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CHAPTER ---------------------------------------------------- 2. ---- Chapter Two Section 2.1 1a+b a,b Based on the direction field, all solutions seem to converge to a specific increasing function. a- b The integrating factor is .a>b oe /$> , and hence Ca>b oe > $ " * /#> - /$> It follows that all solutions converge to the function C" a>b oe > $ " * 2a+b a,b. All slopes eventually become positive, hence all solutions will increase without bound. a- b The integrating factor is .a>b oe /#> , and hence Ca>b oe >$ /#> $ - /#> 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 oe " 4a+b. a- b The integrating factor is .a>b oe /#> , and hence Ca>b oe ># /> # " - /> It is clear that all solutions converge to the specific solution C! a>b oe " . a,b. Based on the direction field, the solutions eventually become oscillatory. a- b The integrating factor is .a>b oe > , and hence the general solution is C a >b oe $-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 oe $=38a#>b # ________________________________________________________________________ page 19 ---------------------------------------------------- CHAPTER 2. ---- a,b. All slopes eventually become positive, hence all solutions will increase without bound. a- b The integrating factor is .a>b oe /B:a ' #.>b oe /#> The differential equation can w be written as /#> C w #/#> C oe $/> , that is, a/#> C b oe $/> Integration of both sides of the equation results in the general solution Ca>b oe $/> - /#> It follows that all solutions will increase exponentially. 6a+b a,b All solutions seem to converge to the function C! a>b oe ! Ca>b oe -9=a>b =38a#>b # # > > > a- b The integrating factor is .a>b oe ># , and hence the general solution is in which - is an arbitrary constant. As > becomes large, all solutions converge to the function C! a>b oe ! 7a+b. ________________________________________________________________________ page 20 ---------------------------------------------------- CHAPTER 2. ---- a,b All solutions seem to converge to the function C! a>b oe ! a- b The integrating factor is .a>b oe /B:a># b, and hence C a>b oe ># /> - /> It is clear that all solutions converge to the function C! a>b oe ! . # # 8a+b a,b All solutions seem to converge to the function C! a>b oe ! a- b Since .a>b oe a" ># b# , the general solution is Ca>b oe c>+8" a>b G d a" ># b# It follows that all solutions converge to the function C! a>b oe ! . 9a+b ________________________________________________________________________ page 21 ---------------------------------------------------- CHAPTER 2. ---- a,b. All slopes eventually become positive, hence all solutions will increase without bound. 10a+b. a- b The integrating factor is .a>b oe /B:^' " .> oe /> # . The differential equation can # w be written as /> # C w /> # C # oe $> /> # # , that is, ^/> # C # oe $> /> # # Integration of both sides of the equation results in the general solution Ca>b oe $> ' - /> # All solutions approach the specific solution C! a>b oe $> ' 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- b First divide both sides of the equation by > . From the resulting standard form, the integrating factor is .a>b oe /B:^ ' " .> oe " > . The differential equation can be > written as C w > C ># oe > /> , that is, a C >bw oe > /> Integration leads to the general solution Ca>b oe >/> - > For - ! , solutions diverge, as implied by the direction field. For the case - oe ! , the specific solution is Ca>b oe >/> , which evidently approaches zero as > p _ . 11a+b. a,b The solutions appear to be oscillatory. ________________________________________________________________________ page 22 ---------------------------------------------------- CHAPTER 2. ---- a- b The integrating factor is .a>b oe /> , and hence Ca>b oe =38a#>b # -9=a#>b - /> It is evident that all solutions converge to the specific solution C! a>b oe =38a#>b # -9=a#>b . 12a+b a- b The integrating factor is .a>b oe /#> . The differential equation can be w written as /> # C w /> # C # oe $># # , that is, ^/> # C # oe $># # Integration of both sides of the equation results in the general solution Ca>b oe $># "#> #% - /> # It follows that all solutions converge to the specific solution C! a>b oe $># "#> #% . a,b. All solutions eventually have positive slopes, and hence increase without bound. 14. The integrating factor is .a>b oe /#> . After multiplying both sides by .a>b, the w equation can be written as ^/2> C oe > Integrating both sides of the equation results in the general solution Ca>b oe ># /#> # - /#> Invoking the specified condition, we require that /# # - /# oe ! . Hence - oe " # , and the solution to the initial value problem is Ca>b oe a># "b/#> # 16. The integrating factor is .a>b oe /B:^' # .> oe ># . Multiplying both sides by .a>b, > w # the equation can be written as a> Cb oe -9=a>b Integrating both sides of the equation results in the general solution Ca>b oe =38a>b ># - ># Substituting > oe 1 and setting the value equal to zero gives - oe ! . Hence the specific solution is Ca>b oe =38a>b ># 17. The integrating factor is .a>b oe /#> , and the differential equation can be written as ^/2> Cw oe " Integrating, we obtain /2> C a>b oe > - Invoking the specified initial condition results in the solution Ca>b oe a> #b/#> 19. After writing the equation in standard 0 orm, we find that the integrating factor is .a>b oe /B:^' % .> oe >% . Multiplying both sides by .a>b, the equation can be written as > ^>% Cw oe > /> Integrating both sides results in >% C a>b oe a> "b/> - Letting > oe " and setting the value equal to zero gives - oe ! Hence the specific solution of the initial value problem is Ca>b oe ^>$ >% /> 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 +! oe " . For + " , the solutions increase without bound. For + " , solutions decrease without bound. a,b The integrating factor is .a>b oe /> # . The general solution of the differential equation is Ca>b oe a)=38a>b %-9=a>bb & - /> # . The solution is sinusoidal as long as - oe ! . The initial value of this sinusoidal solution is +! oe a)=38a!b %-9=a!bb & oe % & a- b See part a,b. 22a+b 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 +! oe " a,b The integrating factor is .a>b oe /> # , and the general solution of the differential equation is Ca>b oe $/> $ - /> # Invoking the initial condition C a!b oe + , the solution may also be expressed as Ca>b oe $/> $ a+ $b /> # Differentiating, follows that C w a!b oe " a+ $b # oe a+ "b # The critical value is evidently +! oe " ________________________________________________________________________ page 24 ---------------------------------------------------- CHAPTER 2. ---- a- b. For +! oe " , the solution is Ca>b oe $/> $ # /> # , which afor large >b is dominated by the term containing /> # is Ca>b oe a)=38a>b %-9=a>bb & - /> # . 23a+b As > p ! , solutions increase without bound if C a"b oe + % , and solutions decrease without bound if Ca"b oe + % a,b. The integrating factor is .a>b oe /B:^' >" .> oe > /> The general solution of the > differential equation is Ca>b oe > /> - /> > . Invoking the specified value C a"b oe + , we have " - oe + / . That is, - oe + / " . Hence the solution can also be expressed as Ca>b oe > /> a+ / "b /> > . For small values of > , the second term is dominant. Setting + / " oe ! , critical value of the parameter is +! oe " / a- b. For + " / , solutions increase without bound. For + " / , solutions decrease without bound. When + oe " / , the solution is C a>b oe > /> , which approaches ! as > p ! . 24a+b. As > p ! , solutions increase without bound if C a"b oe + % , and solutions decrease without bound if Ca"b oe + % ________________________________________________________________________ page 25 ---------------------------------------------------- CHAPTER 2. ---- a,b. Given the initial condition, Ca 1 #b oe + , the solution is Ca>b oe a+ 1# % -9= >b > Since lim -9= > oe " , solutions increase without bound if + % 1# , and solutions decrease without bound if + % 1# Hence the critical value is +! oe % 1# oe ! %&#)%( . > ! a- b For + oe % 1# , the solution is C a>b oe a" -9= >b > , and lim C a>b oe " # . Hence the solution is bounded. > ! 25. The integrating factor is .a>b oe /B:^' " .> oe /> # Therefore general solution is # Ca>b oe c%-9=a>b )=38a>bd & - /> # Invoking the initial condition, the specific solution is Ca>b oe c%-9=a>b )=38a>b * /> # d & . Differentiating, it follows that Setting C w a>b oe ! , the first solution is >" oe " $'%$ , 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" $'%$ )#!!)b. C w a>b oe %=38a>b )-9=a>b % & /> # ` & C ww a>b oe %-9=a>b )=38a>b # #& /> # ` & 26. The integrating factor is .a>b oe /B:^' # .> oe /#> $ , and the differential equation $ can w be written as a/#> $ Cb oe /#> $ > /#> $ # The general solution is C a>b oe #" '> ) - /#> $ . Imposing the initial condition, we have C a>b oe #" '> ) aC! #" )b/#> $ . Since the solution is smooth, the desired intersection will be a point of tangency. Taking the derivative, C w a>b oe $ % a#C! #" %b/#> $ $ Setting C w a>b oe ! , the solution is >" oe $ 68ca#" )C! b *d Substituting into the solution, the respective value at the # stationary point is Ca>" b oe $ * 68 $ * 68a#" )C! b. Setting this result equal to zero, # % ) we obtain the required initial value C! oe a#" * /% $ b ) oe " '%$ 27. The integrating factor is .a>b oe /> 4 , and the differential equation can be written as w a/> 4 Cb oe $ /> 4 # /> 4 -9=a#>b The general solution is Invoking the initial condition, Ca!b oe ! , the specific solution is Ca>b oe "# c)-9=a#>b '%=38a#>bd '& - /> 4 Ca>b oe "# )-9=a#>b '%=38a#>b ()) /> 4 ` '& As > p _ , the exponential term will decay, and the solution will oscillate about an average value of "# , with an amplitude of ) '& ________________________________________________________________________ page 26 ---------------------------------------------------- CHAPTER 2. ---- 29. The integrating factor is .a>b oe /$> # , and the differential equation can be written w as a/$> # Cb oe $> /$> # # /> # The general solution is C a>b oe #> % $ % /> - /$> # Imposing the initial condition, C a>b oe #> % $ % /> aC! "' $b /$> # As > p _ , the term containing /$> # will dominate the solution. Its sign will determine the divergence properties. Hence the critical value of the initial condition is C! oe "' $ The corresponding solution, Ca>b oe #> % $ % /> , will also decrease without bound. Note on Problems 31-34 : Let 1a>b be given, and consider the function Ca>b oe C" a>b 1a>b , in which C" a>b p _ as > p _ . Differentiating, C w a>b oe C"w a>b 1 w a>b . Letting + be a constant, it follows that C w a>b +C a>b oe C"w a>b +C" a>b 1 w a>b +1a>b Note that the hypothesis on the function C" a>b will be satisfied, if C"w a>b +C" a>b oe ! . That is, C" a>b oe - /+> Hence Ca>b oe - /+> 1a>b, which is a solution of the equation C w +C oe 1 w a>b +1a>b For convenience, choose + oe " . 31. Here 1a>b oe $ , and we consider the linear equation C w C oe $ The integrating w factor is .a>b oe /> , and the differential equation can be written as a/> C b oe $/> The general solution is Ca>b oe $ - /> 33. 1a>b oe $ > Consider the linear equation C w C oe " $ > The integrating w factor is .a>b oe /> , and the differential equation can be written as a/> C b oe a# >b/> The general solution is Ca>b oe $ > - /> 34. 1a>b oe % ># Consider the linear equation C w C oe % #> ># The integrating w factor is .a>b oe /> , and the equation can be written as a/> C b oe a% #> ># b/> The general solution is Ca>b oe % ># - /> ________________________________________________________________________ page 27 ---------------------------------------------------- CHAPTER 2. ---- Section 2.2 2. For B " , the differential equation may be written as C .C oe cB# a" B$ bd.B Integrating both sides, with respect to the appropriate variables, we obtain the relation C# # oe " 68k" B$ k - That is, C aBb oe ,, # 68k" B$ k - $ $ 3. The differential equation may be written as C# .C oe =38 B .B Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation C " oe -9= B - That is, aG -9= BbC oe ", in which G is an arbitrary constant. Solving for the dependent variable, explicitly, CaBb oe " aG -9= Bb . 5. Write the differential equation as -9=# #C .C oe -9=# B .B, or =/- # #C .C oe -9=# B .B Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation >+8 #C oe =38 B -9= B B - 7. The differential equation may be written as aC /C b.C oe aB /B b.B Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation C # # / C oe B # # / B - 8. Write the differential equation as a" C# b.C oe B# .B Integrating both sides of the equation, we obtain the relation C C $ $ oe B$ $ - , that is, $C C $ oe B$ G 9a+b. The differential equation is separable, with C# .C oe a" #Bb.B Integration yields C" oe B B# - Substituting B oe ! and C oe " ' , we find that - oe ' Hence the specific solution is C" oe B# B ' . The explicit form is CaBb oe " aB# B 'b a, b a- b. Note that B# B ' oe aB #baB $b . Hence the solution becomes singular at B oe # and B oe $ 10a+b CaBb oe #B #B# % ________________________________________________________________________ page 28 ---------------------------------------------------- CHAPTER 2. ---- 10a,b 11a+b Rewrite the differential equation as B /B .B oe C .C Integrating both sides of the equation results in B /B /B oe C # # - Invoking the initial condition, we obtain - oe " # Hence C # oe #/B #B /B " The explicit form of the solution is CaBb oe #/B #B /B " The positive sign is chosen, since C a!b oe " a, b a- b The function under the radical becomes negative near B oe " ( and B oe ! (' 11a+b Write the differential equation as <# .< oe ) " . ) Integrating both sides of the equation results in the relation <" oe 68 ) - Imposing the condition <a"b oe # , we obtain - oe " # . The explicit form of the solution is <a) b oe # a" # 68 )b ________________________________________________________________________ page 29 ---------------------------------------------------- CHAPTER 2. ---- a, b a- b Clearly, the solution makes sense only if ) ! Furthermore, the solution becomes singular when 68 ) oe " # , that is, ) oe / 13a+b CaBb oe # 68a" B# b % a, b 14a+b. Write the differential equation as C$ .C oe Ba" B# b .B Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation C# # oe " B# - Imposing the initial condition, we obtain - oe $ # Hence the specific solution can be expressed as C# oe $ # " B# The explicit form of the solution is CaBb oe " $ # " B# The positive sign is chosen to " # satisfy the initial condition. ________________________________________________________________________ page 30 ---------------------------------------------------- CHAPTER 2. ---- a, b a- b The solution becomes singular when # " B# oe $ . That is, at B oe ,, & # 15a+b CaBb oe " # B# "& % a, b 16a+b. Rewrite the differential equation as %C$ .C oe BaB# "b.B Integrating both sides of the equation results in C% oe aB# "b# % - Imposing the initial condition, we obtain - oe ! Hence the solution may be expressed as aB# "b# %C % oe ! The explicit form of the solution is CaBb oe aB# "b # The sign is chosen based on C a!b oe " # ________________________________________________________________________ page 31 ---------------------------------------------------- CHAPTER 2. ---- a, b a- b The solution is valid for all B - ` . a, b 17a+b CaBb oe & # B$ /B "$ % a- b. The solution is valid for B " %& This value is found by estimating the root of %B$ %/B "$ oe ! 18a+b. Write the differential equation as a$ %Cb.C oe a/B /B b.B Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation $C #C # oe a/B /B b - Imposing the initial condition, C a!b oe " , we obtain - oe ( Thus, the solution can be expressed as $C #C # oe a/B /B b ( Now by completing the square on the left hand side, #aC $ %b# oe a/B /B b '& ) . Hence the explicit form of the solution is CaBb oe $ % '& "' -9=2 B ________________________________________________________________________ page 32 ---------------------------------------------------- CHAPTER 2. ---- a, b a- b. Note the '& "' -9=2 B ! , as long as kBk # " Hence the solution is valid on the interval # " B # " . 19a+b C aBb oe 1 $ " =38" a$ -9=# Bb $ a, b 20a+b Rewrite the differential equation as C# .C oe +<-=38 B " B# .B Integrating # both sides of the equation results in C$ $ oe a+<-=38 Bb # - Imposing the condition $ $ Ca!b oe !, we obtain - oe ! The explicit form of the solution is CaBb oe # a+<-=38 Bb# $ ________________________________________________________________________ page 33 ---------------------------------------------------- CHAPTER 2. ---- a, b . a- b. Evidently, the solution is defined for " Y B Y " 22. The differential equation can be written as a$C# %b.C oe $B# .B Integrating both sides, we obtain C$ %C oe B$ - Imposing the initial condition, the specific solution is C$ %C oe B$ " Referring back to the differential equation, we find that C w p _ as C p ,,# $ The respective values of the abscissas are B oe " #(' , " &*) Hence the solution is valid for " #(' B " &*) 24. Write the differential equation as a$ #Cb.C oe a# /B b.B Integrating both sides, we obtain $C C # oe #B /B - Based on the specified initial condition, the solution can be written as $C C # oe #B /B " Completing the square, it follows that CaBb oe $ # #B /B "$ % The solution is defined if #B /B "$ % ! , that is, " & Y B Y # aapproximatelyb. In that interval, C w oe ! , for B oe 68 # 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 oe 68 # 26. The differential equation can be written as a" C# b" .C oe #a" Bb.B Integrating both sides of the equation, we obtain +<->+8C oe #B B# - Imposing the given initial condition, the specific solution is +<->+8C oe #B B# Therefore, C aBb oe >+8a#B B# b Observe that the solution is defined as long as 1 # #B B# 1 # It is easy to see that #B B# " Furthermore, #B B# oe 1 # for B oe # ' and ! ' Hence the solution is valid on the interval # ' B ! ' Referring back to the differential ________________________________________________________________________ page 34 ---------------------------------------------------- CHAPTER 2. ---- equation, the solution is stationary at B oe " Since C ww aBb ! on the entire interval of definition, the solution attains a global minimum at B oe " 28a+b. Write the differential equation as C" a% Cb" .C oe >a" >b" .> Integrating both sides of the equation, we obtain 68 kCk 68kC %k oe %> %68k" >k - Taking the exponential of both sides, it follows that kC aC %bk oe G /%> a" >b% It follows that as > p _ , kC aC %bk oe k" % aC %bkp _ . That is, C a>b p % a,b Setting Ca!b oe # , we obtain that G oe " . Based on the initial condition, the solution may be expressed as C aC %b oe /%> a" >b% Note that C aC %b ! , for all > ! Hence C % for all > ! 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 /%> a" >b% oe $** is near > oe # )%% a- b Note the Ca>b oe % is an equilibrium solution. Examining the local direction field, we see that if Ca!b ! , then the corresponding solutions converge to C oe % . Referring back to part a+b, we have C aC %b oe cC! aC! %bd/%> a" >b% , for C! % Setting % > oe # , we obtain C! aC! %b oe a$ /# b C a#b aC a#b %b Now since the function 0 aCb oe C aC %b is monotone for C % and C % , we need only solve the equations C! aC! %b oe $**a$ /# b% and C! aC! %b oe %!"a$ /# b% The respective solutions are C! oe $ ''## and C! oe % %!%# 30a0 b ________________________________________________________________________ page 35 ---------------------------------------------------- CHAPTER 2. ---- 31a- b C " 32a+b. Observe that aB# $C # b #BC oe " ^ B # is homogeneous. $C #B Hence the differential equation a,b. The substitution C oe B @ results in @ B @ w oe aB# $B# @# b #B# @ . The transformed equation is @ w oe a" @# b #B@ This equation is separable, with general solution @# " oe - B In terms of the original dependent variable, the solution is B# C # oe - B$ a- b 33a- b ________________________________________________________________________ page 36 ---------------------------------------------------- CHAPTER 2. ---- C C " 34a+b Observe that a%B $C b #B C oe # B # B ` Hence the differential equation is homogeneous. a,b. The substitution C oe B @ results in @ B @ w oe # @ a# @b. The transformed equation is @ w oe a@# &@ %b # @ B This equation is separable, with general solution a@%b# k@"k oe G B$ In terms of the original dependent variable, the solution is a%B Cb# kBCk oe G a- b 35a- b. 36a+b Divide by B# to see that the equation is homogeneous. Substituting C oe B @ , we obtain B @ w oe a" @b# The resulting differential equation is separable. a,b Write the equation as a" @b# .@ oe B" .B Integrating both sides of the equation, we obtain the general solution " a" @b oe 68kBk - In terms of the original dependent variable, the solution is C oe B cG 68kBkd" B ________________________________________________________________________ page 37 ---------------------------------------------------- CHAPTER 2. ---- a- b a,b Integrating both sides of the transformed equation yields " & 68k" &@# k oe 68kBk - , that is, " &@# oe G kBk& In terms of the original dependent variable, the general solution is &C# oe B# G kBk$ a- b C " $C 37a+b The differential equation can be expressed as C w oe " ^ B # B . Hence the # equation is homogeneous. The substitution C oe B @ results in B @ w oe a" &@# b #@ . #@ " Separating variables, we have "&@# .@ oe B .B C " C " 38a+b The differential equation can be expressed as C w oe $ B # ^ B . Hence the # equation is homogeneous. The substitution C oe B @ results in B @ w oe a@# "b #@, that " is, @##@ .@ oe B .B " a,b Integrating both sides of the transformed equation yields 68k@# "k oe 68kBk - , that is, @# " oe G kBk In terms of the original dependent variable, the general solution is C# oe G B# kBk B# ________________________________________________________________________ page 38 ---------------------------------------------------- CHAPTER 2. ---- a- b ________________________________________________________________________ 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 # " ^" " =38 > oe " " =38 > 9D 738 It leaves the tank at a rate of # U "!! 9D 738 % # # % Hence the differential equation governing the amount of salt at any time is .U " " oe =38 > U &! .> # % The initial amount of salt is U! oe &! 9D The governing ODE is linear, with integrating w factor .a>b oe /> &! Write the equation as ^/> &! U oe /> &! ^ " " =38 > The # % specific solution is Ua>b oe #& "# &=38 > '#&-9= > '$"&! /> &! ` #&!" 9D a, b a- b The amount of salt approaches a steady state, which is an oscillation of amplitude " % about a level of #& 9D 6a+b. The equation governing the value of the investment is .W .> oe < W . The value of the investment, at any time, is given by W a>b oe W! /<> Setting W aX b oe #W! , the required time is X oe 68a#b < a- b Referring to Parta+b, < oe 68a#b X . Setting X oe ) , the required interest rate is to be approximately < oe ) '' % 8a+b. Based on the solution in Eq.a"'b , with W! oe ! , the value of the investments with contributions is given by W a>b oe #& !!!a/<> "b After ten years, person A has WE oe $#& !!!a" ##'b oe $ $! '%! Beginning at age $& , the investments can now be analyzed using the equations WE oe $! '%! / !)> and WF oe #& !!!a/ !)> "b After thirty years, the balances are WE oe $ $$( ($% and WF oe $ #&! &(* a,b For the case < oe (% oe !( , X * * C<= a,b For an unspecified rate < , the balances after thirty years are WE oe $! '%! /$!< and WF oe #& !!!a/$!< "b ________________________________________________________________________ page 40 ---------------------------------------------------- CHAPTER 2. ---- a- b . a. b 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 < oe !* is the annual interest rate. Monthly payments of $ )!! are equivalent to $ * '!! per year The differential equation governing the value of the mortgage is .W .> oe !* W * '!! Given that W! is the original amount borrowed, the debt is W a>b oe W! / !*> "!' ''(a/ !*> "b Setting W a$!b oe ! , it follows that W! oe $ ** &!! . a,b The total payment, over $! years, becomes $ #)) !!! . The interest paid on this purchase is $ ")) &!! . 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 .W .> oe <W 5 . The balance at any time is given by W a>b oe W! /<> 5 a/<> "b < a,b. The solution may also be expressed as W a>b oe W! 5 /<> 5 Note that if the < < withdrawal rate is 5! oe < W! , the balance will remain at a constant level W! 5 a- b. Assuming that 5 5! , W aX! b oe ! for X! oe " 68' 55! " < a. b. If < oe !) and 5 oe #5! , then X! oe ) '' years . a/b. Setting W a>b oe ! and solving for /<> in Parta,b, /<> oe results in 5 oe <W! /<X a/<X "b 5 5<W! Now setting > oe X a0 b In parta/b, let 5 oe "# !!! , < oe !) , and X oe #! . The required investment becomes W! oe $ ""* ("& . 14a+b Let U w oe < U The general solution is Ua>b oe U! /<> Based on the definition of half-life, consider the equation U! # oe U! /&($! < It follows that ________________________________________________________________________ page 41 ---------------------------------------------------- CHAPTER 2. ---- &($! < oe 68 " # , that is, < oe " #!*( , "!% per year. a,b. Hence the amount of carbon-14 is given by Ua>b oe U! /" #!*(,"! > % % a- b Given that UaX b oe U! & , we have the equation " & oe /" #!*(,"! X Solving for the decay time, the apparent age of the remains is approximately X oe "$ $!% '& years . 15. Let T a>b be the population of mosquitoes at any time > . The rate of increase of the mosquito population is <T The population decreases by #! !!! per day. Hence the equation that models the population is given by .T .> oe <T #! !!! . Note that the variable > represents days . The solution is T a>b oe T! /<> #! !!! a/<> "b In the < absence of predators, the governing equation is .T" .> oe <T" , with solution T" a>b oe T! /<> Based on the data, set T" a(b oe #T! , that is, #T! oe T! /(< The growth rate is determined as < oe 68a#b ( oe !**!# per day Therefore the population, including the predation by birds, is T a>b oe # , "!& / !**> #!" **(a/ !**> "b oe oe #!" **( $ "*(( $ / !**> 16a+b. Ca>b oe /B:c# "! > "! #-9= > "!d The doubling-time is 7 # *'$# a,b The differential equation is .C .> oe C "! , with solution C a>b oe C a!b/> "! The doubling-time is given by 7 oe "!68a#b ' *$"& a- b. Consider the differential equation .C .> oe a! & =38 #1> b C & The equation is " " separable, with C .C oe ^! " & =38 #1> .> Integrating both sides, with respect to the appropriate variable, we obtain 68 C oe a1> -9= #1> b "!1 - Invoking the initial condition, the solution is Ca>b oe /B:ca" 1> -9= #1> b "!1d The doubling-time is 7 ' $)!% The doubling-time approaches the value found in parta,b. a. b . 17a+b. The differential equation .C .> oe <a>b C 5 is linear, with integrating factor .a>b oe /B: ' <a>b.>` Write the equation as a. C bw oe 5 .a>b Integration of both ________________________________________________________________________ page 42 ---------------------------------------------------- CHAPTER 2. ---- sides yields the general solution C oe 5 ' .a7 b. 7 C! .a!b` .a>b . In this problem, the integrating factor is .a>b oe /B:ca-9= > >b &d a,b The population becomes extinct, if Ca> b oe ! , for some > oe > . Referring to parta+b, we find that Ca> b oe ! ( > ! /B:ca-9= 7 7 b &d. 7 oe & /" & C- It can be shown that the integral on the left hand side increases monotonically, from zero to a limiting value of approximately & !)*$ . Hence extinction can happen only if & /" & C- & !)*$ , that is, C- ! )$$$ a- b. Repeating the argument in parta,b, it follows that Ca> b oe ! ( > ! /B:ca-9= 7 7 b &d. 7 oe " " & / C- 5 19a+b. Let Ua>b be the volume of carbon monoxide in the room. The rate of increase of CO is a !%ba! "b oe ! !!% 0 >$ 738 The amount of CO leaves the room at a rate of a! "bUa>b "#!! oe Ua>b "#!!! 0 >$ 738 Hence the total rate of change is given by the differential equation .U .> oe ! !!% Ua>b "#!!! This equation is linear and separable, with solution Ua>b oe %) %) /B:a > "#!!!b 0 >$ Note that U! oe ! 0 >$ Hence the concentration at any time is given by Ba>b oe Ua>b "#!! oe Ua>b "# % . a,b. The concentration of CO in the room is Ba>b oe % %/B:a > "#!!!b % A level of ! !!!"# corresponds to ! !"# % . Setting Ba7 b oe ! !"# , the solution of the equation % %/B:a > "#!!!b oe ! !"# is 7 $' minutes . page 43 a. b. Evidently, C- is a linear function of the parameter 5 . Hence extinction can happen only if /" & C- 5 & !)*$ , that is, C- % "''( 5 ________________________________________________________________________ ---------------------------------------------------- CHAPTER 2. ---- 20a+b The concentration is - a>b oe 5 T < a-! 5 T <b/<> Z It is easy to see that - a>p_b oe 5 T < a,b - a>b oe -! /<> Z . The reduction times are X&! oe 68 # Z < and X"! oe 68 "! Z < a- b The reduction times, in years, are XW oe 68a"!ba'& #b "# #!! oe %$! )& XQ oe 68a"!ba"&)b % *!! oe (" % XI oe 68a"!ba"(&b %'! oe ' !& XS oe 68a"!ba#!*b "' !!! oe "( '$ 21a- b 22a+b. The differential equation for the motion is 7 .@ .> oe @ $! 71 Given the initial condition @a!b oe #! m/s , the solution is @a>b oe %% " '% " /B:a > % &b . Setting @a>" b oe ! , the ball reaches the maximum height at >" oe " ')$ sec . Integrating @a>b , the position is given by Ba>b oe $") %& %% " > #)) %& /B:a > % &b Hence the maximum height is Ba>" b oe %& () m . a- b a,b Setting Ba># b oe ! , the ball hits the ground at ># oe & "#) sec . 23a+b The differential equation for the upward motion is 7 .@ .> oe . @# 71 , in which . oe " "$#& . This equation is separable, with . @#7 .@ oe .> Integrating 71 ________________________________________________________________________ page 44 ---------------------------------------------------- CHAPTER 2. ---- both sides and invoking the initial condition, @a>b oe %% "$$ >+8a %#& ### >b Setting @a>" b oe ! , the ball reaches the maximum height at >" oe " *"' sec . Integrating @a>b , the position is given by Ba>b oe "*) (& 68c-9=a! ### > ! %#&bd %) &( Therefore the maximum height is Ba>" b oe %) &' m . a,b The differential equation for the downward motion is 7 .@ .> oe .@# 71 7 This equation is also separable, with 71. @# .@ oe .> For convenience, set > oe ! at the top of the trajectory. The new initial condition becomes @a!b oe ! . Integrating both sides and invoking the initial condition, we obtain 68ca%% "$ @b a%% "$ @bd oe > # #& Solving for the velocity, @a>b oe %% "$a" /> # #& b a" /> # #& b Integrating @a>b , the # position is given by Ba>b oe ** #* 68'/> # #& a" /> # #& b " ")' # To estimate the a- b duration of the downward motion, set Ba># b oe ! , resulting in ># oe $ #(' sec . Hence the total time that the ball remains in the air is >1 ># oe & "*# sec . 24a+b Measure the positive direction of motion downward . Based on Newton's #nd law, the equation of motion is given by 7 .@ ! (& @ 71 , ! > "! oeoe "# @ 71 , > "! .> 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 .@ .> oe @ ( & $# , with initial velocity @a!b oe ! fps This differential equation is separable and linear, with solution @a>b oe #%!a" /> ( & b. Hence @a"!b oe "(' ( fps a,b. Integrating the velocity, with Ba>b oe ! , the distance fallen is given by Hence Ba"!b oe "!(% & ft . Ba>b oe #%! > ")!! /> ( & ")!! . ________________________________________________________________________ page 45 ---------------------------------------------------- CHAPTER 2. ---- a- b For computational purposes, reset time to > oe ! . For the remainder of the motion, the initial value problem is .@ .> oe $#@ "& $# , with specified initial velocity @a!b oe "(' ( fps The solution is given by @a>b oe "& "'" ( /$# > "& As > p _ , @a>b p @P oe "& fps Integrating the velocity, with Ba!b oe "!(% & , the distance fallen after the parachute is open is given by Ba>b oe "& > (& ) /$# > "& ""&! $ To find the duration of the second part of the motion, estimate the root of the transcendental equation "& X (& ) /$# X "& ""&! $ oe &!!! The result is X oe #&' ' sec a. b 25a+b. Measure the positive direction of motion upward . The equation of motion is given by 7.@ .> oe 5 @ 71 . The initial value problem is .@ .> oe 5@ 7 1 , with @a!b oe @! . The solution is @a>b oe 71 5 a@! 71 5 b/5> 7 Setting @a>7 b oe !, the maximum height is reached at time >7 oe a7 5 b68ca71 5 @! b 71d Integrating the velocity, the position of the body is Ba>b oe 71 > 5 "S Hence the maximum height reached is B7 oe Ba>7 b oe 7 @! 7 # 71 5 @! 1S < 68" 5 5 71 7 # 7 @! 5> 7 < 1 " / 5 5 a,b Recall that for $ " , 68a" $ b oe $ " $ # " $ $ " $ % # $ % 26a,b. a- b 5 ! 71 lim 71a5 @!5 b/ 71 5 5> 7 oe lim 5 ! 28a+b. In terms of displacement, the differential equation is 7@ .@ .B oe 5 @ 71 .@ This follows from the chain rule : .@ oe .B .B oe @ .@ . The differential equation is .> .> .> separable, with 7 ! lim ^ 71 @! /5> 7 ` oe ! , since 5 > 7 a5 @ ! 71b/5> 7 oe 1> 7 ! lim /5> 7 oe ! ________________________________________________________________________ page 46 ---------------------------------------------------- CHAPTER 2. ---- Ba@b oe 7@ 7# 1 71 5 @ # 68 5 5 71 The inverse exists, since both B and @ are monotone increasing. In terms of the given parameters, Ba@b oe " #& @ "& $" 68k! !)"' @ "k a,b Ba"!b oe "$ %& meters . The required value is 5 oe ! #% . a- b In parta+b, set @ oe "! m/s and B oe "! meters . 29a+b Let B represent the height above the earth's surface. The equation of motion is Q7 given by 7 .@ oe K aVBb# , in which K is the universal gravitational constant. The .> 7@ .@ Q7 oe 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 @ .@ oe KQ aV Bb# .B Integrating both sides, and invoking the initial condition @a!b oe #1V , the solution is @# oe #KQ aV Bb" #1V #KQ V From elementary physics, it follows that 1 oe KQ V # . Therefore @aBb oe #1 'V V B " a Note that 1 oe () &%& mi/hr# .b a,b We now consider .B .> oe #1 'V V B ". This equation is also separable, with V B .B oe #1 V .> By definition of the variable B, the initial condition is # $ Ba!b oe ! Integrating both sides, we obtain Ba>b oe $ ^ #1 V > # V $ # ` V # $ Setting the distance BaX b V oe #%! !!! , and solving for X , the duration of such a flight would be X %* hours . 32a+b Both equations are linear and separable. The initial conditions are @a!b oe ? -9= E and Aa!b oe ? =38 E . The two solutions are @a>b oe ? -9= E /<> and Aa>b oe 1 < a? =38 E 1 <b/<> ________________________________________________________________________ page 47 ---------------------------------------------------- CHAPTER 2. ---- a,b Integrating the solutions in parta+b, and invoking the initial conditions, the coordinates are Ba>b oe ? -9= Ea" /<> b and < a- b ? Ca>b oe 1> < ^1 ?< =38 E 2<# <# S =38 E 1 <# </<> < a. b Let X be the time that it takes the ball go to $&! ft horizontally. Then from above, /X & oe a? -9= E (!b ? -9= E At the same time, the height of the ball is given by CaX b oe "'! X #'( "#&?=38 E )!! &? =38 E ca? -9= E (!b ? -9= E d Hence E and ? must satisfy the inequality )!!68" ? -9= E (! #'( "#&?=38 E )!! &? =38 E ca? -9= E (!b ? -9= E d "! ? -9= E " # 33a+b Solving equation a3b, C w aBb oe ca5 # C b Cd chosen, since C is an increasing function of B . . The positive answer is a,b. Let C oe 5 # =38# > . Then .C oe #5 # =38 > -9= > .> Substituting into the equation in parta+b, we find that #5 # =38 > -9= > .> -9= > oe .B =38 > Hence #5 # =38# > .> oe .B a- b Letting ) oe #> , we further obtain 5 # =38# ) . ) oe .B Integrating both sides of the # equation and noting that > oe ) oe ! corresponds to the origin, we obtain the solutions Ba) b oe 5 # a) =38 )b # and cfrom parta,bd Ca)b oe 5 # a" -9= )b # a. b. Note that C B oe a" -9= )b a) =38 )b Setting B oe " , C oe # , the solution of the equation a" -9= ) b a) =38 )b oe # is ) " %!" . Substitution into either of the expressions yields 5 # "*$ ________________________________________________________________________ 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! %b , the initial value problem has a unique solution on the interval a! %b 3. The function >+8 > is discontinuous at odd multiples of 1 . Since # 1 initial value problem has a unique solution on the interval ^ 1 $# # 1 # 1 $1 # , the 5. :a>b oe #> a% ># b and 1a>b oe $># a% ># b. These functions are discontinuous at B oe ,,# . The initial value problem has a unique solution on the interval a # #b 6. The function 68 > is defined and continuous on the interval a! _b . Therefore the initial value problem has a unique solution on the interval a! _b. 7. The function 0 a> C b is continuous everywhere on the plane, except along the straight line C oe #> & The partial derivative `0 `C oe (> a#> &C b# has the same region of continuity. 9. The function 0 a> C b is discontinuous along the coordinate axes, and on the hyperbola ># C # oe " . Furthermore, `0 ,," C 68k>C k oe # # C# b `C C a" > a " ># C # b # has the same points of discontinuity. 10. 0 a> C b is continuous everywhere on the plane. The partial derivative `0 `C is also continuous everywhere. 12. The function 0 a> C b is discontinuous along the lines > oe ,,5 1 and C oe " . The partial derivative `0 `C oe -9>a>b a" C b# has the same region of continuity. 14. The equation is separable, with .C C # oe #> .> Integrating both sides, the solution is given by Ca>b oe C! " C! ># For C! ! , solutions exist as long as ># " C! . For C! Y ! , solutions are defined for all > . 15. The equation is separable, with .C C $ oe .> Integrating both sides and invoking the initial condition, Ca>b oe C! #C! > " Solutions exist as long as #C! > " ! , that is, #C! > " . If C! ! , solutions exist for > " #C! . If C! oe ! , then the solution Ca>b oe ! exists for all > . If C! ! , solutions exist for > " #C! . 16. The function 0 a> C b is discontinuous along the straight lines > oe " and C oe ! . The partial derivative `0 `C is discontinuous along the same lines. The equation is ________________________________________________________________________ page 49 ---------------------------------------------------- CHAPTER 2. ---- separable, with C .C oe ># .> a" >$ b Integrating and invoking the initial condition, the # " # solution is Ca>b oe # 68k" >$ k C! ` Solutions exist as long as $ # that is, C! # 68k" >$ k For all C! it can be verified that C! oe ! yields a valid $ solution, even though Theorem # % # does not guarantee one , solutions exists as long as # k" >$ k /B:a $C! #b From above, we must have > " . Hence the inequality $ # may be written as > /B:a $C! #b " It follows that the solutions are valid for " $ # c/B:a $C! #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 >! ! Otherwise, the slopes eventually become negative, and solutions tend to zero . Furthermore, C! oe ! is an equilibrium solution. Note that slopes are zero along the curves C oe ! and >C oe $ . 19. ________________________________________________________________________ page 50 ---------------------------------------------------- CHAPTER 2. ---- For initial conditions a>! C! b satisfying >C $ , the respective solutions all tend to zero . Solutions with initial conditions above or below the hyperbola >C oe $ eventually tend to ,,_ . Also, C! oe ! is an equilibrium solution. 20. Solutions with >! ! all tend to _ . Solutions with initial conditions a>! C! b to the right of the parabola > oe " 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 _ 21. Define C- a>b oe # a> - b$ # ?a> - b, in which ?a>b is the Heaviside step function. $ Note that C- a- b oe C- a!b oe ! and C- ^- a$ #b# $ oe " a,b Let - oe # a$ #b# $ a+b Let - oe " a$ #b# $ a- b Observe that C! a#b oe # a#b$ # , C- a>b # a#b$ # for ! - #, and that C- a#b oe ! for $ $ - # So for any - !, ,,C- a#b - c # #d 26a+b Recalling Eq. a$&b in Section # ", ________________________________________________________________________ page 51 ---------------------------------------------------- CHAPTER 2. ---- " ( .a=b1a=b .= .a>b .a>b " ' .a>b Coe It is evident that C" a>b oe " .a>b " a,b. By definition, .a>b oe /B:^ ' :a>b.>. Hence C"w oe :a>b That is, C"w :a>bC" oe ! and C# a>b oe .a=b1a=b .=. " .a>b oe :a>bC" a- b C#w oe S :a>b That is, C#w :a>bC# oe 1a>b " '> .a>b < ! .a=b1a=b .= " S .a>b <.a>b1a>b oe :a>bC# 1a>b 30. Since 8 oe $, set @ oe C # . It follows that .@ oe #C $ .C and .C oe C# .@ .> .> .> .> $ Substitution into the differential equation yields C# .@ &C oe 5C $ , which further .> results in @ w #&@ oe #5 The latter differential equation is linear, and can be written as w a/#&> b oe #5 The solution is given by @a>b oe #5> /#&> -/#&> Converting back to the original dependent variable, C oe ,,@" # 31. Since 8 oe $, set @ oe C # . It follows that .@ oe #C $ .C and .C oe C# .@ .> .> .> .> C$ .@ $ The differential equation is written as # .> a>-9= > X bC oe 5C , which upon further substitution is @ w #a>-9= > X b@ oe # This ODE is linear, with integrating factor .a>b oe /B:^#' a>-9= > X b.> oe /B:a #>=38 > #X >b The solution is > ! $ $ @a>b oe #/B:a#>=38 > #X >b( /B:a #>=38 7 #X 7 b. 7 - /B:a #>=38 > #X >b Converting back to the original dependent variable, C oe ,,@" # 33. The solution of the initial value problem C"w #C" oe !, C" a!b oe " is C" a>b oe /#> Therefore ya" b oe C" a"b oe /# On the interval a" _b the differential equation is C#w C# oe !, with C# a>b oe -/> Therefore C a" b oe C# a"b oe -/" Equating the limits Ca" b oe Ca" b, we require that - oe /" Hence the global solution of the initial value problem is Ca>b oe oe Note the discontinuity of the derivative C a >b oe oe #/#> , ! > " /"> , > " /#> , ! Y > Y " /"> , > " ________________________________________________________________________ page 52 ---------------------------------------------------- CHAPTER 2. ---- Section 2.5 1. For C! ! , the only equilibrium point is C oe ! . 0 w a!b oe + ! , hence the equilibrium solution 9a>b oe ! is unstable. 2. The equilibrium points are C oe + , and C oe ! . 0 w a + , b ! , therefore the equilibrium solution 9a>b oe + , is asymptotically stable. 3. ________________________________________________________________________ page 53 ---------------------------------------------------- CHAPTER 2. ---- 4. The only equilibrium point is C oe ! . 0 w a!b ! , hence the equilibrium solution 9a>b oe ! is unstable. 5. The only equilibrium point is C oe ! . 0 w a!b ! , hence the equilibrium solution 9a>b oe ! is +=C7:>9>3-+66C stable. 6. ________________________________________________________________________ page 54 ---------------------------------------------------- CHAPTER 2. ---- 7a,b. 8. The only equilibrium point is C oe " . Note that 0 w a"b oe ! , and that C w ! for C " . As long as C! " , the corresponding solution is monotone decreasing. Hence the equilibrium solution 9a>b oe " is semistable. 9. ________________________________________________________________________ page 55 ---------------------------------------------------- CHAPTER 2. ---- 10. The equilibrium points are C oe ! , ,," . 0 w aC b oe " $C # . The equilibrium solution 9a>b oe ! is unstable, and the remaining two are asymptotically stable. 11. 12. The equilibrium points are C oe ! , ,,# . 0 w aC b oe )C %C $ . The equilibrium solutions 9a>b oe # and 9a>b oe # are unstable and asymptotically stable, respectively. The equilibrium solution 9a>b oe ! is semistable. ________________________________________________________________________ page 56 ---------------------------------------------------- CHAPTER 2. ---- 13. The equilibrium points are C oe ! and " . 0 w aC b oe #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! oe O $, " a" $bc" aC O bd > oe 68 < aC O bc" a" $bd " a" $bc" a# $bd 7 oe 68 < a# $bc" a" $bd Setting C oe #C! , That is, 7 oe " 68 % If < oe ! !#& per year, 7 oe && %& years. < " ! c" " d X oe 68 < " c" ! d a,b In Eq. a"$b, set C! O oe ! and C O oe " . As a result, we obtain Given ! oe ! ", " oe ! * and < oe ! !#& per year, 7 oe "(& () years. 16a+b. ________________________________________________________________________ page 57 ---------------------------------------------------- CHAPTER 2. ---- 17. Consider the change of variable ? oe 68aC O b Differentiating both sides with respect to >, ? w oe C w C Substitution into the Gompertz equation yields ? w oe <?, with solution ? oe ?! /<> It follows that 68aC O b oe 68aC! O b/<> That is, C oe /B:68aC! O b/<> ` O a,b Solving for >, a+b. Given O oe )! & , "!' , C! O oe ! #& and < oe ! (" per year, C a#b oe &( &) , "!' . " 68aC O b > oe 68" < 68aC! O b Setting Ca7 b oe ! (&O , the corresponding time is 7 oe # #" years. 19a+b. The rate of increase of the volume is given by rate of flow in rate of flow out. That is, .Z .> oe 5 !+ #12 Since the cross section is constant, .Z .> oe E.2 .> Hence the governing equation is .2 .> oe ^5 !+ #12 E " 5 # a,b Setting .2 .> oe !, the equilibrium height is 2/ oe #1 ^ !+ 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 E. 22a+b. The equilibrium points are at C oe ! and C oe ". Since 0 w aC b oe ! #!C , the equilibrium solution 9 oe ! is unstable and the equilibrium solution 9 oe " is asymptotically stable. a,b. The ODE is separable, with cCa" C bd" .C oe ! .> . Integrating both sides and invoking the initial condition, the solution is Ca>b oe C! /!> " C! C! /!> It is evident that aindependent of C! b 23a+b. Ca>b oe C! /"> > _ lim Ca>b oe ! and > _ lim Ca>b oe " . a,b. From parta+b, .B .> oe ! B C! /"> Separating variables, .B B oe ! C! /"> .> Integrating both sides, the solution is Ba>b oe B! /B:! C! " ^" /"> ` a- b. As > p _ , Ca>b p ! and Ba>b p B! /B:a! C! " b 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! " b 25a+b. Note that 0 aBb oe BcaV V- b + B# d, and 0 w aBb oe aV V- b $+ B# . So if aV V- b ! , the only equilibrium point is B oe ! . 0 w a!b ! , and hence the solution 9a>b oe ! is asymptotically stable. a,b. If aV V- b ! , there are three equilibrium points B oe ! , ,, aV V- b + Now 0 w a!b !, and 0 w ^,, aV V- b + !. Hence the solution 9 oe ! is unstable, and the solutions 9 oe ,, aV V- b + are asymptotically stable. a- b . ________________________________________________________________________ page 59 ---------------------------------------------------- CHAPTER 2. ---- Section 2.6 1. Q aB Cb oe #B $ and R aB C b oe #C # . Since QC oe RB oe ! , the equation is exact. Integrating Q with respect to B , while holding C constant, yields <aB Cb oe oe B# $B 2aC b . Now <C oe 2 w aC b , and equating with R results in the possible function 2aCb oe C # #C . Hence <aB C b oe B# $B C # #C , and the solution is defined implicitly as B# $B C # #C oe - . 2. Q aB Cb oe #B %C and R aB C b oe #B #C . Note that QC RB , and hence the differential equation is not exact. 4. First divide both sides by a#BC #b We now have Q aB C b oe C and R aB C b oe B . Since QC oe RB oe ! , the resulting equation is exact. Integrating Q with respect to B , while holding C constant, results in <aB Cb oe BC 2aC b . Differentiating with respect to C , <C oe B 2 w aC b . Setting <C oe R , we find that 2 w aC b oe ! , and hence 2 aC b oe ! is acceptable. Therefore the solution is defined implicitly as BC oe - . Note that if BC " oe ! , the equation is trivially satisfied. 6. Write the given equation as a+B ,C b.B a,B -C b.C . Now Q aB C b oe +B ,C and R aB C b oe ,B -C . Since QC RB , the differential equation is not exact. 8. Q aB Cb oe /B =38 C $C and R aB C b oe $B /B =38 C . Note that QC RB , and hence the differential equation is not exact. 10. Q aB Cb oe C B 'B and R aB C b oe 68 B #. Since QC oe RB oe " B, the given equation is exact. Integrating R with respect to C , while holding B constant, results in <aB Cb oe C 68 B #C 2 aBb Differentiating with respect to B, <B oe C B 2 w aBb. Setting <B oe Q , we find that 2w aBb oe 'B , and hence 2aBb oe $B# . Therefore the solution is defined implicitly as $B# C 68 B #C oe - . 11. Q aB Cb oe B 68 C BC and R aB C b oe C 68 B BC . Note that QC RB , and hence the differential equation is not exact. 13. Q aB Cb oe #B C and R aB C b oe #C B. Since QC oe RB oe ", the equation is exact. Integrating Q with respect to B , while holding C constant, yields <aB Cb oe oe B# BC 2aC b. Now <C oe B 2 w aC b. Equating <C with R results in 2 w aC b oe #C , and hence 2aCb oe C # . Thus <aB C b oe B# BC C # , and the solution is given implicitly as B# BC C # oe - . Invoking the initial condition C a"b oe $ , the specific solution is B# BC C # oe (. The explicit form of the solution is CaBb oe " 'B #) $B# " # Hence the solution is valid as long as $B# Y #) 16. Q aB Cb oe C /#BC B and R aB C b oe ,B /#BC . Note that QC oe /#BC #BC /#BC , and RB oe , /#BC #,BC /#BC The given equation is exact, as long as , oe " . Integrating ________________________________________________________________________ page 60 ---------------------------------------------------- CHAPTER 2. ---- R with respect to C , while holding B constant, results in <aB Cb oe /#BC # 2aBb Now differentiating with respect to B, <B oe C /#BC 2w aBb. Setting <B oe Q , we find that 2w aBb oe B , and hence 2aBb oe B# # . Conclude that <aB C b oe /#BC # B# # . Hence the solution is given implicitly as /#BC B# oe - . 17. Integrating <C oe R , while holding B constant, yields <aB Cb oe ' R aB C b.C 2 aBb ` Taking the partial derivative with respect to B , <B oe ' `B R aB C b.C 2 w aBb Now ` set <B oe Q aB C b and therefore 2w aBb oe Q aB C b ' `B R aB C b.C . Based on the fact ` that QC oe RB , it follows that `C c2w aBbd oe ! . Hence the expression for 2w aBb can be integrated to obtain 2aBb oe ( Q aB C b.B ( "( 18. Observe that ` `C cQ aBbd ` R aB C b.C .B . `B oe 20. QC oe C " -9= C C # =38 C and RB oe # /B a-9= B =38 Bb C . Multiplying both sides by the integrating factor .aB Cb oe C /B , the given equation can be written as a/B =38 C #C =38 Bb.B a/B -9= C #-9= Bb.C oe ! . Let Q oe .Q and R oe .R . Observe that Q C oe R B , and hence the latter ODE is exact. Integrating R with respect to C , while holding B constant, results in <aB Cb oe /B =38 C #C -9= B 2 aBb Now differentiating with respect to B, <B oe /B =38 C #C =38 B 2 w aBb. Setting <B oe Q , we find that 2w aBb oe ! , and hence 2aBb oe ! is feasible. Hence the solution of the given equation is defined implicitly by /B =38 C #C -9= B oe " . 21. QC oe " and RB oe # . Multiply both sides by the integrating factor .aB C b oe C to obtain C# .B a#BC C # /C b.C oe !. Let Q oe CQ and R oe CR . It is easy to see that Q C oe R B , and hence the latter ODE is exact. Integrating Q with respect to B yields <aB Cb oe BC # 2aC b . Equating <C with R results in 2 w aC b oe C # /C , and hence 2aCb oe /C aC # #C #b. Thus <aB C b oe BC # /C aC # #C #b, and the solution is defined implicitly by BC# /C aC # #C #b oe - . ` `B cR aC bd oe ! 24. The equation .Q .R C w oe ! has an integrating factor if a.Q bC oe a.R bB , that is, .C Q .B R oe .RB .QC . Suppose that RB QC oe V aBQ CR b, in which V is some function depending only on the quantity D oe BC . It follows that the modified form of the equation is exact, if .C Q .B R oe .V aBQ CR b oe V a. BQ . CR b. This relation is satisfied if .C oe a. BbV and .B oe a. CbV . Now consider . oe .aBCb Then the partial derivatives are .B oe .w C and .C oe .w B . Note that .w oe . . .D . Thus . must satisfy .w aD b oe V aD b The latter equation is separable, with . . oe V aD b.D , and .aD b oe ' V aD b.D Therefore, given V oe V aBC b, it is possible to determine . oe .aBC b which becomes an integrating factor of the differential equation. ________________________________________________________________________ page 61 ---------------------------------------------------- CHAPTER 2. ---- 28. The equation is not exact, since RB QC oe #C " . However, aRB QC b Q oe oe a#C "b C is a function of C alone. Hence there exists . oe .aC b , which is a solution of the differential equation .w oe a# " Cb. . The latter equation is separable, with . . . oe # " C . One solution is .aC b oe /B:a#C 68 C b oe /#C C . Now rewrite the given ODE as /#C .B a#B /#C " C b.C oe ! . This equation is exact, and it is easy to see that <aB Cb oe B /#C 68 C . Therefore the solution of the given equation is defined implicitly by B /#C 68 C oe - . 30. The given equation is not exact, since RB QC oe )B$ C $ ' C # . But note that aRB QC b Q oe # C is a function of C alone, and hence there is an integrating factor . oe .aCb. Solving the equation .w oe a# Cb. , an integrating factor is .aCb oe C # Now rewrite the differential equation as a%B$ $Cb.B a$B %C $ b.C oe !. By inspection, <aB Cb oe B% $BC C % , and the solution of the given equation is defined implicitly by B% $BC C % oe - . 32. Multiplying both sides of the ODE by . oe cBC a#B C bd" , the given equation is equivalent to ca$B Cb a#B# BC bd.B caB C b a#BC C # bd.C oe ! Rewrite the differential equation as " # # " " .B " .C oe ! . B #B C C #B C It is easy to see that QC oe RB Integrating Q with respect to B, while keeping C constant, results in <aB Cb oe #68kBk 68k#B C k 2aC b . Now taking the partial derivative with respect to C , <C oe a#B C b" 2 w aC b . Setting <C oe R , we find that 2 w aCb oe " C , and hence 2aC b oe 68 kC k . Therefore <aB Cb oe #68kBk 68k#B C k 68 kC k , and the solution of the given equation is defined implicitly by #B$ C B# C # oe - . ________________________________________________________________________ page 62 ---------------------------------------------------- CHAPTER 2. ---- Section 2.7 2a+b. The Euler formula is C8" oe C8 2a#C8 "b oe a" #2 bC8 2 . 4a+b. The Euler formula is C8" oe a" #2bC8 $2 -9= >8 . a. b The differential equation is linear, with solution Ca>b oe a" /#> b # a. b. The exact solution is Ca>b oe a'-9= > $=38 > ' /#> b & . 5. All solutions seem to converge to 9a>b oe #& * . 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 oe ! 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 > oe ! "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 oe ! is an equilibrium solution. # 12. The iteration formula is C8" oe a" $2bC8 2 >8 C8 . a>! C! b oe a! ! &b 11. The Euler formula is C8" oe C8 $2 C8 &2 . The initial value is C! oe # . $ 14. The iteration formula is C8" oe a" 2 >8 bC8 2 C8 "! a>! C! b oe a! "b 17. The Euler formula is C8" oe C8 The initial point is a>! C! b oe a" #b 18a+b. See Problem 8. 19a+b # 2aC8 #>8 C8 b . $ ># 8 # # a,b. The iteration formula is C8" oe C8 2 C8 2 >8 . The critical value of ! appears to be near !! ! ')"& . For C! !! , the iterations diverge. ________________________________________________________________________ page 65 ---------------------------------------------------- CHAPTER 2. ---- 20a+b. The ODE is linear, with general solution Ca>b oe > - /> . Invoking the specified initial condition, Ca>! b oe C! , we have C! oe >! - />! Hence - oe aC! >! b/>! Thus the solution is given by 9a>b oe aC! >! b/>>! > . a,b. The Euler formula is C8" oe a" 2bC8 2 2 >8 . Now set 5 oe 8 " . a- b. We have C" oe a" 2bC! 2 2 >! oe a" 2 bC! a>" >! b 2 >! . Rearranging the terms, C" oe a" 2baC! >! b >" . Now suppose that C5 oe a" 2b5 aC! >! b >5 , for some 5 " . Then C5" oe a" 2bC5 2 2 >5 . Substituting for C5 , we find that C5" oe a" 2b5" aC! >! b a" 2b >5 2 2 >5 oe a" 2 b5" aC! >! b >5 2 . Noting that >5" oe >5 5 , the result is verified. a. b. Substituting 2 oe a> >! b 8 , with >8 oe > , C8 oe OE" > >! 8 aC! >! b > . 8 pointwise convergence is proved. Taking the limit of both sides, as 8 p _ , and using the fact that lim a" + 8b8 oe /+ , 8 _ 23. The exact solution is 9a>b oe > # /#> . The Euler formula is C8" oe a" #2bC8 2 # 2 >8 . Since C! oe " , C" oe a" #2 b 2 # oe a" #2 b >" # It is easy to show by mathematical induction, that C8 oe a" #2b8 >8 # . For > ! , set 2 oe > 8 and thus >8 oe > . Taking the limit, we find that lim C8 oe lim ca" #> 8b8 > #d oe oe /#> > # . Hence pointwise convergence is proved. 8 _ 8 _ 21. The exact solution is 9a>b oe /> . The Euler formula is C8" oe a" 2bC8 . It is easy to see that C8 oe a" 2b8 C! oe a" 2b8 Given > ! , set 2 oe > 8 . Taking the limit, we find that lim C8 oe lim a" > 8b8 oe /> 8 _ 8 _ ________________________________________________________________________ page 66 ---------------------------------------------------- CHAPTER 2. ---- Section 2.8 2. Let D oe C $ and 7 oe > " . It follows that .D . 7 oe a.D .>ba.> . 7 b oe .D .> . Furthermore, .D .> oe .C .> oe " C $ . Hence .D . 7 oe " aD $b$ . The new initial condition is D a7 oe !b oe ! 3. The approximating functions are defined recursively by 98" a>b oe '! #c98 a=b "d.= . Setting 9! a>b oe ! , 9" a>b oe #> . Continuing, 9# a>b oe #># #> , 9$ a>b oe % >$ #># #> , $ 9% a>b oe # >% % >$ #># #> , . Given convergence, set $ $ > 9a>b oe 9" a>b "c95" a>b 95 a>bd _ oe #> " _ 5oe# 5oe" +5 5 > . 5x Comparing coefficients, +$ $x oe % $ , +% %x oe # $ , . It follows that +$ oe ) , +% oe "', and so on. We find that in general, that +5 oe #5 . Hence 9a>b oe " _ 5oe" #> 5x #5 5 > oe / ". 5. The approximating functions are defined recursively by 98" a>b oe ( c 98 a=b # =d.= . > ! Setting 9! a>b oe ! , 9" a>b oe ># # . Continuing, 9# a>b oe ># # >$ "# ,9$ a>b oe ># # >$ "# >% *' , 9% a>b oe ># # >$ "# >% *' >& *'! , . Given convergence, set ________________________________________________________________________ page 67 ---------------------------------------------------- CHAPTER 2. ---- 9a>b oe 9" a>b "c95" a>b 95 a>bd _ oe ># # " 5oe$ 5oe" _ +5 5 > . 5x Comparing coefficients, +$ $x oe " "# , +% %x oe " *' , +& &x oe " *'! , . We find that +$ oe " # , +% oe " %, +& oe " ) , In general, +5 oe #5" . Hence 9a>b oe " _ #5# a >b 5 5x 5oe# oe % /> # #> % . 6. The approximating functions are defined recursively by 98" a>b oe ( c98 a=b " =d.= . > ! Setting 9! a>b oe ! , 9" a>b oe > ># # , 92 a>b oe > >$ ' , 9$ a>b oe > >% #% , 9% a>b oe oe > >& "#! , . Given convergence, set 9a>b oe 9" a>b "c95" a>b 95 a>bd _ 5oe" # oe > > # ># # >$ '` >$ ' >% #%` oe>!! . 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 oe !. The iterates are given by 9" a>b oe ># # , 92 a>b oe ># # >& "! , 9$ a>b oe ># # >& "! >) )! , 9% a>b oe ># # >& "! >) )! >"" ))! , . Upon inspection, it becomes apparent that 8" " >$ >' a >$ b 98 a>b oe > " # #& #&) # & ) c# $a8 "bd 98" a>b oe ( =# 98 a=b =`.= . > ! # oe ># " 8 5oe" a, b . # & ) c# $a5 "bd a >$ b 5" The iterates appear to be converging. 9a+b. The approximating functions are defined recursively by Set 9! a>b oe !. The first three iterates are given by 9" a>b oe >$ $ , 92 a>b oe >$ $ >( '$ , 9$ a>b oe >$ $ >( '$ #>"" #!(* >"& &*&$& . ________________________________________________________________________ page 69 # 98" a>b oe ( =# 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 oe !. The first three iterates are given by 9" a>b oe > , 92 a>b oe > >% % , 9$ a>b oe > >% % $>( #) $>"! "'! >"$ )$$ . a, b . $ 98" a>b oe ( " 98 a=b`.= . > ! The approximations appear to be diverging. 12a+b. The approximating functions are defined recursively by 98" a>b oe ( " > ! Note that " a#C #b oe above iteration formula by " # 5oe! ! C 5 S ^C( . For computational purposes, replace the ' $=# %= # .= . #a98 a=b "b ________________________________________________________________________ page 70 ---------------------------------------------------- CHAPTER 2. ---- ' " > 5 98" a>b oe ( ^$=# %= #" 98 a=b--.= . # ! 5oe! Set 9! a>b oe !. The first four approximations are given by 9" a>b oe > ># >$ # , 92 a>b oe > ># # >$ ' >% % >& & >' #% , 9$ a>b oe > ># # >% "# $>& #! %>' %& , 9% a>b oe > ># # >% ) (>& '! >' "& a, b . The approximations appear to be converging to the exact solution, 9a>b oe " " #> #># >$ 13. Note that 98 a!b oe ! and 98 a"b oe " , a 8 " . Let + - a! "b . Then 98 a+b oe +8 . Clearly, lim +8 oe ! . Hence the assertion is true. 8 _ 14a+b. 98 a!b oe ! , a 8 " . Let + - ! " . Then 98 a+b oe #8+ /8+ oe #8+ /8+ # # Using l'Hospital's rule, lim #+D /+D oe lim " D/+D oe ! . Hence lim 98 a+b oe ! # # # # " " a,b. '! #8B /8B .B oe /8B ! oe " /8 Therefore, D _ D _ 8 _ 8 _ ! lim ( 98 aBb.B ( lim 98 aBb.B . " " ! 8 _ 15. Let > be fixed, such that a> C" b a> 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" C# b such that 0 a> C" b 0 a> C# b oe 0C a> 0baC" C# b Taking the absolute value of both sides, k0 a> C" b 0 a> C# bk oe k0C a> 0bk kC" C# k Since, by assumption, `0 `C is continuous in H, 0C attains a maximum on any closed and bounded subset of H . ________________________________________________________________________ page 71 ---------------------------------------------------- CHAPTER 2. ---- Hence k0 a> C" b 0 a> C# bk Y O kC" C# k 16. For a sufficiently small interval of >, 98" a>b , 98 a>b - H . Since 0 satisfies a Lipschitz condition, k0 a> 98 a>bb 0 a> 98" a>bbk Y O k98 a>b 98" a>bk Here O oe 7+B k0C k . k> k k>k > 17a+b 9" a>b oe '! 0 a= !b.= . Hence k9" a>bk Y '! k0 a= !bk.= Y '! Q .= oe Q k>k , in which Q is the maximum value of k0 a> C bk on H . > a,b. By definition, 9# a>b 9" a>b oe '! c0 a= 9" a=bb 0 a= !bd.= . Taking the absolute k>k value of both sides, k9# a>b 9" a>bk Y '! kc0 a= 9" a=bb 0 a= !bdk.= . Based on the k>k k>k results in Problems 16 and 17, k9# a>b 9" a>bk Y '! O k9" a=b !k.= Y OQ '! k=k.= . Evaluating the last integral, we obtain k9# a>b 9" a>bk Y Q O k>k# # . a- b. Suppose that for some 3 " . By definition, 93" a>b 93 a>b oe '! c0 a> 93 a=bb 0 a= 93" a=bbd.= . It follows that > k93 a>b 93" a>bk Y Q O 3" k>k3 3x k93" a>b 93 a>bk Y ( Y( Y( k> k k> k k> k ! k0 a= 93 a=bb 0 a= 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" oe Y a 3 "b x a 3 "b x ! Hence, by mathematical induction, the assertion is true. 18a+b. Use the triangle inequality, k+ ,k Y k+k k,k . k98 a>bk Y Q " 8 a,b. For k>k Y 2 , k9" a>bk Y Q 2 , and k98 a>b 98" a>bk Y Q O 8" 2 8 a8 xb . Hence Q 8 aO2b3 " O 3oe" 3 x O 3" 23 3x 3oe" oe ________________________________________________________________________ page 72 ---------------------------------------------------- CHAPTER 2. ---- a- b. The sequence of partial sums in a,b converges to Q ^/O2 " 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 oe '! 0 a= 9a=bb.= and <a>b oe '! 0 a= <a=bb.= Then by linearity of > the integral, 9a>b <a>b oe '! c0 a= 9a=bb 0 a= <a=bbd.= a,b. It follows that k9a>b <a>bk Y '! k0 a= 9a=bb 0 a= <a=bbk.= > a- b. We know that 0 satisfies a Lipschitz condition, based on k`0 `Ck Y O in H. Therefore, > ! k0 a> C" b 0 a> C# bk Y O kC" C# k , k9a>b <a>bk Y ( k0 a= 9a=bb 0 a= <a=bbk.= Y ( O k9a=b <a=bk.= . > ! ________________________________________________________________________ page 73 ---------------------------------------------------- CHAPTER 2. ---- Section 2.9 1. Writing the equation for each 8 ! , C" oe ! * C! , C# oe ! * C" , C$ oe ! * C# and so on, it is apparent that C8 oe a ! *b8 C! . The terms constitute an alternating series, which converge to zero, regardless of C! . 3. Write the equation for each 8 ! , C" oe $ C! , C# oe % # C" , C$ oe & $ C# , . Upon substitution, we find that C# oe a% $b # C" , C$ oe a& % $b a$ #b C! , It can be proved by mathematical induction, that C8 oe " a 8 #b x C! # 8x " oe a8 "ba8 #b C! . # This sequence is divergent, except for C! oe ! . 4. Writing the equation for each 8 ! , C" oe C! , C# oe C" , C$ oe C# , C% oe C$ , and so on, it can be shown that C! C8 oe oe C! , for 8 oe %5 or 8 oe %5 " , for 8 oe %5 # or 8 oe %5 $ The sequence is convergent only for C! oe ! . 6. Writing the equation for each 8 ! , C" oe ! & C! ' C# oe ! & C" ' oe ! &a! & C! 'b ' oe a! &b# C! ' a! &b' C$ oe ! & C# ' oe ! &a! & C" 'b ' oe a! &b$ C! 'c" a! &b ! & # d C8 oe a! &b8 C! "#c" a! &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" oe a" < $&'b C8 . The solution of this difference equation is C8 oe a" < $'&b8 C! , in which C! is the initial balance. At the end of one year, the balance is C$'& oe a" < $'&b$'& C! . Given that < oe !(, C$'& oe a" < $'&b$'& C! oe " !(#& C! . Hence the effective annual yield is a" !(#& C! C! b C! oe ( #& % . 8. Let C8 be the balance at the end of the 8-th month. Then C8" oe a" < "#b C8 #& . As in the previous solutions, we have ________________________________________________________________________ page 74 ---------------------------------------------------- CHAPTER 2. ---- C8 oe 38 "C! a"#b#& < in which 3 oe a" < "#b. Here < is the annual interest rate, given as ) % . Therefore C$' oe a" !!''b$' '"!!! a"#b#& < " #& #& , "3 "3 oe # #)$ '$ dollars. 9. Let C8 be the balance due at the end of the 8-th month. The appropriate difference equation is C8" oe a" < "#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 oe 38 "C! T T , "3 "3 in which 3 oe a" < "#b and C! oe ) !!! To figure out the monthly payment, T , we require that C$' oe ! That is, 336 "C! T T . oe "3 "3 After the specified amounts are substituted, we find the T oe $ #&) "% . 11. Let C8 be the balance due at the end of the 8-th month. The appropriate difference equation is C8" oe a" < "#b C8 T , in which < oe !* and T is the monthly payment. The initial value of the mortgage is C! oe "!! !!! dollars. Then the balance due at the end of the 8-th month is C8 oe 38 "C! T T "3 "3 "#T "#T . < < where 3 oe a" < "#b. In terms of the specified values, C8 oe a! !!(&b8 ""!& Setting 8 oe $!a"#b oe $'! , and C$'! oe ! , we find that T oe )!% '# dollars For the monthly payment corresponding to a #! year mortgage, set 8 oe #%! and C#%! oe ! . 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" oe a" < "#b C8 T , in which < oe ! " and T oe *!! dollars is the maximum monthly payment. Given that the life of the mortgage is #! years, we require that C#%! oe !. The balance due at the end of the 8th month is C8 oe 38 "C! T T "3 "3 In terms of the specified values for the parameters, the solution of ________________________________________________________________________ page 75 ---------------------------------------------------- CHAPTER 2. ---- a !!)$$b#%! "C! is C! oe "!$ '#% '# dollars. 15. "#a"!!!b "#a"!!!b oe ! " ! " 16. For example, take 3 oe $ & and ?! oe " " : ________________________________________________________________________ page 76 ---------------------------------------------------- CHAPTER 2. ---- 19a+b. $# oe a3# 3" b a3$ 3# b oe a$ %%* $b a$ &%% $ %%*b oe % (#'$ . a,b. % diff oe a- b. Assuming a3$ 3# b a3% 3$ b oe $ , 3% $ &'%$ k$ $# k $ , "!! oe k% ''*#% ($'$k % ''*# , "!! " ## % a. b. A period "' solutions appears near 3 $ &'& . " a/b. Note that a38" 38 b oe $8 a38 38" b With the assumption that $8 oe $ , we have a38" 38 b oe $ " a38 38" b, which is of the form C8" oe ! C8 , 8 $ . It follows that a35 35" b oe $ $5 a3$ 3# b for 5 % . Then 38 oe 3" a3# 3" b a3$ 3# b a3% 3$ b a38 38" b oe 3" a3# 3" b a3$ 3# bc" $ " $ # $ $8 d " $ %8 oe 3" a3# 3" b a3$ 3# b" " $ " $ Hence lim 38 oe 3# a3$ 3# b $ " ` Substitution of the appropriate values yields 8 _ 8 _ lim 38 oe $ &'** ________________________________________________________________________ page 77 ---------------------------------------------------- CHAPTER 2. ---- Miscellaneous Problems 1. Linear C oe - B# B$ & 2. Homogeneous +<->+8aC Bb 68 B# C # oe - . 3. Exact B# BC $C C $ oe ! . 4. Linear in BaCb B oe - / C C / C . 5. Exact B# C BC # B oe - . 6. Linear C oe B" a" /"B b . # 7. Let ? oe B# B# C # " oe - /C . 8. Linear C oe a% -9= # -9= Bb B# . 9. Exact B# C B C # oe - . 10. . oe .aBb C # B$ C B# oe - . 11. Exact B$ $ BC /C oe - . 12. Linear C oe - /B /B 68a" /B b . 13. Homogeneous # C B 68 kBk oe - . 14. Exact/Homogeneous B# #BC #C # oe $% . 15. Separable C oe - -9=2# aB #b . 16. Homogeneous S# $ <+<->+8' #C B $ B" 68 kBk oe - . 17. 18. 19. 20. 21. 22. 23. Linear C oe - /$B /#B . Linear/Homogeneous C oe - B# B . . oe .aBb $C #BC $ "!B oe ! . Separable /B /C oe - . Homogeneous /C B 68 kBk oe - . Separable C$ $C B$ $B oe # . Bernoulli " C oe B' B# /#B .B -B . 24. Separable =38# B =38 C oe - . 25. Exact B# C +<->+8aC Bb oe - . 26. . oe .aBb B# #B# C C # oe - . 27. . oe .aBb =38 B -9= #C " =38# B oe - . # 28. Exact #BC BC $ B$ oe - . 29. Homogeneous +<-=38aC Bb 68 kBk oe - . 30. Linear in BaCb BC # 68 kC k oe ! . 31. Separable 32. . oe .aCb B 68 kBk B" C # 68 kC k oe - . B$ C # BC $ oe % . ________________________________________________________________________ page 78

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