solution_manual_to_accompany_boyce_elementary_differential_equations_8e_mathematic87.blogfa.com - CHAPTER 1 Chapter One Section 1.1 1 For C the slopes

# Solution_manual_to_accompany_boyce_elementary_differential_equations_8e_mathematic87.blogfa.com

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This preview shows page 1 out of 756 pages. 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, [email protected] œ 71 # @# .> a,b After a long time, [email protected] # œ 1 @# Þ .> 7 [email protected] .> ¸ ! 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 [email protected] 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 [email protected]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 differentia...
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