solutions for boyce elementary differential equationsFull description
Descrição: solutions for boyce elementary differential equations
Elementary Differential EquationsFull description
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Solutions to Elementary Differential Equations 10th Edition By William E Boyce and Richard C Diprima
gockenbach 2nd editionFull description
DIFFERENTIAL EQUATIONS
Differential Equations by (Shepley L. Ross)Descripción completa
Differential Equations by (Shepley L. Ross)Full description
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Elementary Applied Partial Differential Equations -Haberman R.
Blanchard Differential Equations 3e Solutions Manual
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A set of lecture notes compiled to teach physics students at undergraduate level. Mathematical Methods in Physics Course....supplemented with examples and home work problems.
Descripción: Nonlinear Ordinary Differential Equations by jordan, smith
differential geometryDescripción completa
—————————————————————————— 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
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 $C " . 12.
Note that C w œ ! for C œ ! and C œ & . The two equilibrium solutions are C a>b œ ! and Ca>b œ & . Based on the direction field, C w ! for C & ; thus solutions with initial values greater than & diverge from the solution Ca>b œ & . For ! C &, the slopes are negative, and hence solutions with initial values between ! and & all decrease toward the
—————————————————————————— CHAPTER 1. —— solution Ca>b œ ! . For C ! , the slopes are all positive; thus solutions with initial values less than ! approach the solution Ca>b œ ! . 14.
Observe that C w œ ! for C œ ! and C œ # . The two equilibrium solutions are C a>b œ ! and Ca>b œ # . Based on the direction field, C w ! for C # ; thus solutions with initial values greater than # diverge from Ca>b œ # . For ! C #, the slopes are also positive, and hence solutions with initial values between ! and # all increase toward the solution Ca>b œ # . For C ! , the slopes are all negative; thus solutions with initial values less than ! diverge from the solution Ca>b œ ! .
16. a+b Let Q a>b be the total amount of the drug ain milligramsb in the patient's body at any given time > a2<=b . The drug is administered into the body at a constant rate of &!! 71Î2<Þ The rate at which the drug leaves the bloodstream is given by !Þ%Q a>b Þ Hence the accumulation rate of the drug is described by the differential equation
a, b
.Q œ &!! !Þ% Q .>
a71Î2
Based on the direction field, the amount of drug in the bloodstream approaches the equilibrium level of "#&! 71 aA3>238 + 0 /A 29?<=bÞ 18. a+b Following the discussion in the text, the differential equation is
The direction field is rather complicated. Nevertheless, the collection of points at which the slope field is zero, is given by the implicit equation C$ '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.
—————————————————————————— CHAPTER 1. —— Section 1.2 1a+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
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
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
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
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
—————————————————————————— 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 >0 œ # 68Œ
*!! Þ *!! :!
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 .
7a+b. The general solution is :a>b œ :! /<> . Based on the discussion in the text, time > is measured in months . Assuming " month œ $! days , the hypothesis can be expressed as :! /<†" œ #:! . Solving for the rate constant, < œ 68a#b , with units of per month .
a,b. R days œ R Î$! months . The hypothesis is stated mathematically as :! /
.@ œ 71 .>
in which 1 is the gravitational constant measured in appropriate units. The equation can be ________________________________________________________________________ page 10
—————————————————————————— CHAPTER 1. —— written as .@Î.> œ 1 , with solution @a>b œ 1> @! Þ The object is released with an initial velocity @! .
a,b. Suppose that the object is released from a height of 2 units above the ground. Using the fact that @ œ .BÎ.> , in which B is the downward displacement of the object, we obtain the differential equation for the displacement as .BÎ.> œ 1> @! Þ With the origin placed at the point of release, direct integration results in Ba>b œ 1># Î# @! > . Based on the chosen coordinate system, the object reaches the ground when Ba>b œ 2 . Let > œ X be the time that it takes the object to reach the ground. Then 1X # Î# @! X œ 2 . Using the quadratic formula to solve for X , X œ
@! „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 "!! /
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 . .>
Since the pond is initially free of the chemical, Ua!b œ ! .
a,b. The differential equation can be rewritten as
.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!!!$> Þ Hence, one year after the source is removed, the amount of chemical in the pond is Ua)('!b œ '(!Þ" grams .
—————————————————————————— CHAPTER 1. —— a/b. Letting > be the amount of time after the source is removed, we obtain the equation "! œ *#((Þ(( /!Þ!!!$> 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 solution of the governing differential equation, with the specified initial value, is ; a>b œ &!!! /!Þ# > Þ
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
—————————————————————————— 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 .
—————————————————————————— 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
9. Ca>b œ $> ># Ê C w a>b œ $ #> . Substituting into the differential equation, we have >a$ #>b a$> ># b œ $> #># $> ># œ ># . Hence the given function is a solution.
10. C" a>b œ >Î$ 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 œ /> , 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!
Likewise, C# a>b œ >" Ê C#w a>b œ ># and C#ww a>b œ # >$ . Substituting into the left hand side of the differential equation, we have #># a# >$ b $>a ># b >" œ % >" $ >" >" œ ! . Hence both functions are solutions of the differential equation. 12. C" a>b œ ># Ê C"w a>b œ #>$ and C"ww a>b œ ' >% . Substituting into the left hand side of the differential equation, we have ># a' >% b &>a #>$ b % ># œ ' ># "! ># % ># œ ! . Likewise, C# a>b œ >2 68 > Ê C#w a>b œ >$ #>$ 68 > and C#ww a>b œ & >% ' >% 68 >. Substituting into the left hand side of the equation, we have ># a & >% ' >% 68 >b &>a>$ #>$ 68 >b %a>2 68 >b œ & >2 ' >2 68 > ________________________________________________________________________ page 15
—————————————————————————— CHAPTER 1. —— & >2 "! >2 68 > % >2 68 > œ ! Þ Hence both functions are solutions of the differential equation. 13. Ca>b œ a-9= >b68 -9= > > =38 > Ê C w a>b œ a=38 >b68 -9= > > -9= > and C ww a>b œ a-9= >b68 -9= > > =38 > =/- > . Substituting into the left hand side of the differential equation, we have a a-9= >b68 -9= > > =38 > =/- >b a-9= >b68 -9= > > =38 > œ a-9= >b68 -9= > > =38 > =/- > a-9= >b68 -9= > > =38 > œ =/- > . Hence the function Ca>b is a solution of the differential equation.
15. Let Ca>b œ /<> . Then C ww a>b œ <# /<> , and substitution into the differential equation results in <# /<> # /<> œ !. Since /<> Á ! , we obtain the algebraic equation <# # œ !Þ The roots of this equation are <"ß# œ „ 3È# Þ 17. Ca>b œ /<> Ê C w a>b œ < /<> and C ww a>b œ <# /<> . Substituting into the differential equation, we have <# /<> ' /<> œ ! . Since /<> Á ! , we obtain the algebraic equation <# < ' œ ! , that is, a< #ba< $b œ ! . The roots are <"ß# œ $ , # .
18. Let Ca>b œ /<> . Then C w a>b œ , C ww a>b œ <# /<> and C www a>b œ <$ /<> . Substituting the derivatives into the differential equation, we have <$ /<> $<# /<> # œ ! . Since /<> Á ! , we obtain the algebraic equation <$ $<# #< œ ! Þ By inspection, it follows that b œ >< Ê C w a>b œ < ><" and C ww a>b œ <# . Substituting the derivatives into the differential equation, we have ># c<# d %>a< ><" b % >< œ ! . After some algebra, it follows that < %< >< % >< œ ! . For > Á ! , we obtain the algebraic equation <# &< % œ ! Þ The roots of this equation are <" œ " and <# œ % Þ
21. The order of the partial differential equation is two, since the highest derivative, in fact each one of the derivatives, is of second order. The equation is linear, since the left hand side is a linear function of the partial derivatives. 23. The partial differential equation is fourth order, since the highest derivative, and in fact each of the derivatives, is of order four. The equation is linear, since the left hand side is a linear function of the partial derivatives. 24. The partial differential equation is second order, since the highest derivative of the function ?aBß Cb is of order two. The equation is nonlinear, due to the product ? † ?B on the left hand side of the equation. 25. ?" aBß Cb œ -9= B -9=2 C Ê It is evident that derivatives are
#
` ?" `B#
#
` ?" `C#
` # ?" `B#
œ -9= B -9=2 C and
` # ?" `C#
œ -9= B -9=2 C Þ
œ ! Þ Likewise, given ?# aBß C b œ 68aB# C # b, the second
a,bÞ The path of the particle is a circle, therefore polar coordinates are intrinsic to the problem. The variable < is radial distance and the angle ) is measured from the vertical. Newton's Second Law states that ! F œ 7a Þ In the tangential direction, the equation of motion may be expressed as ! J) œ 7 +) , in which the tangential acceleration, that is,
the linear acceleration along the path is +) œ P . # )Î.># Þ Ð +) is positive in the direction of increasing ) Ñ. Since the only force acting in the tangential direction is the component of weight, the equation of motion is 71 =38 ) œ 7P
.# ) Þ .>#
ÐNote that the equation of motion in the radial direction will include the tension in the rodÑ. a- b. Rearranging the terms results in the differential equation .# ) 1 =38 ) œ ! Þ # .> P
a,bÞ Based on the direction field, all solutions seem to converge to a specific increasing function. a- bÞ The integrating factor is .a>b œ /$> , and hence Ca>b œ >Î$ "Î* /#> - /$> Þ It follows that all solutions converge to the function C" a>b œ >Î$ "Î* Þ 2a+bÞ
a,b. All slopes eventually become positive, hence all solutions will increase without bound. a- bÞ The integrating factor is .a>b œ /#> , and hence Ca>b œ >$ /#> Î$ - /#> Þ It is evident that all solutions increase at an exponential rate. 3a+b
a,b. All solutions seem to converge to the function C! a>b œ " Þ
a- bÞ The integrating factor is .a>b œ /#> , and hence Ca>b œ ># /> Î# " - /> Þ It is clear that all solutions converge to the specific solution C! a>b œ " .
4a+b.
a,b. Based on the direction field, the solutions eventually become oscillatory. a- bÞ The integrating factor is .a>b œ > , and hence the general solution is C a >b œ
$-9=a#>b $ =38a#>b %> # >
in which - is an arbitrary constant. As > becomes large, all solutions converge to the function C" a>b œ $=38a#>bÎ# Þ
a,b. All slopes eventually become positive, hence all solutions will increase without bound.
a- bÞ The integrating factor is .a>b œ /B:a ' #.>b œ /#> Þ The differential equation can w be written as /#> C w #/#> C œ $/> , that is, a/#> C b œ $/> Þ Integration of both sides of the equation results in the general solution Ca>b œ $/> - /#> Þ It follows that all solutions will increase exponentially. 6a+b
a,bÞ All solutions seem to converge to the function C! a>b œ ! Þ
a- bÞ The integrating factor is .a>b œ ># , and hence the general solution is Ca>b œ
-9=a>b =38a#>b # # > > >
in which - is an arbitrary constant. As > becomes large, all solutions converge to the function C! a>b œ ! Þ 7a+b.
a,bÞ All solutions seem to converge to the function C! a>b œ ! Þ
a- bÞ The integrating factor is .a>b œ /B:a># b, and hence C a>b œ ># /> - /> Þ It is clear that all solutions converge to the function C! a>b œ ! . #
#
8a+b
a,bÞ All solutions seem to converge to the function C! a>b œ ! Þ
a- bÞ Since .a>b œ a" ># b# , the general solution is Ca>b œ c>+8" a>b G dÎa" ># b# Þ It follows that all solutions converge to the function C! a>b œ ! . 9a+bÞ
—————————————————————————— CHAPTER 2. —— a,b. All slopes eventually become positive, hence all solutions will increase without bound.
a- bÞ The integrating factor is .a>b œ /B:ˆ' "# .>‰ œ />Î# . The differential equation can w be written as />Î# C w />Î# CÎ# œ $> />Î# Î# , that is, ˆ/>Î# CÎ#‰ œ $> />Î# Î#Þ Integration of both sides of the equation results in the general solution Ca>b œ $> ' - />Î# Þ All solutions approach the specific solution C! a>b œ $> ' Þ
10a+b.
a,b. For C ! , the slopes are all positive, and hence the corresponding solutions increase without bound. For C ! , almost all solutions have negative slopes, and hence solutions tend to decrease without bound. a- bÞ First divide both sides of the equation by > . From the resulting standard form, the integrating factor is .a>b œ /B:ˆ ' "> .>‰ œ "Î> . The differential equation can be written as C w Î> CÎ># œ > /> , that is, a CÎ>bw œ > /> Þ Integration leads to the general solution Ca>b œ >/> - > Þ For - Á ! , solutions diverge, as implied by the direction field. For the case - œ ! , the specific solution is Ca>b œ >/> , which evidently approaches zero as > p _ . 11a+b.
a,bÞ The solutions appear to be oscillatory. ________________________________________________________________________ page 22
—————————————————————————— CHAPTER 2. —— a- bÞ The integrating factor is .a>b œ /> , and hence Ca>b œ =38a#>b # -9=a#>b - /> Þ It is evident that all solutions converge to the specific solution C! a>b œ =38a#>b # -9=a#>b . 12a+bÞ
a,b. All solutions eventually have positive slopes, and hence increase without bound.
a- bÞ The integrating factor is .a>b œ /#> . The differential equation can be w written as />Î# C w />Î# CÎ# œ $># Î# , that is, ˆ/>Î# CÎ#‰ œ $># Î#Þ Integration of both sides of the equation results in the general solution Ca>b œ $># "#> #% - />Î# Þ It follows that all solutions converge to the specific solution C! a>b œ $># "#> #% .
14. The integrating factor is .a>b œ /#> . After multiplying both sides by .a>b, the w equation can be written as ˆ/2> C‰ œ > Þ Integrating both sides of the equation results in the general solution Ca>b œ ># /#> Î# - /#> Þ Invoking the specified condition, we require that /# Î# - /# œ ! . Hence - œ "Î# , and the solution to the initial value problem is Ca>b œ a># "b/#> Î# Þ 16. The integrating factor is .a>b œ /B:ˆ' #> .>‰ œ ># . Multiplying both sides by .a>b, w the equation can be written as a># Cb œ -9=a>b Þ Integrating both sides of the equation results in the general solution Ca>b œ =38a>bÎ># - ># Þ Substituting > œ 1 and setting the value equal to zero gives - œ ! . Hence the specific solution is Ca>b œ =38a>bÎ># Þ
17. The integrating factor is .a>b œ /#> , and the differential equation can be written as ˆ/2> C‰w œ " Þ Integrating, we obtain /2> C a>b œ > - Þ Invoking the specified initial condition results in the solution Ca>b œ a> #b/#> Þ 19. After writing the equation in standard 0 orm, we find that the integrating factor is .a>b œ /B:ˆ' %> .>‰ œ >% . Multiplying both sides by .a>b, the equation can be written as ˆ>% C‰w œ > /> Þ Integrating both sides results in >% C a>b œ a> "b/> - Þ Letting > œ " and setting the value equal to zero gives - œ ! Þ Hence the specific solution of the initial value problem is Ca>b œ ˆ>$ >% ‰/> Þ 21a+b.
The solutions appear to diverge from an apparent oscillatory solution. From the direction field, the critical value of the initial condition seems to be +! œ " . For + " , the solutions increase without bound. For + " , solutions decrease without bound. a,bÞ The integrating factor is .a>b œ />Î# . The general solution of the differential equation is Ca>b œ a)=38a>b %-9=a>bbÎ& - />Î# . The solution is sinusoidal as long as - œ ! . The initial value of this sinusoidal solution is +! œ a)=38a!b %-9=a!bbÎ& œ %Î& Þ a- bÞ See part a,b. 22a+bÞ
All solutions appear to eventually increase without bound. The solutions initially increase or decrease, depending on the initial value + . The critical value seems to be +! œ " Þ a,bÞ The integrating factor is .a>b œ />Î# , and the general solution of the differential equation is Ca>b œ $/>Î$ - />Î# Þ Invoking the initial condition C a!b œ + , the solution may also be expressed as Ca>b œ $/>Î$ a+ $b />Î# Þ Differentiating, follows that C w a!b œ " a+ $bÎ# œ a+ "bÎ# Þ The critical value is evidently +! œ " Þ
—————————————————————————— CHAPTER 2. —— a- b. For +! œ " , the solution is Ca>b œ $/>Î$ # />Î# , which afor large >b is dominated by the term containing />Î# Þ is Ca>b œ a)=38a>b %-9=a>bbÎ& - />Î# . 23a+bÞ
As > p ! , solutions increase without bound if C a"b œ + Þ% , and solutions decrease without bound if Ca"b œ + Þ% Þ
> ‰ a,b. The integrating factor is .a>b œ /B:ˆ' >" > .> œ > / Þ The general solution of the differential equation is Ca>b œ > /> - /> Î> . Invoking the specified value C a"b œ + , we have " - œ + / . That is, - œ + / " . Hence the solution can also be expressed as Ca>b œ > /> a+ / "b /> Î> . For small values of > , the second term is dominant. Setting + / " œ ! , critical value of the parameter is +! œ "Î/ Þ
a- b. For + "Î/ , solutions increase without bound. For + "Î/ , solutions decrease without bound. When + œ "Î/ , the solution is C a>b œ > /> , which approaches ! as > p ! . 24a+b.
As > p ! , solutions increase without bound if C a"b œ + Þ% , and solutions decrease without bound if Ca"b œ + Þ% Þ ________________________________________________________________________ page 25
—————————————————————————— CHAPTER 2. —— a,b. Given the initial condition, Ca 1Î#b œ + , the solution is Ca>b œ a+ 1# Î% -9= >bÎ> Þ Since lim -9= > œ " , solutions increase without bound if + %Î1# , and solutions >Ä!
decrease without bound if + %Î1# Þ Hence the critical value is +! œ %Î1# œ !Þ%&#)%(ÞÞÞ.
a- bÞ For + œ %Î1# , the solution is C a>b œ a" -9= >bÎ> , and lim C a>b œ "Î# . Hence the solution is bounded.
>Ä!
25. The integrating factor is .a>b œ /B:ˆ' "# .>‰ œ />Î# Þ Therefore general solution is Ca>b œ c%-9=a>b )=38a>bdÎ& - />Î# Þ Invoking the initial condition, the specific solution is Ca>b œ c%-9=a>b )=38a>b * />Î# dÎ& . Differentiating, it follows that C w a>b œ %=38a>b )-9=a>b %Þ& />Î# ‘Î& C ww a>b œ %-9=a>b )=38a>b #Þ#& />Î# ‘Î&
Setting C w a>b œ ! , the first solution is >" œ "Þ$'%$ , which gives the location of the first stationary point. Since C ww a>" b ! , the first stationary point in a local maximum. The coordinates of the point are a"Þ$'%$ ß Þ)#!!)b.
26. The integrating factor is .a>b œ /B:ˆ' #$ .>‰ œ /#>Î$ , and the differential equation can w be written as a/#>Î$ Cb œ /#>Î$ > /#>Î$ Î# Þ The general solution is C a>b œ Ð#" '>ÑÎ) - /#>Î$ . Imposing the initial condition, we have C a>b œ Ð#" '>ÑÎ) aC! #"Î)b/#>Î$ . Since the solution is smooth, the desired intersection will be a point of tangency. Taking the derivative, C w a>b œ $Î% a#C! #"Î%b/#>Î$ Î$ Þ Setting C w a>b œ ! , the solution is >" œ $# 68ca#" )C! bÎ*dÞ Substituting into the solution, the respective value at the stationary point is Ca>" b œ $# *% 68 $ *) 68a#" )C! b. Setting this result equal to zero, we obtain the required initial value C! œ a#" * /%Î$ bÎ) œ "Þ'%$ Þ 27. The integrating factor is .a>b œ />Î4 , and the differential equation can be written as w a/>Î4 Cb œ $ />Î4 # />Î4 -9=a#>bÞ The general solution is Ca>b œ "# c)-9=a#>b '%=38a#>bdÎ'& - />Î4 Þ
Invoking the initial condition, Ca!b œ ! , the specific solution is
—————————————————————————— CHAPTER 2. —— 29. The integrating factor is .a>b œ /$>Î# , and the differential equation can be written w as a/$>Î# Cb œ $> /$>Î# # />Î# Þ The general solution is C a>b œ #> %Î$ % /> - /$>Î# Þ Imposing the initial condition, C a>b œ #> %Î$ % /> aC! "'Î$b /$>Î# Þ As > p _ , the term containing /$>Î# will dominate the solution. Its sign will determine the divergence properties. Hence the critical value of the initial condition is C! œ "'Î$Þ The corresponding solution, Ca>b œ #> %Î$ % /> , will also decrease without bound. Note on Problems 31-34 : Let 1a>b be given, and consider the function Ca>b œ C" a>b 1a>b , in which C" a>b p _ as > p _ . Differentiating, C w a>b œ C"w a>b 1 w a>b . Letting + be a constant, it follows that C w a>b +C a>b œ C"w a>b +C" a>b 1 w a>b +1a>bÞ Note that the hypothesis on the function C" a>b will be satisfied, if C"w a>b +C" a>b œ ! . That is, C" a>b œ - /+> Þ Hence Ca>b œ - /+> 1a>b, which is a solution of the equation C w +C œ 1 w a>b +1a>bÞ For convenience, choose + œ " . 31. Here 1a>b œ $ , and we consider the linear equation C w C œ $ Þ The integrating w factor is .a>b œ /> , and the differential equation can be written as a/> C b œ $/> Þ The general solution is Ca>b œ $ - /> Þ
33. 1a>b œ $ > Þ Consider the linear equation C w C œ " $ > ÞThe integrating w factor is .a>b œ /> , and the differential equation can be written as a/> C b œ a# >b/> Þ The general solution is Ca>b œ $ > - /> Þ 34. 1a>b œ % ># Þ Consider the linear equation C w C œ % #> ># ÞThe integrating w factor is .a>b œ /> , and the equation can be written as a/> C b œ a% #> ># b/> Þ The general solution is Ca>b œ % ># - /> Þ
—————————————————————————— CHAPTER 2. —— Section 2.2 2. For B Á " , the differential equation may be written as C .C œ cB# Îa" B$ bd.B Þ Integrating both sides, with respect to the appropriate variables, we obtain the relation C# Î# œ "$ 68k" B$ k - Þ That is, C aBb œ „É #$ 68k" B$ k - Þ
3. The differential equation may be written as C# .C œ =38 B .B Þ Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation C " œ -9= B - Þ That is, aG -9= BbC œ ", in which G is an arbitrary constant. Solving for the dependent variable, explicitly, CaBb œ "ÎaG -9= Bb . 5. Write the differential equation as -9=# #C .C œ -9=# B .B, or =/- # #C .C œ -9=# B .BÞ Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation >+8 #C œ =38 B -9= B B - Þ 7. The differential equation may be written as aC /C b.C œ aB /B b.B Þ Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation C # # / C œ B # # / B - Þ 8. Write the differential equation as a" C# b.C œ B# .B Þ Integrating both sides of the equation, we obtain the relation C C $ Î$ œ B$ Î$ - , that is, $C C $ œ B$ GÞ 9a+b. The differential equation is separable, with C# .C œ a" #Bb.B Þ Integration yields C" œ B B# - Þ Substituting B œ ! and C œ "Î' , we find that - œ ' Þ Hence the specific solution is C" œ B# B ' . The explicit form is CaBb œ "ÎaB# B 'bÞ a, b
a- b. Note that B# B ' œ aB #baB $b . Hence the solution becomes singular at B œ # and B œ $ Þ 10a+bÞ CaBb œ È#B #B# % Þ
11a+bÞ Rewrite the differential equation as B /B .B œ C .C Þ Integrating both sides of the equation results in B /B /B œ C # Î# - Þ Invoking the initial condition, we obtain - œ "Î# Þ Hence C # œ #/B #B /B "Þ The explicit form of the solution is CaBb œ È#/B #B /B " Þ The positive sign is chosen, since C a!b œ "Þ a, bÞ
a- bÞ The function under the radical becomes negative near B œ "Þ( and B œ !Þ(' Þ
11a+bÞ Write the differential equation as <# .< œ ) " . ) Þ Integrating both sides of the equation results in the relation <" œ 68 ) - Þ Imposing the condition
________________________________________________________________________ page 29
—————————————————————————— CHAPTER 2. —— a, bÞ
a- bÞ Clearly, the solution makes sense only if ) ! Þ Furthermore, the solution becomes singular when 68 ) œ "Î# , that is, ) œ È/ Þ 13a+bÞ CaBb œ È# 68a" B# b % Þ a, bÞ
14a+b. Write the differential equation as C$ .C œ Ba" B# b .B Þ Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation C# Î# œ È" B# - Þ Imposing the initial condition, we obtain - œ $Î# Þ Hence the specific solution can be expressed as C# œ $ #È" B# Þ The explicit form of the solution is CaBb œ "ÎÉ$ #È" B# Þ The positive sign is chosen to "Î#
a- bÞ The solution becomes singular when #È" B# œ $ . That is, at B œ „È& Î# Þ 15a+bÞ CaBb œ "Î# ÈB# "&Î% Þ a, bÞ
16a+b. Rewrite the differential equation as %C$ .C œ BaB# "b.B Þ Integrating both sides of the equation results in C% œ aB# "b# Î% - Þ Imposing the initial condition, we obtain - œ ! Þ Hence the solution may be expressed as aB# "b# %C % œ ! Þ The explicit form of the solution is CaBb œ ÈaB# "bÎ# Þ The sign is chosen based on C a!b œ "ÎÈ# Þ
a- b. The solution is valid for B "Þ%& Þ This value is found by estimating the root of %B$ %/B "$ œ ! Þ 18a+b. Write the differential equation as a$ %Cb.C œ a/B /B b.B Þ Integrating both sides of the equation, with respect to the appropriate variables, we obtain the relation $C #C # œ a/B /B b - Þ Imposing the initial condition, C a!b œ " , we obtain - œ (Þ Thus, the solution can be expressed as $C #C # œ a/B /B b (Þ Now by completing the square on the left hand side, #aC $Î%b# œ a/B /B b '&Î) . Hence the explicit form of the solution is CaBb œ $Î% È'&Î"' -9=2 B Þ
a- b. Note the '& "' -9=2 B ! , as long as kBk #Þ" Þ Hence the solution is valid on the interval #Þ" B #Þ" . 19a+bÞ C aBb œ 1Î$ "$ =38" a$ -9=# BbÞ a, bÞ
20a+bÞ Rewrite the differential equation as C# .C œ +<-=38 BÎÈ" B# .B Þ Integrating # both sides of the equation results in C$ Î$ œ a+<-=38 Bb Î# - Þ Imposing the condition
#Î$ $ $ Ca!b œ !, we obtain - œ !Þ The explicit form of the solution is CaBb œ É # a+<-=38 Bb Þ
a- b. Evidently, the solution is defined for " Ÿ B Ÿ " Þ
22. The differential equation can be written as a$C# %b.C œ $B# .B Þ Integrating both sides, we obtain C$ %C œ B$ - Þ Imposing the initial condition, the specific solution is C$ %C œ B$ " Þ Referring back to the differential equation, we find that C w p _ as C p „#ÎÈ$ Þ The respective values of the abscissas are B œ "Þ#(' , "Þ&*) Þ
Hence the solution is valid for "Þ#(' B "Þ&*) Þ 24. Write the differential equation as a$ #Cb.C œ a# /B b.B Þ Integrating both sides, we obtain $C C # œ #B /B - Þ Based on the specified initial condition, the solution can be written as $C C # œ #B /B " Þ Completing the square, it follows that CaBb œ $Î# È#B /B "$Î% Þ The solution is defined if #B /B "$Î% ! , that is, "Þ& Ÿ B Ÿ # aapproximatelyb. In that interval, C w œ ! , for B œ 68 # Þ It can be verified that C ww a68 #b ! . In fact, C ww aBb ! on the interval of definition. Hence the solution attains a global maximum at B œ 68 # Þ 26. The differential equation can be written as a" C# b" .C œ #a" Bb.B Þ Integrating both sides of the equation, we obtain +<->+8C œ #B B# - Þ Imposing the given initial condition, the specific solution is +<->+8C œ #B B# Þ Therefore, C aBb œ >+8a#B B# bÞ Observe that the solution is defined as long as 1Î# #B B# 1Î# Þ It is easy to see that #B B# "Þ Furthermore, #B B# œ 1Î# for B œ #Þ' and !Þ' Þ Hence the solution is valid on the interval #Þ' B !Þ' Þ Referring back to the differential ________________________________________________________________________ page 34
—————————————————————————— CHAPTER 2. —— equation, the solution is stationary at B œ "Þ Since C ww aBb ! on the entire interval of definition, the solution attains a global minimum at B œ " Þ 28a+b. Write the differential equation as C" a% Cb" .C œ >a" >b" .> Þ Integrating both sides of the equation, we obtain 68 kCk 68kC %k œ %> %68k" >k - Þ Taking the exponential of both sides, it follows that kCÎaC %bk œ G /%> Îa" >b% Þ It follows that as > p _ , kCÎaC %bk œ k" %ÎaC %bkp _ . That is, C a>b p % Þ
a,bÞ Setting Ca!b œ # , we obtain that G œ " . Based on the initial condition, the solution may be expressed as CÎaC %b œ /%> Îa" >b% Þ Note that CÎaC %b ! , for all > !Þ Hence C % for all > !Þ Referring back to the differential equation, it follows that C w is always positive. This means that the solution is monotone increasing. We find that the root of the equation /%> Îa" >b% œ $** is near > œ #Þ)%% Þ a- bÞ Note the Ca>b œ % is an equilibrium solution. Examining the local direction field,
we see that if Ca!b ! , then the corresponding solutions converge to C œ % . Referring back to part a+b, we have CÎaC %b œ cC! ÎaC! %bd/%> Îa" >b% , for C! Á % Þ Setting % > œ # , we obtain C! ÎaC! %b œ a$Î/# b C a#bÎaC a#b %bÞ Now since the function 0 aCb œ CÎaC %b is monotone for C % and C % , we need only solve the equations C! ÎaC! %b œ $**a$Î/# b% and C! ÎaC! %b œ %!"a$Î/# b% Þ The respective solutions are C! œ $Þ''## and C! œ %Þ%!%# Þ 30a0 bÞ
" 32a+b. Observe that aB# $C # bÎ#BC œ "# ˆ BC ‰ is homogeneous.
$C #B
Þ Hence the differential equation
a,b. The substitution C œ B @ results in @ B @ w œ aB# $B# @# bÎ#B# @ . The transformed equation is @ w œ a" @# bÎ#B@ Þ This equation is separable, with general solution @# " œ - B Þ In terms of the original dependent variable, the solution is B# C # œ - B$ Þ a- bÞ
—————————————————————————— CHAPTER 2. —— " 34a+bÞ Observe that a%B $C bÎÐ#B CÑ œ # BC # BC ‘ Þ Hence the differential equation is homogeneous.
a,b. The substitution C œ B @ results in @ B @ w œ # @Îa# @b. The transformed equation is @ w œ a@# &@ %bÎÐ# @ÑB Þ This equation is separable, with general solution a@%b# k@"k œ GÎB$ Þ In terms of the original dependent variable, the solution is a%B Cb# kBCk œ GÞ a- bÞ
35a- b.
36a+bÞ Divide by B# to see that the equation is homogeneous. Substituting C œ B @ , we obtain B @ w œ a" @b# Þ The resulting differential equation is separable.
a,bÞ Write the equation as a" @b# .@ œ B" .B Þ Integrating both sides of the equation, we obtain the general solution "Îa" @b œ 68kBk - Þ In terms of the original dependent variable, the solution is C œ B cG 68kBkd" B Þ
" 37a+bÞ The differential equation can be expressed as C w œ "# ˆ BC ‰ #$ BC . Hence the equation is homogeneous. The substitution C œ B @ results in B @ w œ a" &@# bÎ#@ . #@ " Separating variables, we have "&@ # .@ œ B .B Þ
a,bÞ Integrating both sides of the transformed equation yields "& 68k" &@# k œ 68kBk - , that is, " &@# œ GÎkBk& Þ In terms of the original dependent variable, the general solution is &C# œ B# GÎkBk$ Þ a- bÞ
" 38a+bÞ The differential equation can be expressed as C w œ $# BC #" ˆ BC ‰ . Hence the equation is homogeneous. The substitution C œ B @ results in B @ w œ a@# "bÎ#@, that is, @##@" .@ œ B" .B Þ
a,bÞ Integrating both sides of the transformed equation yields 68k@# "k œ 68kBk - , that is, @# " œ G kBkÞ In terms of the original dependent variable, the general solution is C# œ G B# kBk B# Þ
—————————————————————————— CHAPTER 2. —— Section 2.3 5a+b. Let U be the amount of salt in the tank. Salt enters the tank of water at a rate of # "% ˆ" "# =38 >‰ œ "# "% =38 > 9DÎ738 Þ It leaves the tank at a rate of # UÎ"!! 9DÎ738Þ Hence the differential equation governing the amount of salt at any time is .U " " œ =38 > UÎ&! Þ .> # % The initial amount of salt is U! œ &! 9D Þ The governing ODE is linear, with integrating w factor .a>b œ />Î&! Þ Write the equation as ˆ/>Î&! U‰ œ />Î&! ˆ "# "% =38 >‰Þ The specific solution is Ua>b œ #& "#Þ&=38 > '#&-9= > '$"&! />Î&! ‘Î#&!" 9D Þ a, bÞ
a- bÞ The amount of salt approaches a steady state, which is an oscillation of amplitude "Î% about a level of #& 9D Þ 6a+b. The equation governing the value of the investment is .WÎ.> œ < W . The value of the investment, at any time, is given by W a>b œ W! /<> Þ Setting W aX b œ #W! , the required time is X œ 68a#bÎ< Þ
a,bÞ For the case < œ (% œ Þ!( , X ¸ *Þ* C<= Þ
a- bÞ Referring to Parta+b, < œ 68a#bÎX . Setting X œ ) , the required interest rate is to be approximately < œ )Þ'' % Þ 8a+b. Based on the solution in Eq.a"'b , with W! œ ! , the value of the investments with contributions is given by W a>b œ #&ß !!!a/<> "bÞ After ten years, person A has WE œ $#&ß !!!a"Þ##'b œ $ $!ß '%! Þ Beginning at age $& , the investments can now be analyzed using the equations WE œ $!ß '%! /Þ!)> and WF œ #&ß !!!a/Þ!)> "bÞ After thirty years, the balances are WE œ $ $$(ß ($% and WF œ $ #&!ß &(*Þ
a,bÞ For an unspecified rate < , the balances after thirty years are WE œ $!ß '%! /$!< and WF œ #&ß !!!a/$!< "bÞ ________________________________________________________________________ page 40
—————————————————————————— CHAPTER 2. —— a- b .
a. bÞ The two balances can never be equal.
11a+b. Let W be the value of the mortgage. The debt accumulates at a rate of œ Þ!* W *ß '!! Þ Given that W! is the original amount borrowed, the debt is W a>b œ W! /Þ!*> "!'ß ''(a/Þ!*> "bÞ Setting W a$!b œ ! , it follows that W! œ $ **ß &!! . a,bÞ The total payment, over $! years, becomes $ #))ß !!! . The interest paid on this purchase is $ "))ß &!! .
13a+b. The balance increases at a rate of < W $/yr, and decreases at a constant rate of 5 $ per year. Hence the balance is modeled by the differential equation .WÎ.> œ b œ W! /<> 5< a/<> "bÞ a,b. The solution may also be expressed as W a>b œ ÐW! 5< Ñ/<> 5< Þ Note that if the withdrawal rate is 5! œ < W! , the balance will remain at a constant level W! Þ 5 a- b. Assuming that 5 5! , W aX! b œ ! for X! œ "< 68’ 55 “Þ !
a. b. If < œ Þ!) and 5 œ #5! , then X! œ )Þ'' years .
a/b. Setting W a>b œ ! and solving for /<> in Parta,b, /<> œ results in 5 œ
5 5
Now setting > œ X
a0 bÞ In parta/b, let 5 œ "#ß !!! , < œ Þ!) , and X œ #! . The required investment becomes W! œ $ ""*ß ("& . 14a+bÞ Let U w œ < U Þ The general solution is Ua>b œ U! /<> Þ Based on the definition of half-life, consider the equation U! Î# œ U! /&($! < Þ It follows that
—————————————————————————— CHAPTER 2. —— &($! < œ 68Ð"Î#Ñ, that is, < œ "Þ#!*( ‚ "!% per year. a,b. Hence the amount of carbon-14 is given by Ua>b œ U! /"Þ#!*(‚"! > Þ %
a- bÞ Given that UaX b œ U! Î& , we have the equation "Î& œ /"Þ#!*(‚"! X Þ Solving for the decay time, the apparent age of the remains is approximately X œ "$ß $!%Þ'& years . %
15. Let T a>b be the population of mosquitoes at any time > . The rate of increase of the mosquito population is œ variable > represents days . The solution is T a>b œ T! /<> #!ß!!! < a/ "bÞ In the absence of predators, the governing equation is .T" Î.> œ b œ T! /<> Þ Based on the data, set T" a(b œ #T! , that is, #T! œ T! /(< Þ The growth rate is determined as < œ 68a#bÎ( œ Þ!**!# per dayÞ Therefore the population, including the predation by birds, is T a>b œ # ‚ "!& /Þ!**> #!"ß **(a/Þ!**> "b œ œ #!"ß **(Þ$ "*((Þ$ /Þ!**> Þ 16a+b. Ca>b œ /B:c#Î"! >Î"! #-9=Ð>ÑÎ"!dÞ The doubling-time is 7 ¸ #Þ*'$# Þ
a,bÞ The differential equation is .CÎ.> œ CÎ"! , with solution C a>b œ C a!b/>Î"! Þ The doubling-time is given by 7 œ "!68a#b ¸ 'Þ*$"& Þ
a- b. Consider the differential equation .CÎ.> œ a!Þ& =38Ð#1>Ñb CÎ& Þ The equation is separable, with C" .C œ ˆ!Þ" &" =38Ð#1>щ.> Þ Integrating both sides, with respect to the appropriate variable, we obtain 68 C œ a1> -9=Ð#1>ÑbÎ"!1 - Þ Invoking the initial condition, the solution is Ca>b œ /B:ca" 1> -9=Ð#1>ÑbÎ"!1dÞ The doubling-time is 7 ¸ 'Þ$)!% Þ The doubling-time approaches the value found in parta,b. a. b .
—————————————————————————— CHAPTER 8. —— In this problem, 08" œ È>8" C8" Þ Since the ODE is 898linear, an equation solver must be implemented in order to approximate the solution of C8" œ C8
2 *È>8" C8" "* 08 & 08" 08# ‘ #%
at each time step. >8 C8
8 œ "! !Þ& $Þ*'#")'
8 œ #! "Þ! &Þ"!)*!$
8 œ $! "Þ& 'Þ%$"$*!
8 œ %! #Þ! (Þ*#$$)&
a- b. The fourth order backward differentiation formula is C8" œ
—————————————————————————— CHAPTER 8. —— 10a+b. The predictor formula is C8" œ C8
2 a&& 08 &* 08" $( 08# * 08$ bÞ #%
With 08" œ 0 a>8" ß C8" b, the corrector formula is C8" œ C8
2 a* 08" "* 08 & 08" 08# b Þ #%
We use the starting values generated by the Runge-Kutta method À >8 C8
8œ! !Þ! "Þ!
8 œ "! !Þ& "Þ'"#'##
>8 C8
8œ" !Þ!& "Þ!&"#$!
8 œ #! "Þ! #Þ%)!*!*
8œ# !Þ" "Þ"!%)%$
8œ$ !Þ"& "Þ"'!(%!
8 œ $! "Þ& $Þ(%&"%(*
8 œ %! #Þ! &Þ%*&)(#
a,b. With 08" œ 0 a>8" ß C8" b, the fourth order Adams-Moulton formula is C8" œ C8
2 a* 08" "* 08 & 08" 08# b Þ #%
In this problem, 08" œ # >8" /B:a >8" C8" b Þ Since the ODE is 898linear, an equation solver must be implemented in order to approximate the solution of C8" œ C8
The exact solution of the IVP is Ca>b œ a$ ># bÎa' >b Þ
8 œ %! #Þ! "Þ(&!!!"
13. Both Adams methods entail the approximation of 0 a> ß C b, on the interval c>8 ß >8" d, by a polynomial. Approximating 9 w a>b œ T" a>b ´ E , which is a constant polynomial, we have 9a>8" b 9a>8 b œ (
>8"
E .>
œ Ea>8" >8 b œ E2 Þ >8
Setting E œ - 08 a" -b08" , where ! Ÿ - Ÿ " , we obtain the approximation C8" œ C8 2c- 08 a" -b08" dÞ
An appropriate choice of - yields the familiar Euler formula. Similarly, setting E œ - 08 a" -b08" ,
where ! Ÿ - Ÿ " , we obtain the approximation ________________________________________________________________________ page 484
—————————————————————————— CHAPTER 8. —— C8" œ C8 2c- 08 a" -b08" dÞ 14. For a third order Adams-Bashforth formula, we approximate 0 a> ß C b, on the interval c>8 ß >8" d, by a quadratic polynomial using the points a>8# ß C8# b , a>8" ß C8" b and a>8 ß C8 bÞ Let T$ a>b œ E># F> G Þ We obtain the system of equations E>#8# F>8# G œ 08# E>#8" F>8" G œ 08" E>#8 F>8 G œ 08 . For computational purposes, assume that >! œ ! , and >8 œ 82 Þ It follows that 08 #08" 08# # 2# a$ #8b08 a%8 %b08" a" #8b08# Fœ #2 8# $8 # 8# 8 Gœ 08 ˆ#8 8# ‰08" 08# Þ # # Eœ
We then have C8" C8 œ (
>8" >8
E># F> G ‘.>
" " œ E2$ Œ8# 8 F2 # Œ8 G2 , $ #
which yields C8" C8 œ
2 a#$ 08 "' 08" & 08# b Þ "#
15. For a third order Adams-Moulton formula, we approximate 0 a> ß C b, on the interval c>8 ß >8" d, by a quadratic polynomial using the points a>8" ß C8" b , a>8 ß C8 b and a>8" ß C8" bÞ Let T$ a>b œ !># " > # Þ This time we obtain the system of algebraic equations !>#8" " >8" # œ 08" !>#8 " >8 # œ 08 !>#8" " >8" # œ 08" . For computational purposes, again assume that >! œ ! , and >8 œ 82 Þ It follows that
—————————————————————————— CHAPTER 8. —— Section 8.5 1a+b. The general solution of the ODE is Ca>b œ - /> # > Þ Imposing the initial condition, Ca!b œ # , the solution of the IVP is 9" a>b œ # > .
a,b. If instead, the initial condition Ca!b œ #Þ!!" is given, the solution of the IVP is 9# a>b œ !Þ!!" /> # > . We then have 9# a>b 9" a>b œ !Þ!!" /> Þ 3. The solution of the initial value problem is 9a>b œ /"!! > > Þ
a+ß ,bÞ Based on the exact solution, the local truncation error for both of the Euler methods is k/69- k Ÿ
"!% "!! >8 # / 2 Þ #
Hence k/8 k Ÿ &!!! 2# , for all ! >8 " . Furthermore, the local truncation error is greatest near > œ ! . Therefore k/" k Ÿ &!!! 2# !Þ!!!& for 2 !Þ!!!$ Þ Now the truncation error accumulates at each time step. Therefore the actual time step should be much smaller than 2 ¸ !Þ!!!$ . For example, with 2 œ !Þ!!!#& , we obtain the data > œ !Þ!& > œ !Þ"
Euler !Þ!&'$#$ !Þ"!!!%!
B.Euler !Þ!&("'& !Þ"!!!&"
9a>b !Þ!&'($) !Þ"!!!%&
Note that the total number of time steps needed to reach > œ !Þ" is R œ %!! . a- b. Using the Runge-Kutta method, comparisons are made for several values of 2 À 2 œ !Þ" À > œ !Þ!& > œ !Þ"
9a>b !Þ!&'($) !Þ"!!!%&
C8 !Þ!&(%"' !Þ"!!!&&
C8 9 a> 8 b !Þ!!!'() !Þ!!!!"!
> œ !Þ!& > œ !Þ"
9a>b !Þ!&'($) !Þ"!!!%&
C8 !Þ!&'('' !Þ"!!!%'
C8 9 a> 8 b !Þ!!!!#( !Þ!!!!!!%
2 œ !Þ!!& À
6a+b. Using the method of undetermined coefficients, it is easy to show that the general solution of the ODE is Ca>b œ - /-> ># Þ Imposing the initial condition, it follows that - œ ! and hence the solution of the IVP is 9a>b œ ># Þ a,b. Using the Runge-Kutta method, with 2 œ !Þ!" , numerical solutions are generated
The following table shows the calculated value, C" , at the first time step À 9a>b "!%
C" a- œ "b *Þ***** ‚ "!&
C" a- œ "!b *Þ***(* ‚ "!&
C" a- œ #!b *Þ**)$$ ‚ "!&
C" a- œ &!b *Þ*($*' ‚ "!&
a- b. Referring back to the exact solution given in Parta+b, if a nonzero initial condition, say Ca!b œ & , is specified, the solution of the IVP becomes 9& a>b œ & /-> ># .
We then have k9a> b 9& a>bk œ k&k /-> Þ It is evident that for any > ! , ________________________________________________________________________ page 488
This implies that virtually any error introduced early in the calculations will be magnified as - p _ . The initial value problem is inherently unstable.
—————————————————————————— CHAPTER 8. —— Section 8.6 1. In vector notation, the initial value problem can be written as . B BC> " Œ œŒ , xa!b œ Œ Þ .> C %B #C !
