 ß >8" d, by a polynomial. Approximating 9 w a>b œ T" a>b ´ E , which is a constant polynomial, we have 9a>8" b 9a>8 b œ ( >8" Setting E œ - 08 a" -b08" , where ! Ÿ - Ÿ " , we obtain the approximation C8" œ C8 2c- 08 a" -b08" dÞ E œ - 08 a" -b08" , An appropriate choice of - yields the familiar Euler formula. Similarly, setting œ Ea>8" >8 b œ E2 Þ >8 E .> where ! Ÿ - Ÿ " , we obtain the approximation ________________________________________________________________________ page 484 —————————————————————————— CHAPTER 8. —— C8" œ C8 2c- 08 a" -b08" dÞ 14. For a third order Adams-Bashforth formula, we approximate 0 a> ß C b, on the interval c>8 ß >8" d, by a quadratic polynomial using the points a>8# ß C8# b , a>8" ß C8" b and a>8 ß C8 bÞ Let T$ a>b œ E># F> G Þ We obtain the system of equations E># # F>8# G œ 08# 8 E># " F>8" G œ 08" 8 E># F>8 G œ 08 . 8 For computational purposes, assume that >! œ ! , and >8 œ 82 Þ It follows that 08 #08" 08# # 2# a$ #8b08 a%8 %b08" a" #8b08# Fœ #2 8# $8 # 8# 8 Gœ 08 ˆ#8 8# ‰08" 08# Þ # # Eœ We then have C8" C8 œ ( >8" >8 E># F> G ‘.> " " œ E2$ Œ8# 8 F2 # Œ8 G2 , $ # 2 a#$ 08 "' 08" & 08# b Þ "# which yields C8" C8 œ 15. For a third order Adams-Moulton formula, we approximate 0 a> ß C b, on the interval c>8 ß >8" d, by a quadratic polynomial using the points a>8" ß C8" b , a>8 ß C8 b and a>8" ß C8" bÞ Let T$ a>b œ !># " > # Þ This time we obtain the system of algebraic equations !># " " >8" # œ 08" 8 !># " >8 # œ 08 8 !># " " >8" # œ 08" . 8 For computational purposes, again assume that >! œ ! , and >8 œ 82 Þ It follows that ________________________________________________________________________ page 485 —————————————————————————— CHAPTER 8. —— 08" #08 08" # 2# a#8 "b08" %808 a" #8b08" "œ #2 # 8 8 8# 8 #œ 08" ˆ" 8# ‰08 08" Þ # # !œ We then have C8" C8 œ ( >8" >8 !># " > # ‘.> " " œ !2$ Œ8# 8 " 2 # Œ8 # 2 , $ # 2 a&08" )08 08" b Þ "# which results in C8" C8 œ ________________________________________________________________________ page 486 —————————————————————————— CHAPTER 8. —— Section 8.5 1a+b. The general solution of the ODE is Ca>b œ - /> # > Þ Imposing the initial condition, Ca!b œ # , the solution of the IVP is 9" a>b œ # > . a,b. If instead, the initial condition Ca!b œ #Þ!!" is given, the solution of the IVP is 9# a>b œ !Þ!!" /> # > . We then have 9# a>b 9" a>b œ !Þ!!" /> Þ a+ß ,bÞ Based on the exact solution, the local truncation error for both of the Euler methods is k/69- k Ÿ "!% "!! >8 # / 2Þ # 3. The solution of the initial value problem is 9a>b œ /"!! > > Þ Hence k/8 k Ÿ &!!! 2# , for all ! >8 " . Furthermore, the local truncation error is greatest near > œ ! . Therefore k/" k Ÿ &!!! 2# !Þ!!!& for 2 !Þ!!!$ Þ Now the truncation error accumulates at each time step. Therefore the actual time step should be much smaller than 2 ¸ !Þ!!!$ . For example, with 2 œ !Þ!!!#& , we obtain the data > œ !Þ!& > œ !Þ" Euler !Þ!&'$#$ !Þ"!!!%! B.Euler !Þ!&("'& !Þ"!!!&" 9a>b !Þ!&'($) !Þ"!!!%& Note that the total number of time steps needed to reach > œ !Þ" is R œ %!! . a- b. Using the Runge-Kutta method, comparisons are made for several values of 2 À 2 œ !Þ" À > œ !Þ!& > œ !Þ" 2 œ !Þ!!& À > œ !Þ!& > œ !Þ" 9a>b !Þ!&'($) !Þ"!!!%& 9a>b !Þ!&'($) !Þ"!!!%& C8 9 a> 8 b !Þ!!!'() !Þ!!!!"! C8 9 a> 8 b !Þ!!!!#( !Þ!!!!!!% C8 !Þ!&(%"' !Þ"!!!&& C8 !Þ!&'('' !Þ"!!!%' 6a+b. Using the method of undetermined coefficients, it is easy to show that the general solution of the ODE is Ca>b œ - /-> ># Þ Imposing the initial condition, it follows that - œ ! and hence the solution of the IVP is 9a>b œ ># Þ a,b. Using the Runge-Kutta method, with 2 œ !Þ!" , numerical solutions are generated page 487 ________________________________________________________________________ —————————————————————————— CHAPTER 8. —— for various values of - À -œ"À > œ !Þ#& > œ !Þ& > œ !Þ(& > œ "Þ! - œ "! À > œ !Þ#& > œ !Þ& > œ !Þ(& > œ "Þ! - œ #! À > œ !Þ#& > œ !Þ& > œ !Þ(& > œ "Þ! - œ &! À > œ !Þ#& > œ !Þ& > œ !Þ(& > œ "Þ! 9a>b !Þ!'#& !Þ#& !Þ&'#& "Þ! 9a>b !Þ!'#& !Þ#& !Þ&'#& "Þ! 9a>b !Þ!'#& !Þ#& !Þ&'#& "Þ! kC8 9a>8 bk # ‚ "!"" ! ! ! kC8 9a>8 bk #Þ#"& ‚ "!( #Þ*#! ‚ "!' $Þ&(* ‚ "!& %Þ$'# ‚ "!% kC8 9a>8 bk "Þ"! ‚ "!& "Þ'&) ‚ "!$ !Þ#%'!%# $'Þ&"$* kC8 9a>8 bk !Þ"!($!$ #)''*Þ)!% (Þ''!"% ‚ "!* #Þ!%'') ‚ "!"& C8 !Þ!'#%***ÞÞ !Þ#& !Þ&'#& "Þ! C8 !Þ!'#%**)ÞÞ !Þ#%***( !Þ&'#%'% !Þ***&'% C8 !Þ!'#))*ÞÞ !Þ#%)$%# !Þ$"'%&) $&Þ&"$* 9a>b !Þ!'#& !Þ#& !Þ&'#& "Þ! C8 !Þ!%%)!$ÞÞ #)''*Þ&& (Þ''!"% ‚ "!* #Þ!%'') ‚ "!"& The following table shows the calculated value, C" , at the first time step À 9a>b "!% C" a- œ "b *Þ***** ‚ "!& C" a- œ "!b *Þ***(* ‚ "!& C" a- œ #!b *Þ**)$$ ‚ "!& C" a- œ &!b *Þ*($*' ‚ "!& a- b. Referring back to the exact solution given in Parta+b, if a nonzero initial condition, say Ca!b œ & , is specified, the solution of the IVP becomes We then have k9a> b 9& a>bk œ k&k /-> Þ It is evident that for any > ! , ________________________________________________________________________ page 488 9& a>b œ & /-> ># . —————————————————————————— CHAPTER 8. —— lim k9a> b 9& a>bk œ _ Þ -Ä_ This implies that virtually any error introduced early in the calculations will be magnified as - p _ . The initial value problem is inherently unstable. ________________________________________________________________________ page 489 —————————————————————————— CHAPTER 8. —— Section 8.6 1. In vector notation, the initial value problem can be written as .B BC> " Œ œŒ , xa!b œ Œ Þ .> C %B #C ! B8" B8 B8 C8 >8 œ Œ 2Œ Þ C8" C8 %B8 #C8 B8" œ B8 2aB8 C8 >8 b C8" œ C8 2a%B8 #C8 bÞ 8œ# !Þ# "Þ#' !Þ(' 8œ% !Þ% "Þ(("% "Þ%)#% 8œ' !Þ' #Þ&)**" #Þ$(!$ 8œ) !Þ) $Þ)#$(% $Þ'!%"$ 8 œ "! "Þ! &Þ'%#%' &Þ$)))& a+b. The Euler formula is Œ That is, With 2 œ !Þ" , we obtain the values >8 B8 C8 a,b. The Runge-Kutta method uses the following intermediate calculations: k8" œ aB8 C8 >8 ß %B8 #C8 bX X 2" 2# 2 2" 2# k8# œ ”B8 58" C8 58" >8 ß %ŒB8 58" #ŒC8 58" • # # # # # X 2" 2# 2 2" 2# k8$ œ ”B8 58# C8 58# >8 %ŒB8 58# #ŒC8 58# • # # # # # X " # " # k8% œ cB8 258$ C8 258$ >8 2 ß %aB8 258$ b #aC8 258$ bd Þ With 2 œ !Þ# , we obtain the values: >8 B8 C8 8œ" !Þ# "Þ$#%*$ !Þ(&)*$$ 8œ# !Þ% "Þ*$'(* "Þ&(*"* 8œ$ !Þ' #Þ*$%"% #Þ''!** 8œ% !Þ) %Þ%)$") %Þ##'$* 8œ& "Þ! 'Þ)%#$' 'Þ&'%&# a- bÞ With 2 œ !Þ" , we obtain ________________________________________________________________________ page 490 —————————————————————————— CHAPTER 8. —— >8 B8 C8 8œ# !Þ# "Þ$#%)* !Þ(&*&"' 8œ% !Þ% "Þ*$'* "Þ&(*** 8œ' !Þ' #Þ*$%&* #Þ''#!" 8œ) !Þ) %Þ%)%## %Þ##()% 8 œ "! "Þ! 'Þ)%%% 'Þ&'')% The exact solution of the IVP is # " # Ba>b œ /#> /$> > * $ * ) $> # " Ca>b œ /#> / > Þ * $ * 3a+b. The Euler formula is Œ B8" B8 >8 B8 C8 " œ Œ 2Œ Þ C8" C8 B8 B8" œ B8 2a >8 B8 C8 "b C8" œ C8 2aB8 bÞ That is, With 2 œ !Þ" , we obtain the values >8 B8 C8 8œ# !Þ# !Þ&)# "Þ") 8œ% !Þ% !Þ""(*'* "Þ#($%% 8œ' !Þ' !Þ$$'*"# "Þ#($)# 8œ) !Þ) !Þ($!!!( "Þ")&(# 8 œ "! "Þ! "Þ!#"$% "Þ!#$(" a,b. The Runge-Kutta method uses the following intermediate calculations: k8" œ a >8 B8 C8 "ß B8 bX X 2 2" 2# 2" k8# œ ” Œ>8 ŒB8 58" ŒC8 58" " ß B8 58" • # # # # X 2 2" 2# 2" k8$ œ ” Œ>8 ŒB8 58# ŒC8 58# " ß B8 58# • # # # # " # "X k8% œ c a>8 2baB8 258$ b aC8 258$ b " ß B8 258$ d Þ ________________________________________________________________________ page 491 —————————————————————————— CHAPTER 8. —— With 2 œ !Þ# , we obtain the values: >8 B8 C8 8œ" !Þ# !Þ&')%&" "Þ"&((& 8œ# !Þ% !Þ"!*((' "Þ##&&' 8œ$ !Þ' !Þ$##!) "Þ#!$%( 8œ% !Þ) !Þ')"#*' "Þ"!"'# 8œ& "Þ! !Þ*$()&# !Þ*$()&# a- bÞ With 2 œ !Þ" , we obtain >8 B8 C8 8œ# !Þ# !Þ&')%& "Þ"&((& 8œ% !Þ% !Þ"!*(($ "Þ##&&( 8œ' !Þ' !Þ$##!)" "Þ#!$%( 8œ) !Þ) !Þ')"#*" "Þ"!"'" 8 œ "! "Þ! !Þ*$()%" !Þ*$()% 4a+b. The Euler formula gives B8" œ B8 2aB8 C8 B8 C8 b C8" œ C8 2a$B8 #C8 B8 C8 bÞ 8œ' !Þ' !Þ&"*$# !Þ!$#"!#& 8œ) !Þ) !Þ&*%$#% !Þ$#')!" 8 œ "! "Þ! !Þ&))#() !Þ&(&%& With 2 œ !Þ" , we obtain the values >8 B8 C8 a,b. Given 8œ# !Þ# !Þ"*) !Þ'") 8œ% !Þ% !Þ$()(*' !Þ#)$#* 0 a>ß Bß C b œ B C B C 1a>ß Bß C b œ $B #C B C , the Runge-Kutta method uses the following intermediate calculations: k8" œ c0 a>8 ß B8 ß C8 bß 1a>8 ß B8 ß C8 bdX 2 2 # 2 2 # 2 2 # " # " # k8# œ ”0 Œ>8 ß B8 58" ß C8 58" ß 1Œ>8 ß B8 58" ß C8 58" • # # # # # # 2 2 2 X " # " # k8$ œ ”0 Œ>8 ß B8 58# ß C8 58# ß 1Œ>8 ß B8 58# ß C8 58# • " # " # k8% œ 0 ˆ>8 2ß B8 258$ ß C8 258$ ‰ß 1ˆ>8 2ß B8 258$ß C8 258$ ‰‘ Þ X 2 # 2 # 2 # X ________________________________________________________________________ page 492 —————————————————————————— CHAPTER 8. —— With 2 œ !Þ# , we obtain the values: >8 B8 C8 8œ" !Þ# !Þ"*'*!% !Þ'$!*$' 8œ# !Þ% !Þ$(#'%$ !Þ#*)))) 8œ$ !Þ' !Þ&!"$!# !Þ!"""%#* 8œ% !Þ) !Þ&'"#(! !Þ#))*%$ 8œ& "Þ! !Þ&%(!&$ !Þ&!)$!$ a- bÞ With 2 œ !Þ" , we obtain >8 B8 C8 8œ# !Þ# !Þ"*'*$& !Þ'$!*$* 8œ% !Þ% !Þ$(#')( !Þ#*))'' 8œ' !Þ' !Þ&!"$%& !Þ!""#")% 8œ) !Þ) !Þ&'"#*# !Þ#)*!( 8 œ "! "Þ! !Þ&%(!$" !Þ&!)%#( 5a+b. The Euler formula gives B8" œ B8 2cB8 a" !Þ& B8 !Þ& C8 bd C8" œ C8 2cC8 a !Þ#& !Þ& B8 bdÞ 8œ% !Þ% #Þ$%""* "Þ'("#" 8œ' !Þ' "Þ*!#$' "Þ*("&) 8œ) !Þ) "Þ&''!# #Þ#$)*& 8 œ "! "Þ! "Þ#*(') #Þ%'($# With 2 œ !Þ" , we obtain the values >8 B8 C8 a,b. Given 8œ# !Þ# #Þ*'##& "Þ$%&$) 0 a>ß Bß C b œ Ba" !Þ& B !Þ& C b 1a>ß Bß C b œ C a !Þ#& !Þ& Bb , the Runge-Kutta method uses the following intermediate calculations: k8" œ c0 a>8 ß B8 ß C8 bß 1a>8 ß B8 ß C8 bdX 2 2 # 2 2 # 2 2 # " # " # k8# œ ”0 Œ>8 ß B8 58" ß C8 58" ß 1Œ>8 ß B8 58" ß C8 58" • # # # # # # 2 2 2 X " # " # k8$ œ ”0 Œ>8 ß B8 58# ß C8 58# ß 1Œ>8 ß B8 58# ß C8 58# • " # " # k8% œ 0 ˆ>8 2ß B8 258$ ß C8 258$ ‰ß 1ˆ>8 2ß B8 258$ß C8 258$ ‰‘ Þ X 2 # 2 # 2 # X ________________________________________________________________________ page 493 —————————————————————————— CHAPTER 8. —— With 2 œ !Þ# , we obtain the values: >8 B8 C8 8œ" !Þ# $Þ!'$$* "Þ$%)&) 8œ# !Þ% #Þ%%%*( "Þ')'$) 8œ$ !Þ' "Þ**"" #Þ!!!$' 8œ% !Þ) "Þ'$)") #Þ#(*)" 8œ& "Þ! "Þ$&&& #Þ&"(& a- bÞ With 2 œ !Þ" , we obtain >8 B8 C8 8œ# !Þ# $Þ!'$"% "Þ$%)** 8œ% !Þ% #Þ%%%'& "Þ')'** 8œ' !Þ' "Þ**!(& #Þ!!"!( 8œ) !Þ) "Þ'$()" #Þ#)!&( 8 œ "! "Þ! "Þ$&&"% #Þ&")#( 6a+b. The Euler formula gives B8" œ B8 2c/B:a B8 C8 b -9= B8 d C8" œ C8 2c=38aB8 $ C8 bdÞ 8œ% !Þ% "Þ)##$% #Þ$'(*" 8œ' !Þ' #Þ#"(#) #Þ&$$#* 8œ) !Þ) #Þ'""") #Þ')('$ 8 œ "! "Þ! #Þ**&& #Þ)$$&% With 2 œ !Þ" , we obtain the values >8 B8 C8 8œ# !Þ# "Þ%#$)' #Þ")*&( a,b. The Runge-Kutta method uses the following intermediate calculations: k8" œ c0 a>8 ß B8 ß C8 bß 1a>8 ß B8 ß C8 bdX 2 # 2 2 # 2 2 # 2 " # " # k8# œ ”0 Œ>8 ß B8 58" ß C8 58" ß 1Œ>8 ß B8 58" ß C8 58" • 2 # 2 # 2 # X " # " # k8$ œ ”0 Œ>8 ß B8 58# ß C8 58# ß 1Œ>8 ß B8 58# ß C8 58# • # # # # # # 2 2 2 X " # " # k8% œ 0 ˆ>8 2ß B8 258$ ß C8 258$ ‰ß 1ˆ>8 2ß B8 258$ß C8 258$ ‰‘ Þ X ________________________________________________________________________ page 494 —————————————————————————— CHAPTER 8. —— With 2 œ !Þ# , we obtain the values: >8 B8 C8 8œ" !Þ# "Þ%"&"$ #Þ")'** 8œ# !Þ% "Þ)"#!) #Þ$'#$$ 8œ$ !Þ' #Þ#!'$& #Þ&#&) 8œ% !Þ) #Þ&*)#' #Þ'(*% 8œ& "Þ! #Þ*()!' #Þ)#%)( a- bÞ With 2 œ !Þ" , we obtain >8 B8 C8 8œ# !Þ# "Þ%"&"$ #Þ")'** 8œ% !Þ% "Þ)"#!* #Þ$'#$$ 8œ' !Þ' #Þ#!'$& #Þ&#&)" 8œ) !Þ) #Þ&*)#' #Þ'(*%" 8 œ "! "Þ! #Þ*()!' #Þ)#%)) 7. The Runge-Kutta method uses the following intermediate calculations: k8" œ cB8 % C8 ß B8 C8 dX 2 # 2 # 2 # 2 # " # " # k8# œ ”B8 58" %ŒC8 58" ß ŒB8 58" C8 58" • 2 # 2 # 2 # 2 # X " # " # k8$ œ ”B8 58# %ŒC8 58# ß ŒB8 58# C8 58# • " # " # k8% œ B8 258$ %ˆC8 258$ ‰ß ˆB8 258$ ‰ C8 258$ ‘ Þ X X Using 2 œ !Þ!% , we obtain the following values: >8 B8 C8 8œ& !Þ# "Þ$#!% !Þ#&!)& 8 œ "! !Þ% "Þ**&# !Þ''#%& 8 œ "& !Þ' $Þ#**# "Þ$(&# 8 œ #! !Þ) &Þ($'# #Þ'%$& 8 œ #& "Þ! "!Þ##( %Þ*#*% The exact solution is given by /> /$> /> /$> 9a>b œ , <a>b œ , # % and the associated tabulated values: >8 9a>8 b <a>8 b 8œ& !Þ# "Þ$#!% !Þ#&!)& 8 œ "! !Þ% "Þ**&# !Þ''#%& 8 œ "& !Þ' $Þ#**# "Þ$(&# 8 œ #! !Þ) &Þ($'# #Þ'%$& 8 œ #& "Þ! "!Þ##( %Þ*#*% 8. Let C œ B w . The second order ODE can be transformed into the first order system ________________________________________________________________________ page 495 —————————————————————————— CHAPTER 8. —— Bw œ C C w œ > $B ># C , with initial conditions Ba!b œ " , Ca!b œ # . Given 0 a>ß Bß C b œ C 1a>ß Bß C b œ > $B ># C , X X the Runge-Kutta method uses the following intermediate calculations: k8" œ C8 ß >8 $ B8 ># C8 ‘ 8 k8# œ ”C8 k 8$ k 8% 2# 2 2" 2# 58" ß 1Œ>8 ß B8 58" ß C8 58" • # # # # X 2# 2 2" 2# œ ”C8 58# ß 1Œ>8 ß B8 58# ß C8 58# • # # # # # " # ‰‘X œ C8 258$ ß 1ˆ>8 2ß B8 258$ ß C8 258$ Þ With 2 œ !Þ" , we obtain the following values: >8 B8 C8 9. The predictor formulas are B8" œ B8 C8" 2 a&& 08 &* 08" $( 08# * 08$ b #% 2 œ C8 a&& 18 &* 18" $( 18# * 18$ bÞ #% 2 a* 08" "* 08 & 08" 08# b #% 2 œ C8 a* 18" "* 18 & 18" 18# bÞ #% 8œ& !Þ& "Þ&%$ "Þ"%(%$ 8 œ "! "Þ! !Þ!(!(& "Þ$))& With 08" œ B8" % C8" and 18" œ B8" C8" , the corrector formulas are B8" œ B8 C8" ________________________________________________________________________ page 496 —————————————————————————— CHAPTER 8. —— We use the starting values from the exact solution À >8 B8 C8 8œ! ! "Þ! !Þ! 8œ" !Þ" "Þ"#))$ !Þ""!&( 8œ# !Þ# "Þ$#!%# !Þ#&!)%( 8œ$ !Þ$ "Þ'!!#" !Þ%#*'*' One time step using the predictor-corrector method results in the approximate values: >8 B8 C8 8 œ %a:</b !Þ% "Þ**%%& !Þ''#!'% 8 œ %a-9<b !Þ% "Þ**&#" !Þ''#%%# ________________________________________________________________________ page 497 —————————————————————————— CHAPTER 9. —— Chapter Nine Section 9.1 2a+bÞ Setting x œ 0 /<> results in the algebraic equations Œ &< $ For a nonzero solution, we must have ./>aA < Ib œ <# ' < ) œ ! . The roots of the characteristic equation are <" œ # and <# œ % . For < œ #, the system of equations reduces to $0" œ 0# . The corresponding eigenvector is 0 a"b œ a" ß $bX Þ Substitution of < œ % results in the single equation 0" œ 0# . A corresponding eigenvector is 0 a#b œ a" ß "bX Þ a,b. The eigenvalues are real and positive, hence the critical point is an unstable node. a-ß . b. 0" ! " Œ œ Œ . "< 0# ! ________________________________________________________________________ page 498 —————————————————————————— CHAPTER 9. —— 3a+b. Solution of the ODE requires analysis of the algebraic equations For a nonzero solution, we must have ./>aA < Ib œ <# " œ ! . The roots of the characteristic equation are <" œ " and <# œ " . For < œ " , the system of equations reduces to 0" œ 0# . The corresponding eigenvector is 0 a"b œ a" ß "bX Þ Substitution of < œ " results in the single equation $ 0" 0# œ ! . A corresponding eigenvector is 0 a#b œ a" ß $bX Þ a-ß . b. a,b. The eigenvalues are real, with <" <# ! . Hence the critical point is a saddle. Œ #< $ 0" ! " Œ œ Œ . #< 0# ! ________________________________________________________________________ page 499 —————————————————————————— CHAPTER 9. —— 5a+b. The characteristic equation is given by º "< " & œ <# # < # œ ! Þ $<º The equation has complex roots <" œ " 3 and <# œ " 3. For < œ " 3, the components of the solution vector must satisfy 0" a# 3b0# œ ! . Thus the corresponding eigenvector is 0 a"b œ a# 3 ß "bX Þ Substitution of < œ " 3 results in the single equation 0" a# 3b0# œ ! . A corresponding eigenvector is 0 a#b œ a# 3 ß "bX Þ a,b. The eigenvalues are complex conjugates, with negative real part. Hence the origin is a stable spiral. ________________________________________________________________________ page 500 —————————————————————————— CHAPTER 9. —— a-ß . b. 6a+b. Solution of the ODEs is based on the analysis of the algebraic equations For a nonzero solution, we require that ./>aA < Ib œ <# " œ !. The roots of the characteristic equation are < œ „3 . Setting < œ 3 , the equations are equivalent to 0" a# 3b0# œ ! . The eigenvectors are 0 a"b œ a# 3 ß "bX and 0 a#b œ a# 3 ß "bX Þ a,b. The eigenvalues are purely imaginary. Hence the critical point is a center. Œ #< " 0" ! & Œ œ Œ . #< 0# ! ________________________________________________________________________ page 501 —————————————————————————— CHAPTER 9. —— a-ß . b. 7a+b. Setting x œ 0 /<> results in the algebraic equations Œ $< % a,b. The eigenvalues are complex conjugates, with positive real part. Hence the origin is an unstable spiral. For a nonzero solution, we require that ./>aA < Ib œ <# #< & œ !. The roots of the characteristic equation are < œ " „ #3 . Substituting < œ " #3 , the two equations reduce to a" 3b0" 0# œ ! . The two eigenvectors are 0 a"b œ a" ß " 3bX and 0 a#b œ a" ß " 3bX Þ 0" ! # œ . " < Œ 0# Œ ! ________________________________________________________________________ page 502 —————————————————————————— CHAPTER 9. —— a-ß . b. 8a+b. The characteristic equation is given by º "< " & œ a< "ba< !Þ#&b œ ! , $<º with roots <" œ " and <# œ !Þ#& . For < œ ", the components of the solution vector must satisfy 0# œ ! . Thus the corresponding eigenvector is 0 a"b œ a" ß !bX Þ Substitution of < œ !Þ#& results in the single equation !Þ(& 0" 0# œ ! . A corresponding eigenvector is 0 a#b œ a% ß $bX Þ a,b. The eigenvalues are real and both negative. Hence the critical point is a stable page 503 ________________________________________________________________________ —————————————————————————— CHAPTER 9. —— nodeÞ a-ß . b. 9a+b. Solution of the ODEs is based on the analysis of the algebraic equations For a nonzero solution, we require that ./>aA < Ib œ <# # < " œ !. The single root of the characteristic equation is < œ " . Setting < œ " , the components of the solution vector must satisfy 0" # 0# œ ! . A corresponding eigenvector is 0 œ a # ß "b X Þ a,b. Since there is only one linearly independent eigenvector, the critical point is an ?8=>+,6/, improper node. Œ $< " 0" ! % Œ œ Œ . "< 0# ! ________________________________________________________________________ page 504 —————————————————————————— CHAPTER 9. —— a-ß . bÞ 10a+b. The characteristic equation is given by "< º & # œ <# * œ ! Þ "<º The equation has complex roots <"ß# œ „ $3. For < œ $3, the components of the solution vector must satisfy & 0" a" $3b0# œ ! . Thus the corresponding eigenvector is 0 a"b œ a" $3 ß &bX Þ Substitution of < œ $3 results in & 0" a" $3b0# œ ! . A corresponding eigenvector is 0 a#b œ a" $3 ß &bX Þ a,b. The eigenvalues are purely imaginary, hence the critical point is a center. ________________________________________________________________________ page 505 —————————————————————————— CHAPTER 9. —— a-ß . bÞ a,bÞ Since there are two linearly independent eigenvectors, the critical point is a stable proper node. a-ß . b. 11a+b. The characteristic equation is a< "b# œ ! , with double root < œ " . It is easy to see that the two linearly independent eigenvectors are 0 a"b œ a" ß !bX and 0 a#b œ a! ß "bX Þ ________________________________________________________________________ page 506 —————————————————————————— CHAPTER 9. —— 12a+b. Setting x œ 0 /<> results in the algebraic equations #< Œ *Î& For a nonzero solution, we require that ./>aA < Ib œ <# < &Î# œ !. The roots of the characteristic equation are < œ "Î# „ $3Î# . Substituting < œ "Î# $3Î# , the equations reduce to a$ $3b0" & 0# œ ! . Therefore the two eigenvectors are 0 a"b œ a& ß $ $3bX and 0 a#b œ a& ß $ $3bX Þ a,b. Since the eigenvalues are complex, with positive real part, the critical point is an unstable spiral. a-ß . b. 0" ! &Î# Œ œ Œ . "< 0# ! ________________________________________________________________________ page 507 —————————————————————————— CHAPTER 9. —— 14. Setting x w œ 0 , that is, # Œ" # " xœŒ , # " we find that the critical point is x! œ a "ß !bX Þ With the change of dependent variable, x œ x! u , the differential equation can be written as .u # œŒ " .> " u. # The critical point for the transformed equation is the origin. Setting u œ 0 /<> results in the algebraic equations Œ #< " 0" ! " Œ œ Œ . #< 0# ! For a nonzero solution, we require that ./>aA < Ib œ <# %< $ œ !. The roots of the characteristic equation are < œ $ , " . Hence the critical point is a stable node. 15. Setting x w œ 0 , that is, " Œ# " " x œ Œ , " & we find that the critical point is x! œ a #ß "bX Þ With the change of dependent variable, x œ x! u , the differential equation can be written as .u " œŒ # .> " u. " The characteristic equation is ./>aA < Ib œ <# #< $ œ !, with complex conjugate roots < œ " „ 3È# Þ Since the real parts of the eigenvalues are negative, the critical point is a stable spiral. 16. The critical point x! satisfies the system of equations ________________________________________________________________________ page 508 —————————————————————————— CHAPTER 9. —— ! Œ$ ! " x œ Œ Þ ! # It follows that B! œ # Î$ and C ! œ !Î" . Using the transformation, x œ x! u , the differential equation can be written as .u ! œŒ $ .> " u. ! The characteristic equation is ./>aA < Ib œ <# " $ œ !Þ Since " $ ! , the roots are purely imaginary, with < œ „ 3È"$ Þ Hence the critical point is a center. 20Þ The system of ODEs can be written as .x + œ Œ "" +#" .> +"# x. +## The characteristic equation is <# : < ; œ !Þ The roots are given by <"ß# œ : „ È:# %; : „ È? œ . # # The results can be verified using Table *Þ"Þ" . 21a+b. If ; ! and : ! , then the roots are either complex conjugates with negative real parts, or both real and negative. a,b. If ; ! and : œ ! , then the roots are purely imaginary. a- b. If ; ! , then the roots are real, with <" † <# ! . If : ! , then either the roots are real, with <" † <# ! or the roots are complex conjugates with positive real parts. ________________________________________________________________________ page 509 —————————————————————————— CHAPTER 9. —— Section 9.2 2. The differential equations can be combined to obtain a related ODE .C #C œ Þ .B B The equation is separable, with .C # .B œ Þ C B The solution is given by C œ G B# Þ Note that the system is uncoupled, and hence we also have B œ B! /> and C œ C! /#> . In order to determine the direction of motion along the trajectories, observe that for positive initial conditions, B will decrease, whereas C will increase. 4. The trajectories of the system satisfy the ODE .C ,B œ Þ .B +C The equation is separable, with +C .C œ ,B .B Þ Hence the trajectories are given by , B# + C # œ G # , in which G is arbitrary. Evidently, the trajectories are ellipses. Invoking the initial condition, we find that G # œ +, . The system of ODEs can also be written as .x ! œŒ , .> + x. ! Using the methods in Chapter (, it is easy to show that B œ È+ -9= È+, > C œ È, =38 È+, > Þ ________________________________________________________________________ page 510 —————————————————————————— CHAPTER 9. —— 5a+b. The critical points are given by the solution set of the equations Ba" Cb œ ! Ca" #Bb œ ! Þ Note that for positive initial conditions, B will 38crease, whereas C will ./crease. Clearly, a! ß !b is a solution. If B Á ! , then C œ " and B œ "Î# . Hence the critical points are a! ß !b and a "Î# ß "b. a, b . a- b. Based on the phase portrait, all trajectories starting near the origin diverge. Hence the critical point a! ß !b is unstable. Examining the phase curves near the critical point a "Î# ß "b, ________________________________________________________________________ page 511 —————————————————————————— CHAPTER 9. —— the equilibrium point has the properties of a saddle, and hence it is unstable. 6a+b. The critical points are solutions of the equations There are two equilibrium points, Š "ÎÈ$ ß "Î#‹ and Š"ÎÈ$ ß "Î#‹Þ a, bÞ " #C œ ! " $B# œ ! Þ a- b. Locally, the trajectories near the point Š "ÎÈ$ ß "Î#‹ resemble the behavior near a saddle. Hence the critical point is unstable. Near the point Š"ÎÈ$ ß "Î#‹, the solutions are periodic. Therefore the second critical point is stable. 8a+b. The critical points are solutions of the equations aB C ba" B C b œ ! Ba# Cb œ ! Þ If B œ C , then B œ C œ ! or B œ C œ # Þ If B œ " C , then B œ ! and C œ " , or B œ $ and C œ # Þ It follows that the critical points are a! ß !bß a # ß #bß a! ß "b ________________________________________________________________________ page 512 —————————————————————————— CHAPTER 9. —— and a$ ß #bÞ a, bÞ a- b. Near the origin, the trajectories resemble those of a saddle, and hence it is unstable. Near the critical point a! ß "b , the trajectories resemble those of a stable spiral. Hence the equilibrium point is asymptotically stable. Based on the global phase portrait, it is evident that the other critical points are nodes. Closer examination reveals that the point a # ß #b is asymptotically stable, whereas the point a$ ß #b is unstable. ________________________________________________________________________ page 513 —————————————————————————— CHAPTER 9. —— 9a+b. The critical points are given by the solution set of the equations Clearly, a! ß !b is a critical point. If B œ # C , then it follows that C aC #b œ " . The additional critical points are Š" È# ß " È# ‹ and Š" È# ß " È# ‹Þ a, b . C a# B C b œ ! B C #BC œ ! Þ a- b. The behavior near the origin is that of a stable spiral. Hence the point a! ß !b is asymptotically stable. At the critical point Š" È# ß " È# ‹, the trajectories resemble those near a saddle. Hence the critical point is unstable. ________________________________________________________________________ page 514 —————————————————————————— CHAPTER 9. —— Near the point Š" È# ß " È# ‹, the trajectories resemble those near a saddle. Hence the critical point is also unstable. 10a+bÞ The critical points are solutions of the equations a# BbaC Bb œ ! Cˆ# B B# ‰ œ ! Þ The origin is evidently a critical point. If B œ # , then C œ ! . If B œ C , then either C œ ! or B œ C œ " or B œ C œ # . Hence the other critical points are a # ß !b, a " ß "b and a# ß #b. ________________________________________________________________________ page 515 —————————————————————————— CHAPTER 9. —— a, b . a- b. Based on the global phase portrait, the critical points a! ß !b and a # ß !b have the characteristics of a saddle. Hence these points are unstable. The behavior near the remaining two critical points resembles those near a stable spiral. Hence the critical points a " ß "b and a# ß #b are asymptotically stable. 11a+b. The critical points are given by the solution set of the equations Ba" #Cb œ ! C B# C # œ ! Þ a, b . If B œ ! , then either C œ ! or C œ " . If C œ "Î# , then B œ „"Î# . Hence the critical points are at a! ß !b, a! ß "b, a "Î# ß "Î#b and a"Î# ß "Î#b. a- b. The trajectories near the critical points a "Î# ß "Î#b and a"Î# ß "Î#b are closed curves. Hence the critical points have the characteristics of a center, which is stable. The trajectories near the critical points a! ß !b and a!ß "b resemble those near a saddle. Hence these critical points are unstable. ________________________________________________________________________ page 516 —————————————————————————— CHAPTER 9. —— 13a+b. The critical points are solutions of the equations a# BbaC Bb œ ! a% BbaC Bb œ ! Þ If C œ B , then either B œ C œ ! or B œ C œ % Þ If B œ # , then C œ # . If B œ C , then C œ # or C œ ! . Hence the critical points are at a! ß !b, a% ß %b and a # ß #b. a, b . a- b. The critical point at a% ß %b is evidently a stable spiral, which is asymptotically stable. Closer examination of the critical point at a! ß !b reveals that it is a saddle, which is unstable. The trajectories near the critical point a # ß #b resemble those near an unstable node. ________________________________________________________________________ page 517 —————————————————————————— CHAPTER 9. —— 14a+b. The critical points consist of the solution set of the equations It is easy to see that the only critical point is at a! ß !b. a, b . Cœ! ˆ" B C B œ ! Þ #‰ a- b. The origin is an unstable spiral. 16a+b. The trajectories are solutions of the differential equation .C %B œ , .B C which can also be written as %B .B C .C œ ! . Integrating, we obtain %B# C # œ G # . Hence the trajectories are ellipses. ________________________________________________________________________ page 518 —————————————————————————— CHAPTER 9. —— a, b . Based on the differential equations, the direction of motion on each trajectory is clockwise. 17a+b. The trajectories of the system satisfy the ODE which can also be written as a#B Cb.B C.C œ ! . This differential equation is homogeneous. Setting C œ B @aBb, we obtain @B that is, B .@ # @ @# œ Þ .B @ .@ # œ ", .B @ .C #B C œ , .B C The resulting ODE is separable, with solution B$ a@ "ba@ #b# œ G . Reverting back to the original variables, the trajectories are level curves of L aB ß C b œ aB C baC #Bb# Þ ________________________________________________________________________ page 519 —————————————————————————— CHAPTER 9. —— a, b . The origin is a saddle. Along the line C œ #B , solutions increase without bound. Along the line C œ B , solutions converge toward the origin. 18a+b. The trajectories are solutions of the differential equation which is homogeneous. Setting C œ B @aBb, we obtain @B that is, B .@ " @# œ Þ .B "@ .@ B B@ œ , .B B B@ .C BC œ , .B BC The resulting ODE is separable, with solution +<->+8a@b œ 68kBkÈ" @# Þ Reverting back to the original variables, the trajectories are level curves of L aB ß C b œ +<->+8aCÎBb 68ÈB# C # Þ ________________________________________________________________________ page 520 —————————————————————————— CHAPTER 9. —— a, b . The origin is a stable spiral. 20a+b. The trajectories are solutions of the differential equation which can also be written as a#BC# ' BCb.B a#B# C $B# %C b.C œ ! . The resulting ODE is exact, with `L `L œ #BC# ' BC and œ #B# C $B# %C . `B `C .C #BC # ' BC œ# , .B #B C $B# %C Integrating the first equation, we find that L aB ß C b œ B# C # $B# C 0 aC b. It follows that `L œ #B# C $B# 0 w aC b. `C Comparing the two partial derivatives, we obtain 0 aCb œ #C # - . Hence L aB ß C b œ B# C # $B# C #C # Þ ________________________________________________________________________ page 521 —————————————————————————— CHAPTER 9. —— a, b . The associated direction field shows the direction of motion along the trajectories. 22a+b. The trajectories are solutions of the differential equation which can also be written as a' B B$ b.B ' C.C œ ! . The resulting ODE is exact, with `L `L œ ' B B$ and œ 'C. `B `C `L œ 0 w aC b . `C .C ' B B$ , œ .B 'C Integrating the first equation, we have L aB ß Cb œ $B# B% Î% 0 aC b. It follows that Comparing the two partial derivatives, we conclude that 0 aCb œ $C # - . Hence ________________________________________________________________________ page 522 —————————————————————————— CHAPTER 9. —— L aB ß C b œ $B# a, b . B% $C # Þ % ________________________________________________________________________ page 523 —————————————————————————— CHAPTER 9. —— Section 9.3 1. Write the system in the form x w œ Ax gaxb. In this case, it is evident that That is, gaxb œ a C # ß B# bX Þ Using polar coordinates, lgaxbl œ <# È=38% ) -9=% ) and lxl œ < . Hence lgaxbl œ lim <È=38% ) -9=% ) œ ! , < Ä ! lxl <Ä! lim .B " Œ œ Œ" .> C B C# ! Œ # Þ # Œ C B and the system is almost linear. The origin is an isolated critical point of the linear system .B " Œ œ Œ" .> C B ! Œ Þ # C The characteristic equation of the coefficient matrix is <# < # œ ! , with roots <" œ " and <# œ # . Hence the critical point is a saddle, which is unstable. 2. The system can be written as .B " Œ œŒ % .> C B #BC " Þ " Œ C Œ B# C # Following the discussion in Example $ , we note that J aB ß Cb œ B C #BC and KaB ß Cb œ %B C B# C # . Both of the functions J and K are twice differentiable, hence the system is almost linear. Furthermore, JB œ " #C , JC œ " #B , KB œ % #B , KC œ " #C . The origin is an isolated critical point, with J B a ! ß !b Œ K a ! ß !b B J C a ! ß !b " œ KC a! ß !b Œ % " . " The characteristic equation of the associated linear system is <# # < & œ ! , with complex conjugate roots <"ß# œ " „ #3 Þ The origin is a stable spiral, which is asymptotically stable. 5a+b. The critical points consist of the solution set of the equations As shown in Prob. "$ of Section *Þ# , the only critical points are at a! ß !b, a% ß %b and a # ß #b . ________________________________________________________________________ page 524 a# BbaC Bb œ ! a% BbaC Bb œ ! Þ —————————————————————————— CHAPTER 9. —— a,ß - b. First note that J aB ß C b œ a# BbaC Bb and KaB ß C b œ a% BbaC Bb . The Jacobian matrix of the vector field is JœŒ JB a B ß C b KB aB ß Cb J C aB ß C b # #B C œ Œ % C #B KC a B ß C b Ja! ß !b œ Œ # % # , % #B . % B At the origin, the coefficient matrix of the linearized system is with eigenvalues <" œ " È"( and <# œ " È"( . The eigenvalues are real, with opposite sign. Hence the critical point is a saddle, which is unstable. At the equilibrium point a # ß #b, the coefficient matrix of the linearized system is Ja # ß #b œ Œ % ' ! , ' with eigenvalues <" œ % and <# œ ' . The eigenvalues are real, unequal and positive, hence the critical point is an unstable node. At the point a% ß %b, the coefficient matrix of the linearized system is Ja% ß %b œ Œ ' ) ' , ! with complex conjugate eigenvalues <"ß# œ $ „ 3È$* . The critical point is a stable spiral, which is asymptotically stable. Based on Table *Þ$Þ" , the nonlinear terms do not affect the stability and type of each critical point. a. b . 7a+bÞ The critical points are solutions of the equations ________________________________________________________________________ page 525 —————————————————————————— CHAPTER 9. —— "C œ! aB CbaB Cb œ ! Þ The first equation requires that C œ " . Based on the second equation, B œ „ " . Hence the critical points are a " ß "b and a" ß "b. JB a B ß C b KB aB ß Cb J C aB ß C b ! œ KC aB ß Cb Œ #B ! # a,ß - b. J aB ß C b œ " C and KaB ß C b œ B# C # . The Jacobian matrix of the vector field is JœŒ " . #C At the critical point a " ß "b, the coefficient matrix of the linearized system is with eigenvalues <" œ " È$ and <# œ " È$ . The eigenvalues are real, with opposite sign. Hence the critical point is a saddle, which is unstable. At the equilibrium point a" ß "b, the coefficient matrix of the linearized system is Ja" ß "b œ Œ ! # " , # Ja " ß "b œ Œ " , # with complex conjugate eigenvalues <"ß# œ " „ 3 . The critical point is a stable spiral, which is asymptotically stable. a. b . Based on Table *Þ$Þ" , the nonlinear terms do not affect the stability and type of each critical point. 8a+b. The critical points are given by the solution set of the equations ________________________________________________________________________ page 526 —————————————————————————— CHAPTER 9. —— Ba" B Cb œ ! Ca# C $Bb œ ! Þ If B œ ! , then either C œ ! or C œ # . If C œ ! , then B œ ! or B œ " . If C œ " B , then either B œ "Î# or B œ ". If C œ # $B , then B œ ! or B œ "Î# . Hence the critical points are at a! ß !b, a! ß #b, a" ß !b and a"Î# ß "Î#bÞ a,ß - b. Note that J aB ß C b œ B B# BC and KaB ß C b œ a#C C # $BC bÎ% . The Jacobian matrix of the vector field is JœŒ JB a B ß C b KB aB ß Cb J C aB ß C b " #B C œ Œ $CÎ% KC a B ß C b Ja! ß !b œ Œ " ! ! " # B . "Î# CÎ# $BÎ% At the origin, the coefficient matrix of the linearized system is , with eigenvalues <" œ " and <# œ "Î# . The eigenvalues are real and both positive. Hence the critical point is an unstable node. At the equilibrium point a! ß #b, the coefficient matrix of the linearized system is Ja! ß #b œ Œ " $ # ! , " # with eigenvalues <" œ " and <# œ "Î# . The eigenvalues are both negative, hence the critical point is a stable node. At the point a" ß !b, the coefficient matrix of the linearized system is Ja" ß !b œ Œ " ! " , " % with eigenvalues <" œ " and <# œ "Î% . Both of the eigenvalues are negative, and hence the critical point is a stable node. At the critical point a"Î# ß "Î#b, the coefficient matrix of the linearized system is Ja"Î# ß "Î#b œ " # $ ) with eigenvalues <" œ &Î"' È&( Î"' and <# œ &Î"' È&( Î"' . The eigenvalues are real, with opposite sign. Hence the critical point is a saddle, which is unstable. " # , " ) ________________________________________________________________________ page 527 —————————————————————————— CHAPTER 9. —— a. b . Based on Table *Þ$Þ" , the nonlinear terms do not affect the stability and type of each critical point. 9a+b. Based on Prob. ) , in Section *Þ# , the critical points are at a! ß !bß a # ß #bß a! ß "b and a$ ß #bÞ a,ß - bÞ First note that J aB ß C b œ aB C ba" B C b and KaB ß C b œ Ba# C b . The Jacobian matrix of the vector field is JœŒ #B " #C " #C . B At the origin, the coefficient matrix of the linearized system is ________________________________________________________________________ page 528 —————————————————————————— CHAPTER 9. —— Ja! ß !b œ Œ " , ! " # with eigenvalues <" œ " and <# œ # . The eigenvalues are real, with opposite sign. Hence the critical point is a saddle, which is unstable. At the critical point a! ß "b, the coefficient matrix of the linearized system is Ja! ß "b œ Œ " $ " , ! with complex conjugate eigenvalues <"ß# œ "Î# „ 3È"" Î# . The critical point is a stable spiral, which is asymptotically stable. At the point a # ß #b, the coefficient matrix of the linearized system is Ja # ß #b œ Œ & ! & , # with eigenvalues <" œ # and <# œ & . The eigenvalues are unequal and negative, hence the critical point is a stable node. At the point a$ ß #b, the coefficient matrix of the linearized system is Ja$ ß #b œ Œ & ! & , $ with eigenvalues <" œ $ and <# œ & . The eigenvalues are unequal and positive, hence the critical point is an unstable node. a. b . Based on Table *Þ$Þ" , the nonlinear terms do not affect the stability and type of each critical point. ________________________________________________________________________ page 529 —————————————————————————— CHAPTER 9. —— 11a+b. The critical points are solutions of the equations Substitution of C œ BÎaB #b into the first equation results in $B% "$B$ #)B# #!B œ ! . One root of the resulting equation is B œ ! . The only other real root of the equation is Bœ "Î$ "Î$ " "$•. ”Š#)( ")È#!"* ‹ )$Š#)( ")È#!"* ‹ * #B C BC $ œ ! B #C BC œ ! Þ a,ß - b. J aB ß C b œ B B# BC and KaB ß C b œ a#C C # $BC bÎ% . The Jacobian matrix of the vector field is JœŒ JB a B ß C b KB aB ß Cb J C aB ß C b # C$ œŒ "C KC a B ß C b # " " , # " $BC # . #B Hence the critical points are a! ß !b and a "Þ"*$%&ÞÞÞ ß "Þ%(*(ÞÞÞbÞ At the origin, the coefficient matrix of the linearized system is Ja! ß !b œ Œ with eigenvalues <" œ È& and <# œ È& . The eigenvalues are real and of opposite sign. Hence the critical point is a saddle, which is unstable. At the equilibrium point a "Þ"*$%&ÞÞÞ ß "Þ%(*(ÞÞÞb, the coefficient matrix of the linearized system is Ja "Þ"*$%& ß "Þ%(*(b œ Œ "Þ#$** #Þ%(*( 'Þ)$*$ , !Þ)!'& with complex conjugate eigenvalues <"ß# œ "Þ!#$# „ %Þ""#& 3 . The critical point is a stable spiral, which is asymptotically stable. ________________________________________________________________________ page 530 —————————————————————————— CHAPTER 9. —— a. b . In both cases, the nonlinear terms do not affect the stability and type of the critical point. 12a+b. The critical points are given by the solution set of the equations a" Bb=38 C œ ! " B -9= C œ ! Þ If B œ " , then we must have -9= C œ # , which is impossible. Therefore =38 C œ ! , which implies that C œ 81 , 8 œ ! ß „ " ß # ß ÞÞÞ Þ Based on the second equation, It follows that the critical points are located at a! ß #5 1b and a# ß a#5 "b1b , where 5 œ ! ß „ " ß # ß ÞÞÞ Þ a,ß - b. Given that J aB ß Cb œ a" Bb=38 C and KaB ß C b œ " B -9= C , the Jacobian matrix of the vector field is JœŒ =38 C " a" Bb-9= C . =38 C " , ! B œ " -9= 81 Þ At the critical points a! ß #5 1b, the coefficient matrix of the linearized system is Ja! ß #5 1b œ Œ ! " with purely complex eigenvalues <"ß# œ „ 3 . The critical points of the associated linear systems are centers, which are stable. Note that Theorem *Þ$Þ# does not provide a definite conclusion regarding the relation between the nature of the critical points of the nonlinear systems and their corresponding linearizations. At the points a# ß a#5 "b1b, the coefficient matrix of the linearized system is Jc# ß a#5 "b1d œ Œ ! " $ , ! ________________________________________________________________________ page 531 —————————————————————————— CHAPTER 9. —— with eigenvalues <" œ È$ and <# œ È$ . The eigenvalues are real, with opposite sign. Hence the critical points of the associated linear systems are saddles, which are unstable. a. b . As asserted in Theorem *Þ$Þ# , the trajectories near the critical points a# ß a#5 "b1b resemble those near a saddle. Upon closer examination, the critical points a! ß #5 1b are indeed centers. 13a+b. The critical points are solutions of the equations ________________________________________________________________________ page 532 —————————————————————————— CHAPTER 9. —— B C# œ ! C B# œ ! Þ Substitution of C œ B# into the first equation results in with real roots B œ ! , " . Hence the critical points are at a! ß !b and a" ß "b. B B% œ ! , a,ß - b. In this problem, J aB ß C b œ B C # and KaB ß C b œ C B# . The Jacobian matrix of the vector field is JœŒ " #B #C . " ! , " At the origin, the coefficient matrix of the linearized system is Ja! ß !b œ Œ " ! with repeated eigenvalues <" œ " and <# œ " . It is easy to see that the corresponding eigenvectors are linearly independent. Hence the critical point is an unstable proper node. Theorem *Þ$Þ# does not provide a definite conclusion regarding the relation between the nature of the critical point of the nonlinear system and the corresponding linearization. At the critical point a" ß "b, the coefficient matrix of the linearized system is Ja" ß "b œ Œ " # # , " with eigenvalues <" œ $ and <# œ " . The eigenvalues are real, with opposite sign. Hence the critical point is a saddle, which is unstable. a. b . Closer examination reveals that the critical point at the origin is indeed a proper node. ________________________________________________________________________ page 533 —————————————————————————— CHAPTER 9. —— 14a+b. The critical points are given by the solution set of the equations " BC œ ! B C$ œ ! Þ After multiplying the second equation by C , it follows that C œ „ " . Hence the critical points of the system are at a" ß "b and a " ß "b. a,ß - b. Note that J aB ß C b œ " BC and KaB ß C b œ B C $ . The Jacobian matrix of the vector field is JœŒ C " B . $C# At the critical point a" ß "b, the coefficient matrix of the linearized system is Ja" ß "b œ Œ " " " , $ with eigenvalues <" œ # and <# œ # . The eigenvalues are real and equal. It is easy to show that there is only one linearly independent eigenvector. Hence the critical point is a stable improper node. Theorem *Þ$Þ# does not provide a definite conclusion regarding the relation between the nature of the critical point of the nonlinear system and the corresponding linearization. At the point a " ß "b, the coefficient matrix of the linearized system is Ja " ß "b œ Œ " " " , $ with eigenvalues <" œ " È& and <# œ " È& . The eigenvalues are real, with opposite sign. Hence the critical point of the associated linear system is a saddle, which is unstable. ________________________________________________________________________ page 534 —————————————————————————— CHAPTER 9. —— a. b . Closer examination reveals that the critical point at a" ß "b is indeed a stable improper node, which is asymptotically stable. 15a+b. The critical points are given by the solution set of the equations #B C BˆB# C # ‰ œ ! B C CˆB# C # ‰ œ ! Þ It is clear that the origin is a critical point. Solving the first equation for C , we find that Cœ " „È" )B# %B% Þ #B Substitution of these relations into the second equation results in two equations of the form 0" aBb œ ! and 0# aBb œ ! . Plotting these functions, we note that only 0" aBb œ ! has real roots given by B ¸ „ !Þ$$!(' Þ It follows that the additional critical points are at a !Þ$$!(' ß "Þ!*#%b and a !Þ$$!(' ß "Þ!*#%bÞ ________________________________________________________________________ page 535 —————————————————————————— CHAPTER 9. —— a,ß - b. Given that J aB ß Cb œ #B C BˆB# C # ‰ KaB ß Cb œ B C C ˆB# C # ‰, JœŒ # $B# C # " #BC Ja! ß !b œ Œ " #BC . " B# $C # " , " the Jacobian matrix of the vector field is At the critical point a! ß !b, the coefficient matrix of the linearized system is with complex conjugate eigenvalues <"ß# œ Š $ „ 3È$ ‹Î# . Hence the critical point is a stable spiral, which is asymptotically stable. At the point a !Þ$$!(' ß "Þ!*#%b, the coefficient matrix of the linearized system is Ja !Þ$$!(' ß "Þ!*#%b œ Œ $Þ&#"' !Þ#(($& !Þ#(($& , #Þ')*& # " a. b . with eigenvalues <" œ $Þ&!*# and <# œ #Þ'((" . The eigenvalues are real, with opposite sign. Hence the critical point of the associated linear system is a saddle, which is unstable. Identical results hold for the point at a !Þ$$!(' ß "Þ!*#%b . A closer look at the origin reveals a spiral: ________________________________________________________________________ page 536 —————————————————————————— CHAPTER 9. —— Near the point a !Þ$$!(' ß "Þ!*#%b the nature of the critical point is evident: Based on Table *Þ$Þ" , the nonlinear terms do not affect the stability and type of each critical point. 16a+b. The critical points are solutions of the equations C Bˆ" B# C # ‰ œ ! B Cˆ" B# C # ‰ œ ! Þ Multiply the first equation by C and the second equation by B . The difference of the two equations gives B# C # œ ! . Hence the only critical point is at the origin. a,ß - b. With J aB ß C b œ C Ba" B# C # b and KaB ß C b œ B C a" B# C # b, the Jacobian matrix of the vector field is JœŒ " $B# C # " #BC Ja! ß !b œ Œ " #BC . " B# $C # " , " At the origin, the coefficient matrix of the linearized system is " " with complex conjugate eigenvalues <"ß# œ " „ 3 . Hence the origin is an unstable ________________________________________________________________________ page 537 —————————————————————————— CHAPTER 9. —— spiral. a. bÞ 17a+b. The Jacobian matrix of the vector field is JœŒ ! " 'B# " . ! ! , " At the origin, the coefficient matrix of the linearized system is Ja! ß !b œ Œ " ! with eigenvalues <" œ " and <# œ " . The eigenvalues are real, with opposite sign. Hence the critical point is a saddle point. a,b. The trajectories of the linearized system are solutions of the differential equation .C B œ, .B C which is separable. Integrating both sides of the equation B .B C .C œ ! , the solution is B# C # œ G . The trajectories consist of a family of hyperbolas. ________________________________________________________________________ page 538 —————————————————————————— CHAPTER 9. —— It is easy to show that the general solution is given by Ba>b œ -" /> -# /> and Ca>b œ -" /> -# /> Þ The only bounded solutions consist of those for which -" œ ! . In that case, Ba>b œ -# /> œ C a>b. a- b. The trajectories of the given system are solutions of the differential equation .C B #B$ œ , .B C which can also be written as aB #B$ b.B C .C œ ! . The resulting ODE is exact, with `L `L œ B #B$ and œ C. `B `C Integrating the first equation, we find that L aB ß C b œ B# Î# B% Î# 0 aC b. It follows that `L œ 0 w aC b . `C Comparing the partial derivatives, we obtain 0 aCb œ C # Î# - . Hence the solutions are level curves of the function L aB ß Cb œ B# Î# B% Î# C # Î# Þ The trajectories approaching to, or diverging from, the origin are no longer straight lines. ________________________________________________________________________ page 539 —————————————————————————— CHAPTER 9. —— 19a+b. The solutions of the system of equations consist of the points a„ 81 ß !b , 8 œ ! ß " ß # ß â Þ The functions J aB ß C b œ C and KaB ß Cb œ =# =38 B are analytic on the entire plane. It follows that the system is almost linear near each of the critical points. a,b. The Jacobian matrix of the vector field is JœŒ # Cœ! = =38 B œ ! # ! " . = -9= B ! " , ! At the origin, the coefficient matrix of the linearized system is Ja! ß !b œ Œ ! =# with purely complex eigenvalues <"ß# œ „ 3= . Hence the origin is a center. Since the eigenvalues are purely complex, Theorem *Þ$Þ# gives no definite conclusion about the critical point of the nonlinear system. Physically, the critical point corresponds to the state ) œ ! , ) w œ ! Þ That is, the rest configuration of the pendulum. a- b. At the critical point a1 ß !b, the coefficient matrix of the linearized system is Ja1 ß !b œ Œ ! =# " , ! with eigenvalues <"ß# œ „ = . The eigenvalues are real and of opposite sign. Hence the critical point is a saddle. Theorem *Þ$Þ# asserts that the critical point for the nonlinear system is also a saddle, which is unstable. This critical point corresponds to the state ) œ 1 , ) w œ ! Þ That is, the upright rest configuration. ________________________________________________________________________ page 540 —————————————————————————— CHAPTER 9. —— a. b. Let =# œ " . The following is a plot of the phase curves near a! ß !bÞ The local phase portrait shows that the origin is indeed a center. a/ b . It should be noted that the phase portrait has a periodic pattern, since ) œ B 79. #1 . 20a+b. The trajectories of the system in Problem "* are solutions of the differential equation .C =# =38 B œ , .B C which can also be written as =# =38 B .B C .C œ ! . The resulting ODE is exact, with `L `L œ =# =38 B and œ C. `B `C Integrating the first equation, we find that L aB ß C b œ =# -9= B 0 aC b. It follows that ________________________________________________________________________ page 541 —————————————————————————— CHAPTER 9. —— `L œ 0 w aC b . `C Comparing the partial derivatives, we obtain 0 aCb œ C # Î# G . Hence the solutions are level curves of the function Adding an arbitrary constant, say =# , to the function L aB ß C b does not change the nature of the level curves. Hence the trajectories are can be written as "# C =# a" -9= Bb œ - , # in which - is an arbitrary constant. a,b. Multiplying by 7P# and reverting to the original physical variables, we obtain " .) 7P# Œ 7P# =# a" -9= ) b œ 7P# - Þ # .> # L aB ß C b œ =# -9= B C # Î# Þ Since =# œ 1ÎP , the equation can be written as " .) 7P# Œ 71Pa" -9= ) b œ I , # .> # in which I œ 7P# - . a- b. The absolute velocity of the point mass is given by @ œ P . )Î.> . The kinetic energy of the mass is X œ 7@# Î# . Choosing the rest position as the datum, that is, the level of zero potential energy, the gravitational potential energy of the point mass is Z œ 71Pa" -9= ) bÞ It follows that the total energy, X Z , is constant along the trajectories. 21a+b. E œ !Þ#& ________________________________________________________________________ page 542 —————————————————————————— CHAPTER 9. —— Since the system is undamped, and Ca!b œ ! , the amplitude is !Þ#& . The period is estimated at 7 ¸ $Þ"' . a, b . E œ !Þ& E œ "Þ! E œ "Þ& E œ #Þ! V !Þ& "Þ! "Þ& #Þ! 7 $Þ#! $Þ$& $Þ'$ %Þ"( a- bÞ Since the system is conservative, the amplitude is equal to the initial amplitude. On ________________________________________________________________________ page 543 —————————————————————————— CHAPTER 9. —— the other hand, the period of the pendulum is a monotone increasing function of the initial position E . It appears that as E p ! , the period approaches 1, the period of the corresponding linear pendulum a#1Î=b. a. b . The pendulum is released from rest, at an inclination of % 1 radians from the vertical. Based on conservation of energy, the pendulum will swing past the lower equilibrium position a) œ #1b and come to rest, momentarily, at a maximum rotational displacement of )7+B œ $1 a% 1b œ %1 % . The transition between the two dynamics occurs at E œ 1 , that is, once the pendulum is released beyond the upright configuration. 24a+b. It is evident that the origin is a critical point of each system. Furthermore, it is easy to see that the corresponding linear system, in each case, is given by .B œC .> .C œ BÞ .> The eigenvalues of the coefficient matrix are <"ß# œ „ 3 . Hence the critical point of the ________________________________________________________________________ page 544 —————————————————————————— CHAPTER 9. —— linearized system is a center. a,b. Using polar coordinates, it is also easy to show that lgaxbl œ !. < Ä ! lxl lim Alternatively, the nonlinear terms are analytic in the entire plane. Hence both systems are almost linear near the origin. a- b. For system a33b, note that B .B .C C œ BC B# ˆB# C # ‰ BC C # ˆB# C # ‰Þ .> .> Converting to polar coordinates, and differentiating the equation <# œ B# C # with respect to > , we find that < .< .B .C # œB C œ ˆB# C # ‰ œ <% Þ .> .> .> # That is, < w œ <$ Þ It follows that <# œ "Îa#> - b, where - œ "Î<! . Since < p ! as > p ! , regardless of the value of <! , the origin is an +=C7:>9>3-+66C =>+,6/ equilibrium point. On the other hand, for system a3b, < .< .B .C # œB C œ ˆB# C # ‰ œ <% Þ .> .> .> - #> That is, < w œ <$ Þ Solving the differential equation results in <# œ a#> - b# . Imposing the initial condition <a!b œ <! , we obtain a specific solution <# œ # <! . # # <! > " # Since the solution becomes unbounded as > p "Î#<! , the critical point is unstable. 25. The characteristic equation of the coefficient matrix is <# " œ ! , with complex roots <"ß# œ „ 3 . Hence the critical point at the origin is a center. The characteristic equation of the perturbed matrix is <# # % < " %# œ ! , with complex conjugate roots <"ß# œ % „ 3 . As long as % Á ! , the critical point of the perturbed system is a spiral point. Its stability depends on the sign of % Þ 26. The characteristic equation of the coefficient matrix is a< "b# œ ! , with roots ________________________________________________________________________ page 545 —————————————————————————— CHAPTER 9. —— <" œ <# œ " . Hence the critical point is an asymptotically stable node. On the other hand, the characteristic equation of the perturbed system is <# #< " % œ ! , with roots <"ß# œ " „È % . If % ! , then <"ß# œ " „ 3È% are complex roots. The critical point is a stable spiral. If % ! , then <"ß# œ " „ Èk%k are real and both negative ak%k ¥ "b. The critical point remains a stable node. 27a. b. Set 5 œ =38a!Î#b œ =38aEÎ#b and 1ÎP œ % . ________________________________________________________________________ page 546 —————————————————————————— CHAPTER 9. —— Section 9.4 1a+b. a,bÞ The critical points are solutions of the system of equations The four critical points are a! ß !b, a! ß #b, a"Þ& ß !b and a!Þ) ß "Þ%b. a- b. The Jacobian matrix of the vector field is JœŒ BÎ# . # $BÎ% #C ! . # Ba"Þ& B !Þ& C b œ ! Ca# C !Þ(& Bb œ ! Þ At the critical point a! ß !b, the coefficient matrix of the linearized system is Ja! ß !b œ Œ $Î# ! $Î# #B CÎ# $CÎ% The eigenvalues and eigenvectors are " ! <" œ $Î# , 0 a"b œ Œ ; <# œ # , 0 a#b œ Œ Þ ! " The eigenvalues are positive, hence the origin is an unstable node. At the critical point a! ß #b, the coefficient matrix of the linearized system is Ja! ß #b œ Œ "Î# $Î# ! . # The eigenvalues and eigenvectors are ________________________________________________________________________ page 547 —————————————————————————— CHAPTER 9. —— <" œ "Î# , 0 a"b œ Œ " ! a#b ; <# œ # , 0 œ Œ Þ !Þ' " The eigenvalues are of opposite sign. Hence the critical point is a saddle, which is unstable. At the critical point a"Þ& ß !b, the coefficient matrix of the linearized system is Ja"Þ& ß !b œ Œ "Þ& ! !Þ(& . !Þ)(& The eigenvalues and eigenvectors are " !Þ$"&(* <" œ "Þ& , 0 a"b œ Œ ; <# œ !Þ)(& , 0 a#b œ Œ Þ ! " The eigenvalues are of opposite sign. Hence the critical point is also a saddle, which is unstable. At the critical point a!Þ) ß "Þ%b, the coefficient matrix of the linearized system is Ja!Þ) ß "Þ%b œ Œ !Þ) "Þ!& !Þ% . "Þ% The eigenvalues and eigenvectors are <" œ "" È&" "! "! " " "" È&" , 0 a"b œ $È&" ; <# œ , 0 a#b œ $È&" Þ "! "! % % The eigenvalues are both negative. Hence the critical point is a stable node, which is asymptotically stable. a.ß /b. a0 b. Except for initial conditions lying on the coordinate axes, almost all trajectories ________________________________________________________________________ page 548 —————————————————————————— CHAPTER 9. —— converge to the stable node at a!Þ) ß "Þ%b. 2a+b. a,bÞ The critical points are the solution set of the system of equations The four critical points are a! ß !b, a! ß %b, a"Þ& ß !b and a" ß "b. a- b. The Jacobian matrix of the vector field is JœŒ BÎ# . # $BÎ# C ! . # Ba"Þ& B !Þ& C b œ ! Ca# !Þ& C "Þ& Bb œ ! Þ $Î# #B CÎ# $CÎ# Ja! ß !b œ Œ At the origin, the coefficient matrix of the linearized system is $Î# ! The eigenvalues and eigenvectors are " ! <" œ $Î# , 0 a"b œ Œ ; <# œ # , 0 a#b œ Œ Þ ! " The eigenvalues are positive, hence the origin is an unstable node. At the critical point a! ß %b, the coefficient matrix of the linearized system is Ja! ß %b œ Œ "Î# ' ! . # The eigenvalues and eigenvectors are ________________________________________________________________________ page 549 —————————————————————————— CHAPTER 9. —— <" œ "Î# , 0 a"b œ Œ " ! a#b ; <# œ # , 0 œ Œ Þ % " The eigenvalues are both negative, hence the critical point a! ß %b is a stable node, which is asymptotically stable. At the critical point a$Î# ß !b, the coefficient matrix of the linearized system is Ja$Î# ß !b œ Œ $Î# ! $Î% . "Î% The eigenvalues and eigenvectors are " $ <" œ $Î# , 0 a"b œ Œ ; <# œ "Î% , 0 a#b œ Œ Þ ! & The eigenvalues are both negative, hence the critical point is a stable node, which is asymptotically stable. At the critical point a "ß "b, the coefficient matrix of the linearized system is Ja" ß "b œ Œ " $Î# "Î# . "Î# The eigenvalues and eigenvectors are <" œ $ È"$ % The eigenvalues are of opposite sign, hence a "ß "b is a saddle, which is unstable. a.ß /b. , 0 a"b œ "È"$ # " ; <# œ $ È"$ % ! , 0 a#b œ "È"$ Þ # a0 b. Trajectories approaching the critical point a" ß "b form a separatrix. Solutions on either side of the separatrix approach either a! ß %b or a"Þ& ß !b. ________________________________________________________________________ page 550 —————————————————————————— CHAPTER 9. —— 4a+b. a,bÞ The critical points are solutions of the system of equations The four critical points are a! ß !b, a! ß $Î%b, a$ ß !b and a# ß "Î#b. a- b. The Jacobian matrix of the vector field is JœŒ $Î# B C CÎ) Ja! ß !b œ Œ The eigenvalues and eigenvectors are " ! <" œ $Î# , 0 a"b œ Œ ; <# œ $Î% , 0 a#b œ Œ Þ ! " The eigenvalues are positive, hence the origin is an unstable node. At the critical point a! ß $Î%b, the coefficient matrix of the linearized system is Ja! ß $Î%b œ Œ $Î% $Î$# ! . $Î% B . $Î% BÎ) #C ! . $Î% Ba"Þ& !Þ& B C b œ ! Ca!Þ(& C !Þ"#& Bb œ ! Þ At the origin, the coefficient matrix of the linearized system is $Î# ! The eigenvalues and eigenvectors are <" œ $Î% , 0 a"b œ Œ "' ! a#b ; <# œ $Î% , 0 œ Œ Þ " " The eigenvalues are of opposite sign, hence the critical point a! ß $Î%b is a saddle, which is unstable. ________________________________________________________________________ page 551 —————————————————————————— CHAPTER 9. —— At the critical point a$ ß !b, the coefficient matrix of the linearized system is Ja$ ß !b œ Œ $Î# ! $ . $Î) The eigenvalues and eigenvectors are " ) <" œ $Î# , 0 a"b œ Œ ; <# œ $Î) , 0 a#b œ Œ Þ ! & The eigenvalues are of opposite sign, hence the critical point a! ß $Î%b is a saddle, which is unstable. At the critical point a# ß "Î#b, the coefficient matrix of the linearized system is Ja# ß "Î#b œ Œ " "Î"' # . "Î# $ È$ % The eigenvalues and eigenvectors are <" œ $ È$ % The eigenvalues are negative, hence the critical point a# ß "Î#b is a stable node, which is asymptotically stable. a.ß /bÞ , 0 a"b œ "È$ ) " ; <# œ ! , 0 a#b œ "È$ Þ ) a0 b. Except for initial conditions along the coordinate axes, almost all solutions converge to the stable node a# ß "Î#b. ________________________________________________________________________ page 552 —————————————————————————— CHAPTER 9. —— 7. It follows immediately that # # a5" \ 5# ] b# %5" 5# \] œ 5" \ # #5" 5# \] 5# ] # %5" 5# \] œ a5" \ 5# ] b# Þ Since all parameters and variables are positive, it follows that Hence the radicand in Eq.a$*b is nonnegative. a5" \ 5# ] b# %a5" 5# !" !# b\] ! Þ 10a+b. The critical points consist of the solution set of the equations Ba%" 5" B !" C b œ ! Ca%# 5# C !# Bb œ ! Þ If B œ ! , then either C œ ! or C œ %# Î5# . If %" 5" B !" C œ ! , then solving for B results in B œ a%" !" C bÎ5" . Substitution into the second equation yields Based on the hypothesis, it follows that a5" % # % " !# bC œ ! Þ So if 5" % # % " !# Á ! , then C œ ! , and the critical points are located at a! ß !b, a! ß %# Î5# b and a%" Î5" ß !bÞ a5" 5# !" !# bC# a5" % # % " !# bC œ ! Þ For the case 5" % # % " !# œ !, C can be arbitrary. From the relation B œ a%" !" C bÎ5" , we conclude that all points on the line 5" B !" C œ %" are critical points, in addition to the point a! ß !b. a,b. The Jacobian matrix of the vector field is JœŒ %" #5" B !" C !# C Ja! ß !b œ Œ !" B . %# #5# C !# B ! , %# At the origin, the coefficient matrix of the linearized system is %" ! with eigenvalues <" œ %" and <# œ %# . Since both eigenvalues are positive, the origin is an unstable node. At the point a! ß %# Î5# b, the coefficient matrix of the linearized system is with eigenvalues <" œ a%" !# 5" %# bÎ!# and <# œ %# . If 5" %# %" !# ! , then both eigenvalues are negative. Hence the point a! ß %# Î5# b is a stable node, which is asymptotically stable. If 5" %# %" !# ! , then the eigenvalues are of opposite sign. Hence the point a! ß %# Î5# b is a saddle, which is unstable. ________________________________________________________________________ page 553 Ja! ß %# Î5# b œ Œ a%" !# 5" %# bÎ!# %# !# Î5# ! , %# —————————————————————————— CHAPTER 9. —— At the point a%" Î5" ß !b, the coefficient matrix of the linearized system is with eigenvalues <" œ a5" %# %" !# bÎ5" and <# œ %" . If 5" %# %" !# ! , then the eigenvalues are of opposite sign. Hence the point a%" Î5" ß !b is a saddle, which is unstable. If 5" %# %" !# ! , then both eigenvalues are negative. In that case the point a%" Î5" ß !b is a stable node, which is asymptotically stable. a- b. As shown in Part a+b, when 5" %# %" !# œ ! , the set of critical points consists of a! ß !b and all of the points on the straight line 5" B !" C œ %" . Based on Part a,b, the origin is still an unstable node. Setting C œ a%" 5" BbÎ!" , the Jacobian matrix of the vector field, along the given straight line, is JœŒ 5" B !# a%" 5" BbÎ!" !" B . !# B %" !# Î5" Ja%" Î5" ß !b œ Œ %" ! %" !" Î5" , a5" %# %" !# bÎ5" The characteristic equation of the matrix is <# ” # %" !# !# 5" B 5" B •< œ ! . 5" # Using the given hypothesis, a%" !# !# 5" B 5" BbÎ5" œ % # !# B 5" B. Hence the characteristic equation can be written as <# c% # !# B 5" Bd< œ ! . First note that ! Ÿ B Ÿ %" Î5" . Since the coefficient in the quadratic equation is linear, and % # !# B 5" B œ œ % # at B œ ! % " at B œ %" Î5" , it follows that the coefficient is positive for ! Ÿ B Ÿ %" Î5" . Therefore, along the straight line 5" B !" C œ %" , one eigenvalue is zero and the other one is negative. Hence the continuum of critical points consists of stable nodes, which are asymptotically stable. 11a+b. The critical points are solutions of the system of equations Ba" B Cb $ + œ ! Ca!Þ(& C !Þ& Bb $ , œ ! Þ B œ B! B" $ B# $ # â C œ C! C " $ C # $ # â . Substitution of the series expansions results in ________________________________________________________________________ page 554 Assume solutions of the form —————————————————————————— CHAPTER 9. —— B! a" B! C! b aB" #B" B! B! C" B" C! +b$ â œ ! C! a!Þ(& C! !Þ& B! b a!Þ(& C" #C! C" B" C! Î# B! C" Î# , b$ â œ ! Þ a,b. Taking a limit as $ p ! , the equations reduce to the original system of equations. It follows that B! œ C! œ !Þ& . a- b. Setting the coefficients of the linear terms equal to zero, we find that C" Î# B" Î# + œ ! B" Î% C" Î# , œ ! , with solution B" œ %+ %, and C" œ #+ %, . a. b. Consider the +, - parameter space . The collection of points for which , + represents an increase in the level of species ". At points where , + , B" $ ! Þ Likewise, the collection of points for which #, + represents an increase in the level of species # . At points where #, + , C" $ ! Þ It follows that if , + #, , the level of both species will increase. This condition is represented by the wedge-shaped region on the graph. Otherwise, the level of one species will increase, whereas the level of the other species will simultaneously decrease. Only for + œ , œ ! will both populations remain the same. 13a+b. The critical points consist of the solution set of the equations Setting C œ ! , the second equation becomes BaB !Þ"&baB #b œ ! , with roots B œ ! , !Þ"& and # . Hence the critical points are located at a! ß !b, a!Þ"& ß !b and a# ß !b. The Jacobian matrix of the vector field is C œ! # C BaB !Þ"&baB #b œ ! Þ ________________________________________________________________________ page 555 —————————————————————————— CHAPTER 9. —— JœŒ " . # ! $B %Þ$ B !Þ$ # At the origin, the coefficient matrix of the linearized system is Ja! ß !b œ Œ with eigenvalues <"ß# œ ! !Þ$ " , # Regardless of the value of # , the eigenvalues are real and of opposite sign. Hence a! ß !b is a saddle, which is unstable. At the critical point a!Þ"& ß !b, the coefficient matrix of the linearized system is Ja!Þ"& ß !b œ Œ ! !Þ#((& " , # # " „ È#& # # $! . # "! with eigenvalues <"ß# œ # " „ È"!! # # """ . # #! If "!! # # """ ! , then the eigenvalues are real. Furthermore, since <" <# œ !Þ#((& , both eigenvalues will have the same sign. Therefore the critical point is a node, with its stability dependent on the sign of # . If "!! # # """ ! , the eigenvalues are complex conjugates. In that case the critical point a!Þ"& ß !b is a spiral, with its stability dependent on the sign of # . At the critical point a# ß !b, the coefficient matrix of the linearized system is Ja# ß !b œ Œ ! $Þ( " , # with eigenvalues <"ß# œ Regardless of the value of # , the eigenvalues are real and of opposite sign. Hence a# ß !b is a saddle, which is unstable. # " „ È#& # # $(! . # "! ________________________________________________________________________ page 556 —————————————————————————— CHAPTER 9. —— a, b . It is evident that for # œ !Þ) , the critical point a!Þ"& ß !b is a stable spiral. Closer examination shows that for # œ "Þ& , the critical point a!Þ"& ß !b is a stable node. a- b. Based on the phase portraits in Part a,b, it is apparent that the required value of # satisfies !Þ) # "Þ& . Using the initial condition Ba!b œ # and C a!b œ !Þ!" , it is possible to solve the initial value problem for various values of # . A reasonable first guess is # œ È"Þ"" . This value marks the change in qualitative behavior of the critical ________________________________________________________________________ page 557 —————————————————————————— CHAPTER 9. —— point a!Þ"& ß !b. Numerical experiments show that the solution remains positive for # ¸ "Þ#! . ________________________________________________________________________ page 558 —————————————————————————— CHAPTER 9. —— Section 9.5 1a+b. a,bÞ The critical points are solutions of the system of equations The two critical points are a! ß !b and a!Þ& ß $b. a- b. The Jacobian matrix of the vector field is JœŒ $Î# CÎ# C Ba"Þ& !Þ& C b œ ! Ca !Þ& Bb œ ! Þ At the critical point a! ß !b, the coefficient matrix of the linearized system is Ja! ß !b œ Œ $Î# ! ! . "Î# BÎ# . "Î# B The eigenvalues and eigenvectors are " ! <" œ $Î# , 0 a"b œ Œ ; <# œ "Î# , 0 a#b œ Œ Þ ! " The eigenvalues are of opposite sign, hence the origin is a saddle, which is unstable. At the critical point a!Þ& ß $b, the coefficient matrix of the linearized system is Ja!Þ& ß $b œ Œ ! $ "Î% . ! The eigenvalues and eigenvectors are <" œ 3 È$ # , 0 a"b œ Œ È$ " " , 0 a#b œ Œ È Þ ; <# œ 3 # 3È $ # #3 $ ________________________________________________________________________ page 559 —————————————————————————— CHAPTER 9. —— The eigenvalues are purely imaginary. Hence the critical point is a center, which is stable. a.ß /b. a0 b. Except for solutions along the coordinate axes, almost all trajectories are closed curves about the critical point a!Þ& ß $b. 2a+b. a,b. The critical points are the solution set of the system of equations The two critical points are a! ß !b and a!Þ& ß #b. a- b. The Jacobian matrix of the vector field is JœŒ " CÎ# CÎ# Ba" !Þ& C b œ ! Ca !Þ#& !Þ& Bb œ ! Þ At the critical point a! ß !b, the coefficient matrix of the linearized system is ________________________________________________________________________ page 560 BÎ# . "Î% BÎ# —————————————————————————— CHAPTER 9. —— Ja! ß !b œ Œ The eigenvalues and eigenvectors are " ! <" œ " , 0 a"b œ Œ ; <# œ "Î% , 0 a#b œ Œ Þ ! " The eigenvalues are of opposite sign, hence the origin is a saddle, which is unstable. At the critical point a!Þ& ß #b, the coefficient matrix of the linearized system is Ja!Þ& ß #b œ Œ ! " "Î% . ! ! . "Î% " ! The eigenvalues and eigenvectors are <" œ 3Î# , 0 a"b œ Œ " " a#b ; <# œ 3Î# , 0 œ Œ Þ #3 #3 The eigenvalues are purely imaginary. Hence the critical point is a center, which is stable. a.ß /b. a0 b. Except for solutions along the coordinate axes, almost all trajectories are closed curves about the critical point a!Þ& ß #b. ________________________________________________________________________ page 561 —————————————————————————— CHAPTER 9. —— 4a+b. a,b. The critical points are the solution set of the system of equations The three critical points are a! ß !b, a*Î) ß !b and a" ß "Î%b. a- b. The Jacobian matrix of the vector field is JœŒ BÎ# . " B Ba*Î) B CÎ#b œ ! Ca " Bb œ ! Þ At the critical point a! ß !b, the coefficient matrix of the linearized system is Ja! ß !b œ Œ *Î) ! ! . " *Î) #B CÎ# C The eigenvalues and eigenvectors are " ! <" œ *Î) , 0 a"b œ Œ ; <# œ " , 0 a#b œ Œ Þ ! " The eigenvalues are of opposite sign, hence the origin is a saddle, which is unstable. At the critical point a*Î) ß !b, the coefficient matrix of the linearized system is Ja*Î) ß !b œ Œ *Î) ! *Î"' . "Î) The eigenvalues and eigenvectors are <" œ * " " * , 0 a"b œ Œ ; <# œ , 0 a#b œ Œ Þ ) ! ) #! ________________________________________________________________________ page 562 —————————————————————————— CHAPTER 9. —— The eigenvalues are of opposite sign, hence the critical point a*Î) ß !b is a saddle, which is unstable. At the critical point a" ß "Î%b, the coefficient matrix of the linearized system is Ja" ß "Î%b œ Œ " "Î% "Î# . ! The eigenvalues and eigenvectors are <" œ # È# % , 0 a"b œ # È# # È# # È# ; <# œ , 0 a#b œ Þ % " " The eigenvalues are both negative. Hence the critical point is a stable node, which is asymptotically stable. a.ß /b. a0 b. Except for solutions along the coordinate axes, all solutions converge to the critical point a" ß "Î%b. 5a+b. ________________________________________________________________________ page 563 —————————————————————————— CHAPTER 9. —— a,b. The critical points are solutions of the system of equations The four critical points are a! ß !b, a"Î# ß !b, a# ß !b and a$Î# ß &Î$b. a- b. The Jacobian matrix of the vector field is JœŒ " &B $B# $CÎ"! C Ja! ß !b œ Œ $BÎ"! . $Î# B Bˆ " #Þ& B !Þ$ C B# ‰ œ ! Ca "Þ& Bb œ ! Þ At the critical point a! ß !b, the coefficient matrix of the linearized system is " ! ! . $Î# The eigenvalues and eigenvectors are " ! <" œ " , 0 a"b œ Œ ; <# œ $Î# , 0 a#b œ Œ Þ ! " The eigenvalues are both negative, hence the critical point a! ß !b is a stable node, which is asymptotically stable. At the critical point a"Î# ß !b, the coefficient matrix of the linearized system is Ja"Î# ß !b œ Œ $Î% ! $Î#! . " The eigenvalues and eigenvectors are <" œ $ " $ , 0 a"b œ Œ ; <# œ " , 0 a#b œ Œ Þ % ! $& The eigenvalues are of opposite sign, hence the critical point a"Î# ß !b is a saddle, which is unstable. At the critical point a# ß !b, the coefficient matrix of the linearized system is Ja# ß !b œ Œ $ ! $Î& . "Î# The eigenvalues and eigenvectors are " ' <" œ $ , 0 a"b œ Œ ; <# œ "Î# , 0 a#b œ Œ Þ ! $& The eigenvalues are of opposite sign, hence the critical point a# ß !b is a saddle, which is unstable. ________________________________________________________________________ page 564 —————————————————————————— CHAPTER 9. —— At the critical point a$Î# ß &Î$b, the coefficient matrix of the linearized system is Ja$Î# ß &Î$b œ Œ Î *3 $È$* Ñ Ï %! $Î% &Î$ *Î#! . ! The eigenvalues and eigenvectors are <" œ $ 3È$* ) ,0 a"b œ " The eigenvalues are complex conjugates. Hence the critical point a$Î# ß &Î$b is a stable spiral, which is asymptotically stable. a.ß /b. Ò ; <# œ $ 3È$* ) ,0 a#b œ Î *3 $È$* Ñ Ï %! " Ò Þ a0 b. The single solution curve that converges to the node at a"Î# ß !b is a separatrix. Except for initial conditions on the coordinate axes, trajectories on either side of the separatrix converge to the node at a! ß !b or the stable spiral at a$Î# ß &Î$b. 6. Given that > is measured from the time that B is a maximum, we have -O -9=ˆÈ+- >‰ # # + +C œ O Ê =38ˆÈ+- >‰ Þ ! !! Bœ The period of oscillation is evidently X œ #1ÎÈ+- Þ Both populations oscillate about a mean value. The following is based on the properties of the -9= and =38 functions The prey population aBb is maximum at > œ ! and > œ X . It is a minimum at > œ X Î# Þ Its rate of increase is greatest at > œ $ X Î%. The rate of decrease of the prey population is greatest at > œ X Î% . The predator population aCb is maximum at > œ X Î% . It is a minimum at > œ $ X Î% Þ ________________________________________________________________________ page 565 —————————————————————————— CHAPTER 9. —— The rate of increase of the predator population is greatest at > œ ! and > œ X . The rate of decrease of the predator population is greatest at > œ X Î# . In the following example, the system in Problem # is solved numerically with the initial conditions Ba!b œ !Þ( and Ca!b œ # . The critical point of interest is at a!Þ& ß #b. Since + œ " and - œ "Î% , it follows that the period of oscillation is X œ % 1 Þ a,b. The estimated period appears to agree with the graphic in Figure *Þ&Þ$ . Ba!b œ $Þ& (Þ#' Ba!b œ %Þ! (Þ#* Ba!b œ %Þ& (Þ$% Ba!b œ &Þ! (Þ%# 8a+b. The period of oscillation for the linear system is X œ #1ÎÈ+- Þ In system a#b, + œ " and - œ !Þ(& . Hence the period is estimated as X œ #1ÎÈ!Þ(& ¸ (Þ#&&# . a- b. The critical point of interest is at a$ ß #bÞ The system is solved numerically, with Ca!b œ # and Ba!b œ $Þ& ß %Þ! ß %Þ& ß &Þ! Þ The resulting periods are shown in the table: X The actual amplitude steadily increases as the amplitude increases. ________________________________________________________________________ page 566 —————————————————————————— CHAPTER 9. —— 9. The system .B C œ + BŠ" ‹ .> # .C B œ , CŠ " ‹ .> $ is solved numerically for various values of the parameters. The initial conditions are Ba!b œ & , Ca!b œ # . a+b. + œ " and , œ " À The period is estimated by observing when the trajectory becomes a closed curve. In this case, X ¸ 'Þ%& . a,b. + œ $ and + œ "Î$ , with , œ " À For + œ $ , X ¸ $Þ'* . For + œ "Î$ , X ¸ ""Þ%% . ________________________________________________________________________ page 567 —————————————————————————— CHAPTER 9. —— a- bÞ , œ $ and , œ "Î$ , with + œ " À For , œ $ , X ¸ $Þ)# . For , œ "Î$ , X ¸ ""Þ!' . a. b. It appears that if one of the parameters is fixed, the period varies inversely with the other parameter. Hence one might postulate the relation Xœ 5 Þ 0 a+ ß , b 10a+b. Since X œ #1ÎÈ+- , we first note that ( EX E -9=ˆÈ+- > 9‰.> œ ( EX E =38ˆÈ+- > 9‰.> œ ! . Hence " EX " EX + + Bœ ( .> œ and C œ ( .> œ . XE # # XE ! ! a,b. One way to estimate the mean values is to find a horizontal line such that the area above the line is approximately equal to the area under the line. From Figure *Þ&Þ$ , it appears that B ¸ $Þ#& and C ¸ #Þ! . In Example " , + œ " , - œ !Þ(& , ! œ !Þ& and # œ !Þ#& . Using the result in Part a+b, B œ $ and C œ # Þ .B C œ BŠ" ‹ .> # .C $B œ CŒ .> %% a- b. The system is solved numerically for various initial conditionsÞ Ba!b œ $ and Ca!b œ #Þ& À ________________________________________________________________________ page 568 —————————————————————————— CHAPTER 9. —— Ba!b œ $ and Ca!b œ $Þ! À Ba!b œ $ and Ca!b œ $Þ& À ________________________________________________________________________ page 569 —————————————————————————— CHAPTER 9. —— Ba!b œ $ and Ca!b œ %Þ! À It is evident that the mean values increase as the amplitude increases. That is, the mean values increase as the initial conditions move farther from the critical point. 12. The system of equations in model a"b is given by .B œ Ba+ ! C b .> .C œ C a - # Bb . .> Based on the hypothesis, let the death rate of the insect population and the predators be : B and ; C , respectively. The modified system of equations becomes .B œ Ba+ ! C b : B .> .C œ C a - # Bb ; C , .> in which : ! , ; ! . The critical points are solutions of the system of equations Ba+ : ! C b œ ! C a - ; # Bb œ ! Þ since a- ; bÎ# -Î# , the equilibrium level of the insect population has increased. On the other hand, since a+ :bÎ! +Î! , equilibrium level of the predators has decreased. Indeed, the introduction of insecticide creates a potential to significantly affect the predator population a+ ¸ :b. It is easy to see that the critical points are now at a! ß !b and Š -; ß +: ‹Þ Furthermore, # ! ________________________________________________________________________ page 570 —————————————————————————— CHAPTER 9. —— Section 9.6 2. We consider the function Z aBß Cb œ + B# - C # . The rate of change of Z along any trajectory is Þ .B .C Z œ ZB ZC .> .> "$ œ #+BŒ B #BC # #-C ˆ C $ ‰ # œ +B% %+B# C # #-C % Þ Let ? œ B# , @ œ C # , ! œ + , " œ %+ , and # œ #- . We then have If + ! and - ! , then Z aBß Cb is positive definite. Furthermore, ! ! . Recall that Theorem *Þ'Þ% asserts that if %!# " # œ )+- "' +# ! , then the function ! ?# " ?@ # @# Þ is negative definite. Hence if - #+ , then Z aBß C b is negative definite. One such example is Z aBß Cb œ B# $ C # Þ It follows from Theorem *Þ'Þ" that the origin is an asymptotically stable critical point. 4. Given Z aBß Cb œ + B# - C # , the rate of change of Z along any trajectory is Þ .B .C Z œ ZB ZC .> .> ˆB$ C $ ‰ #-C ˆ#BC # %B# C #C $ ‰ œ #+B œ #+ B% a%- #+bBC $ )- B# C # %- C % Þ Þ Z œ %- B% )- B# C # %- C % %- B% %- C % Þ +B% %+B# C # #-C % œ ! ?# " ?@ # @# . Setting + œ #- , Þ As long as + œ #- ! , the function Z aBß C b is positive definite and Z aBß C b is also positive definite. It follows from Theorem *Þ'Þ# that a! ß !b is an unstable critical point. 5. Given Z aBß Cb œ - aB# C # b , the rate of change of Z along any trajectory is Þ .B .C Z œ ZB ZC .> .> œ #- BcC B0 aBß C bd #-C c B C0 aBß C bd œ #- ˆB# C # ‰0 aBß C b Þ If - ! , then Z aBß Cb is positive definite. Furthermore, if 0 aBß C b is positive in some Þ neighborhood of the origin, then Z aBß Cb is negative definite. Theorem *Þ'Þ" asserts that ________________________________________________________________________ page 571 —————————————————————————— CHAPTER 9. —— the origin is an asymptotically stable critical point. On the other hand, if 0 aBß C b is negative in some neighborhood of the origin, then Z aBßÞCb and Z aBß Cb are both positive definite. It follows from Theorem *Þ'Þ# that the origin is an unstable critical point. 9a+b. Letting B œ ? and C œ ? w , we obtain the system of equations .B œC .> .C œ 1aBb C Þ .> Since 1a!b œ ! , it is evident that a! ß !b is a critical point of the system. Consider the function B "# Z aBß Cb œ C ( 1a=b.= Þ # ! It is clear that Z a!ß !b œ ! . Since 1a?b is an odd function in a neighborhood of ? œ ! , ( 1a=b.= ! for B ! , B ! and ( 1a=b.= œ ( 1a=b.= ! for B ! . B ! ! B Therefore Z aBß Cb is positive definite. The rate of change of Z along any trajectory is Þ .B .C Z œ ZB ZC .> .> œ 1aBb † aCb C c 1aBb Cd œ C# Þ Þ It follows that Z aBß Cb is only negative semidefinite Þ Hence the origin is a stable critical point. a,b. Given Z aBß Cb œ B "# " C C =38aBb ( =38a=b.= , # # ! It is easy to see that Z a! ß !b œ ! . The rate of change of Z along any trajectory is ________________________________________________________________________ page 572 —————————————————————————— CHAPTER 9. —— Þ .B .C Z œ ZB ZC .> .> C " œ ’=38 B -9= B“aC b ”C =38 B•c =38 B Cd # # " " C œ C# -9= B =38# B =38 B C # Þ # # # For 1Î# B 1Î# , we can write =38 B œ B ! B$ Î' and -9= B œ " " B# Î# , in which ! œ !aBb , " œ " aBb Þ Note that ! ! ß " " . Then # Þ C# " B# " ! B$ C ! B$ # Z aB ß C b œ " ŒB ŒB Œ C Þ # # # ' # ' Using polar coordinates, Þ <# Z a< ß ) b œ c" =38 ) -9= ) 2 a< ß ) bd # <# " œ ”" =38 #) 2a< ß ) b• . # # It is easy to show that So if < is sufficiently small, then k2a< ß ) bk "Î# and ¸ " =38 #) 2a< ß )b¸ " Þ Hence # Þ Z aB ß Cb is negative definite. Z aBß Cb œ "# " C C =38aBb " -9= B . # # B# B% # . # #% k 2 a< ß ) b k Ÿ "# " < <% . # (# Now we show that Z aBß Cb is positive definite. Since 1a?b œ =38 ? , This time we set -9= B œ " Note that ! # " for 1Î# B 1Î# . Converting to polar coordinates, Z a<ß ) b œ <# <# <# $ % ”" =38 ) -9= ) =38 ) -9= ) # -9= ) • # "# #% <# " <# <# œ ”" =38 #) =38 ) -9=$ ) # -9=% ) •Þ # # "# #% ________________________________________________________________________ page 573 —————————————————————————— CHAPTER 9. —— Now <# <# " =38 ) -9=$ ) # -9=% ) for < " . "# #% ) Z a<ß ) b <# ( " $ <# !. ” =38 #) • #) # "' It follows that when < ! , Therefore Z aBß Cb is indeed positive definite, and by Theorem *Þ'Þ" , the origin is an asymptotically stable critical point. 12a+b. We consider the linear system w Let Z aBß Cb œ EB# FBC GC # , in which Eœ B +"" Œ œ Œ+ C #" B +"# Œ . +## C and ? œ a+"" +## ba+"" +## +"# +#" b. Based on the hypothesis, the coefficients E and F are negative. Therefore, except for the origin, Z aBß C b is negative on each of the coordinate axes. Along each trajectory, Þ Z œ a#EB FC ba+"" B +"# Cb a#GC FBba+#" B +## C b œ B# C # Þ Þ Hence Z aBß Cb is negative definite. Theorem *Þ'Þ# asserts that the origin is an unstable critical point. a,b. We now consider the system B J " aB ß C b +"# Œ Œ , +## C K" aB ß C b # # +#" +## a+"" +## +"# +#" b #? +"# +## +"" +#" Fœ ? # # +"" +"# a+"" +## +"# +#" b Gœ , #? in which J" aB ß CbÎ< p ! and K" aB ß C bÎ< p ! as < p ! . Let in which Bw +"" Œ œ Œ+ C #" Z aBß Cb œ EB# FBC GC # , ________________________________________________________________________ page 574 —————————————————————————— CHAPTER 9. —— # # +#" +## a+"" +## +"# +#" b #? +"# +## +"" +#" Fœ ? # # + +"# a+"" +## +"# +#" b G œ "" , #? Eœ and ? œ a+"" +## ba+"" +## +"# +#" b. Based on the hypothesis, E , F ! . Except for the origin, Z aBß Cb is positive on each of the coordinate axes. Along each trajectory, Þ Z œ B# C # a#EB FC bJ" aB ß Cb a#GC FBbK" aB ß C b . Converting to polar coordinates, for < Á ! , Þ Z œ <# <a#E-9= ) F=38 ) b J" <a#G=38 ) F-9= )b K" J" K" œ <# <# ”a#E-9= ) F=38 ) b a#G=38 ) F-9= )b •Þ < < Since the system is almost linear, there is an V such that ºa#E-9= ) F=38 ) b and hence a#E-9= ) F=38 ) b for < V . It follows that Þ " Z <# # Þ as long as ! < V . Hence Z is positive definite on the domain H œ ˜ aB ß C b l B # C # V # ™ . J" K" " a#G=38 ) F-9= )b < < # J" K" " a#G=38 ) F-9= )b º , < < # By Theorem *Þ'Þ# , the origin is an unstable critical point. ________________________________________________________________________ page 575 —————————————————————————— CHAPTER 9. —— Section 9.7 3. The equilibrium solutions of the ODE .< œ <a< "ba< $b .> are given by <" œ ! , <# œ " and <$ œ $ . Note that .< ! for ! < " and < $ ; .> .< ! for " < $ . .> < œ ! corresponds to an unstable critical point. The equilibrium solution <# œ " is asymptotically stable, whereas the equilibrium solution <$ œ $ is unstable. Since the critical values are isolated, a limit cycle is given by < œ " , ) œ > >! which is asymptotically stable. Another periodic solution is found to be < œ $ , ) œ > >! which is unstable. 5. The equilibrium solutions of the ODE .< œ =38 1< .> are given by < œ 8 , 8 œ ! ß " ß # ß â . Based on the sign of < w in the neighborhood of each critical value, the equilibrium solutions < œ #5 , 5 œ " ß # ß â correspond to unstable periodic solutions, with ) œ > >! . The equilibrium solutions < œ #5 " , 5 œ ! ß " ß # ß â correspond to stable limit cycles, with ) œ > >! . The solution < œ ! represents an unstable critical point. 10. Given J aB ß Cb œ +"" B +"# C and KaB ß C b œ +#" B +## C , it follows that JB KC œ +"" +## . Based on the hypothesis, JB KC is either positive or negative on the entire plane. By Theorem *Þ(Þ# , the system cannot have a nontrivial periodic solution. 12. Given that J aB ß Cb œ #B $C BC # and KaB ß C b œ C B$ B# C , J B KC œ " B # C # . Since JB KC ! on the entire plane, Theorem *Þ(Þ# asserts that the system cannot have a nontrivial periodic solution. ________________________________________________________________________ page 576 —————————————————————————— CHAPTER 9. —— 14a+b. Based on the given graphs, the following table shows the estimated values: . œ !Þ# . œ "Þ! . œ &Þ! X ¸ 'Þ#* X ¸ 'Þ'' X ¸ ""Þ'! a,bÞ The initial conditions were chosen as Ba!b œ # , Ca!b œ ! Þ X ¸ 'Þ$) . X ¸ (Þ'& . ________________________________________________________________________ page 577 —————————————————————————— CHAPTER 9. —— X ¸ )Þ)' . X ¸ "!Þ#& . a- b. The period, X , appears to be a quadratic function of . . 15a+b. Setting B œ ? and C œ ? w , we obtain the system of equations .B œC .> .C " œ B .Œ" C # C . .> $ a,b. Evidently, C œ ! . It follows that B œ ! . Hence the only critical point of the system is at a! ß !b. The components of the vector field are infinitely differentiable everywhere. Therefore the system is almost linear. ________________________________________________________________________ page 578 —————————————————————————— CHAPTER 9. —— The Jacobian matrix of the vector field is JœŒ ! " " . . .C # " , . At the critical point a! ß !b, the coefficient matrix of the linearized system is Ja! ß !b œ Œ ! " with eigenvalues <"ß# œ . "È # . % Þ „ ## If . œ ! , the equation reduces to the ODE for a simple harmonic oscillator. For the case ! . # , the eigenvalues are complex, and the critical point is an unstable spiral. For . # , the eigenvalues are real, and the origin is an unstable node. a- b. The initial conditions were chosen as Ba!b œ # , Ca!b œ ! Þ E ¸ #Þ"' and X ¸ 'Þ'& . a. bÞ ________________________________________________________________________ page 579 —————————————————————————— CHAPTER 9. —— E ¸ #Þ!! and X ¸ 'Þ$! . E ¸ #Þ!% and X ¸ 'Þ$) . E ¸ #Þ' and X ¸ (Þ'# . ________________________________________________________________________ page 580 —————————————————————————— CHAPTER 9. —— E ¸ %Þ$( and X ¸ ""Þ'" . a/ b . . œ !Þ# . œ !Þ& . œ "Þ! . œ #Þ! . œ &Þ! E #Þ!! #Þ!% #Þ"' #Þ' %Þ$( X 'Þ$! 'Þ$) 'Þ'& (Þ'# ""Þ'" ________________________________________________________________________ page 581 —————————————————————————— CHAPTER 9. —— Section 9.8 6. < œ #) , with initial point a& ß & ß &b: < œ #) , with initial point a&Þ!" ß & ß &b: ________________________________________________________________________ page 582 —————————————————————————— CHAPTER 9. —— 7. < œ #) À ________________________________________________________________________ page 583 —————————————————————————— CHAPTER 9. —— 9a+b. < œ "!! , initial point a & ß "$ ß && b À The period appears to be X ¸ "Þ"# . a,b. < œ **Þ*% , initial point a & ß "$ ß && b À The periodic trajectory appears to have split into two strands, indicative of a perioddoubling. Closer examination reveals that the peak values of D a>b are slightly different: ________________________________________________________________________ page 584 —————————————————————————— CHAPTER 9. —— < œ **Þ( , initial point a & ß "$ ß && b À ________________________________________________________________________ page 585 —————————————————————————— CHAPTER 9. —— a- b. < œ **Þ' , initial point a & ß "$ ß && b À The strands again appear to have split. Closer examination reveals that the peak values of D a>b are different: ________________________________________________________________________ page 586 —————————————————————————— CHAPTER 9. —— 10a+b. < œ "!!Þ& , initial point a & ß "$ ß && b À ________________________________________________________________________ page 587 ————————————————————————— CHAPTER 9. —— < œ "!!Þ( , initial point a & ß "$ ß && b À ________________________________________________________________________ page 588 —————————————————————————— CHAPTER 9. —— a,b. < œ "!!Þ) , initial point a & ß "$ ß && b À < œ "!!Þ)" , initial point a & ß "$ ß && b À The strands of the periodic trajectory are beginning to split apart. ________________________________________________________________________ page 589 —————————————————————————— CHAPTER 9. —— < œ "!!Þ)# , initial point a & ß "$ ß && b À ________________________________________________________________________ page 590 —————————————————————————— CHAPTER 9. —— < œ "!!Þ)$ , initial point a & ß "$ ß && b À ________________________________________________________________________ page 591 —————————————————————————— CHAPTER 9. —— < œ "!!Þ)% , initial point a & ß "$ ß && b À ________________________________________________________________________ page 592 —————————————————————————— CHAPTER 10. —— Chapter Ten Section 10.1 1. The general solution of the ODE is CaBb œ -" -9= B -# =38 B Þ Imposing the first boundary condition, it is necessary that -" œ ! . Therefore CaBb œ -# =38 B . Taking its derivative, C w aBb œ -# -9= B . Imposing the second boundary condition, we require that -# -9= 1 œ " . The latter equation is satisfied only if -# œ " . Hence the solution of the boundary value problem is CaBb œ =38 B . 4. The general solution of the differential equation is CaBb œ -" -9= B -# =38 B Þ It follows that C w aBb œ -" =38 B -# -9= B Þ Imposing the first boundary condition, we find that -# œ " . Therefore CaBb œ -" -9= B =38 B Þ Imposing the second boundary condition, we require that -" -9= P =38 P œ ! . If -9= P Á ! , that is, as long as P Á a#5 "b1Î# , with 5 an integer, then -" œ >+8P . The solution of the boundary value problem is CaBb œ >+8P -9= B =38 B Þ If -9= P œ ! , the boundary condition results in =38 P œ ! . The latter two equations are inconsistent, which implies that the BVP has no solution. 5. The general solution of the homogeneous differential equation is CaBb œ -" -9= B -# =38 B Þ Using any of a number of methods, including the method of undetermined coefficients, it is easy to show that a particular solution is ] aBb œ B . Hence the general solution of the given differential equation is CaBb œ -" -9= B -# =38 B B Þ The first boundary condition requires that -" œ ! . Imposing the second boundary condition, it is necessary that -# =38 1 1 œ ! Þ The resulting equation has no solution. We conclude that the boundary value problem has no solution. 6. Using the method of undetermined coefficients, it is easy to show that the general solution of the ODE is CaBb œ -" -9=È# B -# =38È# B BÎ# Þ Imposing the first boundary condition, we find that -" œ ! . The second boundary condition requires that -# =38È# 1 1Î# œ ! Þ It follows that -# œ 1Î#=38È# 1 Þ Hence the solution of the boundary value problem is 1 B CaBb œ =38È# B Þ # #=38È# 1 8. The general solution of the homogeneous differential equation is Using the method of undetermined coefficients, a particular solution is ] aBb œ =38 BÎ$ . ________________________________________________________________________ page 593 CaBb œ -" -9= #B -# =38 #B Þ —————————————————————————— CHAPTER 10. —— Hence the general solution of the given differential equation is " CaBb œ -" -9= #B -# =38 #B =38 B Þ $ The first boundary condition requires that -" œ ! . The second boundary requires that -# =38 #1 " =38 1 œ ! Þ The latter equation is valid for all values of -# . Therefore the $ solution of the boundary value problem is " CaBb œ -# =38 #B =38 B Þ $ 9. Using the method of undetermined coefficients, it is easy to show that the general solution of the ODE is CaBb œ -" -9= #B -# =38 #B -9= BÎ$ Þ It follows that C w aBb œ #-" =38 #B #-# -9= #B =38 BÎ$ . Imposing the first boundary condition, we find that -# œ ! . The second boundary condition requires that " #-" =38 #1 =38 1 œ ! Þ $ The resulting equation is satisfied for all values of -" Þ Hence the solution is the family of functions " CaBb œ -" -9= #B -9= B Þ $ 10. The general solution of the differential equation is Its derivative is C w aBb œ È$ -" =38È$ B È$ -# -9=È$ B =38 BÎ# . The first boundary condition requires that -# œ ! . Imposing the second boundary condition, we obtain È$ -" =38È$ 1 œ ! . It follows that -" œ ! . Hence the solution of the BVP is CaBb œ -9= BÎ# . 12. Assuming that - ! , we can set - œ .# . The general solution of the differential equation is so that C w aBb œ .-" =38 .B .-# -9= .B . Imposing the first boundary condition, it follows that -# œ ! . Therefore CaBb œ -" -9= .B . The second boundary condition requires that -" -9= .1 œ ! . For a nontrivial solution, it is necessary that -9= .1 œ ! , that is, .1 œ a#8 "b1Î# , with 8 œ "ß #ß â . Therefore the eigenvalues are -8 œ a#8 "b# , 8 œ "ß #ß â Þ % CaBb œ -" -9= .B -# =38 .B , " CaBb œ -" -9=È$ B -# =38È$ B -9= B Þ # ________________________________________________________________________ page 594 —————————————————————————— CHAPTER 10. —— The corresponding eigenfunctions are given by C8 œ -9= a#8 "bB , 8 œ "ß #ß â Þ # Assuming that - ! , we can set - œ .# . The general solution of the differential equation is so that C w aBb œ .-" =382 .B .-# -9=2 .B . Imposing the first boundary condition, it follows that -# œ ! . Therefore CaBb œ -" -9=2 .B . The second boundary condition requires that -" -9=2 .1 œ ! , which results in -" œ ! . Hence the only solution is the trivial solution. Finally, with - œ ! , the general solution of the ODE is CaBb œ -" B -# . CaBb œ -" -9=2 .B -# =382 .B , It is easy to show that the boundary conditions require that -" œ -# œ ! . Therefore all of the eigenvalues are positive. 13. Assuming that - ! , we can set - œ .# . The general solution of the differential equation is so that C w aBb œ .-" =38 .B .-# -9= .B . Imposing the first boundary condition, it follows that -# œ ! . The second boundary condition requires that -" =38 .1 œ ! . For a nontrivial solution, we must have .1 œ 81 , 8 œ "ß #ß â . It follows that the eigenvalues are -8 œ 8# , 8 œ "ß #ß â , and the corresponding eigenfunctions are C8 œ -9= 8B , 8 œ "ß #ß â Þ Assuming that - ! , we can set - œ .# . The general solution of the differential equation is so that C w aBb œ .-" =382 .B .-# -9=2 .B . Imposing the first boundary condition, it follows that -# œ ! . The second boundary condition requires that -" =382 .1 œ ! . The latter equation is satisfied only for -" œ ! . Finally, for - œ ! , the solution is CaBb œ -" B -# . Imposing the boundary conditions, we find that CaBb œ -# . Therefore - œ ! is also an eigenvalue, with corresponding eigenfunction C! aBb œ " Þ CaBb œ -" -9=2 .B -# =382 .B , CaBb œ -" -9= .B -# =38 .B , ________________________________________________________________________ page 595 —————————————————————————— CHAPTER 10. —— 14. It can be shown, as in Prob. "# , that - ! . Setting - œ .# , the general solution of the resulting ODE is with C w aBb œ .-" =38 .B .-# -9= .B . Imposing the first boundary condition, we find that -# œ ! . Therefore CaBb œ -" -9= .B . The second boundary condition requires that -" -9= .P œ ! . For a nontrivial solution, it is necessary that -9= .P œ ! , that is, . œ a#8 "b1Îa#Pb , with 8 œ "ß #ß â . Therefore the eigenvalues are a#8 "b# 1# -8 œ , 8 œ "ß #ß â Þ %P# CaBb œ -" -9= .B -# =38 .B , The corresponding eigenfunctions are given by C8 œ -9= a#8 "b1B , 8 œ "ß #ß â Þ #P 16. Assuming that - ! , we can set - œ .# . The general solution of the differential equation is The first boundary condition requires that -" œ ! . Therefore CaBb œ -# =382 .B and C w aBb œ -# -9=2 .B Þ Imposing the second boundary condition, it is necessary that -# -9=2 .P œ ! Þ The latter equation is valid only for -# œ ! . The only solution is the trivial solution. Assuming that - ! , we set - œ .# . The general solution of the resulting ODE is Imposing the first boundary condition, we find that -" œ ! . Hence CaBb œ -# =38 .B and C w aBb œ -# -9= .B . In order to satisfy the second boundary condition, it is necessary that -# -9= .P œ ! . For a nontrivial solution, . œ a#8 "b1Îa#Pb , with 8 œ "ß #ß â . Therefore the eigenvalues are a#8 "b# 1# -8 œ , 8 œ "ß #ß â Þ %P# a#8 "b1B , 8 œ "ß #ß â Þ #P CaBb œ -" -9= .B -# =38 .B . CaBb œ -" -9=2 .B -# =382 .B . The corresponding eigenfunctions are given by C8 œ =38 Finally, for - œ ! , the general solution is linear. Based on the boundary conditions, it follows that CaBb œ ! . Therefore all of the eigenvalues are negative. ________________________________________________________________________ page 596 —————————————————————————— CHAPTER 10. —— 17a+b. Setting - œ .# , write the general solution of the ODE C ww .# C œ ! as Imposing the boundary conditions Ca!b œ Ca1b œ ! , we obtain the system of equations 5" 5 # œ ! 5" /3.1 5 # / 3.1 œ ! Þ The system has a nontrivial solution if and only if the coefficient matrix is singular. Set the determinant equal to zero to obtain / 3.1 / 3.1 œ ! . a,b. Let . œ / 35 . Then 3.1 œ 3/1 51 , and the previous equation can be written as /51 / 3/1 /51 / 3/1 œ ! . Using Euler's relation, /3/1 œ -9= /1 3 =38 /1 , we obtain /51 a-9= / 3 =38 / b /51 a-9= / 3 =38 / b œ ! . a/51 /51 b-9= /1 œ ! a/51 /51 b=38 /1 œ ! Þ CaBb œ 5" /3.B 5 # / 3.B Þ Equating the real and imaginary parts of the equation, a- b. Based on the second equation, / œ 8 , 8 − ˆ . Since -9= 81 Á ! , it follows that /51 œ /51 , or /#51 œ " . Hence 5 œ ! , and . œ 8 , 8 − ˆ . ________________________________________________________________________ page 597 —————————————————————————— CHAPTER 10. —— Section 10.2 1. The period of the function =38 !B is X œ #1Î! . Therefore the function =38 &B has period X œ #1Î& . 2. The period of the function - 9= !B is also X œ #1Î! . Therefore the function -9= #1B has period X œ #1Î#1 œ " . 4. Based on Prob. " , the period of the function =38 1BÎP is X œ #1Îa1ÎPb œ #P . 6. Let X ! and consider the equation aB X b# œ B# Þ It follows that #X B X # œ ! and #B X œ ! . Since the latter equation is not an identity, the function B# cannot be periodic with finite period. 8. The function is defined on intervals of length a#8 "b a#8 "b œ # . On any two consecutive intervals, 0 aBb is identically equal to " on one of the intervals and alternates between " and " on the other. It follows that the period is X œ % . 9. On the interval P B #P , a simple shift to the right results in 0 aBb œ aB #Pb œ #P B . On the interval $P B #P , a simple shift to the left results in 0 aBb œ aB #Pb œ #P B . 11. The next fundamental period to the left is on the interval #P B ! . Hence the interval P B ! is the second half of a fundamental period. A simple shift to the left results in 0 aBb œ P aB #Pb œ P B . 12. First note that -9= and -9= 71B 81 B " a7 8 b 1 B a7 8 b 1 B -9= œ ”-9= -9= • P P # P P 71B 81 B " a8 7b1B a7 8 b 1 B =38 œ ”=38 =38 •Þ P P # P P ________________________________________________________________________ page 598 —————————————————————————— CHAPTER 10. —— It follows that ( P -9= P 71B 81 B "P a7 8 b 1 B a7 8 b 1 B -9= .B œ ( ”-9= -9= •.B P P # P P P " P =38ca7 8b1BÎPd =38ca7 8b1BÎPd P œ œ º #1 78 78 P œ !, as long as 7 8 and 7 8 are not zero. For the case 7 œ 8 , 81B # "P #81B ‹ .B œ ( ”" -9= ( Š-9= •.B P # P P P P " =38a#81BÎPb P B œ º # #81ÎP P œ P. œ Likewise, ( P -9= P a8 7b1B a7 8 b 1 B 71B 81 B "P =38 .B œ ( ”=38 =38 •.B P P # P P P " P -9=ca8 7b1BÎPd -9=ca7 8b1BÎPd P œ œ º #1 78 78 P œ !, as long as 7 8 and 7 8 are not zero. For the case 7 œ 8 , 71B 81B "P #81B =38 .B œ ( =38 .B ( -9= P P # P P P " -9=a#81BÎPb P œ œ º # #81ÎP P œ !. P ________________________________________________________________________ page 599 —————————————————————————— CHAPTER 10. —— 14a+b. For P œ " , a,b. The Fourier coefficients are calculated using the Euler-Fourier formulas: +! œ "P ( 0 aBb.B P P "! œ ( .B P P œ "Þ For 8 ! , +8 œ "P 81 B .B ( 0 aBb-9= P P P "! 81 B œ ( -9= .B P P P œ !Þ Likewise, ,8 œ "P 81B .B ( 0 aBb=38 P P P "! 81B œ ( =38 .B P P P " a "b 8 œ Þ 81 It follows that ,#5 œ ! and ,#5" œ #Îca#5 "b1d, 5 œ "ß #ß $ß â . Therefore the Fourier series for the given function is 0 aBb œ " #_ " a#5 "b1B " =38 Þ # 1 5 œ " #5 " P ________________________________________________________________________ page 600 —————————————————————————— CHAPTER 10. —— 16a+b. a,b. The Fourier coefficients are calculated using the Euler-Fourier formulas: +! œ œ ( aB "b.B ( a" Bb.B ! " " ! "P ( 0 aBb.B P P œ "Þ For 8 ! , +8 œ "P 81B .B ( 0 aBb-9= P P P ! " œ ( aB "b-9= 81B .B ( a" Bb-9= 81B .B " It follows that + #5 œ ! and + #5" œ %Îa#5 "b# 1# ‘, 5 œ "ß #ß $ß â . Likewise, "P 81B ,8 œ ( 0 aBb=38 .B P P P ! " " a "b 8 œ # Þ 8# 1 # ! œ ( aB "b=38 81B .B ( a" Bb=38 81B .B " ! œ !Þ Therefore the Fourier series for the given function is " %_ " 0 aBb œ # " -9=a#5 "b1B Þ # 1 5 œ " a#5 "b# ________________________________________________________________________ page 601 —————————————————————————— CHAPTER 10. —— 17a+b. For P œ " , a,b. The Fourier coefficients are calculated using the Euler-Fourier formulas: +! œ "P ( 0 aBb.B P P "! "P œ ( aB Pb.B ( P .B P P P! œ $PÎ# Þ For 8 ! , +8 œ "P 81 B .B ( 0 aBb-9= P P P "! 81B "P 81 B œ ( aB Pb-9= .B ( P -9= .B P P P P! P Pa" -9= 81b œ Þ 8# 1 # Likewise, ,8 œ "P 81B .B ( 0 aBb=38 P P P "! 81B "P 81 B œ ( aB Pb=38 .B ( P =38 .B P P P P! P P -9= 81 œ Þ 81 Note that -9= 81 œ a "b8 . It follows that the Fourier series for the given function is ________________________________________________________________________ page 602 —————————————————————————— CHAPTER 10. —— 0 aBb œ 18a+b. $P P_ # a#8 "b1B a "b8 1 81 B # "– -9= =38 Þ # % 1 8 œ " a#8 "b P 8 P— a,b. The Fourier coefficients are calculated using the Euler-Fourier formulas: "P +! œ ( 0 aBb.B P P "" œ ( B .B # " œ !Þ For 8 ! , +8 œ "P 81 B .B ( 0 aBb-9= P P P "" 81 B œ ( B -9= .B # " P œ !Þ Likewise, ,8 œ "P 81 B .B ( 0 aBb=38 P P P "" 81B œ ( B =38 .B # " P # 81 81 œ # # Š# =38 81 -9= ‹Þ 81 # # ________________________________________________________________________ page 603 —————————————————————————— CHAPTER 10. —— Therefore the Fourier series for the given function is 0 aBb œ " ” _ 8œ" % 8# 1 # =38 81 # 81 81B -9= •=38 Þ # 81 # # 19a+b. a,b. The Fourier cosine coefficients are given by "P 81 B +8 œ ( 0 aBb-9= .B P P P "! 81B "# 81 B œ ( -9= .B ( -9= .B # # # #! # œ !Þ "P 81B .B ( 0 aBb=38 P P P "! 81 B "# 81 B œ ( =38 .B ( =38 .B # # # #! # " -9= 81 œ# Þ 81 %_ " a#8 "b1B " =38 Þ 1 8 œ " #8 " # The Fourier sine coefficients are given by ,8 œ Therefore the Fourier series for the given function is 0 aBb œ ________________________________________________________________________ page 604 —————————————————————————— CHAPTER 10. —— a- b . 20a+b. a,b. The Fourier cosine coefficients are given by "P 81 B +8 œ ( 0 aBb-9= .B P P P œ ( B -9= 81B.B " " œ !Þ The Fourier sine coefficients are given by ________________________________________________________________________ page 605 —————————————————————————— CHAPTER 10. —— "P 81B ,8 œ ( 0 aBb=38 .B P P P œ ( B =38 81B.B " " œ # -9= 81 Þ 81 Therefore the Fourier series for the given function is # _ a "b 8 0 aBb œ " =38 81B Þ 1 8œ" 8 a- b . ________________________________________________________________________ page 606 —————————————————————————— CHAPTER 10. —— 22a+b. a,b. The Fourier cosine coefficients are given by "P +! œ ( 0 aBb.B P P "! "# œ ( aB #b.B ( a# #Bb.B # # #! œ ", and for 8 ! , "P 81B +8 œ ( 0 aBb-9= .B P P P "! 81 B "# 81 B œ ( aB #b-9= .B ( a# #Bb-9= .B # # # #! # a" -9= 81b œ' Þ 8# 1 # The Fourier sine coefficients are given by ,8 œ "P 81 B .B ( 0 aBb=38 P P P "! 81B "# 81 B œ ( aB #b=38 .B ( a# #Bb=38 .B # # # #! # -9= 81 œ# Þ 81 Therefore the Fourier series for the given function is ________________________________________________________________________ page 607 —————————————————————————— CHAPTER 10. —— " "# _ " a#8 "b1B # _ a "b8 81 B " " 0 aBb œ # =38 Þ # -9= # 1 8 œ " a#8 "b # 1 8œ" 8 # a- b . 23a+b. a,b. The Fourier cosine coefficients are given by ________________________________________________________________________ page 608 —————————————————————————— CHAPTER 10. —— "P +! œ ( 0 aBb.B P P "! B "# " œ ( Š ‹.B ( Œ#B B# .B # # # #! # œ ""Î' , and for 8 ! , "P 81B +8 œ ( 0 aBb-9= .B P P P "! B 81 B "# " 81 B œ ( Š ‹-9= .B ( Œ#B B# -9= .B # # # # #! # # a& -9= 81b œ Þ 8# 1 # The Fourier sine coefficients are given by ,8 œ "P 81B .B ( 0 aBb=38 P P P "! B 81 B "# " 81 B œ ( Š ‹=38 .B ( Œ#B B# =38 .B # # # # #! # # % a% 8# 1# b-9= 81 œ Þ 8$ 1 $ 0 aBb œ "" " _ c a "b 8 & d 81 B #" -9= # "# 1 8 œ " 8 # Therefore the Fourier series for the given function is " _ c% a% 8# 1# ba "b8 d 81 B " =38 Þ $ $ 1 8œ" 8 # ________________________________________________________________________ page 609 —————————————————————————— CHAPTER 10. —— a- b . 24a+b. a,b. The Fourier cosine coefficients are given by +! œ "P ( 0 aBb.B P P "$ œ ( B# a$ Bb.B $! œ *Î% , and for 8 ! , ________________________________________________________________________ page 610 —————————————————————————— CHAPTER 10. —— "P 81B +8 œ ( 0 aBb-9= .B P P P "$ 81B œ ( B# a$ Bb-9= .B $! $ a' ' -9= 81 8# 1# -9= 81b œ #( Þ 8% 1 % The Fourier sine coefficients are given by "P 81B ,8 œ ( 0 aBb=38 .B P P P "$ 81B œ ( B# a$ Bb=38 .B $! $ " # -9= 81 œ &% Þ 8$ 1 $ Therefore the Fourier series for the given function is _ * 'c" a "b8 d a "b8 81 B "” 0 aBb œ #( # # • -9= ) 8% 1 % 81 $ 8œ" &% _ c" #a "b8 d 81 B $" =38 Þ $ 1 8œ" 8 $ a- bÞ ________________________________________________________________________ page 611 —————————————————————————— CHAPTER 10. —— 26. It is evident that k/7 aBbk is greatest at B œ „ $ . Increasing the number of terms in the partials sums, we find that if 7 #( , then k/7 aBbk Ÿ !Þ" , for all B − c $ ß $d . ________________________________________________________________________ page 612 —————————————————————————— CHAPTER 10. —— Graphing the partial sum =#( aBb, the convergence is as predicted: 28. Let B œ X + , for some + − c! ß X d . First note that for any value of 2 , 0 aB 2b 0 aBb œ 0 aX + 2 b 0 aX +b œ 0 a+ 2b 0 a+b Þ 0 w aBb œ lim 0 aB 2b 0 aBb 2Ä! 2 0 a+ 2b 0 a+b œ lim 2Ä! 2 w œ 0 a+ b . Since 0 is differentiable, That is, 0 w a+ X b œ 0 w a+b . By induction, it follows that 0 w a+ X b œ 0 w a+b for every value of + . On the other hand, if 0 aBb œ " -9= B , then the function J aBb œ ( B ! c" -9= >d.> œ B =38 B is not periodic, unless its definition is restricted to a specific interval. 29a+b. Based on the hypothesis, the vectors v" , v# and v$ are a basis for ‘$ . Given any vector u − ‘$ , it can be expressed as a linear combination u œ +" v" +# v# +$ v$ . Taking the inner product of both sides of this equation with v3 , we have u † v 3 œ a + " v " + # v # + $ v $ b † v3 œ +3 v3 † v3 , ________________________________________________________________________ page 613 —————————————————————————— CHAPTER 10. —— since the basis vectors are mutually orthogonal. Hence u † v3 +3 œ , 3 œ "ß #ß $ . v3 † v3 Recall that u † v3 œ ? @3 -9= ) , in which ) is the angle between u and v3 . Therefore +3 œ ? -9= ) Þ @3 a,b. Assuming that a Fourier series converges to a periodic function, 0 aBb , 0 aBb œ _ _ +! 9! aBb " +7 97 aBb " ,7 <7 aBb Þ # 7œ" 7œ" Here ? -9= ) is interpreted as the magnitude of the projection of u in the direction of v3 . Taking the inner product, defined by a? ß @ b œ ( P P ?aBb@aBb.B , of both sides of the series expansion with the specified trigonometric functions, we have _ _ +! " +7 a97 ß 98 b " ,7 a<7 ß 98 b a0 ß 98 b œ a9! ß 98 b # 7œ" 7œ" for 8 œ !ß "ß #ß â . a- b. It also follows that a0 ß < 8 b œ _ _ +! a9! ß <8 b " +7 a97 ß <8 b " ,7 a<7 ß <8 b # 7œ" 7œ" for 8 œ "ß #ß â . Based on the orthogonality conditions, and a<7 ß 98 b œ P $78 . Note that a9! ß 9! b œ #P . Therefore +! œ and +8 œ a97 ß 98 b œ P $78 , a<7 ß <8 b œ P $78 , #a0 ß 9! b "P œ ( 0 aBb9! aBb.B a9! ß 9! b P P a0 ß 98 b "P œ ( 0 aBb98 aBb.B , 8 œ "ß #ß â . a98 ß 98 b P P ________________________________________________________________________ page 614 —————————————————————————— CHAPTER 10. —— Likewise, ,8 œ a0 ß < 8 b "P œ ( 0 aBb<8 aBb.B , 8 œ "ß #ß â . a< 8 ß < 8 b P P ________________________________________________________________________ page 615 —————————————————————————— CHAPTER 10. —— Section 10.3 1a+b. The given function is assumed to be periodic with #P œ # . The Fourier cosine coefficients are given by "P +! œ ( 0 aBb.B P P ! " œ ( a "b.B ( a"b.B " œ !, and for 8 ! , ! "P 81 B +8 œ ( 0 aBb-9= .B P P P ! " œ ( -9= 81B .B ( -9= 81B .B " ! œ !Þ The Fourier sine coefficients are given by ,8 œ "P 81 B .B ( 0 aBb=38 P P P ! " œ ( =38 81B .B ( =38 81B .B " ! " -9= 81 œ# Þ 81 Therefore the Fourier series for the specified function is 0 aBb œ %_ " " =38 a#8 "b1B . 1 8 œ " #8 " ________________________________________________________________________ page 616 —————————————————————————— CHAPTER 10. —— a, b . The function is piecewise continuous on each finite interval. The points of discontinuity are at integer values of B . At these points, the series converges to k0 aB b 0 aB bk œ ! . 3a+b. The given function is assumed to be periodic with X œ #P . The Fourier cosine coefficients are given by +! œ "P ( 0 aBb.B P P "! "P œ ( aP Bb.B ( aP Bb.B P P P! œ P, and for 8 ! , +8 œ "P 81 B .B ( 0 aBb-9= P P P "! 81 B "P 81 B œ ( aP Bb-9= .B ( aP Bb-9= .B P P P P! P " -9= 81 œ #P Þ 8# 1 # The Fourier sine coefficients are given by ,8 œ "P 81B .B ( 0 aBb=38 P P P "! 81 B "P 81 B œ ( aP Bb=38 .B ( aP Bb=38 .B P P P P! P œ !Þ ________________________________________________________________________ page 617 —————————————————————————— CHAPTER 10. —— Therefore the Fourier series of the specified function is 0 aBb œ a,b. For P œ " , P %P _ " a#8 "b1B #" . # -9= # 1 8 œ " a#8 "b P 5a+b. The given function is assumed to be periodic with #P œ #1 . The Fourier cosine coefficients are given by "P +! œ ( 0 aBb.B P P " 1Î# œ( a"b.B 1 1Î# œ ", and for 8 ! , +8 œ "P 81 B .B ( 0 aBb-9= P P P " 1Î# œ( a"b-9= 8B .B 1 1Î# # 81 œ =38Š ‹Þ 81 # Note that 0 aBb is continuous. Based on Theorem "!Þ$Þ" , the series converges to the continuous function 0 aBb. The Fourier sine coefficients are given by ________________________________________________________________________ page 618 —————————————————————————— CHAPTER 10. —— "P 81B ,8 œ ( 0 aBb=38 .B P P P " 1Î# a"b=38 8B .B œ( 1 1Î# œ !Þ Observe that =38Š ! , 8 œ #5 81 ‹œœ 5 +1 # a "b , 8 œ #5 " 0 aBb œ a, b . , 5 œ "ß #ß â . Therefore the Fourier series of the specified function is " # _ a "b 8 " -9= a#8 "bB . # 1 8 œ " #8 " The given function is piecewise continuous, with discontinuities at odd multiples of 1Î# . At B. œ a#5 "b1Î# , 5 œ !ß "ß #ß â , the series converges to k0 aB. b 0 aB. bk œ "Î# . 6a+b. The given function is assumed to be periodic with #P œ # . The Fourier cosine coefficients are given by +! œ "P ( 0 aBb.B P P " ! œ ( B# .B œ "Î$ , and for 8 ! , ________________________________________________________________________ page 619 —————————————————————————— CHAPTER 10. —— "P 81 B +8 œ ( 0 aBb-9= .B P P P œ ( B# -9= 81B .B " ! # -9= 81 œ Þ 8# 1 # The Fourier sine coefficients are given by "P 81B ,8 œ ( 0 aBb=38 .B P P P œ ( B# =38 81B .B " ! œ # # -9= 81 8# 1# -9= 81 Þ 8$ 1 $ Therefore the Fourier series for the specified function is " # _ a "b 8 0 aBb œ # " -9= 81B ' 1 8 œ " 8# "” _ 8œ" #c" a "b8 d a "b8 •=38 81B . 8$ 1 $ 81 a, b . The given function is piecewise continuous, with discontinuities at the odd integers . ________________________________________________________________________ page 620 —————————————————————————— CHAPTER 10. —— At B. œ #5 " , 5 œ !ß "ß #ß â , the series converges to k0 aB. b 0 aB. bk œ "Î# . 8a+b. As shown in Problem "' of Section "!Þ# , 0 aBb œ a, b . " %_ " #" -9=a#8 "b1B Þ # 1 8 œ " a#8 "b# ________________________________________________________________________ page 621 —————————————————————————— CHAPTER 10. — a- b . 9a+b. As shown in Problem #! of Section "!Þ# , # _ a "b 8 0 aBb œ " =38 81B Þ 1 8œ" 8 a, b . ________________________________________________________________________ page 622 —————————————————————————— CHAPTER 10. —— a- bÞ The given function is discontinuous at B œ „ " . At these points, the series will converge to a value of zero. The error can never be made arbitrarily small. 10a+b. As shown in Problem ## of Section "!Þ# , 0 aBb œ " "# _ " a#8 "b1B # _ a "b8 81 B " #" -9= =38 Þ # # 1 8 œ " a#8 "b # 1 8œ" 8 # a, b . ________________________________________________________________________ page 623 —————————————————————————— CHAPTER 10. —— a- b. The given function is discontinuous at B œ „ # . At these points, the series will converge to a value of " . The error can never be made arbitrarily small. 11a+b. As shown in Problem ' , above , " # _ a "b 8 0 aBb œ # " -9= 81B ' 1 8 œ " 8# "” _ 8œ" #c" a "b8 d a "b8 •=38 81B . 8$ 1 $ 81 a, bÞ ________________________________________________________________________ page 624 —————————————————————————— CHAPTER 10. —— a- b. The given function is piecewise continuous, with discontinuities at the odd integers . At B. œ #5 " , 5 œ !ß "ß #ß â , the series converges to k0 aB. b 0 aB. bk œ "Î# . At these points the error can never be made arbitrarily small. 13. The solution of the homogenous differential equation is C- a>b œ -" -9= => -# =38 => Þ Given that =# Á 8# , we can use the method of undetermined coefficients to find a particular solution ] a>b œ =# " =38 8> . 8# " =38 8> . 8# Hence the general solution of the ODE is Ca>b œ -" -9= => -# =38 => =# Imposing the initial conditions, we obtain the equations -" œ ! 8 = -# # œ !. = 8# It follows that -# œ 8Îc=a=# 8# bdÞ The solution of the IVP is Ca>b œ =# " 8 =38 8> =38 => . # # 8# b 8 =a= If =# œ 8# , then the forcing function is also one of the fundamental solutions of the ODE. The method of undetermined coefficients may still be used, with a more elaborate trial solution. Using the method of variation of parameters, we obtain ________________________________________________________________________ page 625 —————————————————————————— CHAPTER 10. —— ] a>b œ -9= 8>( =38# 8> -9= 8> =38 8> .> =38 8>( .> 8 8 =38 8> 8> -9= 8> œ Þ #8# Ca>b œ -" -9= 8> -# =38 8> > -9= 8> . #8 In this case, the general solution is Invoking the initial conditions, we obtain -" œ ! and -# œ "Î#8# . Therefore the solution of the IVP is C a >b œ " > =38 8> -9= 8> . # #8 #8 16. Note that the function 0 a>b and the function given in Problem ) have the same Fourier series. Therefore 0 a >b œ " %_ " #" -9=a#8 "b1> Þ # 1 8 œ " a#8 "b# C- a>b œ -" -9= => -# =38 => Þ The solution of the homogeneous problem is Using the method of undetermined coefficients, we assume a particular solution of the form ] a>b œ E! " E8 -9= 81> . _ 8œ" Substitution into the ODE and equating like terms results in E! œ "Î#=# and +8 E8 œ # . = 8# 1 # It follows that the general solution is Ca>b œ -" -9= => -# =38 => " %_ -9=a#8 "b1> #" Þ # #= 1 8 œ " a#8 "b# =# a#8 "b# 1# ‘ Setting Ca!b œ " , we find that " %_ -9=a#8 "b1> -" œ " # # " Þ #= 1 8 œ " a#8 "b# =# a#8 "b# 1# ‘ ________________________________________________________________________ page 626 —————————————————————————— CHAPTER 10. —— Invoking the initial condition C w a!b œ ! , we obtain -# œ ! . Hence the solution of the initial value problem is " " %_ -9=a#8 "b1> -9= => Ca>b œ -9= => # -9= => # # " Þ #= #= 1 8 œ " a#8 "b# =# a#8 "b# 1# ‘ 17. Let _ +! 81 B 81 B " ’+8 -9= 0 aBb œ ,8 =38 “Þ # 8œ" P P Squaring both sides of the equation, we formally have k0 aBbk# œ " ’-78 -9= 7Á8 _ _ # +! 81 B 81 B 81 B 81 B # # " ’+8 -9=# ,8 =38# ,8 =38 “ +! " ’+8 -9= “ % P P P P 8œ" 8œ" 71B 81B =38 “Þ P P Integrating both sides of the last equation, and using the orthogonality conditions, ( P P k0 aBbk# .B œ ( œ _ # +! # # P " +8 P ,8 P‘Þ # 8œ" _ # P P +! 81 B 81 B # # .B " –( +8 -9=# .B ( ,8 =38# .B— P P P P % 8 œ " P P Therefore, _ # "P +! # # # " ˆ+8 ,8 ‰Þ ( k0 aBbk .B œ P P # 8œ" 19a+b. As shown in the Example, the Fourier series of the function 0 aBb œ œ !, P, PB! ! B P, is given by 0 aBb œ Setting P œ " , P #P _ " a#8 "b1B " =38 Þ # 1 8 œ " #8 " P " #_ " 0 aBb œ " =38a#8 "b1B Þ # 1 8 œ " #8 " ________________________________________________________________________ page 627 —————————————————————————— CHAPTER 10. —— It follows that " 1 " =38a#8 "b1B œ ”0 aBb •. #8 " # # 8œ" _ " a33b a,b. Given that and subtracting Eq.a33b from Eq.a3b, we find that 1aBb œ " _ 8œ"" a#8 "b# #8 " =38a#8 "b1B , a 3b _ 1 " #8 " " 1aBb ”0 aBb • œ =38a#8 "b1B # # # 8 œ " " a#8 "b " _ " =38a#8 "b1B Þ #8 " 8œ" Based on the fact that " a#8 "b 1aBb œ #8 " # " " , œ #8 " a#8 "b" a#8 "b# ‘ and the fact that we can combine the two series, it follows that _ 1 " =38a#8 "b1B 0 aBb • " . ” #‘ # # 8 œ " a#8 "b " a#8 "b ________________________________________________________________________ page 628 —————————————————————————— CHAPTER 10. —— Section 10.4 1. Since the function contains only odd powers of B , the function is odd. 2. Since the function contains both odd and even powers of B , the function is neither even nor odd. 4. We have =/- B œ "Î-9= B . Since the quotient of two even functions is even, the function is even. 5. We can write kBk$ œ kBk † kBk# œ kBk † B# . Since both factors are even, it follows that the function is even. 8. P œ # . ________________________________________________________________________ page 629 —————————————————————————— CHAPTER 10. —— 9. P œ # . 11. P œ # . ________________________________________________________________________ page 630 —————————————————————————— CHAPTER 10. —— 12. P œ " . 16. P œ # . For an odd extension of the function, the cosine coefficients are zero. The sine coefficients are given by ________________________________________________________________________ page 631 —————————————————————————— CHAPTER 10. —— #P 81B ,8 œ ( 0 aBb=38 .B P! P " # 81B 81 B œ ( B =38 .B ( =38 .B # # ! " # =38 8#1 81 -9= 81 œ# Þ 8# 1 # Observe that =38Š Likewise, " , 8 œ #5 -9= 81 œ œ " , 8 œ #5 " 0 aBb œ , 5 œ "ß #ß â . ! , 8 œ #5 81 ‹œœ 5 +1 # a "b , 8 œ #5 " , 5 œ "ß #ß â . Therefore the Fourier sine series of the specified function is "_" # _ #a "b8" a#8 "b1 a#8 "b1B " =38 81B # " =38 . # 1 8œ" 8 1 8œ" # a#8 "b 17. P œ 1 . For an even extension of the function, the sine coefficients are zero. The cosine coefficients are given by +! œ #P ( 0 aBb.B P! #1 œ ( a"b.B 1! œ #, ________________________________________________________________________ page 632 —————————————————————————— CHAPTER 10. —— and for 8 ! , +8 œ #P 81B .B ( 0 aBb-9= P! P #1 œ ( a"b-9= 8B .B 1! œ !Þ The even extension of the given function is a constant function. As expected, the Fourier cosine series is +! 0 aBb œ œ ". # 19. P œ $1 . For an odd extension of the function, the cosine coefficients are zero. The sine coefficients are given by ,8 œ #P 81B .B ( 0 aBb=38 P! P #1 $1 # 8B # 8B œ =38 .B ( ( # =38 .B $1 1 $ $1 #1 $ 81 #81 # -9= 81 -9= $ -9= $ œ # Þ 81 Therefore the Fourier sine series of the specified function is #_" 81 #81 8B 0 aBb œ " ”-9= -9= # -9= 81• =38 . 1 8œ" 8 $ $ $ ________________________________________________________________________ page 633 —————————————————————————— CHAPTER 10. —— 21. Extend the function over the interval c P ß Pd as 0 aBb œ œ B P, P B, PŸB! ! Ÿ B Ÿ P. Since the extended function is even, the sine coefficients are zero. The cosine coefficients are given by +! œ #P ( 0 aBb.B P! #P œ ( aP Bb.B P! œ P, and for 8 ! , +8 œ #P 81B .B ( 0 aBb-9= P! P #P 81B œ ( aP Bb-9= .B P! P " -9= 81 œ #P Þ 8# 1 # Therefore the Fourier cosine series of the extended function is 0 aBb œ P %P _ " a#8 "b1B #" . # -9= # 1 8 œ " a#8 "b P In order to compare the result with Example " of Section "!Þ# , set P œ # . The cosine series converges to the function graphed below: This function is a shift of the function in Example " of Section "!Þ# Þ ________________________________________________________________________ page 634 —————————————————————————— CHAPTER 10. —— 22. Extend the function over the interval c P ß Pd as with 0 a!b œ ! . Since the extended function is odd, the cosine coefficients are zero. The sine coefficients are given by #P 81B ,8 œ ( 0 aBb=38 .B P! P #P 81B œ ( aP Bb=38 .B P! P #P œ Þ 81 Therefore the Fourier cosine series of the extended function is #P _ " 81 B " =38 0 aBb œ . 1 8œ" 8 P 0 aBb œ œ B P, P B, PŸB! ! B Ÿ P, ________________________________________________________________________ page 635 —————————————————————————— CHAPTER 10. —— Setting P œ # , for example, the series converges to the function graphed below: 23a+b. P œ #1 . For an even extension of the function, the sine coefficients are zero. The cosine coefficients are given by +! œ #P ( 0 aBb.B P! "1 œ ( B .B 1! œ 1Î# , and for 8 ! , +8 œ #P 81B .B ( 0 aBb-9= P! P "1 8B œ ( B -9= .B 1! # # -9=ˆ 8#1 ‰ 81=38ˆ 8#1 ‰ # œ# Þ 8# 1 Therefore the Fourier cosine series of the given function is 0 aBb œ Observe that =38Š Likewise, ________________________________________________________________________ page 636 1 #_ 1 81 # 81 8B " ” =38 # Š-9= "‹•-9= . % 1 8œ" 8 # 8 # # ! , 8 œ #5 81 ‹œœ 5 +1 # a "b , 8 œ #5 " , 5 œ "ß #ß â . —————————————————————————— CHAPTER 10. —— -9=Š a, b . 81 a "b5 , 8 œ #5 ‹œœ # ! , 8 œ #5 " , 5 œ "ß #ß â . a- b . 24a+b. P œ 1 . For an odd extension of the function, the cosine coefficients are zero. Note that 0 aBb œ B on ! Ÿ B 1 . The sine coefficients are given by ,8 œ #P 81B .B ( 0 aBb=38 P! P #1 œ ( B =38 8B .B 1! # -9= 81 œ Þ 8 Therefore the Fourier sine series of the given function is ________________________________________________________________________ page 637 —————————————————————————— CHAPTER 10. —— 0 aBb œ # " _ a "b 8 =38 8B. 8 8œ" a, b . a- b . 26a+b. P œ % . For an even extension of the function, the sine coefficients are zero. The cosine coefficients are given by #P +! œ ( 0 aBb.B P! "% œ ( ˆB# #B‰.B #! œ )Î$ , and for 8 ! , ________________________________________________________________________ page 638 —————————————————————————— CHAPTER 10. —— #P 81B +8 œ ( 0 aBb-9= .B P! P "% 81 B œ ( ˆB# #B‰-9= .B #! % " $ -9= 81 œ "' Þ 8# 1 # Therefore the Fourier cosine series of the given function is % "' _ " $a "b8 81 B 0 aBb œ # " -9= . # $ 1 8œ" 8 % a, b . a- b . ________________________________________________________________________ page 639 —————————————————————————— CHAPTER 10. —— 27a+b. a,b. P œ $ . For an even extension of the function, the cosine coefficients are given by +! œ #P ( 0 aBb.B P! #$ œ ( a$ Bb.B $! œ $, and for 8 ! , +8 œ #P 81B .B ( 0 aBb-9= P! P #$ 81B œ ( a$ Bb-9= .B $! $ " -9= 81 œ' Þ 8# 1 # Therefore the Fourier cosine series of the given function is 1aBb œ $ ' _ " a "b 8 81 B #" -9= . # # 1 8œ" 8 $ For an odd extension of the function, the sine coefficients are given by #P 81 B ,8 œ ( 0 aBb=38 .B P! P #$ 81 B œ ( a$ Bb=38 .B $! $ ' œ Þ 81 Therefore the Fourier sine series of the given function is ________________________________________________________________________ page 640 —————————————————————————— CHAPTER 10. —— '_" 81B 2aBb œ " =38 . 1 8œ" 8 $ a- b. For the even extension: For the odd extension: a. b. Since the even extension is continuous, the series converges uniformly. On the other hand, the odd extension is discontinuous. Gibbs' phenomenon results in a finite error for all values of 8 . ________________________________________________________________________ page 641 —————————————————————————— CHAPTER 10. —— 29a+b. a,b. P œ # . For an even extension of the function, the cosine coefficients are given by +! œ #P ( 0 aBb.B P! # %B# %B $ œ( ” •.B % ! œ &Î' , and for 8 ! , +8 œ #P 81B .B ( 0 aBb-9= P! P # %B# %B $ 81 B œ( ” .B •-9= % # ! " $ -9= 81 œ% Þ 8# 1 # & % _ " $a "b 8 81 B #" -9= . # "# 1 8 œ " 8 # Therefore the Fourier cosine series of the given function is 1aBb œ For an odd extension of the function, the sine coefficients are given by #P 81B ,8 œ ( 0 aBb=38 .B P! P # %B# %B $ 81 B œ( ” .B •=38 % # ! $# $8# 1# &8# 1# -9= 81 $# -9= 81 œ Þ # 8$ 1 $ Therefore the Fourier sine series of the given function is ________________________________________________________________________ page 642 —————————————————————————— CHAPTER 10. —— " _ $#a" -9= 81b 8# 1# a$ & -9= 81b 81 B 2aBb œ $ " =38 . $ #1 8 œ " 8 # a- b. For the even extension: For the 9.. extension: a. b. Since the even extension is continuous, the series converges uniformly. On the other hand, the odd extension is discontinuous. Gibbs' phenomenon results in a finite error for all values of 8 . ________________________________________________________________________ page 643 —————————————————————————— CHAPTER 10. —— 30a+b. a,b. P œ $ . For an even extension of the function, the cosine coefficients are given by +! œ #P ( 0 aBb.B P! #$ œ ( ˆB$ &B# &B "‰.B $! œ "Î# , and for 8 ! , +8 œ #P 81B .B ( 0 aBb-9= P! P #$ 81B œ ( ˆB$ &B# &B "‰-9= .B $! $ "'# "& 8# 1# ' 8# 1# -9= 81 "'# -9= 81 œ# Þ 8% 1 % Therefore the Fourier cosine series of the given function is 1aBb œ " # _ "'#a" -9= 81b $ 8# 1# a& # -9= 81b 81 B %" -9= . % 1 8œ" 8% $ #P 81 B .B ( 0 aBb=38 P! P #$ 81B œ ( ˆB$ &B# &B "‰=38 .B $! $ *! 8# 1# # 8# 1# -9= 81 (# -9= 81 œ# Þ 8$ 1 $ For an odd extension of the function, the sine coefficients are given by ,8 œ Therefore the Fourier sine series of the given function is ________________________________________________________________________ page 644 —————————————————————————— CHAPTER 10. —— # _ ")a& % -9= 81b 8# 1# a" # -9= 81b 81 B 2aBb œ $ " =38 . $ 1 8œ" 8 $ a- b. For the even extension: For the odd extension: a. b. Since the even extension is continuous, the series converges uniformly. On the other hand, the odd extension is discontinuous. Gibbs' phenomenon results in a finite error for all values of 8 ; particularly at B œ „ $ . 33. Let 0 aBb be a differentiable even function. For any B in its domain, 0 a B 2b 0 a Bb œ 0 aB 2 b 0 aBb . It follows that ________________________________________________________________________ page 645 —————————————————————————— CHAPTER 10. —— 0 w a Bb œ lim 0 a B 2 b 0 a Bb 2Ä! 2 0 aB 2b 0 aBb œ lim 2Ä! 2 0 aB 2b 0 aBb œ lim . 2Ä! a 2b 0 aB $ b 0 aBb 2Ä! $ 0 aB $ b 0 aBb œ lim $ Ä ! $ w œ 0 aBb . Setting 2 œ $ , we have 0 w a Bb œ lim Therefore 0 w a Bb œ 0 w aBb . If 0 aBb is a differentiable odd function, for any B in its domain, 0 a B 2b 0 a Bb œ 0 aB 2 b 0 aBb . 0 w a Bb œ lim 0 a B 2 b 0 a Bb 2Ä! 2 0 aB 2b 0 aBb œ lim 2Ä! 2 0 aB 2b 0 aBb œ lim . 2Ä! a 2b 0 aB $ b 0 aBb 2Ä! $ 0 aB $ b 0 aBb œ lim $ Ä ! $ w œ 0 aBb . It follows that Setting 2 œ $ , we have 0 w a Bb œ lim Therefore 0 w a Bb œ 0 w aBb . 36. From Example " of Section "!Þ# , the function 0 aBb œ œ B, B, #ŸB! ! Ÿ B #, aP œ #b has a convergent Fourier series ________________________________________________________________________ page 646 —————————————————————————— CHAPTER 10. —— )_ " a#8 "b1B " . # -9= # 1 8 œ " a#8 "b # 0 aBb œ " Since 0 aBb is continuous, the series converges everywhere. In particular, at B œ ! , we have ! œ 0 a!b œ " It follows immediately that _ 1# " " " " œ" œ " # # # â. # ) $ & ( 8 œ " a#8 "b )_ " " . 1# 8 œ " a#8 "b# 40. Since one objective is to obtain a Fourier series containing only cosine terms, any extension of 0 aBb should be an even function. Another objective is to derive a series containing only the terms -9= First note that the functions -9= 81B , 8 œ "ß #ß â P a#8 "b1B , 8 œ "ß #ß â . #P are symmetric about B œ P . Indeed, ________________________________________________________________________ page 647 —————————————————————————— CHAPTER 10. —— 81a#P Bb 81 B œ -9=Š#81 ‹ P P 81B œ -9=Š ‹ P 81B œ -9= . P -9= It follows that if 0 aBb is extended into aP ß #Pb as an antisymmetric function about B œ P, that is, 0 a#P Bb œ 0 aBb for ! Ÿ B Ÿ #P , then ( #P ! 0 aBb-9= 81B .B œ ! . P This follows from the fact that the integrand is antisymmetric function about B œ P. Now extend the function 0 aBb to obtain 0 aBb œ œ µ Finally, extend the resulting function into a #P ß !b as an even function, and then as a periodic function of period %P. By construction, the Fourier series will contain only cosine terms. We first note that #P µ # +! œ ( 0 aBb .B #P ! "P " #P œ ( 0 aBb.B ( 0 a#P Bb.B P! PP P " "P œ ( 0 aBb.B ( 0 a?b.? P! P! œ !. 0 aBb, ! Ÿ B P 0 a#P Bb, P B #P . For 8 ! , +8 œ #P µ # 81B .B ( 0 aBb-9= #P ! #P "P 81B " #P 81 B œ ( 0 aBb-9= .B ( 0 a#P Bb-9= .B . P! #P PP #P For the second integral, let ? œ #P B . Then -9= 81B 81a#P ?b 81 ? œ -9= œ a "b8 -9= #P #P #P ________________________________________________________________________ page 648 —————————————————————————— CHAPTER 10. —— and therefore ( Hence #P P 0 a#P Bb-9= P 81B 81 ? .B œ a "b8 ( 0 a?b-9= .? . #P #P ! " a "b 8 P 81 B +8 œ .B . ( 0 aBb-9= P #P ! #P a#5 "b1B œ ( 0 aBb-9= .B , for 5 œ "ß #ß â . P! #P 0 aBb œ " +#8" -9= _ 8œ! It immediately follows that +8 œ ! for 8 œ #5 , 5 œ !ß "ß #ß â , and +#5" The associated Fourier series representation converges almost everywhere on a #P ß #Pb and hence on a! ß Pb. a#8 "b1B #P For example, if 0 aBb œ B for ! Ÿ B Ÿ P œ " , the graph of the extended function is: ________________________________________________________________________ page 649 —————————————————————————— CHAPTER 10. —— Section 10.5 1. We consider solutions of the form ?aB ß >b œ \ aBbX a>b. Substitution into the partial differential equation results in B\ ww X \X w œ ! . Divide both sides of the differential equation by the product \X to obtain B so that B \ ww Xw œ . \ X \ ww Xw œ!ß \ X Since both sides of the resulting equation are functions of different variables, each must be equal to a constant, say - . We obtain the ordinary differential equations B\ ww -\ œ ! and X w -X œ ! . 2. In order to apply the method of separation of variables, we consider solutions of the form ?aB ß >b œ \ aBbX a>b. Substituting the assumed form of the solution into the partial differential equation, we obtain > \ ww X B \X w œ ! . Divide both sides of the differential equation by the product B>\X to obtain \ ww Xw œ!ß B\ >X so that \ ww Xw œ . B\ >X Since both sides of the resulting equation are functions of different variables, it follows that \ ww Xw œ œ -. B\ >X Therefore \ aBb and X a>b are solutions of the ordinary differential equations \ ww -B \ œ ! and X w -> X œ ! . ________________________________________________________________________ page 650 —————————————————————————— CHAPTER 10. —— 4. Assume that the solution of the PDE has the form ?aB ß >b œ \ aBbX a>b. Substitution into the partial differential equation results in Divide both sides of the differential equation by the product <aBb\X to obtain c:aBb\ w d w X ww œ!ß <aBb\ X c:aBb\ w d w X ww œ . <aBb\ X c:aBb\ w d -<aBb\ œ ! and X ww -X œ ! . w c:aBb\ w d X <aBb\ X ww œ ! . w that is, Since both sides of the resulting equation are functions of different variables, each must be equal to a constant, say - . We obtain the ordinary differential equations 6. We consider solutions of the form ?aB ß Cb œ \ aBb] aCb. Substitution into the partial differential equation results in \ ww ] \] ww B\] œ ! . Divide both sides of the differential equation by the product \] to obtain \ ww ] ww Bœ!ß \ ] that is, \ ww ] ww Bœ Þ \ ] Since both sides of the resulting equation are functions of different variables, it follows that \ ww ] ww Bœ œ -. \ ] We obtain the ordinary differential equations \ ww aB -b\ œ ! and ] ww -] œ ! . 7. The heat conduction equation, "!! ?BB œ ?> , and the given boundary conditions are homogeneous. We consider solutions of the form ?aB ß >b œ \ aBbX a>b. Substitution into the partial differential equation results in ________________________________________________________________________ page 651 —————————————————————————— CHAPTER 10. —— "!! \ ww X œ \X w . Divide both sides of the differential equation by the product \X to obtain \ ww Xw œ Þ \ "!! X Since both sides of the resulting equation are functions of different variables, it follows that \ ww Xw œ œ -. \ "!! X Therefore \ aBb and X a>b are solutions of the ordinary differential equations \ ww - \ œ ! and X w "!!- X œ ! . The general solution of the spatial equation is \ œ -" -9= -"Î# B -# =38 -"Î# B . In order to satisfy the homogeneous boundary conditions, we require that -" œ ! , and -"Î# œ 81 . Hence the eigenfunctions are \8 œ =38 81B , with associated eigenvalues -8 œ 8# 1# . We thus obtain the family of equations X w "!!-8 X œ ! . Solution are given by X8 œ /"!!-8 > Þ Hence the fundamental solutions of the PDE are ?8 aB ß >b œ /"!!8 1 > =38 81B , ## which yield the general solution ? aB ß >b œ " -8 /"!!8 1 > =38 81B Þ _ ## Finally, the initial condition ?aB ß !b œ =38 #1B =38 &1B must be satisfied. Therefore is it necessary that " -8 =38 81B œ =38 #1B =38 &1B . _ 8œ" 8œ" It follows from the othogonality conditions that -# œ -& œ " , with all other -8 œ ! . Therefore the solution of the given heat conduction problem is ? aB ß >b œ /%!!1 > =38 #1B /#&!!1 > =38 &1B . # # ________________________________________________________________________ page 652 —————————————————————————— CHAPTER 10. —— 9. The heat conduction problem is formulated as ?BB œ ?> , ?a! ß >b œ ! , ?aB ß !b œ &! , ! B %! , > ! à ?a%! ß >b œ ! , > ! à ! B %! . Assume a solution of the form ?aB ß >b œ \ aBbX a>b . Following the procedure in this section, we obtain the eigenfunctions \8 œ =38 81BÎ%! , with associated eigenvalues -8 œ 8# 1# /"'!! . The solutions of the temporal equations are X 8 œ / - 8 > Þ Hence the general solution of the given problem is ? aB ß >b œ " -8 /8 1 >Î"'!! =38 _ ## The coefficients -8 are the Fourier sine coefficients of ?aB ß !b œ &! . That is, -8 œ #P 81B .B ( 0 aBb=38 P! P & %! 81B œ ( =38 .B #! %! " -9= 81 œ "!! Þ 81 8œ" 81B Þ %! The sine series of the initial condition is &! œ "!! _ " -9= 81 81 B " =38 . 1 8œ" 8 %! Therefore the solution of the given heat conduction problem is "!! _ " -9= 81 8# 1# >Î"'!! 81 B " ? a B ß >b œ / =38 Þ 1 8œ" 8 %! 11. Refer to Prob. * for the formulation of the problem. In this case, the initial condition is given by Ú !ß ? aB ß !b œ Û &! ß Ü !ß ! Ÿ B "! ß "! Ÿ B Ÿ $! ß $! B Ÿ %! . All other data being the same, the solution of the given problem is ________________________________________________________________________ page 653 —————————————————————————— CHAPTER 10. —— ?aB ß >b œ " -8 /8 1 >Î"'!! =38 _ ## The coefficients -8 are the Fourier sine coefficients of ?aB ß !b . That is, -8 œ #P 81B .B ( 0 aBb=38 P! P & $! 81 B œ ( =38 .B # "! %! -9= 8%1 -9= $81 % œ "!! Þ 81 8œ" 81B Þ %! Therefore the solution of the given heat conduction problem is ?aB ß >b œ "!! _ -9= 8%1 -9= $81 8# 1# >Î"'!! 81B % " / =38 Þ 1 8œ" 8 %! 12. Refer to Prob. * for the formulation of the problem. In this case, the initial condition is given by ?aB ß !b œ B , ! B %! . _ All other data being the same, the solution of the given problem is ? aB ß >b œ " -8 /8 1 >Î"'!! =38 ## The coefficients -8 are the Fourier sine coefficients of ?aB ß !b œ B . That is, #P 81B -8 œ ( 0 aBb=38 .B P! P " %! 81B œ .B ( B =38 #! ! %! -9= 81 œ )! Þ 81 8œ" 81B Þ %! Therefore the solution of the given heat conduction problem is ? a B ß >b œ )! _ a "b8" 8# 1# >Î"'!! 81 B " / =38 Þ 1 8œ" 8 %! ________________________________________________________________________ page 654 —————————————————————————— CHAPTER 10. —— 13. Substituting B œ #! , into the solution, we have ? a#! ß >b œ We can also write "!! _ " -9= 81 8# 1# >Î"'!! 81 " / =38 Þ 1 8œ" 8 # #!! _ a "b5" a#5"b# 1# >Î"'!! " ? a#! ß >b œ / Þ 1 5 œ " #5 " #!! _ a "b5" a#5"b# 1# Î$#! " ? a#! ß &b œ / Þ 1 5 œ " #5 " a "b8" #!! a#5"b# 1# Î$#! / Þ 1a#5 "b Therefore, Let E5 œ It follows that kE5 k !Þ!!& for 5 * . So for 8 œ #5 " "(, the summation is unaffected by additional terms. For > œ #! , ? a#! ß #!b œ Let #!! _ a "b5" a#5"b# 1# Î)! " / Þ 1 5 œ " #5 " It follows that kE5 k !Þ!!$ for 5 & . So for 8 œ #5 " *, the summation is unaffected by additional terms. For > œ )! , ? a#! ß )!b œ Let E5 œ #!! _ a "b5" a#5"b# 1# Î#! " / Þ 1 5 œ " #5 " a "b8" #!! a#5"b# 1# Î#! / Þ 1a#5 "b a "b8" #!! a#5"b# 1# Î)! E5 œ / Þ 1a#5 "b It follows that kE5 k !Þ!!!!& for 5 $ . So for 8 œ #5 " &, the summation is unaffected by additional terms. ________________________________________________________________________ page 655 —————————————————————————— CHAPTER 10. —— The series solution converges faster as > increases. 14a+b. The solution of the given heat conduction problem is "!! _ " -9= 81 8# 1# >Î"'!! 81 B " ? a B ß >b œ / =38 Þ 1 8œ" 8 %! Setting > œ &ß "!ß #!ß %!ß "!!ß #!! À a,b. Setting B œ &ß "!ß "&ß #! À ________________________________________________________________________ page 656 —————————————————————————— CHAPTER 10. —— a- b. Surface plot of ?aB ß >b À a. b. ! Ÿ ?aB ß >b Ÿ " for > '(& =/- . 16a+b. The solution of the given heat conduction problem is "!! _ -9= 8%1 -9= $81 8# 1# >Î"'!! 81B % " ?aB ß >b œ / =38 Þ 1 8œ" 8 %! ________________________________________________________________________ page 657 —————————————————————————— CHAPTER 10. —— Setting > œ &ß "!ß #!ß %!ß "!!ß #!! À a,bÞ Setting B œ &ß "!ß "&ß #! À a- b. Surface plot of ?aB ß >b À ________________________________________________________________________ page 658 —————————————————————————— CHAPTER 10. —— a. b. ! Ÿ ?aB ß >b Ÿ " for > '"& =/- . 17a+b. The solution of the given heat conduction problem is ? a B ß >b œ )! _ a "b8" 8# 1# >Î"'!! 81 B " / =38 Þ 1 8œ" 8 %! Setting > œ &ß "!ß #!ß %!ß "!!ß #!! À a,b. Analyzing the individual plots, we find that the 'hot spot' varies with time: > B2 & $$ "! $" #! #* %! #' "!! ## #!! #" ________________________________________________________________________ page 659 —————————————————————————— CHAPTER 10. —— Location of the 'hot spot', B2 , versus time À Evidently, the location of the greatest temperature migrates to the center of the rod. a- bÞ Setting B œ &ß "!ß "&ß #! À a. b. Surface plot of ?aB ß >b À ________________________________________________________________________ page 660 —————————————————————————— CHAPTER 10. —— a/b. ! Ÿ ?aB ß >b Ÿ " for > &#& =/- . 19. The solution of the given heat conduction problem is ? a B ß >b œ Setting B œ "! -7 , #!! _ " -9= 81 8# 1# !# >Î%!! 81 " ? a"! ß >b œ / =38 Þ 1 8œ" 8 # A two-term approximation is given by ? a"! ß >b ¸ From Table "!Þ&Þ" À silver aluminum cast iron !# "Þ(" !Þ)' !Þ"# %!! ## ## ’$ /1 ! >Î%!! /*1 ! >Î%!! “Þ $1 #!! _ " -9= 81 8# 1# !# >Î%!! 81 B " / =38 Þ 1 8œ" 8 #! ________________________________________________________________________ page 661 —————————————————————————— CHAPTER 10. —— a+b. !# œ "Þ(" À a,b. !# œ !Þ)' À a- b. !# œ !Þ"# À ________________________________________________________________________ page 662 —————————————————————————— CHAPTER 10. —— 21a+b. Given the partial differential equation in which + , , , and - are constants, set ?aB ß >b œ /$ > AaB ß >b. Substitution into the PDE results in + /$ > ABB ,ˆ$ /$ > A /$ > A> ‰ - /$ > A œ ! . + ABB , A> a- , $ b A œ ! . + ?BB , ?> - ? œ ! , Dividing both sides of the equation by /$ > , we obtain As long as , Á ! , choosing $ œ -Î, yields + ABB A> œ ! , , which is the heat conduction equation with dependent variable A . 23. The heat conduction equation in polar coordinates is given by !# ”?<< " " ?< # ?)) • œ ?> . < < We consider solutions of the form ?a< ß ) ß >b œ V a<b@a)bX a>b. Substitution into the PDE results in !# ”V ww @X "w " V @X # V @ ww X • œ V @X w . < < Dividing both sides of the equation by the factor V @X , we obtain V ww " Vw " @ ww Xw # œ#. V <V <@ !X Since both sides of the resulting differential equation depend on different variables, each side must be equal to a constant, say - . That is, V ww " Vw " @ ww Xw # œ # œ -# . V <V <@ !X It follows that X w !# -# X œ ! , and V ww " Vw " @ ww # œ -# . V <V <@ Multiplying both sides of this differential equation by <# , we find that <# V ww Vw @ ww < œ -# < # , V V @ ________________________________________________________________________ page 663 —————————————————————————— CHAPTER 10. —— which can be written as <# V ww Vw @ ww < -# < # œ . V V @ Once again, since both sides of the resulting differential equation depend on different variables, each side must be equal to a constant. Hence <# V ww Vw @ ww < -# <# œ .# and œ .# . V V @ <# V ww <V w ˆ-# <# .# ‰V œ ! The resulting ordinary equations are @ ww .# @ œ ! X w !# -# X œ ! Þ ________________________________________________________________________ page 664 —————————————————————————— CHAPTER 10. —— Section 10.6 1. The steady-state solution, @aBb, satisfies the boundary value problem The general solution of the ODE is @aBb œ EB F . Imposing the boundary conditions, we have @aBb œ %! "! $B B "! œ "! Þ &! & @ ww aBb œ ! , ! B &! , @a!b œ "! , @a&!b œ %! . 2. The steady-state solution, @aBb, satisfies the boundary value problem @ ww aBb œ ! , ! B %! , @a!b œ $! , @a%!b œ #! . @aBb œ #! $! &B B $! œ $! Þ %! % The solution of the ODE is linear. Imposing the boundary conditions, we have 4. The steady-state solution is also a solution of the boundary value problem given by @ ww aBb œ ! , ! B P , and the conditions @ w a!b œ ! , @aPb œ X . The solution of the ODE is @aBb œ EB F . The boundary condition @ w a!b œ ! requires that E œ ! . The other condition requires that F œ X Þ Hence @aBb œ X Þ 5. As in Prob. % , the steady-state solution has the form @aBb œ EB F . The boundary condition @a!b œ ! requires that F œ ! . The boundary condition @ w aPb œ ! requires that E œ ! . Hence @aBb œ ! Þ 6. The steady-state solution has the form @aBb œ EB F . The first boundary condition, @a!b œ X , requires that F œ X . The other boundary condition, @ w aPb œ ! , requires that E œ ! . Hence @aBb œ X Þ 8. The steady-state solution, @aBb, satisfies the differential equation @ ww aBb œ ! , along with the boundary conditions @a!b œ X , @ w aPb @aPb œ ! . @ w aPb @aPb œ E EP X . @aBb œ XB X . "P The general solution of the ODE is @aBb œ EB F . The boundary condition @ w a!b œ ! requires that F œ X . It follows that @aBb œ EB X , and The second boundary condition requires that E œ X Îa" Pb . Therefore ________________________________________________________________________ page 665 —————————————————————————— CHAPTER 10. —— 10a+b. Based on the symmetry of the problem, consider only left half of the bar. The steady-state solution satisfies the ODE @ ww aBb œ ! , along with the boundary conditions @a!b œ ! and @a&!b œ "!! . The solution of this boundary value problem is @aBb œ #B . It follows that the steady-state temperature is the entire rod is given by 0 aBb œ œ #B , ! Ÿ B Ÿ &! #!! #B , &! Ÿ B Ÿ "!! Þ a,b. The heat conduction problem is formulated as !# ?BB œ ?> , ?a! ß >b œ #! , ?aB ß !b œ 0 aBb , ! B "!! , > ! à ?a"!! ß >b œ ! , > ! à ! B "!! . First express the solution as ?aB ß >b œ 1aBb AaB ß >b , where 1aBb œ BÎ& #! and A satisfies the heat conduction problem !# ABB œ A> , ! B "!! , > ! à A a! ß > b œ ! , Aa"!! ß >b œ ! , > ! à AaB ß !b œ 0 aBb 1aBb , ! B "!! . Based on the results in Section "!Þ& , AaB ß >b œ " -8 /8 1 ! >Î"!!!! =38 _ ### in which the coefficients -8 are the Fourier sine coefficients of 0 aBb 1aBb. That is, #P 81B -8 œ ( c0 aBb 1aBbd=38 .B P! P " "!! 81B œ c0 aBb 1aBbd=38 .B ( &! ! "!! #! =38 8#1 81 œ %! Þ 8# 1 # 8œ" 81B , "!! Finally, the thermal diffusivity of copper is "Þ"% -7# Î=/- . Therefore the temperature distribution in the rod is ? aB ß >b œ #! B %! _ #! =38 8#1 81 "Þ"% 8# 1# >Î"!!!! 81B #" / =38 Þ & 1 8œ" 8# "!! ________________________________________________________________________ page 666 —————————————————————————— CHAPTER 10. —— a- b. > œ & ß "!ß #!ß %! =/- À > œ "!!ß #!!ß $!!ß &!! =/- À ________________________________________________________________________ page 667 —————————————————————————— CHAPTER 10. —— a. b. The steady-state temperature of the center of the rod will be 1a&!b œ "!°G . Using a one-term approximation, ? aB ß >b ¸ "! )!! %! 1 "Þ"% 1# >Î"!!!! / Þ 1# Numerical investigation shows that "! ?a&! ß >b "" for > $(&& =/- . ________________________________________________________________________ page 668 —————————————————————————— CHAPTER 10. —— 11a+bÞ The heat conduction problem is formulated as ?BB œ ?> , ?a! ß >b œ $! , ?aB ß !b œ 0 aBb , in which the initial condition is given by 0 aBb œ Ba'! BbÎ$! . Express the solution as ?aB ß >b œ @aBb AaB ß >b , where @aBb œ $! B and A satisfies the heat conduction problem ABB œ A> , ! B $! , > ! à A a ! ß >b œ ! , Aa$! ß >b œ ! , > ! à AaB ß !b œ 0 aBb @aBb , ! B $! . As shown in Section "!Þ& , AaB ß >b œ " -8 /8 1 >Î*!! =38 _ ## ! B $! , > ! à ?a$! ß >b œ ! , > ! à ! B $! , in which the coefficients -8 are the Fourier sine coefficients of 0 aBb @aBb. That is, -8 œ #P 81B .B ( c0 aBb 1aBbd=38 P! P " $! 81B œ .B ( c0 aBb 1aBbd=38 "& ! $! #a" -9= 81b 8# 1# a" -9= 81b œ '! Þ 8$ 1 $ 8œ" 81B , $! Therefore the temperature distribution in the rod is '! _ #a" -9= 81b 8# 1# a" -9= 81b 8# 1# >Î*!! 81B ? aB ß >b œ $! B $ " / =38 Þ 8$ 1 8œ" $! a,b. > œ & ß "!ß #!ß %! =/- À ________________________________________________________________________ page 669 —————————————————————————— CHAPTER 10. —— > œ &!ß (&ß "!!ß #!! =/- À ________________________________________________________________________ page 670 —————————————————————————— CHAPTER 10. —— a- b . Based on the heat conduction equation, the rate of change of the temperature at any given point is proportional to the concavity of the graph of ? versus B , that is, ?BB . Evidently, near > œ '! , the concavity of ?aB ß >b changes. 13a+b. The heat conduction problem is formulated as ?BB œ % ?> , ?B a! ß >b œ ! , ?aB ß !b œ 0 aBb , ! B %! , > ! à ?B a%! ß >b œ ! , > ! à ! B %! , in which the initial condition is given by 0 aBb œ Ba'! BbÎ$! . ? a B ß >b œ As shown in the discussion on rods with insulated ends, the solution is given by _ -! 81B ### " -8 /8 1 ! >Î"'!! -9= , # %! 8œ" where -8 are the Fourier cosine coefficients. In this problem, ________________________________________________________________________ page 671 —————————————————————————— CHAPTER 10. —— #P -! œ ( 0 aBb.B P! " %! Ba'! Bb œ .B ( #! ! $! œ %!!Î* , and for 8 " , #P 81 B -8 œ ( 0 aBb-9= .B P! P " %! Ba'! Bb 81B œ -9= .B ( #! ! $! %! "'!a$ -9= 81b œ Þ $8# 1# Therefore the temperature distribution in the rod is #!! "'! _ a$ -9= 81b 8# 1# >Î'%!! 81 B ? a B ß >b œ #" / -9= . # * $1 8 œ " 8 %! a,b. > œ &!ß "!!ß "&!ß #!! =/- À ________________________________________________________________________ page 672 —————————————————————————— CHAPTER 10. —— > œ %!ß '!!ß )!!ß "!!! =/- À a- b. Since >Ä_ lim /8 1 >Î'%!! -9= ## 81B œ! %! for each B , it follows that the steady-state temperature is ?_ œ #!!Î* . ________________________________________________________________________ page 673 —————————————————————————— CHAPTER 10. —— a. b. We first note that ? a%! ß >b œ #!! "'! _ a "b8 a$ -9= 81b 8# 1# >Î'%!! #" / . 8# * $1 8 œ " For large values of > , an approximation is given by Numerical investigation shows that ##Þ## ?a%! ß >b #$Þ## for > "&&! =/- . 16a+b. The heat conduction problem is formulated as ?BB œ ?> , ?a! ß >b œ ! , ?aB ß !b œ 0 aBb , ! B $! , > ! à ?B a$! ß >b œ ! , > ! à ! B $! , ? a%! ß >b ¸ #!! $#! 1# >Î'%!! #/ . * $1 in which the initial condition is given by 0 aBb œ $! B. Based on the results of Prob. "&, the solution is given by ? aB ß >b œ " -8 /a#8"b 1 >Î$'!! =38 _ ## 8œ" 81B , '! in which -8 œ #P a#8 "b1B .B ( 0 aBb=38 P! '! " $! a#8 "b1B œ .B ( a$! Bb=38 "& ! '! # -9= 81 a#8 "b1 œ "#! Þ a#8 "b# 1# ________________________________________________________________________ page 674 —————————————————————————— CHAPTER 10. —— Therefore the solution of the heat conduction problem is ? aB ß >b œ "#! " _ # -9= 81 a#8 "b1 a#8 "b # 1 # /a#8"b 1 >Î$'!! =38 ## 8œ" 81 B . '! a,b. > œ "!ß #!ß $!ß %! =/- À > œ %!ß '!ß )!ß "!! =/- À ________________________________________________________________________ page 675 —————————————————————————— CHAPTER 10. —— > œ "!!ß "&!ß #!!ß #&! =/- À ________________________________________________________________________ page 676 —————————————————————————— CHAPTER 10. —— a- b . The location of B2 moves from B œ ! to B œ $! . a. b . 17a+b. The heat conduction problem is formulated as ?BB œ ?> , ! B $! , > ! à ?a! ß >b œ %! , ?B a$! ß >b œ ! , > ! à ?aB ß !b œ $! B , ! B $! , @ ww œ ! , @a!b œ %! and @ w a$!b œ ! . ?aB ß >b œ %! AaB ß >b , The steady-state temperature satisfies the boundary value problem It easy to see we must have @aBb œ %! . Express the solution as ________________________________________________________________________ page 677 —————————————————————————— CHAPTER 10. —— in which A satisfies the heat conduction problem ABB œ A> , ! B $! , > ! à A a ! ß >b œ ! , AB a$! ß >b œ ! , > ! à AaB ß !b œ "! B , ! B $! . As shown in Prob. "&, the solution is given by A aB ß >b œ " -8 /a#8"b 1 >Î$'!! =38 _ ## 8œ" 81B , '! in which -8 œ a#8 "b1B #P .B ( 0 aBb=38 P! '! a#8 "b1B " $! œ .B ( a "! Bb=38 "& ! '! ' -9= 81 a#8 "b1 œ %! Þ a#8 "b# 1# _ Therefore the solution of the original heat conduction problem is ? aB ß >b œ %! %! " a,b. > œ "!ß $!ß &!ß (! =/- À ' -9= 81 a#8 "b1 a#8 "b # 1 # /a#8"b 1 >Î$'!! =38 ## 8œ" 81 B . '! ________________________________________________________________________ page 678 —————————————————————————— CHAPTER 10. —— > œ "!!ß #!!ß $!!ß %!! =/- À a- b. Observe the concavity of the curves. Note also that the temperature at the insulated end tends to the value of the fixed temperature at the boundary B œ ! . 18. Setting - œ .# , the general solution of the ODE \ ww .# \ œ ! is The boundary conditions C w a!b œ C w aPb œ ! lead to the system of equations . 5" / 3.P \ aBb œ 5" /3.B 5 # / 3.B Þ . 5" . 5 # œ ! .5 # / 3.P œ ! Þ a‡b If . œ ! , then the solution of the ODE is \ œ EB F . The boundary conditions require that \ œ F . If . Á ! , then the system algebraic equations has a nontrivial solution if and only if the coefficient matrix is singular. Set the determinant equal to zero to obtain / 3.P / 3.P œ ! . ________________________________________________________________________ page 679 —————————————————————————— CHAPTER 10. —— Let . œ / 35 . Then 3.P œ 3/ P 5 P , and the previous equation can be written as / 5 P / 3/ P / 5 P / 3 / P œ ! . Using Euler's relation, /3/ P œ -9= / P 3 =38 / P , we obtain /5P a-9= / 3 =38 / b /5P a-9= / 3 =38 / b œ ! . ˆ/5P /5P ‰-9= / P œ ! ˆ/5P /5P ‰=38 / P œ ! Þ Equating the real and imaginary parts of the equation, Based on the second equation, / P œ 81 , 8 − ˆ . Since -9= 8P Á ! , it follows that /5P œ /5P , or /#5P œ " . Hence 5 œ ! , and . œ 81ÎP , 8 − ˆ . Note that if 5 Á !, then the last two equations have no solution. It follows that the system of equations a‡b has no nontrivial solutions. 20a+b. Consider solutions of the form ?aB ß >b œ \ aBbX a>b. Substitution into the partial differential equation results in !# \ ww X œ X w . Divide both sides of the differential equation by the product \X to obtain \ ww Xw œ#. \ !X Since both sides of the resulting equation are functions of different variables, each must be equal to a constant, say - . We obtain the ordinary differential equations \ ww -\ œ ! and X w -!# X œ ! . Invoking the first boundary condition, ?a! ß >b œ \ a!bX a>b œ ! . At the other boundary, ?B aP ß >b # ?aP ß >b œ c\ w aPb # \ aPbdX a>b œ ! . \ a!b œ ! and \ w aPb # \ aPb œ ! . Since these conditions are valid for all > ! , it follows that ________________________________________________________________________ page 680 —————————————————————————— CHAPTER 10. —— a,bÞ We consider the boundary value problem \ ww -\ œ ! , ! B P à \ a!b œ ! , \ w aPb # \ aPb œ ! . a‡b Assume that - is real, with - œ .# . The general solution of the ODE is \ aBb œ -" -9=2a.Bb -# =382a.Bb . The first boundary condition requires that -" œ !Þ Imposing the second boundary condition, If -# Á ! , then .-9=2a.Pb # =382a.Pb œ ! , which can also be written as If # œ . , then it follows that -9=2a.Pb œ =382a.Pb, and hence . œ !. If # Á ., then /.P œ /.P again implies that . œ ! . For the case . œ ! , the general solution is \ aBb œ EB F . Imposing the boundary conditions, we have F œ ! and If # œ "ÎP , then \ aBb œ EB is a solution of a‡b. Otherwise E œ ! . a- b. Let - œ .# , with . ! . The general solution of a‡b is \ aBb œ -" -9=a.Bb -# =38a.Bb . -# .-9=a.Pb # -# =38a.Pb œ ! . .-9=a.Pb # =38a.Pb œ ! . a‡‡b E # EP œ ! . a. # b/.P a. # b/.P œ ! . -# .-9=2a.Pb # -# =382a.Pb œ ! . The first boundary condition requires that -" œ !Þ From the second boundary condition, For a nontrivial solution, we must have a. b. The last equation can also be written as >+8 .P œ The eigenvalues - obtained from the solutions of a‡‡b, which are infinite in number. In the graph below, we assume # P œ " . . Þ # ________________________________________________________________________ page 681 —————————————————————————— CHAPTER 10. —— For # P œ " À Denote the nonzero solutions of a‡‡b by ." , .# , .$ ,â . a/b. We can in principle calculate the eigenvalues -8 œ .# . Hence the associated 8 eigenfunctions are \8 œ =38 .8 B . Furthermore, the solutions of the temporal equations are X8 œ /B:a !# .# >bÞ The fundamental solutions of the heat conduction problem 8 are given as ?8 aB ß >b œ /! .8 > =38 .8 B , ## which lead to the general solution ? aB ß >b œ " -8 /! .8 > =38 .8 B . _ ## 8œ" ________________________________________________________________________ page 682 —————————————————————————— CHAPTER 10. —— Section 10.7 2a+b. The initial velocity is zero. Therefore the solution, as given by Eq. a#!b, is ?aB ß >b œ " -8 =38 _ 8œ" in which the coefficients are the Fourier sine coefficients of 0 aBb. That is, #P 81B -8 œ ( 0 aBb=38 .B P! P 81 B 81 + > -9= , P P PÎ% $PÎ% P # %B 81 B 81 B %P %B 81 B œ –( =38 .B ( =38 .B ( =38 .B— P! P P P P P PÎ% $PÎ% œ) =38 81Î% =38 $81Î% Þ 8# 1 # )_ 81 $81 81B 81+ > ?aB ß >b œ # " ”=38 =38 -9= . •=38 1 8œ" % % P P Therefore the displacement of the string is given by a,b. With + œ " and P œ "! , )_ 81 $81 81B 81> ?aB ß >b œ # " ”=38 =38 -9= . •=38 1 8œ" % % "! "! ________________________________________________________________________ page 683 —————————————————————————— CHAPTER 10. —— a- b . ________________________________________________________________________ page 684 —————————————————————————— CHAPTER 10. —— a. b . 3a+b. The initial velocity is zero. As given by Eq. a#!b, the solution is ?aB ß >b œ " -8 =38 _ 8œ" in which the coefficients are the Fourier sine coefficients of 0 aBb. That is, 81 B 81 + > , -9= P P ________________________________________________________________________ page 685 —————————————————————————— CHAPTER 10. —— #P 81B -8 œ ( 0 aBb=38 .B P! P # P )BaP Bb# 81 B œ( =38 .B $ P! P P # -9= 81 œ $# Þ 8$ 1 $ Therefore the displacement of the string is given by $# _ # -9= 81 81 B 81 + > ?aB ß >b œ $ " =38 -9= . $ 1 8œ" 8 P P a,b. With + œ " and P œ "! , $# _ # -9= 81 81 B 81 > ?aB ß >b œ $ " =38 -9= . $ 1 8œ" 8 "! "! ________________________________________________________________________ page 686 —————————————————————————— CHAPTER 10. —— a- b . ________________________________________________________________________ page 687 —————————————————————————— CHAPTER 10. —— a. b . 4a+b. As given by Eq. a#!b, the solution is _ 8œ" in which the coefficients are the Fourier sine coefficients of 0 aBb. That is, ?aB ß >b œ " -8 =38 81 B 81 + > -9= , P P ________________________________________________________________________ page 688 —————————————————————————— CHAPTER 10. —— #P 81B -8 œ ( 0 aBb=38 .B P! P # PÎ#" 81B œ( =38 .B P PÎ#" P =38 8#1 =38 81 P œ% Þ 81 Therefore the displacement of the string is given by %_" 81 81 81B 81+ > ?aB ß >b œ " ’=38 =38 “=38 -9= . 1 8œ"8 # P P P a,b. With + œ " and P œ "! , %_" 81 81 81B 81> ?aB ß >b œ " ’=38 =38 “=38 -9= . 1 8œ"8 # "! "! "! ________________________________________________________________________ page 689 —————————————————————————— CHAPTER 10. —— a- b . ________________________________________________________________________ page 690 —————————————————————————— CHAPTER 10. —— a. b . 5a+b. The initial displacement is zero. Therefore the solution, as given by Eq. a$%b, is ?aB ß >b œ " 5 8 =38 _ 8œ" in which the coefficients are the Fourier sine coefficients of ?> aB ß !b œ 1aBb. It follows that 81 B 81 + > =38 , P P ________________________________________________________________________ page 691 —————————————————————————— CHAPTER 10. —— P # 81 B 58 œ .B ( 1aBb=38 81 + ! P œ PÎ# P # #B 81 B #aP Bb 81 B =38 .B ( =38 .B— –( 81 + ! P P P P PÎ# œ )P =38 81Î# Þ 8$ 1 $ + )P _ " 81 81 B 81 + > " $ =38 =38 =38 . +1 $ 8 œ " 8 # P P Therefore the displacement of the string is given by ?aB ß >b œ a,b. With + œ " and P œ "! , )! _ " 81 81B 81> ?aB ß >b œ $ " $ =38 =38 =38 . 1 8œ"8 # "! "! ________________________________________________________________________ page 692 —————————————————————————— CHAPTER 10. —— a- b . ________________________________________________________________________ page 693 —————————————————————————— CHAPTER 10. —— a. b . 7a+b. The initial displacement is zero. As given by Eq. a$%b, the solution is ?aB ß >b œ " 5 8 =38 _ 8œ" in which the coefficients are the Fourier sine coefficients of ?> aB ß !b œ 1aBb. It follows 81B 81 + > =38 , P P ________________________________________________________________________ page 694 —————————————————————————— CHAPTER 10. —— that 58 œ P # 81B .B ( 1aBb=38 81 + ! P P # )BaP Bb# 81 B œ =38 .B ( $ 81 + ! P P # -9= 81 œ $#P Þ 8% 1% + Therefore the displacement of the string is given by ?aB ß >b œ $#P _ # -9= 81 81 B 81 + > " =38 =38 . % % +1 8 œ " 8 P P a,b. With + œ " and P œ "! , ?aB ß >b œ $#! _ # -9= 81 81 B 81 > " =38 =38 . 1% 8 œ " 8% "! "! ________________________________________________________________________ page 695 —————————————————————————— CHAPTER 10. —— a- b . ________________________________________________________________________ page 696 —————————————————————————— CHAPTER 10. —— a. b . 8a+b. As given by Eq. a$%b, the solution is _ 8œ" in which the coefficients are the Fourier sine coefficients of ?> aB ß !b œ 1aBb. It follows that ?aB ß >b œ " 5 8 =38 81 B 81 + > =38 , P P ________________________________________________________________________ page 697 —————————————————————————— CHAPTER 10. —— P # 81B 58 œ .B ( 1aBb=38 81+ ! P PÎ#" # 81B œ =38 .B ( 81+ PÎ#" P =38 8#1 =38 81 P œ %P Þ # 1# + 8 Therefore the displacement of the string is given by %P _ " 81 81 81 B 81 + > " # ’=38 ?aB ß >b œ =38 “=38 =38 . # +1 8 œ " 8 # P P P a,b. With + œ " and P œ "! , %! _ " 81 81 81B 81> ?aB ß >b œ # " # ’=38 =38 “=38 =38 . 1 8œ"8 # "! "! "! ________________________________________________________________________ page 698 —————————————————————————— CHAPTER 10. —— a- b . ________________________________________________________________________ page 699 —————————————————————————— CHAPTER 10. —— a. b . 11a+b. As shown in Prob. * , the solution is ?aB ß >b œ " -8 =38 _ 8œ" in which the coefficients are the Fourier sine coefficients of 0 aBb. It follows that a#8 "b1B a#8 "b1+ > -9= , #P #P ________________________________________________________________________ page 700 —————————————————————————— CHAPTER 10. —— a#8 "b1B #P -8 œ ( 0 aBb=38 .B P! #P a#8 "b1B # P )BaP Bb# œ( =38 .B $ P! P #P $-9= 81 a#8 "b1 œ &"# Þ a#8 "b% 1% Therefore the displacement of the string is given by ?aB ß >b œ &"# _ $-9= 81 a#8 "b1 a#8 "b1B a#8 "b1+ > " =38 -9= . % % 1 8œ" #P #P a#8 "b Note that the period is X œ %PÎ+ . a,b. With + œ " and P œ "! , &"# _ $-9= 81 a#8 "b1 a#8 "b1B a#8 "b1> ?aB ß >b œ % " =38 -9= . % 1 8œ" #! #! a#8 "b ________________________________________________________________________ page 701 —————————————————————————— CHAPTER 10. —— a- b . ________________________________________________________________________ page 702 —————————————————————————— CHAPTER 10. —— a. b . 12. The wave equation is given by +# Setting = œ BÎP , we have `? `? .= " `? œ œ . `B `= .B P `= It follows that ` #? ` #? œ #. `B# `> ________________________________________________________________________ page 703 —————————————————————————— CHAPTER 10. —— ` #? " ` #? œ # #. `B# P `= Likewise, with 7 œ +>ÎP , `? + `? ` #? +# ` # ? œ and œ # #. `> P `7 `># P `7 Substitution into the original equation results in ` #? ` #? œ . `=# `7 # 15. The given specifications are P œ & ft , X œ &! lb , and weight per unit length # œ !Þ!#' lb/ft . It follows that 3 œ # Î$#Þ# œ )!Þ(& ‚ "!& slugs/ft . a+b. The transverse waves propagate with a speed of + œ ÈX Î3 œ #%) ft/sec . a,b. The natural frequencies are =8 œ 81+ÎP œ %*Þ) 18 rad/sec Þ a- b. The new wave speed is + œ ÈaX ?X bÎ3 . For a string with fixed ends, the natural modes are proportional to the functions 81B Q8 aBb œ =38 , P which are independent of + . 19. The solution of the wave equation +# @BB œ @>> in an infinite one-dimensional medium subject to the initial conditions @aB ß !b œ 0 aBb , @> aB ß !b œ ! , _ B _ @ aB ß >b œ The solution of the wave equation +# ABB œ A>> , on the same domain, subject to the initial conditions AaB ß !b œ ! , A> aB ß !b œ 1aBb , _ B _ " c0 aB +>b 0 aB +>bd. # is given by is given by ________________________________________________________________________ page 704 —————————————————————————— CHAPTER 10. —— A aB ß >b œ " B+> 1 a 0 b. 0 . ( #+ B+> Let ?aB ß >b œ @aB ß >b AaB ß >b . Since the PDE is linear, it is easy to see that ?aB ß >b is a solution of the wave equation +# ?BB œ ?>> . Furthermore, we have ?aB ß !b œ @aB ß !b AaB ß !b œ 0 aBb and Hence ?aB ß >b is a solution of the general wave propagation problem. 20. The solution of the specified wave propagation problem is ?aB ß >b œ " -8 =38 _ 8œ" ?> aB ß !b œ @> aB ß !b A> aB ß !b œ 1aBb . 81 B 81 + > -9= . P P Using a standard trigonometric identity, =38 81 B 81 + > " 81 B 81 + > 81 B 81 + > -9= œ ”=38Œ =38Œ • P P # P P P P " 81 81 œ ’=38 aB +>b =38 aB +>b“Þ # P P "_ 81 81 ?aB ß >b œ " -8 ’=38 aB +>b =38 aB +>b“ . # 8œ" P P Assuming that the series can be split up, _ "_ 81 81 ?aB ß >b œ – " -8 =38 aB +>b " -8 =38 aB +>b— . # 8œ" P P 8œ" We can therefore also write the solution as Comparing the solution to the one given by Eq. a#)b, we can infer that 2aBb œ " -8 =38 _ 8œ" 81B . P 21. Let 2a0b be a #P-periodic function defined by 2 a0 b œ œ Set ?aB ß >b œ " c2aB +>b 2 aB +>bd. Assuming the appropriate differentiability # ________________________________________________________________________ page 705 0 a0 b , ! Ÿ 0 Ÿ P à 0 a 0b , P Ÿ 0 Ÿ ! . —————————————————————————— CHAPTER 10. —— conditions on 2 , `? " œ c2 w aB +>b 2 w aB +>bd `B # and ` #? " œ c2 ww aB +>b 2 ww aB +>bd. `B# # Likewise, ` #? +# œ c2 ww aB +>b 2 ww aB +>bd. `># # It follows immediately that +# ` #? ` #? œ #. `B# `> Let > ! . Checking the first boundary condition, ?a! ß >b œ " " c2a +>b 2 a+>bd œ c 2 a+>b 2 a+>bd œ ! . # # " c2aP +>b 2 aP +>bd # " œ c 2a+> Pb 2 a+> PbdÞ # Checking the other boundary condition, ?aP ß >b œ Since 2 is #P-periodic, 2a+> Pb œ 2 a+> P #Pb. Therefore ?aP ß >b œ ! . Furthermore, for ! Ÿ B Ÿ P , Hence ?aB ß >b is a solution of the problem. ?aB ß !b œ " c2aBb 2aBbd œ 2aBb œ 0 aBbÞ # 23. Assuming that we can differentiate term-by-term, _ `? -8 8 81 B 81 + > œ 1+ " =38 =38 `> P P P 8œ" and _ `? -8 8 81 B 81 + > œ 1" -9= -9= Þ `B P P P 8œ" Formally, ________________________________________________________________________ page 706 —————————————————————————— CHAPTER 10. —— _ # `? # 81 B 81 + > # # " -8 8 =38# Š ‹ =38# Œ œ1+ `> P P P 8œ" 1# +# " J87 aBß >b _ 8Á7 and _ `? -8 8 # 81 B 81 + > œ 1# " Š -9=# ‹ -9=# Œ `B P P P 8œ" # 1# " K87 aBß >b , _ 8Á7 in which J87 aBß >b and K87 aBß >b contain products of the natural modes and their derivatives. Based on the orthogonality of the natural modes, `? P _ -8 8 # 81 + > .B œ 1# +# " Š ‹ =38# (Œ `> # 8œ" P P ! P # and _ # `? 81 + > # P " -8 8 Þ Š ‹ -9=# ( Œ .B œ 1 `B # 8œ" P P ! P # Recall that +# œ X Î3 . It follows that ( –3 Œ P ! `? `? X P _ -8 8 # 81 + > "Š X Œ —.B œ 1# ‹ =38# `> `B # 8œ" P P # # 1# Therefore, X P _ -8 8 # 81 + > "Š . ‹ -9=# # 8œ" P P " `? " `? X _ ## "8 - . 3Œ X Œ —.B œ 1# (– # `> # `B %P 8 œ " 8 ! P # # ________________________________________________________________________ page 707 —————————————————————————— CHAPTER 11. —— Chapter Eleven Section 11.1 1. Since the right hand sides of the ODE and the boundary conditions are all zero, the boundary value problem is homogeneous. 3. The right hand side of the ODE is nonzero. Therefore the boundary value problem is nonhomogeneous. 6. The ODE can also be written as C ww -ˆ" B# ‰C œ ! . Although the second boundary condition has a more general form, the boundary value problem is homogeneous. 7. First assume that - œ ! . The general solution of the ODE is CaBb œ -" B -# . The boundary condition at B œ ! requires that -# œ ! . Imposing the second condition, - " a 1 "b - # œ ! . It follows that -" œ -# œ ! . Hence there are no nontrivial solutions. Suppose that - œ .# . In this case, the general solution of the ODE is CaBb œ -" -9=2 .B -# =382 .B . The first boundary condition requires that -" œ ! . Imposing the second condition, -" a-9=2 .1 .=382 .1b -# a=382 .1 .-9=2 .1b œ ! . -# a>+82 .1 .b œ ! . The two boundary conditions result in Since the only solution of the equation >+82 .1 . œ ! is . œ ! , we have -# œ ! . Hence there are no nontrivial solutions. Let - œ .# , with . ! . Then the general solution of the ODE is CaBb œ -" -9= .B -# =38 .B . Imposing the boundary conditions, we obtain -" œ ! and -" a-9= .1 .=38 .1b -# a=38 .1 .-9= .1b œ ! . -9= .1 œ ! Ê =38 .1 œ ! , For a nontrivial solution of the ODE, we require that =38 .1 .-9= .1 œ ! . Note that which is false. It follows that >+8 .1 œ . . From a plot of 1 >+8 1. and 1. , ________________________________________________________________________ page 720 —————————————————————————— CHAPTER 11. —— we find that there is a sequence of solutions, ." ¸ !Þ()(' , .# ¸ "Þ'("' , â ; For large values of 8, 1 1 .8 ¸ a#8 "b . # Therefore the eigenfunctions are 98 aBb œ =38 .8 B , with corresponding eigenvalues -" ¸ !Þ'#!% , -# ¸ #Þ(*%$ , â Þ a#8 "b# -8 ¸ . % Asymptotically, 8. With - œ ! , the general solution of the ODE is CaBb œ -" B -# . Imposing the two boundary conditions, -" œ ! and #-" -# œ ! . It follows that -" œ -# œ ! . Hence there are no nontrivial solutions. Setting - œ .# , the general solution of the ODE is CaBb œ -" -9=2 .B -# =382 .B . The first boundary condition requires that -# œ ! . Imposing the second condition, -" a-9=2 . .=382 .b -# a=382 . .-9=2 .b œ ! . -" a" . >+82 .b œ ! . The two boundary conditions result in Since . >+82 . ! , it follows that -" œ ! , and there are no nontrivial solutions. Let - œ .# , with . ! . Then the general solution of the ODE is CaBb œ -" -9= .B -# =38 .B . Imposing the boundary conditions, we obtain -# œ ! and ________________________________________________________________________ page 721 —————————————————————————— CHAPTER 11. —— -" a-9= . .=38 .b -# a=38 . .-9= .b œ ! . For a nontrivial solution of the ODE, we require that -9= . .=38 . œ ! . First note that -9= . œ ! Ê . œ ! or =38 . œ ! . Therefore we find that " . >+8 . œ ! . From a plot of . >+8 . , there is a sequence of solutions, ." ¸ !Þ)'!$ , .# ¸ $Þ%#&' , â ; For large 8, Therefore the eigenfunctions are 98 aBb œ -9= .8 B , with corresponding eigenvalues -" ¸ !Þ(%!# , -# ¸ ""Þ($%* , â Þ - 8 ¸ a8 "b # 1 # . Asymptotically, . 8 ¸ a8 "b 1 . 12. First note that T aBb œ " , UaBb œ #B and V aBb œ - . Based on Prob. "" , the integrating factor is a solution of the ODE The differential equation is first order linear, with solution .aBb œ - /B:a B# b. It then follows that the Hermite equation can be written as ’/B C w “ - /B C œ ! . # . w aBb œ #B .aBb . w # 14. For the Laguerre equation, T aBb œ B , UaBb œ " B and V aBb œ - . Using the result of Prob. "", the integrating factor is a solution of the ODE The general solution of . w aBb œ .aBb is .aBb œ - /B . Therefore the Laguerre equation can be written as ________________________________________________________________________ page 722 B . w aBb œ B .aBb . —————————————————————————— CHAPTER 11. —— cB /B C w d - /B C œ ! . w 15. For the Chebyshev equation, T aBb œ " B# , UaBb œ B and V aBb œ !# . The integrating factor is a solution of the ODE ˆ" B# ‰. w aBb œ B .aBb . .. B œ . . " B# The general solution of the resulting ODE is .aBb œ . Èk" B# k The differential equation is separable, with Recall that the Chebyshev equation is typically defined for kBk Ÿ " . Therefore it can also be written as ’È" B# C w “ w È" B# !# C œ !. 16. We consider solutions of the form ?aB ß >b œ \ aBbX a>b . Substitution into the PDE results in \X ww - \X w 5 \X œ !# \ ww X . Dividing both sides of the equation by \X , we obtain \X ww \X w \ ww X , 5 œ !# \X \X \X that is, ww X ww Xw #\ œ! 5. X X \ Since both sides of the resulting equation are functions of different variables, each must be equal to a constant, say - . Therefore we obtain two ordinary differential equations !# \ ww a- 5 b\ œ ! and X ww -X w -X œ ! . 17a+b. Setting C œ =aBb? , we have C w œ = w ? = ?w and C ww œ = ww ? #= w ? w = ?ww . Substitution into the given ODE results in = ww ? #= w ? w = ?ww #a= w ? = ?w b a" -b=? œ ! . ________________________________________________________________________ page 723 —————————————————————————— CHAPTER 11. —— Collecting the various terms, = ?ww a#= w #=b? w c= ww #= w a" -b=d? œ ! . The second term on the left vanishes as long as = w œ = . a,b. With =aBb œ /B , the transformed differential equation can be written as Since the boundary conditions are homogeneous, we also have ?a!b œ ?a"b œ ! . It now follows that the eigenfunctions are ?8 œ =38 È-8 B , with corresponding eigenvalues Therefore the eigenfunctions for the original problem are 98 aBb œ /B =38 81B , with corresponding eigenvalues " -8 œ " 8# 1 # . a- b. The given equation is a second order constant coefficient differential equation. The characteristic equation is with roots <"ß# œ " „ È - . <# #< a" -b œ ! , - 8 œ 8# 1 # . ?ww -? œ ! . If - œ ! , then the general solution is C œ -" /B -# B/B . Imposing the two boundary conditions, we find that -" œ -# œ ! , and hence there are no nontrivial solutions. If - ! , then the general solution is C œ -" /B:Š" È - ‹B -# /B:Š" È - ‹B . It again follows that -" œ -# œ ! , and hence there are no nontrivial solutions. Therefore - ! , and the general solution is C œ -" /B -9= È- B -# /B =38 È- B . Invoking the boundary conditions, we have -" œ ! and -# /=38È- œ ! . For a nontrivial solution, È- œ 81 . 19. First write the differential equation as C ww a" -bC w -C œ ! , which is a second order constant coefficient differential equation. The characteristic equation is ________________________________________________________________________ page 724 —————————————————————————— CHAPTER 11. —— < # a" - b< - œ ! , with roots <" œ " and <# œ - . For - Á " , the general solution is C œ -" /B -# /-B . Imposing the boundary conditions, we require that -" -# œ ! and -" /" -# /- œ ! . For a nontrivial solution, it follows that /" œ /- , and hence - œ " , which is contrary to the assumption. If - œ " , then the general solution is C œ -" /B -# B/B . The boundary conditions require that -" œ ! and -" /" -# /" œ ! . Hence there are no nontrivial solutions. 21. Suppose that - œ ! . In that case the general solution is C œ -" B -# . The boundary conditions require that -" #-# œ ! and -" -# œ ! . We find that -" œ -# œ ! , and hence there are no nontrivial solutions. a+b. Let - œ .# , with . ! . Then the general solution of the ODE is CaBb œ -" -9= .B -# =38 .B . The boundary conditions require that #-" . -# œ ! and -" -9= . -# =38 . œ ! . These equations have a nonzero solution only if #=38 . . -9= . œ ! , which can also be written as #>+8 . . œ ! . Based on the graph, the positive roots of the determinantal equation are ________________________________________________________________________ page 725 —————————————————————————— CHAPTER 11. —— 1 ." ¸ %Þ#(%) , .# ¸ (Þ&*'& ,â ; for large 8, .8 ¸ a#8 "b . # Therefore the eigenvalues are -" ¸ ")Þ#($) , -# ¸ &(Þ(!(& ,â ; for large 8, -8 ¸ a#8 "b# CaBb œ -" -9=2 .B -# =382 .B . 1# . % a,b. Setting - œ .# ! , the general solution of the ODE is Imposing the boundary conditions, we obtain the equations #-" . -# œ ! and -" -9=2 . -# =382 . œ ! . These equations have a nonzero solution only if #=382 . . -9=2 . œ ! . The latter equation is satisfied only for . œ ! and . œ „ "Þ*"&! . Hence the only negative eigenvalue is -" œ $Þ''($ . 24. Based on the physical problem, - œ 7=# ÎIM ! . Let - œ .% . The characteristic equation is <% .% œ ! , with roots <"ß# œ „ .3 , <$ œ . and <% œ . . Hence the general solution is CaBb œ -" -9=2 .B -# =382 .B -$ -9= .B -% =38 .B . a+b. Simply supported on both ends À Ca!b œ C ww a!b œ ! à C aPb œ C ww aPb œ ! Þ Invoking the boundary conditions, we obtain the system of equations -" - $ œ ! -" - $ œ ! -" -9=2 .P -# =382 .P -$ -9= .P -% =38 .P œ ! # -" . -9=2 .P -# .# =382 .P -$ .# -9= .P -% .# =38 .P œ ! . The determinantal equation is .% =382 .P =38 .P œ ! . The nonzero roots are .8 œ 81ÎP , 8 œ "ß #ß â . The first two equations result in -" œ -$ œ ! . The last two equations, -# =382 81 -% =38 81 œ ! -# =382 81 -% =38 81 œ ! , imply that -# œ ! . Therefore the eigenfunctions are 98 œ =38 .8 B , with corresponding eigenvalues -8 œ 8% 1% ÎP% . ________________________________________________________________________ page 726 —————————————————————————— CHAPTER 11. —— a,b. Simply supported À Ca!b œ C ww a!b œ ! à clamped À C aPb œ C w aPb œ ! Þ Invoking the boundary conditions, we obtain the system of equations -" - $ œ ! -" - $ œ ! -" -9=2 .P -# =382 .P -$ -9= .P -% =38 .P œ ! -" .=382 .P -# .-9=2 .P -$ .=38 .P -% .-9= .P œ ! . The determinantal equation is #.$ =382 .P -9= .P #.$ -9=2 .P =38 .P œ ! . Based on numerical analysis, ." ¸ $Þ*#''ÎP and .# ¸ (Þ!')'ÎP . The first two equations result in -" œ -$ œ ! . The last two equations, -# =382 .8 P -% =38 .8 P œ ! -# -9=2 .8 P -% -9= .8 P œ ! , imply that -# œ Therefore the eigenfunctions are 98 œ =38 .8 P =382 .8 B =38 .8 B , =382 .8 P =38 .8 P -% . =382 .8 P with corresponding eigenvalues -8 œ .% . 8 a- b. Clamped À Ca!b œ C w a!b œ ! à free À C ww aPb œ C www aPb œ ! Þ Invoking the boundary conditions, we obtain the system of equations ________________________________________________________________________ page 727 —————————————————————————— CHAPTER 11. —— -" - $ œ ! . -# . - % œ ! # # # -" . -9=2 .P -# . =382 .P -$ . -9= .P -% .# =38 .P œ ! -" .$ =382 .P -# .$ -9=2 .P -$ .$ =38 .P -% .$ -9= .P œ ! . The determinantal equation is " -9=2 .P -9= .P œ ! . The first two nonzero roots are ." ¸ "Þ)(&"ÎP and .# ¸ %Þ'*%"ÎP . With -$ œ -" and -% œ -# , the system of equations reduce to -" a-9=2 .8 P -9= .8 Pb -# a=382 .8 P =38 .8 Pb œ ! -" a=382 .8 P =38 .8 Pb -# a-9=2 .8 P -9= .8 Pb œ ! . Let E8 œ a-9=2 .8 P -9= .8 PbÎa=382 .8 P =38 .8 Pb . The eigenfunctions are given by 98 aBb œ -9=2 .8 B -9= .8 B E8 a=38 .8 B =382 .8 Bb , with corresponding eigenvalues -8 œ .% . 8 25a+b. Assume that the solution has the form ?aB ß >b œ \ aBbX a>b . Substitution into the PDE results in I ww \ X œ \X ww . 3 Dividing both sides of the equation by \X , we obtain I \ ww X \X ww œ , 3 \X \X that is, ________________________________________________________________________ page 728 —————————————————————————— CHAPTER 11. —— \ ww 3 X ww œ . \ IX Since both sides of the resulting equation are functions of different variables, each must be equal to a constant, say - . Therefore we obtain two ordinary differential equations \ ww -\ œ ! and X ww I X œ !. 3 a,b. Given that ?a! ß >b œ \ a!bX a>b for > ! , it follows that \ a!b œ ! . The second boundary condition can be expressed as From the result in Part a+b, IE\ w aPbX a>b 7\ aPbX ww a>b œ ! , > ! . IE\ w aPbX a>b -7 I \ aPbX a>b œ ! , > ! . 3 Since the condition is to be satisfied for all > ! , we arrive at the boundary condition 7 \ w aPb \ aPb œ ! . 3E a- b. If - œ ! , the general solution of the spatial equation is \ aBb œ -" B -# . The boundary condition require that -" œ -# œ !. Hence there are no nontrivial solutions. If - œ .# ! , then the general solution is \ aBb œ -" -9=2 .B -# =382 .B . The first boundary condition implies that -" œ !. The second boundary condition requires that 7 -# -9=2 .P -# . =382 .P œ ! . 3E The solution is nontrivial only if 3E . 7 Since . >+82 .P ! , there are no nontrivial solutions. . >+82 .P œ Let - œ .# ! . The general solution of the spatial equation is \ aBb œ -" -9= .B -# =38 .B . ________________________________________________________________________ page 729 —————————————————————————— CHAPTER 11. —— The first boundary condition implies that -" œ !. The second boundary condition requires that 7 -# -9= .P -# . =38 .P œ ! . 3E For a nontrivial solution, it is necessary that 7 -9= .P . =38 .P œ ! , 3E or >+8 .P œ For the case a7Î3EPb œ !Þ& , 3E . 7. we find that ." P ¸ "Þ!('* and .# P ¸ $Þ'%$' . Therefore the eigenfunctions are given by 98 aBb œ =38 .8 B . The corresponding eigenvalues are solutions of -9= È-8 P PÈ -8 =38 È-8 P œ ! . # The first two eigenvalues are approximated as -" ¸ "Þ"&*(ÎP# and -# ¸ "$Þ#('ÎP# . ________________________________________________________________________ page 730 —————————————————————————— CHAPTER 11. —— Section 11.2 2. Based on the boundary conditions, - ! . The general solution of the ODE is The boundary condition C w a!b œ ! requires that -# œ ! . Imposing the second boundary condition, we find that -" -9=È- œ ! . So for a nontrivial solution, È- œ a#8 "b1Î# , 8 œ "ß #ß â Þ Therefore the eigenfunctions are given by 98 aBb œ 58 -9= a#8 "b1B . # # CaBb œ -" -9=È-B -# =38È-B . In this problem, <aBb œ " , and the normalization condition is # 58 ( ! " a#8 "b1B ”-9= • .B œ " . # # It follows that 58 œ # . Therefore the normalized eigenfunctions are 98 aBb œ È# -9= a#8 "b1B , 8 œ "ß #ß â Þ # 3. Based on the boundary conditions, - !. For - œ !, the eigenfunction is 9! aBb œ 5 ! Þ Set 5! œ " . With - ! , the general solution of the ODE is Invoking the boundary conditions, we require that -# œ ! and -" È- =38È- œ ! . Since - ! , the eigenvalues are -8 œ 8# 1# , 8 œ "ß #ß â , with corresponding eigenfunctions 98 aBb œ 58 -9= 81B . " CaBb œ -" -9=È-B -# =38È-B . The normalization condition is # 58 ( -9=# 81B .B œ " . ! # It follows that 58 œ # . Therefore the normalized eigenfunctions are 9! aBb œ " , and 98 aBb œ È# -9= 81B , 8 œ "ß #ß â Þ 4. From Prob. ) in Section ""Þ" , the eigenfunctions are 98 aBb œ 58 -9= È-8 B , in which -9=È-8 È-8 =38È-8 œ ! . The normalization condition is ________________________________________________________________________ page 731 —————————————————————————— CHAPTER 11. —— # 58 ( ! " -9=# È-8 B .B œ " . -9= È-8 =38 È-8 È-8 . #È-8 First note that # ( -9= È-8 B .B œ " ! Based on the determinantal equation, -9= È-8 =38 È-8 È-8 " =38# È-8 œ # #È-8 $ -9= #È-8 œ . % % $ -9= #È-8 # -9= È-8 B Therefore # 58 œ and the normalized eigenfunctions are given by 98 aBb œ É$ -9= #È-8 . 6. As shown in Prob. ", the normalized eigenfunctions are 98 aBb œ È# =38 Based on Eq. a$%b, with <aBb œ " , the coefficients in the eigenfunction expansion are given by -7 œ ( 0 aBb97 aBb.B " a#8 "b1B , 8 œ "ß #ß âÞ # œ È#( =38 " ! œ a#7 "b1 #È# ! a#7 "b1B .B # . Therefore we obtain the formal expansion "œ #È# _ " a#8 "b1B " =38 . 1 8 œ " #8 " # ________________________________________________________________________ page 732 —————————————————————————— CHAPTER 11. —— 8. We consider the normalized eigenfunctions 98 aBb œ È# =38 a#8 "b1B , 8 œ "ß #ß âÞ # Based on Eq. a$%b, with <aBb œ " , the coefficients in the eigenfunction expansion are given by -7 œ ( 0 aBb97 aBb.B " œ È #( œ ! "Î# #È# a#7 "b1 ”" -9= •. a#7 "b1 % ! =38 a#7 "b1B .B # Therefore we obtain the formal expansion 0 aBb œ #È# _ a#8 "b1 a#8 "b1B " ”" -9= . •=38 1 8œ" % # 9. The normalized eigenfunctions are Based on Eq. a$%b, with <aBb œ " , the coefficients in the eigenfunction expansion are given by -7 œ ( 0 aBb97 aBb.B " " a#7 "b1B a#7 "b1B .B È#( =38 .B # # ! "Î# ) 71 71 œ # # ’=38 # -9= # “ . a#7 "b 1 98 aBb œ È# =38 a#8 "b1B , 8 œ "ß #ß âÞ # œ È #( ! "Î# #B =38 Therefore the formal expansion of the given function is 0 aBb œ ) _ =38 8#1 -9= 8#1 a#8 "b1B " =38 . # 1# 8 œ " a#8 "b # # -9= È-8 B 11. From Prob. %, the normalized eigenfunctions are given by 98 aBb œ É$ -9= #È-8 , ________________________________________________________________________ page 733 —————————————————————————— CHAPTER 11. —— in which the eigenvalues satisfy - 9=È-8 È-8 =38È-8 œ !. Based on Eq. a$%b, the coefficients in the eigenfunction expansion are given by -7 œ ( 0 aBb97 aBb.B " ! œ œ È# ˆ# -9= È-7 "‰ -7 !7 É$ -9= #È-7 # ( B -9= È-7 B.B " ! , in which !7 œ É" =38# È-7 Þ 12. The normalized eigenfunctions are given by 98 aBb œ in which the eigenvalues satisfy - 9=È-8 È-8 =38È-8 œ !. Based on Eq. a$%b, the coefficients in the eigenfunction expansion are given by -7 œ ( 0 aBb97 aBb.B " ! É$ -9= #È-8 # -9= È-8 B , œ É$ -9= #È-7 -7 !7 # œ È# ˆ" -9= È-7 ‰ ( a" Bb-9= È-7 B.B " ! , in which !7 œ É" =38# È-7 Þ 13. We consider the normalized eigenfunctions 98 aBb œ in which the eigenvalues satisfy -9=È-8 È-8 =38È-8 œ !. The coefficients in the eigenfunction expansion are given by É$ -9= #È-8 # -9= È-8 B , ________________________________________________________________________ page 734 —————————————————————————— CHAPTER 11. —— -8 œ ( 0 aBb98 aBb.B " ! œ œ in which !8 œ É" =38# È-8 Þ È# =38ˆÈ-8 Î#‰ , È -8 !8 É$ -9= #È-8 # ( "Î# ! -9= È-8 B.B 15. The differential equation can be written as with :aBb œ " B# and ; aBb œ " . The boundary conditions are homogeneous and separated. Hence the BVP is self-adjoint. 16. Since the boundary conditions are not separated, the inner product is computed: Given ? and @, sufficiently smooth and satisfying the boundary conditions, aPc?dß @b œ ( c? ww @ ?@d.B " ! " ! " ! ˆ" B# ‰C w ‘ w C œ ! , œ ? w @ ¹ ( c? w @ w ?@d.B œ c? w @ ?@ w d ¹ a?ß Pc@dbÞ " ! Based on the given boundary conditions, ? w a"b@a"b ? w a!b@a!b œ ?a!b@a"b #?a"b@a!b ?a"b@ w a"b ?a!b@ w a!b œ ?a"b@a!b #?a!b@a"b Þ c? w @ ?@ w d ¹ œ ?a"b@a!b ?a!b@a"b , " ! Since the BVP is not self-adjoint. 18. The differential equation can be written as w with :aBb œ " , ; aBb œ ! , and <aBb œ " . The boundary conditions are homogeneous and separated. Hence the BVP is self-adjoint. cC w d œ - C , ________________________________________________________________________ page 735 —————————————————————————— CHAPTER 11. —— 19. If +# œ ! , then ? w a"b@a"b ?a"b@ w a"b œ ,# w ,# ? a"b@ w a"b ? w a"b@ w a"b œ ! , ," ," and since ?a!b œ @a!b œ ! , If ,# œ ! , then ?a"b œ @a"b œ ! implies that Furthermore, ? w a!b@a!b ?a!b@ w a!b œ ! . ? w a"b@a"b ?a"b@ w a"b œ ! . +# w +# ? a!b@ w a!b ? w a!b@ w a!b œ ! . +" +" ? w a!b@a!b ?a!b@ w a!b œ Clearly, the results are also true if +# œ ,# œ ! . 20. Suppose that 9" aBb and 9# aBb are linearly independent eigenfunctions associated with an eigenvalue - . The Wronskian is given by Each of the eigenfunctions satisfies the boundary condition +" Ca!b +# C w a!b œ ! . If either +" œ ! or +# œ ! , then clearly [ a9" , 9# ba!b œ ! . On the other hand, if +# is not equal to zero, then [ a9" , 9# ba!b œ 9" a!b9#w a!b 9# a!b9"w a!b +" +" œ 9" a!b9# a!b 9# a!b9" a!b +# +# œ !. [ a9" , 9# baBb œ 9" aBb9#w aBb 9# aBb9"w aBb. By Theorem $Þ$Þ# , [ a9" , 9# baBb œ ! for all ! Ÿ B Ÿ " . Based on Theorem $Þ$Þ$ , 9" aBb and 9# aBb must be linearly dependent. Hence - must be a simple eigenvalue. 22. We consider the operator PcCd œ c:aBbC w d ; aBbC w on the interval ! B " , together with the boundary conditions +" Ca!b +# C w a!b œ ! , ," C a"b ,# C w a"b œ ! . Let ? œ 9 3< and @ œ 0 3( . If ? and @ both satisfy the boundary conditions, then the real and imaginary parts also satisfy the same boundary conditions. Using the inner product a? ß @b œ ( ?aBb@aBb.B , " ! ________________________________________________________________________ page 736 —————————————————————————— CHAPTER 11. —— w aPc?d ß @b œ ( c:aBb? w d @ ; aBb?@‘.B " ! " w œ ( ˜ c:aBba9 w 3< w bd @ ; aBb?@™.B ! " ! " œ :aBba9 w 3< w b@¹ ( e:aBba9 w 3< w b @ w ; aBb?@f.B Þ ! Integrating by parts, again, w w w w ww ( e:aBba9 3< b@ f.B œ a9 3<b:aBb@ ¹ ( ˜c:aBb@ d ?™.B . " " ! " ! ! Collecting the boundary terms, :aBbca9 w 3< w b@ a9 3<b@ w d¹ œ :aBbca9 w 3< w ba0 3(b a9 3<ba0 w 3( w bd¹ " ! " ! . The real part is given by :aBbca9 w 0 < w (b a90 w <( w bd¹ œ :aBbca9 w 0 90 w b a< w ( <( w bd¹ " ! w w " ! w " ! " ! œ :aBbc9 0 90 d¹ :aBbc< ( <( w d¹ Þ Since 9 , < , 0 and ( satisfy the boundary conditions, it follows that :aBbca9 w 0 < w (b a90 w <( w bd¹ œ ! . " ! Similarly, the imaginary part also vanishes. That is, :aBbca< w 0 <0 w b a9 w ( 9( w bd¹ œ ! . " ! Therefore w aPc?d ß @b œ ( ˜ c:aBb@ w d ? ; aBb?@™.B " The result follows from the fact that a? ß Pc@db œ a? ß Pc@db . 24. Based on the physical problem, - œ T ÎIM ! . Let - œ .# . The characteristic equation is <% .# <# œ ! , with roots <"ß# œ ! , <$ œ .3 and <% œ .3 . Hence the general solution is CaBb œ -" -# B -$ -9= .B -% =38 .B . œ a P c @ d ß ?b œ a ? ß P c@ d b . ! ________________________________________________________________________ page 737 —————————————————————————— CHAPTER 11. —— a+b. Simply supported on both ends À Ca!b œ C ww a!b œ ! à C aPb œ C ww aPb œ ! Þ Invoking the boundary conditions, we obtain the system of equations -" - $ œ ! -$ œ ! -$ -9= .P -% =38 .P œ ! -" -# P -$ -9= .P -% =38 .P œ ! . The determinantal equation is =38 .P œ ! . The nonzero roots are .8 œ 81ÎP , 8 œ "ß #ß â . Therefore the eigenfunctions are 98 œ =38 .8 B , with corresponding eigenvalues -8 œ 8# 1# ÎP# . Hence the smallest eigenvalue is -" œ 1# ÎP# . a,b. Simply supported À Ca!b œ C ww a!b œ ! à clamped À C aPb œ C w aPb œ ! Þ Invoking the boundary conditions, we obtain the system of equations -" - $ œ ! -$ œ ! -# -$ .=38 .P -% .-9= .P œ ! -" -# P -$ -9= .P -% =38 .P œ ! . The determinantal equation is .P -9= .P =38 .P œ ! . It follows that the eigenfunctions are given by 98 aBb œ =38È-8 B ŠÈ-8 -9=È-8 P‹B , and the eigenvalues satisfy the equation PÈ-8 -9= È-8 P =38 È-8 P œ ! . The smallest eigenvalue is estimated as -" ¸ a%Þ%*$%b# ÎP# Þ a- b. Clamped À Ca!b œ C w a!b œ ! à clamped À C aPb œ C w aPb œ ! Þ Invoking the boundary conditions, we obtain the system of equations -" - $ œ ! -# .- % œ ! -" -# P -$ -9= .P -% =38 .P œ ! -# -$ .=38 .P -% .-9= .P œ ! . The determinantal equation is # #-9= .P œ .P =38 .P Þ It follows that the eigenfunctions are given by ________________________________________________________________________ page 738 —————————————————————————— CHAPTER 11. —— 98 aBb œ " -9=È-8 B , and the eigenvalues satisfy the equation # #-9= È-8 P œ È-8 P =38 È-8 P Þ The smallest eigenvalue is -" œ a#1b# ÎP# Þ 26. As shown is Prob. #& , the general solution is CaBb œ -" -# B -$ -9= .B -% =38 .B . -# œ ! -" - $ œ ! -# .- % œ ! -$ -9= .P -% =38 .P œ ! . For a nontrivial solution, it is necessary that -9= .P œ ! . We find that -# œ -% œ ! , and hence the eigenfunctions are given by The corresponding eigenvalues are -8 œ a#8 "b# 1# Î%P# , 8 œ "ß #ß â . The smallest eigenvalue is -" œ 1# Î%P# Þ 98 aBb œ " -9=È-8 B . Imposing the boundary conditions, we obtain the system of equations ________________________________________________________________________ page 739 —————————————————————————— CHAPTER 11. —— Section 11.3 4. The eigensystem of the associated homogeneous problem is given in Prob. "" of Section ""Þ# . The normalized eigenfunctions are 98 aBb œ É" =38# È-8 È# -9= È-8 B , in which the eigenvalues satisfy -9=È-8 È-8 =38È-8 œ !. Rewrite the given differential equation as C ww œ #C B . Since . œ # Á -8 , the formal solution of the nonhomogeneous problem is CaBb œ " _ -8 98 aBb, - # 8œ" 8 in which -8 œ ( 0 aBb98 aBb.B " ! œ œ É" =38# È-8 È# È#ˆ# -9= È-8 "‰ ! ( B -9= È-8 B .B " -8 É" =38# È-8 . Therefore we obtain the formal expansion CaBb œ # " _ È#ˆ# -9= È-8 "‰ -9= È-8 B -8 a-8 #bˆ" =38# È-8 ‰ . 8œ" 5. The solution follows that in Prob. " , except that the coefficients are given by -8 œ ( 0 aBb98 aBb.B " œ È #( œ% ! "Î# ! È# =38a81Î#b 8# 1 # #B =38 81B .B È#( . " "Î# a# #Bb =38 81B .B Therefore the formal solution is CaBb œ ) " _ =38a81Î#b =38 81B . 8# 1# a8# 1# #b 8œ" ________________________________________________________________________ page 740 —————————————————————————— CHAPTER 11. —— 6. The differential equation can be written as C ww œ .C 0 aBb. Note that ; aBb œ ! and <aBb œ " . As shown in Prob. " in Section ""Þ# , the normalized eigenfunctions are with associated eigenvalues -8 œ a#8 "b# 1# Î% . Based on Theorem ""Þ$Þ" , the formal solution is given by CaBb œ È# " _ 98 aBb œ È# =38 a#8 "bB , # -8 a#8 "bB =38 , a-8 .b # 8œ" " as long as . Á -8 Þ The coefficients in the series expansion are computed as -8 œ È#( 0 aBb=38 ! a#8 "bB .B . # 7. As shown in Prob. " in Section ""Þ# , the normalized eigenfunctions are 98 aBb œ È# -9= a#8 "bB , # with associated eigenvalues -8 œ a#8 "b# 1# Î% . Based on Theorem ""Þ$Þ" , the formal solution is given by CaBb œ È# " _ -8 a#8 "bB -9= , a-8 .b # 8œ" " as long as . Á -8 Þ The coefficients in the series expansion are computed as a#8 "bB -8 œ È#( 0 aBb-9= .B . # ! È# -9= È-8 B 9. The normalized eigenfunctions are 98 aBb œ The eigenvalues satisfy -9=È-8 È-8 =38È-8 œ !. Based on Theorem ""Þ$Þ" , the formal solution is given by CaBb œ È# " _ 8 œ " a-8 É" =38# È-8 . .bÉ" =38# È-8 -8 -9= È-8 B , as long as . Á -8 Þ The coefficients in the series expansion are computed as ________________________________________________________________________ page 741 —————————————————————————— CHAPTER 11. —— È# ( 0 aBb-9= È-8 B .B . " ! -8 œ É" =38# È-8 13. The differential equation can be written as C ww œ 1# C -9= 1B + . Note that . œ 1# and 0 aBb œ -9= 1B + . Furthermore, . œ 1# is an eigenvalue corresponding to the eigenfunction 9" aBb œ È# =38 1B . A solution exists only if 0 aBb and 9" aBb are orthogonal. Since ( a-9= 1B +b=38 1B .B œ #+Î1 , " ! there exists a solution as long as + œ ! . In that case, the ODE is The complementary solution is C- aBb œ -" -9= 1B -# =38 1B . A particular solution is ] aBb œ EB -9= 1B FB =38 1B . Using the method of undetermined coefficients, we find that E œ ! and F œ "Î#1 . Therefore the general solution is B CaBb œ -" -9= 1B -# =38 1B =38 1B . #1 The boundary conditions require that -" œ ! . Hence the solution of the boundary value problem is B CaBb œ -# =38 1B =38 1B . #1 15. Let CaBb œ 9" aBb 9# aBb . It follows that PcCd œ Pc9" d Pc9# d œ 0 aBb. Also, +" Ca!b +# C w a!b œ +" 9" a!b +" 9# a!b +# 9"w a!b +# 9#w a!b œ +" 9" a!b +# 9"w a!b +" 9# a!b +# 9#w a!b œ !. C ww 1# C œ -9= 1B . Similarly, the boundary condition at B œ " is satisfied as well. 16. The complementary solution is C- aBb œ -" -9= 1B -# =38 1B . A particular solution is ] aBb œ E FB . Using the method of undetermined coefficients, we find that E œ ! and F œ " . Therefore the general solution is CaBb œ -" -9= 1B -# =38 1B B . CaBb œ -9= 1B -# =38 1B B . Imposing the boundary conditions, we find that -" œ " . Therefore the solution of the BVP is Now attempt to solve the problem as shown in Prob. "& . Let BVP-1 be given by ________________________________________________________________________ page 742 —————————————————————————— CHAPTER 11. —— ? ww 1# ? œ 1# B , ?a!b œ ! , ?a"b œ ! Þ The general solution of the ODE is ?aBb œ -" -9= 1B -# =38 1B B . The boundary conditions require that -" œ ! and -" " œ ! . We find that BVP-1 has no solution. Let BVP-2 be given by @ ww 1# @ œ ! , @ a !b œ " , @ a " b œ ! Þ The general solution of the ODE is @aBb œ -" -9= 1B -# =38 1B . Imposing the boundary conditions, we obtain -" œ " and -" œ ! . Thus BVP-2 has no solution. 17. Setting CaBb œ ?aBb @aBb, substitution results in ? ww @ ww :aBbc? w @ w d ; aBbc? @d œ ? ww :aBb? w ; aBb? @ ww :aBb@ w ; aBb@ Þ ? ww :aBb? w ; aBb? œ c@ ww :aBb@ w ; aBb@dÞ 1aBb œ a+ ,b:aBb a+ ,bB ; aBb + ; aBb Þ Since the left hand side of the equation is zero, Furthermore, ?a!b œ Ca!b @a!b œ ! and ?a"b œ Ca"b @a"b œ ! . The simplest function having the assumed properties is @aBb œ a, +bB + . In this case, 20. The associated homogeneous PDE is ?> œ ?BB , ! B " , with ?B a! ß >b œ !, ?B a"ß >b ?a"ß >b œ ! and ?aB ß !b œ " B. Applying the method of separation of variables, we obtain the eigenvalue problem \ ww -\ œ ! , with boundary conditions \ w a!b œ ! and \ w a"b \ a"b œ !. It was shown in Prob. % , in Section ""Þ# , that the normalized eigenfunctions are 98 aBb œ where -9=È-8 È-8 =38È-8 œ !. We assume a solution of the form ?aB ß >b œ ",8 a>b98 aBb . _ 8 œ" É" =38# È-8 È# -9= È-8 B , ________________________________________________________________________ page 743 —————————————————————————— CHAPTER 11. —— Substitution into the given PDE results in w ww ",8 a>b98 aBb œ ",8 a>b98 aBb /> _ _ 8 œ" 8 œ" _ œ "-8 ,8 a>b98 aBb /> , 8 œ" that is, w "c,8 a>b -8 ,8 a>bd98 aBb œ /> Þ _ 8 œ" We now note that "œ" _ 8 œ" È-8 É" È# =38È-8 =38# È-8 98 aBb . Therefore / > œ ""8 /> 98 aBb , _ 8 œ" in which "8 œ È# =38È-8 ΔÈ-8 É" =38# È-8 • . Combining these results, w ",8 a>b -8 ,8 a>b "8 /> ‘98 aBb œ ! Þ _ 8 œ" Since the resulting equation is valid for ! B " , it follows that w ,8 a>b -8 ,8 a>b œ "8 /> , 8 œ "ß #ß â . Prior to solving the sequence of ODEs, we establish the initial conditions. These are obtained from the expansion ?aB ß !b œ " B œ "!8 98 aBb , _ in which !8 œ È# ˆ" -9=È-8 ‰Î”-8 É" =38# È-8 •Þ That is, ,8 a!b œ !8 . Therefore the solutions of the first order ODEs are ,8 a>b œ "8 ˆ/> /-8 > ‰ !8 /-8 > , 8 œ "ß #ß â . a - 8 "b 8 œ" Hence the solution of the boundary value problem is ________________________________________________________________________ page 744 —————————————————————————— CHAPTER 11. —— ?aB ß >b œ "– _ 8 œ" "8 ˆ/> /-8 > ‰ !8 /-8 > —98 aBb Þ a - 8 "b 21. Based on the boundary conditions, the normalized eigenfunctions are given by 98 aBb œ È# =38 81B , _ with associated eigenvalues -8 œ 8# 1# Þ We now assume a solution of the form ?aB ß >b œ ",8 a>b98 aBb . 8 œ" Substitution into the given PDE results in w ww ",8 a>b98 aBb œ ",8 a>b98 aBb " k" #Bk _ _ 8 œ" 8 œ" _ œ "-8 ,8 a>b98 aBb " k" #Bk, 8 œ" that is, w "c,8 a>b -8 ,8 a>bd98 aBb œ " k" #BkÞ _ 8 œ" It was shown in Prob. & that " k" #Bk œ "% _ 8 œ" È# =38a81Î#b 8# 1# 98 aBb . Substituting on the right hand side and collecting terms, we obtain w "–,8 a>b -8 ,8 a>b % _ 8 œ" È# =38a81Î#b 8# 1 # —98 aBb œ !Þ Since the resulting equation is valid for ! B " , it follows that w ,8 a>b 8# 1# ,8 a>b œ % È# =38a81Î#b 8# 1# , 8 œ "ß #ß â . Based on the given initial condition, we also have ,8 a!b œ ! , for 8 œ "ß #ß â . The solutions of the first order ODEs are ,8 a>b œ % È# =38a81Î#b 8% 1% Š" /8 1 > ‹, 8 œ "ß #ß â . ## Hence the solution of the boundary value problem is ________________________________________________________________________ page 745 —————————————————————————— CHAPTER 11. —— ) _ =38a81Î#b ## ?aB ß >b œ % " Š" /8 1 > ‹=38 81B Þ % 1 8 œ" 8 23a+b. Let ?aB ß >b be a solution of the boundary value problem and @aBb be a solution of the related BVP. Substituting for ?aB ß >b œ AaB ß >b @aBb, we have <aBb?> œ <aBbA> and Hence AaB ß >b is a solution of the homogeneous PDE The required boundary conditions are c:aBb?B dB ; aBb? J aBb œ c:aBbAB dB ; aBbA c:aBb@ w d ; aBb@ J aBb œ c:aBbAB dB ; aBbA J aBb J aBb œ c:aBbAB dB ; aBbA . w <aBbA> œ c:aBbAB dB ; aBbA . Aa! ß >b œ ?a! ß >b @a!b œ ! , Aa" ß >b œ ?a" ß >b @a"b œ ! . The associated initial condition is AaB ß !b œ ?aB ß !b @aBb œ 0 aBb @aBb. a,b. Let @aBb be a solution of the ODE w and satisfying the boundary conditions @ w a!b 2" @a!b œ X" , @ w a"b 2# @a"b œ X# . If AaB ß >b œ ?aB ß >b @aBb, then it is easy to show the A satisfies the PDE and initial condition given in Part a+b. Furthermore, AB a! ß >b 2" Aa! ß >b œ ?B a! ß >b @ w a!b 2" ?a! ß >b 2" @a!b œ ?B a! ß >b 2" ?a! ß >b @ w a!b 2" @a!b œ !. c:aBb@ w d ; aBb@ œ J aBb , Similarly, the other boundary condition is also homogeneous. ________________________________________________________________________ page 746 —————————————————————————— CHAPTER 11. —— 25. In this problem, J aBb œ 1# -9= 1B . First find a solution of the boundary value problem The general solution is @aBb œ EB F -9= 1B . Imposing the initial conditions, the solution of the related BVP is @aBb œ -9= 1B . Now let AaB ß >b œ ?aB ß >b -9= 1B . It follows that AaB ß >b satisfies the homogeneous boundary value problem, and the initial condition AaB ß !b œ -9=a$1BÎ#b -9= 1B a -9= 1Bb œ -9=a$1BÎ#b . We now seek solutions of the homogeneous problem of the form AaB ß >b œ ",8 a>b98 aBb , _ 8 œ" @ ww œ 1# -9= 1B , @ w a!b œ ! , @a"b œ " . in which 98 aBb œ È# -9= a#8 "b1BÎ# are the normalized eigenfunctions of the homogeneous problem and -8 œ a#8 "b# 1# Î% , with 8 œ "ß #ß â . Substitution into the PDE for A, we have w ww ",8 a>b98 aBb œ ",8 a>b98 aBb _ _ 8 œ" 8 œ" _ œ "-8 ,8 a>b98 aBb . 8 œ" Since the latter equation is valid for ! B " , it follows that with ,8 a>b œ ,8 a!b/B:a -8 >b. Hence _ 8 œ" w ,8 a>b -8 ,8 a>b œ ! , 8 œ "ß #ß â , AaB ß >b œ ",8 a!b/B:a -8 >b98 aBb . Imposing the initial condition, we require that _ It is evident that all of the coefficients are zero, except for ,# a!b œ "ÎÈ# . Therefore AaB ß >b œ /B:ˆ *1# >Î%‰-9= $1B , # È# ",8 a!b-9= a#8 "b1B œ -9= $1B . # # 8 œ" and the solution of the original BVP is ?aB ß >b œ /B:ˆ *1# >Î%‰-9= $1B -9= 1B . # ________________________________________________________________________ page 747 —————————————————————————— CHAPTER 11. —— 26a+b. Let ?aB ß >b œ \ aBbX a>b. Substituting into the homogeneous form of a3b, <aBb\X ww œ c:aBb\ w d X ; aBb\X . w Now divide both sides of the resulting equation by \X to obtain X ww c:aBb\ w d w ; aBb œ œ -. X <aBb\ <aBb w It follows that Since the boundary conditions a33b are valid for all > ! , we also have \ w a!b 2" \ a!b œ ! , \ w a"b 2# \ a"b œ ! . a,b. Let -8 and 98 aBb denote the eigenvalues and eigenfunctions of the BVP in Part a+b. Assume a solution, of the PDE a3b, of the form ?aB ß >b œ ",8 a>b98 aBb Þ _ 8 œ" c:aBb\ w d ; aBb\ œ -<aBb\ . Substituting into a3b, ww ww <aBb ",8 a>b98 œ ",8 a>b˜c:aBb98 d ; aBb98 ™ J aB ß >b _ _ 8 œ" œ ",8 a>bc -8 <aBb98 d J aB ß >b . 8 œ" 8 œ" _ Rearranging the terms, ww <aBb "c,8 a>b -8 ,8 a>bd98 œ J aB ß >b , _ 8 œ" or ww "c,8 a>b -8 ,8 a>bd98 œ _ 8 œ" J a B ß >b . <aBb Now expand the right hand side in terms of the eigenfunctions. That is, write _ J a B ß >b œ "#8 a>b98 aBb , <aBb 8 œ" in which ________________________________________________________________________ page 748 —————————————————————————— CHAPTER 11. —— #8 a>b œ ( <aBb " ! " ! œ ( J aB ß >b98 aBb.B , 8 œ "ß #ß â . J a B ß >b 98 aBb.B <aBb Combining these results, we have _ 8 œ" ww "c,8 a>b -8 ,8 a>b #8 a>bd98 œ ! . It follows that w In order to solve this sequence of ODEs, we require initial conditions ,8 a!b and ,8 a!b Þ Note that w ?aB ß !b œ ",8 a!b98 aBb and ?> aB ß !b œ ",8 a!b98 aBb . _ _ 8 œ" 8 œ" ww ,8 a>b -8 ,8 a>b œ #8 a>b , 8 œ "ß #ß â . Based on the given initial conditions, w 0 aBb œ ",8 a!b98 aBb and 1aBb œ ",8 a!b98 aBb . _ _ 8 œ" 8 œ" w Hence ,8 a!b œ !8 and ,8 a!b œ "8 , the expansion coefficients for 0 aBb and 1aBb in terms of the eigenfunctions, 98 aBb . 27a+b. Since the eigenvectors are orthogonal, they form a basis. Given any vector b , b œ ",3 0 a3b Þ 8 3œ" Taking the inner product, with 0 a4b , of both sides of the equation, we have ˆb ß 0 a4b ‰ œ ,4 ˆ0 a4b ß 0 a4b ‰Þ a,b. Consider solutions of the form x œ "+3 0 a3b Þ 8 3œ" Substituting into Eq. a3b, and using the above form of b , 8 8 3œ" 3œ" "+3 A0 a3b ". +3 0 a3b œ ",3 0 a3b Þ 8 3œ" It follows that ________________________________________________________________________ page 749 —————————————————————————— CHAPTER 11. —— "c+3 -3 . +3 ,3 d0 a3b œ 0 Þ 8 3œ" Since the eigenvectors are linearly independent, +3 -3 . +3 ,3 œ ! , for 3 œ "ß #ß âß 8 . That is, +3 œ ,3 Îa-3 .b , 3 œ "ß #ß âß 8 . xœ" 8 Assuming that the eigenvectors are normalized, the solution is given by ab ß 0 a3b b a3b 0, -3 . 3œ" as long as . is not equal to one of the eigenvalues. 29. First write the ODE as C ww C œ 0 aBb . A fundamental set of solutions of the homogeneous equation is given by The Wronskian is equal to [ c-9= B ß =38 Bd œ " . Applying the method of variation of parameters, a particular solution is ] aBb œ C" aBb?" aBb C# aBb?# aBb , B C" œ -9= B and C# œ =38 B . in which ?" aBb œ ( =38a=b0 a=b.= and ?# aBb œ ( -9=a=b0 a=b.= . B ! ! Therefore the general solution is C œ 9aBb œ -" -9= B -# =38 B -9= B( =38a=b0 a=b.= =38 B( -9=a=b0 a=b.=. B B ! ! Imposing the boundary conditions, we must have -" œ ! and -# =38 " -9= "( =38a=b0 a=b.= =38 "( -9=a=b0 a=b.= œ ! . " " ! ! It follows that -# œ and " " ( =38a" =b0 a=b.= , =38 " ! ________________________________________________________________________ page 750 —————————————————————————— CHAPTER 11. —— B =38 B " 9aBb œ ( =38a" =b0 a=b.= ( =38aB =b0 a=b.=. =38 " ! ! Using standard identities, =38 B † =38a" =b =38 " † =38aB =b œ =38 = † =38a" Bb . =38 B † =38a" =b =38 = † =38a" Bb =38aB =b œ . =38 " =38 " B ! " =38 = † =38a" Bb =38 B † =38a" =b 0 a=b.= ( 0 a=b.= =38 " =38 " B Therefore Splitting up the first integral, we obtain 9aBb œ ( œ ( KaB ß =b0 a=b.= , " ! in which K aB ß =b œ =38 =†=38a"Bb =38 " =38 B†=38a"=b =38 " ß ß !Ÿ=ŸB B Ÿ = Ÿ "Þ 31. The general solution of the homogeneous problem is By inspection, it is easy to see that C" aBb œ " satisfies the BC C w a!b œ ! and that the function C# aBb œ " B satisfies the BC C a"b œ ! . The Wronskian of these solutions is [ cC" ß C# d œ " . Based on Prob. $! , with :aBb œ " , the Green's function is given by K aB ß =b œ œ a" Bb ß a " =b ß !Ÿ=ŸB B Ÿ = Ÿ "Þ C œ -" -# B . Therefore the solution of the given BVP is 9aBb œ ( a" Bb0 a=b.= ( a" =b0 a=b.= . B " ! B 32. The general solution of the homogeneous problem is We find that C" aBb œ B satisfies the BC Ca!b œ ! . Imposing the boundary condition C œ -" -# B . ________________________________________________________________________ page 751 —————————————————————————— CHAPTER 11. —— Ca"b C w a"b œ ! , we must have -" #-# œ ! . Hence choose C# aBb œ # B . The Wronskian of these solutions is [ cC" ß C# d œ # . Based on Prob. $! , with :aBb œ " , the Green's function is given by K aB ß =b œ œ =aB #bÎ# ß Ba= #bÎ# ß !Ÿ=ŸB B Ÿ = Ÿ "Þ Therefore the solution of the given BVP is "B "" 9aBb œ ( =aB #b0 a=b.= ( Ba= #b0 a=b.= . #! #B 34. The general solution of the homogeneous problem is By inspection, it is easy to see that C" aBb œ B satisfies the BC Ca!b œ ! and that the function C# aBb œ " satisfies the BC C w a"b œ ! . The Wronskian of these solutions is [ cC" ß C# d œ " . Based on Prob. $! , with :aBb œ " , the Green's function is given by K aB ß =b œ œ =ß Bß !Ÿ=ŸB B Ÿ = Ÿ "Þ C œ -" -# B . Therefore the solution of the given BVP is 9aBb œ ( =0 a=b.= ( B0 a=b.= . B " ! B 35a+b. We proceed to show that if the expression given by Eq. a3@b is substituted into the integral of Eq. a333b, then the result is the solution of the nonhomogeneous problem. As long as we can interchange the summation and integration, C œ 9aBb œ ( KaB ß = ß .b0 a=b.= " œ" 93 aBb " ( 0 a=b93 a=b.= . - . ! 8œ" 3 ! _ Note that ( 0 a=b93 a=b.= œ -3 . " ! Therefore C œ 9aBb œ " _ -3 93 aBb , -3 . 8œ" ________________________________________________________________________ page 752 —————————————————————————— CHAPTER 11. —— as given by Eq. a"$b in the text. It is assumed that the eigenfunctions are normalized and -3 Á . . a,b. For any fixed value of B, KaB ß = ß .b is a function of = and the parameter . . With appropriate assumptions on K , we can write the eigenfunction expansion KaB ß = ß .b œ "+3 aB ß .b93 a=b . _ 3œ" Since the eigenfunctions are orthonormal with respect to <aBb, " ! ( KaB ß = ß .b<a=b93 a=b.= œ +3 aB ß .b . C3 aBb œ ( KaB ß = ß .b<a=b93 a=b.= . " ! Now let Based on the association 0 aBb œ <aBb93 aBb , it is evident that PcC3 d œ . <aBbC3 aBb <aBb93 aBb . C3 aBb œ " ,35 95 aBb . _ 5œ" In order to evaluate the left hand side, we consider the eigenfunction expansion It follows that PcC3 d œ " ,35 Pc95 d _ 5œ" _ 5œ" œ " ,35 -5 <aBb95 aBb . Therefore <aBb " ,35 -5 95 aBb œ . <aBb" ,35 95 aBb <aBb93 aBb , _ _ 5œ" 5œ" and since <aBb Á ! , 5œ" " ,35 -5 95 aBb œ . " ,35 95 aBb 93 aBb . _ _ 5œ" Rearranging the terms, we find that ________________________________________________________________________ page 753 —————————————————————————— CHAPTER 11. —— 93 aBb œ " ,35 a-5 .b95 aBb . _ 5œ" Since the eigenfunctions are linearly independent, ,35 a-5 .b œ $35 , and thus C3 aBb œ " _ " $35 95 aBb œ 93 aBb . - . -3 . 5œ" 5 + 3 aB ß . b œ " 93 aBb , -3 . 93 aBb93 a=b . -3 . 3œ" _ We conclude that which verifies that K aB ß = ß .b œ " 36. First note that . # CÎ.=# œ ! for = Á B . On the interval ! = B , the solution of the ODE is C" a=b œ -" -# = . Given that C a!b œ ! , we have C" a=b œ -# = . On the interval B = " , the solution is C# a=b œ ." .# = . Imposing the condition C a"b œ ! , we have C# a=b œ ." a" =b. Assuming continuity of the solution, at = œ B , which gives -# œ ." a" BbÎB . Next, integrate both sides of the given ODE over an infinitesimal interval containing = œ B À ( It follows that B B .# C .= œ ( $ a= Bb.= œ " . .=# B -# B œ ." a" Bb , B and hence -# a ." b œ " . Solving for the two coefficients, we obtain -# œ " B and ." œ B . Therefore the solution of the BVP is given by C a =b œ œ =a" Bb ß Ba" =b ß !Ÿ=ŸB B Ÿ = Ÿ ", C w aB b C w aB b œ " , which is identical to the Green's function in Prob. #). ________________________________________________________________________ page 754 —————————————————————————— CHAPTER 11. —— Section 11.4 1. Let 98 aBb œ N! ˆÈ-8 B‰ be the eigenfunctions of the singular problem Let 9aBb be a solution of the given BVP, and set _ 8œ! aB C w b œ -BC , ! B " , C ß C w bounded as B p ! , C a"b œ ! . w 9aBb œ " ,8 98 aBb . a‡b Then Substituting a‡b, we obtain _ 8œ! aB 9 w b œ . B9 0 aBb 0 aBb œ . B9 B . B w _ _ in which the -8 are the expansion coefficients of 0 aBbÎB for B ! . That is, -8 œ " " 0 aBb 98 aBb.B (B # m98 aBbm ! B " " œ ( 0 aBb98 aBb.B . m98 aBbm# ! " ,8 -8 B 98 aBb œ .B " ,8 98 aBb B " -8 98 aBb , 8œ! 8œ! It follows that if B Á ! , " c-8 ,8 a-8 .bd98 aBb œ ! . _ 8œ! As long as . Á -8 , linear independence of the eigenfunctions implies that -8 , 8 œ "ß #ß â . ,8 œ -8 . Therefore a formal solution is given by 9aBb œ " _ in which È-8 are the positive roots of N! aBb œ ! . -8 N! ŠÈ-8 B‹, -8 . 8œ! ________________________________________________________________________ page 755 —————————————————————————— CHAPTER 11. —— 3a+b. Setting > œ È- B , it follows that .C .C .#C .#C œ Èand œ- # . .B .> .B# .> . > È .C 5 # È- œ È- > C , .> È.> > . .C 5# Œ> œ >CÞ .> .> > The given ODE can be expressed as Èor An equivalent form is given by ># which is known as a Bessel equation of order 5 . A bounded solution is N5 a>b . .C .C > ˆ># 5 # ‰C œ ! , .> .> a,b. N5 ŠÈ- B‹ satisfies the boundary condition at B œ ! . Imposing the other boundary condition, it is necessary that N5 ŠÈ- ‹ œ ! . Therefore the eigenvalues are given by -8 , 8 œ "ß #â , where È-8 are the positive zeroes of N5 aBb . The eigenfunctions of the BVP are 98 aBb œ N5 ˆÈ-8 B‰Þ a- b. The BVP is a singular Sturm-Liouville problem with PcCd œ aB C w b w 5# C and <aBb œ " . B " We note that -8 ( B 98 aBb97 aBb.B œ ( Pc98 d 97 aBb.B " ! œ ( 98 aBbPc97 d.B " ! " ! ! œ -7 ( B 98 aBb97 aBb.B Þ Therefore a-8 -7 b( B 98 aBb97 aBb.B œ ! Þ " ! ________________________________________________________________________ page 756 —————————————————————————— CHAPTER 11. —— So for 8 Á 7 , we have -8 Á -7 and ( B 98 aBb97 aBb.B œ ! Þ " ! a. b. Consider the expansion Multiplying both sides of equation by B 94 aBb and integrating from ! to " , and using the orthogonality of the eigenfunction, ( B 0 aBb94 aBb.B œ " +8 ( B 94 aBb98 aBb.B " _ " ! 0 aBb œ " +8 98 aBb Þ _ 8œ! œ +4 ( B 94 aBb94 aBb.B Þ " ! 8œ! ! Therefore +4 œ ( B 0 aBb94 aBb.BÎ( Bc94 aBbd# .B , 4 œ "ß #ß â Þ " " ! ! a/b. Let 9aBb be a solution of the given BVP, and set _ where 98 aBb œ N5 ˆÈ-8 B‰. Then 9aBb œ " ,8 98 aBb , 8œ! a‡b Substituting a‡b, we obtain _ 8œ! Pc9d œ . B9 0 aBb 0 aBb œ . B9 B . B _ _ in which the -8 are the expansion coefficients of 0 aBbÎB for B ! . That is, -8 œ " " 0 aBb 98 aBb.B (B # m98 aBbm ! B " " œ ( 0 aBbN5 ŠÈ-8 B‹.B . mN5 ˆÈ-8 B‰m# ! " ,8 -8 B 98 aBb œ .B " ,8 98 aBb B " -8 98 aBb , 8œ! 8œ! ________________________________________________________________________ page 757 —————————————————————————— CHAPTER 11. —— It follows that if B Á ! , " c-8 ,8 a-8 .bdN5 ŠÈ-8 B‹ œ ! . _ 8œ! As long as . Á -8 , linear independence of the eigenfunctions implies that -8 , 8 œ "ß #ß â . ,8 œ -8 . Therefore a formal solution is given by 9aBb œ " _ -8 N5 ŠÈ-8 B‹. -8 . 8œ! 5a+b. Setting - œ !# in Prob. "& of Section ""Þ" , the Chebyshev equation can also be written as ’È" B# C w “ œ w C. È" B# Note that :aBb œ È" B# , ; aBb œ ! , and <aBb œ "ÎÈ" B# , hence both boundary points are singular. a,b. Observe that :a" &b œ È#& &# and :a " &b œ È#& &# Þ It follows that if ?aBb and @aBb satisfy the boundary conditions a333b, then & Ä ! lim :a" &bc? w a" &b@a" &b ?a" &b@ w a" &bd œ ! and a- b. For 8 Á ! , Therefore Eq. a"(b is satisfied and the boundary value problem is self-adjoint. 8# ( " X! aBb X8 aBb .B œ ( X! aBb PcX8 d.B " È" B# " " " " & Ä ! lim :a " &bc? w a " &b@a " &b ?a " &b@ w a " &bd œ ! . œ ( PcX! d X8 aBb.B œ !, since PcX! d œ ! † X! œ ! . Otherwise, ________________________________________________________________________ page 758 —————————————————————————— CHAPTER 11. —— " X8 aBb X7 aBb 8( .B œ ( PcX8 d X7 aBb.B " È" B# " # " œ ( X8 aBbPcX7 d.B " œ 7# ( Therefore ˆ8# 7# ‰( So for 8 Á 7 , ( " X8 aBb X7 aBb .B Þ " È" B# " X8 aBb X7 aBb .B œ ! Þ " È" B# " X8 aBb X7 aBb .B œ ! Þ " È" B# " ________________________________________________________________________ page 759 —————————————————————————— CHAPTER 11. —— Section 11.5 3. The equations relating to this problem are given by Eqs. a#b to a"(b in the text. Based on the boundary conditions, the eigenfunctions are 98 aBb œ N! a-8 <b and the associated eigenvalues -" ß -# ß â are the positive zeroes of N! a-b . The general solution has the form ?a< ß >b œ " c-8 N! a-8 <b -9= -8 +> 58 N! a-8 <b =38 -8 +>d . _ 8œ" The initial conditions require that ?a< ß !b œ " -8 N! a-8 <b œ 0 a<b _ 8œ" and ?> a< ß !b œ " +-8 58 N! a-8 <b œ 1a<b . _ 8œ" The coefficients -8 and 58 are obtained from the respective eigenfunction expansions. That is, -8 œ and " " 58 œ ( <1a<bN! a-8 <b.< , +-8 mN! a-8 <bm# ! " " ( <0 a<bN! a-8 <b.< mN! a-8 <bm# ! in which mN! a-8 <bm# œ ( <cN! a-8 <bd# .< " ! for 8 œ "ß #ß â . 8. A more general equation was considered in Prob. #$ of Section "!Þ& . Assuming a solution of the form ?a< ß >b œ V a<bX a>b, substitution into the PDE results in !# ”V ww X "w V X • œ VX w . < Dividing both sides of the equation by the factor VX , we obtain ________________________________________________________________________ page 760 —————————————————————————— CHAPTER 11. —— V ww " Vw Xw œ#. V <V !X Since both sides of the resulting differential equation depend on different variables, each side must be equal to a constant, say -# . That is, V ww " Vw Xw œ # œ -# . V <V !X It follows that X w !# -# X œ ! , and V ww " Vw œ -# , V <V which can be written as <# V ww < V w -# <# V œ ! . Introducing the variable 0 œ -< , the last equation can be expressed as 0# V ww 0 V w 0# V œ ! , which is the Bessel equation of order zero. The temporal equation has solutions which are multiples of X a>b œ /B:a !# -# >b. The general solution of the Bessel equation is V a<b œ ," N! a-8 <b ,# ]! a-8 <b . Since the steady state temperature will be zero, all solutions must be bounded, and hence we set ,# œ ! . Furthermore, the boundary condition ?a" ß >b œ ! requires that V a"b œ ! and hence N! a-b œ ! . It follows that the eigenfunctions are 98 aBb œ N! a-8 <b , with the associated eigenvalues -" ß -# ß â , which are the positive zeroes of N! a-b . Therefore # the fundamental solutions of the PDE are ?8 a< ß >b œ N! a-8 <b/B:a !# -8 >b, and the general solution has the form # ?a< ß >b œ " -8 N! a-8 <b/B:ˆ !# -8 >‰ . _ 8œ" The initial condition requires that ?a< ß !b œ " -8 N! a-8 <b œ 0 a<b. _ 8œ" The coefficients in the general solution are obtained from the eigenfunction expansion of 0 a<b. That is, " " -8 œ ( <0 a<bN! a-8 <b.< , mN! a-8 <bm# ! in which mN! a-8 <bm# œ ( <cN! a-8 <bd# .< " ! a8 œ "ß #ß â b. ________________________________________________________________________ page 761 —————————————————————————— CHAPTER 11. —— Section 11.6 1. The sine expansion of 0 aBb œ " , on ! B " , is given by 0 aBb œ # " _ " -9= 71 =38 71B , 71 7œ" with partial sums W8 aBb œ # " 8 " -9= 71 =38 71B . 71 7œ" The mean square error in this problem is V8 œ ( k" W8 aBbk# .B . " ! Several values are shown in the Table : 8 V8 & !Þ!'( "! !Þ!% "& !Þ!#' #! !Þ!# Further numerical calculation shows that V8 !Þ!# for 8 #" . 0 aBb œ # " _ 3a+b. The sine expansion of 0 aBb œ Ba" Bb , on ! B " , is given by " -9= 71 =38 71B , 71 7œ" with partial sums W8 aBb œ % " 8 a,ß - b. The mean square error in this problem is " ! " -9= 71 =38 71B . 7$ 1$ 7œ" V8 œ ( k Ba" Bb W8 aBbk# .B . ________________________________________________________________________ page 762 —————————————————————————— CHAPTER 11. —— We find that V" œ !Þ!!!!%) . The graphs of 0 aBb and W" aBb are plotted below : 6a+b. The function is bounded on intervals not containing B œ ! , so for & ! , "Î# .B œ # #È& Þ ( 0 aBb.B œ ( B " " & & Hence the improper integral is evaluated as "Î# .B œ # Þ ( 0 aBb.B œ lim ( B " " ! &Ä! & On the other hand, 0 # aBb œ "ÎB for B Á ! , and " # " & & " ( 0 aBb.B œ ( B .B œ 68È& Þ Therefore the improper integral does not exist. a,b. Since 0 # aBb ´ " , it is evident that the Riemann integral of 0 # aBb exists. Let ________________________________________________________________________ page 763 —————————————————————————— CHAPTER 11. —— TR œ e! œ B" ß B# ß âß BR " œ "f be a partition of c! ß "d into equal subintervals. We can always choose a rational point, 03 , in each of the subintervals so that the Riemann sum V a0" ß 0# ß âß 0R b œ " 0 a08 b R 8œ" " œ ". R Likewise, can always choose an irrational point, (3 , in each of the subintervals so that the Riemann sum V a(" ß (# ß âß (R b œ " 0 a(8 b R 8œ" It follows that 0 aBb is not Riemann integrable. " œ ". R 8. With T! aBb œ " and T" aBb œ B, the normalization conditions are satisfied. Using the usual inner product on c " ß "d, and hence the polynomials are also orthogonalÞ Let T# aBb œ +# B# +" B +! . The normalization condition requires that +# +" +! œ " . For orthogonality, we need # # ( ˆ+# B +" B +! ‰.B œ ! and ( Bˆ+# B +" B +! ‰.B œ ! . " " " " ( T! aBbT" aBb.B œ ! " " It follows that +# œ $Î# , +" œ ! and +! œ "Î# . Hence T# aBb œ a$B# "bÎ# . Now let T$ aBb œ +$ B$ +# B# +" B +! . The coefficients must be chosen so that +$ +# +" +! œ " and the orthogonality conditions ( T3 aBbT4 aBb.B œ ! a3 Á 4b " " are satisfied. Solution of the resulting algebraic equations leads to +$ œ &Î# , +# œ ! , +1 œ $Î# and +! œ ! . Therefore T$ aBb œ a&B$ $BbÎ# . 11. The implied sequence of coefficients is +8 œ " , 8 " . Since the limit of these coefficients is not zero, the series cannot be an eigenfunction expansion. 13. Consider the eigenfunction expansion 0 aBb œ "+3 93 aBbÞ _ 3œ" Formally, ________________________________________________________________________ page 764 —————————————————————————— CHAPTER 11. —— ## 0 aBb œ "+3 93 aBb # "+3 +4 93 aBb94 aBb Þ _ # 3œ" 3Á4 Integrating term-by-term, # # ( <aBb0 aBb.B œ "( +3 <aBb93 aBb.B # "( +3 +4 <aBb93 aBb94 aBb.B " _ # " " ! œ 3œ" ! _ " # # "+3 ( 93 aBb.B , ! 3œ" 3Á4 ! since the eigenfunctions are orthogonal. Assuming that they are also normalized, # ( <aBb0 aBb.B œ "+3 . " _ # ! 3œ" ________________________________________________________________________ page 765 Math 219, Homework 2 Due date: 23.11.2005, Wednesday This homework concerns two (fictitious) design problems about the solar car “MES¸ e” of the METU Robotics Society, which won the Formula-G trophy in September 2005. Just for the purposes of this homework, assume that they want to modify the car, and they are asking for your help on two issues. 1. The first problem is about the shock absorbing system of the car. We may model the shock absorber as a single linear spring. This question concerns how to adjust the damping coefficient in order to meet certain requirements. (a) It is known that when the pilot, weighing 80kg , gets into the car seat, the shock absorber is compressed by 5cm. From this data, compute the spring constant k (in kg/sec2 ). (b) The car (without the pilot) weighs 240kg . Write a differential equation which governs the vertical motion of the car (this could for instance describe the vertical displacement when the car goes over a speed bump). (Hint: Check “Damped free vibrations” from Boyce,Di Prima, section 3.8). (c) It is required that, when the car goes over a speed bump of 5cm high, the vertical displacement x(t) should approach the equilibrium point 0 in a way that for t ≥ 1sec, |x(t)| ≤ 1cm. For several values of the damping coefficient and for this initial condition, sketch the graph of the solution curve for 0 ≤ t ≤ 2sec. Finally, decide which values of the damping coefficient are allowable in order to meet this requirement. 2. The second problem is about the power supply of the motor. The panels convert solar energy to electrical energy and store it in the accumulator. Assume that the accumulator provides a voltage of E (t) = 48 cos(ωt) Volts to the system. The frequency ω is adjusted by the gas pedal. The circuit can be modeled as a series L − C circuit. Take L = 0.05Henry and C = 10−6 Farad. (a) Write a differential equation for the current I (t) through the voltage source (Hint: Check “Electric Circuits” from Boyce,Di Prima, section 3.8. We are assuming R = 0). (b) Solve this differential equation for I (t) in terms of ω using the method of undetermined coefficients. (c) Currents over 20 Amperes may harm the accumulator. Using your result from part (b), find out which frequencies (in Hertz) should not be allowed. For several values of ω graph I (t) for 0 ≤ t ≤ 0.2sec using ODE Architect, and denote which of these are potentially harmful and which are not. Math 219, Homework 4 Due date: 30.12.2005, Wednesday Suppose that K > 0, and f (t) is defined as 1 0 if 4n ≤ t < 4n + 1 otherwise f (t) = where n runs through the set of integers. (a) Determine the Fourier series for f (t). (b) Consider the differential equation d2 x + x = f (t). dt2 By using ODE Architect, solve (and graph the solutions of) this equation for as many values of K as possible between 0.5 and 10 for 0 ≤ t ≤ 100.(You can enter f (t) in ODE Architect using the built in command SqW ave(t, L, K ) and then set L = 4 ∗ K , and assign K definite values on the lines below). Record the maximum values of x(t) for each of these K ’s, and plot a K vs. max(x(t)) graph by hand. (c) Which values of K result in a resonance in the system? (Hint: you should find 5 such values). Can you relate these values to the Fourier series terms? Please print the resonance graphs. (d) What do the Fourier coefficients correspond to on the resonance graphs? We wish you a happy new year and success on your finals! Math 219, Homework 3 Due date: 9.12.2005, Friday 1. Consider the initial value problem d2 x dx + + x = u4 (t), dt2 dt y (0) = y (0) = 0 (a) Solve this initial value problem using the Laplace transform. dx dt with respect to t (You can use the function Step(t, 4) to create a unit step function with discontinuity at t = 4). (b) Use ODE Architect to solve the equation, and graph the solution. Also graph (c) Discuss how the graphs agree with the solutions in (a): in particular determine dx (if any) all the points where x(t) and are discontinuous, behavior of these two dt functions for t → ∞, their maxima and minima. 2. Write each of the following systems of differential equations in matrix form, find the eigenvalues and eigenvectors of the coefficient matrices, and using these, find all solutions of each system. Also, graph the phase portraits (x − y graph) using ODE Architect. Please use a scale which includes the point (0, 0), and graph several solutions in order to clearly observe the behavior around (0, 0). Also, place arrows on the solution curves which indicate the direction of increasing t, and make sure that solution curves along the eigenvector directions are graphed if there are any real eigenvectors. (a) dx = 2x − y dt dy = 3x + 3y dt (b) dx = −x + y dt dy = 3x − 4y dt (c) dx = 2 x + 3y dt dy = 5x + 5 y dt (d) dx = −4x + 3y dt dy = −3x + 2y dt (e) dx = −x − 3y dt dy = 2x + y dt ...
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This note was uploaded on 02/08/2011 for the course MATH 421 taught by Professor Staff during the Spring '08 term at Rutgers.

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