solution-manual-elementary-differential-equations-and-boundary-value-problems-william-e-boyce
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—————————————————————————— 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
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—————————————————————————— CHAPTER 1. ——
the equilibrium solution Ca>b œ "Î# .
6.
For C # , the slopes are :9=3tive, and hence the solutions increase. For C # ,
the slopes are negative, and hence the solutions decrease. All solutions diverge away
from
the equilibrium solution Ca>b œ # .
8. For all solutions to approach the equilibrium solution Ca>b œ #Î$ , we must have
C w ! for C #Î$ , and C w ! for C #Î$ . The required rates are satisfied by the
differential equation C w œ # $C .
9. For solutions other than Ca>b œ # to diverge from C œ # , C a>b must be an increasing
function for C # , and a decreasing function for C # . The simplest differential
equation
whose solutions satisfy these criteria is C w œ C # .
10. For solutions other than Ca>b œ "Î$ to diverge from C œ "Î$ , we must have C w !
for C "Î$ , and C w ! for C "Î$ . The required rates are satisfied by the differential
equation C w œ $C " .
12.
Note that C w œ ! for C œ ! and C œ & . The two equilibrium solutions are C a>b œ ! and
Ca>b œ & . Based on the direction field, C w ! for C & ; thus solutions with initial
values greater than & diverge from the solution Ca>b œ & . For ! C &, the slopes are
negative, and hence solutions with initial values between ! and & all decrease toward the
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—————————————————————————— 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<b Þ
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
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—————————————————————————— CHAPTER 1. ——
7
.@
œ 71 # @#
.>
or equivalently,
a,b After a long time,
.@
#
œ 1 @# Þ
.>
7
.@
.>
¸ ! Þ Hence the object attains a terminal velocity given by
@_ œ Ê
71
Þ
#
a- b Using the relation # @_# œ 71 , the required drag coefficient is # œ !Þ!%!) 51Î=/- Þ
a. b
19.
All solutions appear to approach a linear asymptote aA3>2 =69:/ /;?+6 >9 "b. It is easy to
verify that Ca>b œ > $ is a solution.
20.
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—————————————————————————— CHAPTER 1. ——
All solutions approach the equilibrium solution Ca>b œ ! Þ
23.
All solutions appear to diverge from the sinusoid Ca>b œ
$
È#
=38Ð> 1% Ñ " ,
which is also a solution corresponding to the initial value Ca!b œ &Î# .
25.
All solutions appear to converge to Ca>b œ ! . First, the rate of change is small. The
slopes
eventually increase very rapidly in magnitude.
26.
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—————————————————————————— CHAPTER 1. ——
The direction field is rather complicated. Nevertheless, the collection of points at which
the slope field is zero, is given by the implicit equation C$ 'C œ #># Þ The graph of
these points is shown below:
The y-intercepts of these curves are at C œ ! , „È' . It follows that for solutions with
initial values C È' , all solutions increase without bound. For solutions with initial
values in the range C È' and ! C È' , the slopes remain negative, and
hence
these solutions decrease without bound. Solutions with initial conditions in the range
È' C ! initially increase. Once the solutions reach the critical value, given by
the equation C$ 'C œ #># , the slopes become negative and remain negative. These
solutions eventually decrease without bound.
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—————————————————————————— 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
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—————————————————————————— CHAPTER 1. ——
.C
œ .> Þ
C&
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
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—————————————————————————— CHAPTER 1. ——
All solutions appear to diverge from the equilibrium solution Ca>b œ & .
3a+b. Rewrite the differential equation as
.C
œ .> ,
, +C
which is valid for C Á , Î+. Integrating both sides results in +" 68k, +C k œ > -" , or
equivalently, , +C œ - /+> . Hence the general solution is C a>b œ a, - /+> bÎ+ Þ
Note that if C œ ,Î+ , then .CÎ.> œ ! , and C a>b œ ,Î+ is an equilibrium solution.
a, b
a3b As + increases, the equilibrium solution gets closer to Ca>b œ ! , from above.
Furthermore, the convergence rate of all solutions, that is, + , also increases.
a33b As , increases, then the equilibrium solution C a>b œ ,Î+ also becomes larger. In
this case, the convergence rate remains the same.
a333b If + and , both increase abut ,Î+ œ constantb, then the equilibrium solution
Ca>b œ ,Î+ remains the same, but the convergence rate of all solutions increases.
5a+b. Consider the simpler equation .C" Î.> œ +C" . As in the previous solutions, rewrite the equation as
.C"
œ + .> Þ
C"
Integrating both sides results in C" a>b œ - /+> Þ
a,bÞ Now set Ca>b œ C" a>b 5 , and substitute into the original differential equation. We
find that
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—————————————————————————— 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 :! /<N/30 œ #:!
.
It follows that <R Î$! œ 68a#b , and hence the rate constant is given by < œ $! 68a#bÎR Þ
The units are understood to be per month .
9a+b. Assuming no air resistance, with the positive direction taken as downward,
Newton's
Second Law can be expressed as
7
.@
œ 71
.>
in which 1 is the gravitational constant measured in appropriate units. The equation can
be
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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 "!! /<X , in which < œ Þ"*(*'/week . Taking the
natural logarithm of both sides, we find that X œ $Þ&!"% weeks or X œ #%Þ&" .+C s .
11. The general solution of the differential equation .UÎ.> œ < U is Ua>b œ U! /<> ,
in which U! œ Ua!b is the initial amount of the substance. Let 7 be the time that it takes
the substance to decay to one-half of its original amount , U! . Setting > œ 7 in the
solution,
we have !Þ& U! œ U! /<7 . Taking the natural logarithm of both sides, it follows that
<7 œ 68a!Þ&b or <7 œ 68 # Þ
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—————————————————————————— 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!!!$> Þ
Hence, one year after the source is removed, the amount of chemical in the pond is
Ua)('!b œ '(!Þ" grams .
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a/b. Letting > be the amount of time after the source is removed, we obtain the equation
"! œ *#((Þ(( /!Þ!!!$> 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
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—————————————————————————— 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 .
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—————————————————————————— 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 >
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& >2 "! >2 68 > % >2 68 > œ ! Þ Hence both functions are solutions of the
differential equation.
13. Ca>b œ a-9= >b68 -9= > > =38 > Ê C w a>b œ a=38 >b68 -9= > > -9= > and
C ww a>b œ a-9= >b68 -9= > > =38 > =/- > . Substituting into the left hand side of the
differential equation, we have a a-9= >b68 -9= > > =38 > =/- >b a-9= >b68 -9= >
> =38 > œ a-9= >b68 -9= > > =38 > =/- > a-9= >b68 -9= > > =38 > œ =/- > .
Hence the function Ca>b is a solution of the differential equation.
15. Let Ca>b œ /<> . Then C ww a>b œ <# /<> , and substitution into the differential equation
results in <# /<> # /<> œ !. Since /<> Á ! , we obtain the algebraic equation <# # œ !Þ
The roots of this equation are <"ß# œ „ 3È# Þ
17. Ca>b œ /<> Ê C w a>b œ < /<> and C ww a>b œ <# /<> . Substituting into the differential
equation, we have <# /<> </<> ' /<> œ ! . Since /<> Á ! , we obtain the algebraic
equation <# < ' œ ! , that is, a< #ba< $b œ ! . The roots are <"ß# œ $ , # .
18. Let Ca>b œ /<> . Then C w a>b œ </<> , C ww a>b œ <# /<> and C www a>b œ <$ /<> . Substituting
the derivatives into the differential equation, we have <$ /<> $<# /<> #</<> œ ! . Since
/<> Á ! , we obtain the algebraic equation <$ $<# #< œ ! Þ By inspection, it follows
that <a< "ba< #b œ ! . Clearly, the roots are <" œ ! , <# œ " and <$ œ # Þ
20. Ca>b œ >< Ê C w a>b œ < ><" and C ww a>b œ <a< "b><# . Substituting the derivatives
into the differential equation, we have ># c<a< "b><# d %>a< ><" b % >< œ ! . After
some algebra, it follows that <a< "b>< %< >< % >< œ ! . For > Á ! , we obtain the
algebraic equation <# &< % œ ! Þ The roots of this equation are <" œ " and <# œ % Þ
21. The order of the partial differential equation is two, since the highest derivative, in
fact each one of the derivatives, is of second order. The equation is linear, since the left
hand side is a linear function of the partial derivatives.
23. The partial differential equation is fourth order, since the highest derivative, and in
fact each of the derivatives, is of order four. The equation is linear, since the left hand
side is a linear function of the partial derivatives.
24. The partial differential equation is second order, since the highest derivative of the
function ?aBß Cb is of order two. The equation is nonlinear, due to the product ? † ?B on
the left hand side of the equation.
25. ?" aBß Cb œ -9= B -9=2 C Ê
It is evident that
derivatives are
#
` ?"
`B#
#
` ?"
`C#
` # ?"
`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
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—————————————————————————— CHAPTER 1. ——
` # ?2
#
%B#
œ
`B#
B# C #
aB# C # b#
` # ?2
#
%C#
œ
`C#
B# C #
aB# C # b#
Adding the partial derivatives,
` # ?2
` # ?2
#
%B#
#
%C #
œ
`B#
`C#
B# C #
B# C #
aB# C # b#
aB# C # b#
%
%aB# C# b
œ #
B C#
aB# C # b#
œ !.
Hence ?# aBß Cb is also a solution of the differential equation.
27. Let ?" aBß >b œ =38 -B =38 -+> . Then the second derivatives are
` # ?"
œ -# =38 -B =38 -+>
#
`B
` # ?"
œ -# +# =38 -B =38 -+>
`>#
#
It is easy to see that +# ``B?#" œ
` # ?"
`>#
. Likewise, given ?# aBß >b œ =38aB +>b , we have
` # ?#
œ =38aB +>b
`B#
` # ?#
œ +# =38aB +>b
#
`>
Clearly, ?# aBß >b is also a solution of the partial differential equation.
28. Given the function ?aBß >b œ È1Î> /B Î%! > , the partial derivatives are
#
#
È1Î> /B# Î%!# >
È1Î> B# /B# Î%!# >
?BB œ
#! # >
%! % > #
#
#
#
!
B
Î%
>
È1> /
È1 B /B# Î%!# >
?> œ
#>#
%!# ># È>
It follows that !# ?BB œ ?> œ
È1 ˆ#!# >B# ‰/B# Î%!# >
%! # > # È >
.
Hence ?aBß >b is a solution of the partial differential equation.
________________________________________________________________________
page 17
—————————————————————————— CHAPTER 1. ——
29a+b.
a,bÞ The path of the particle is a circle, therefore polar coordinates are intrinsic to the
problem. The variable < is radial distance and the angle ) is measured from the vertical.
Newton's Second Law states that ! F œ 7a Þ In the tangential direction, the equation of
motion may be expressed as ! J) œ 7 +) , in which the tangential acceleration, that is,
the linear acceleration along the path is +) œ P . # )Î.># Þ Ð +) is positive in the direction
of increasing ) Ñ. Since the only force acting in the tangential direction is the component
of weight, the equation of motion is
71 =38 ) œ 7P
.# )
Þ
.>#
ÐNote that the equation of motion in the radial direction will include the tension in the
rodÑ.
a- b. Rearranging the terms results in the differential equation
.# )
1
=38 ) œ ! Þ
#
.>
P
________________________________________________________________________
page 18
—————————————————————————— CHAPTER 2. ——
Chapter Two
Section 2.1
1a+bÞ
a,bÞ Based on the direction field, all solutions seem to converge to a specific increasing
function.
a- bÞ The integrating factor is .a>b œ /$> , and hence Ca>b œ >Î$ "Î* /#> - /$> Þ
It follows that all solutions converge to the function C" a>b œ >Î$ "Î* Þ
2a+bÞ
a,b. All slopes eventually become positive, hence all solutions will increase without
bound.
a- bÞ The integrating factor is .a>b œ /#> , and hence Ca>b œ >$ /#> Î$ - /#> Þ It is
evident that all solutions increase at an exponential rate.
3a+b
________________________________________________________________________
page 18
—————————————————————————— CHAPTER 2. ——
a,b. All solutions seem to converge to the function C! a>b œ " Þ
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
5a+b.
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page 19
—————————————————————————— CHAPTER 2. ——
a,b. All slopes eventually become positive, hence all solutions will increase without
bound.
a- bÞ The integrating factor is .a>b œ /B:a ' #.>b œ /#> Þ The differential equation
can
w
be written as /#> C w #/#> C œ $/> , that is, a/#> C b œ $/> Þ Integration of both
sides of the equation results in the general solution Ca>b œ $/> - /#> Þ It follows that
all solutions will increase exponentially.
6a+b
a,bÞ All solutions seem to converge to the function C! a>b œ ! Þ
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.
________________________________________________________________________
page 20
—————————————————————————— CHAPTER 2. ——
a,bÞ All solutions seem to converge to the function C! a>b œ ! Þ
a- bÞ The integrating factor is .a>b œ /B:a># b, and hence C a>b œ ># /> - /> Þ It is
clear that all solutions converge to the function C! a>b œ ! .
#
#
8a+b
a,bÞ All solutions seem to converge to the function C! a>b œ ! Þ
a- bÞ Since .a>b œ a" ># b# , the general solution is Ca>b œ c>+8" a>b G dÎa" ># b# Þ
It follows that all solutions converge to the function C! a>b œ ! .
9a+bÞ
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page 21
—————————————————————————— 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.
________________________________________________________________________
page 23
—————————————————————————— CHAPTER 2. ——
The solutions appear to diverge from an apparent oscillatory solution. From the
direction
field, the critical value of the initial condition seems to be +! œ " . For + " , the
solutions increase without bound. For + " , solutions decrease without bound.
a,bÞ The integrating factor is .a>b œ />Î# . The general solution of the differential
equation is Ca>b œ a)=38a>b %-9=a>bbÎ& - />Î# . The solution is sinusoidal as long
as - œ ! . The initial value of this sinusoidal solution is
+! œ a)=38a!b %-9=a!bbÎ& œ %Î& Þ
a- bÞ See part a,b.
22a+bÞ
All solutions appear to eventually increase without bound. The solutions initially
increase
or decrease, depending on the initial value + . The critical value seems to be +! œ " Þ
a,bÞ The integrating factor is .a>b œ />Î# , and the general solution of the differential
equation is Ca>b œ $/>Î$ - />Î# Þ Invoking the initial condition C a!b œ + , the
solution
may also be expressed as Ca>b œ $/>Î$ a+ $b />Î# Þ Differentiating, follows that
C w a!b œ " a+ $bÎ# œ a+ "bÎ# Þ The critical value is evidently +! œ " Þ
________________________________________________________________________
page 24
—————————————————————————— CHAPTER 2. ——
a- b. For +! œ " , the solution is Ca>b œ $/>Î$ # />Î# , which afor large >b is
dominated by the term containing />Î# Þ
is Ca>b œ a)=38a>b %-9=a>bbÎ& - />Î# .
23a+bÞ
As > p ! , solutions increase without bound if C a"b œ + Þ% , and solutions decrease
without bound if Ca"b œ + Þ% Þ
>
‰
a,b. The integrating factor is .a>b œ /B:ˆ' >"
> .> œ > / Þ The general solution of the
differential equation is Ca>b œ > /> - /> Î> . Invoking the specified value C a"b œ + ,
we have " - œ + / . That is, - œ + / " . Hence the solution can also be expressed as
Ca>b œ > /> a+ / "b /> Î> . For small values of > , the second term is dominant.
Setting + / " œ ! , critical value of the parameter is +! œ "Î/ Þ
a- b. For + "Î/ , solutions increase without bound. For + "Î/ , solutions decrease
without bound. When + œ "Î/ , the solution is C a>b œ > /> , which approaches ! as > p !
.
24a+b.
As > p ! , solutions increase without bound if C a"b œ + Þ% , and solutions decrease
without bound if Ca"b œ + Þ% Þ
________________________________________________________________________
page 25
—————————————————————————— CHAPTER 2. ——
a,b. Given the initial condition, Ca 1Î#b œ + , the solution is Ca>b œ a+ 1# Î% -9= >bÎ>
Þ
Since lim -9= > œ " , solutions increase without bound if + %Î1# , and solutions
>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
Ca>b œ "# )-9=a#>b '%=38a#>b ()) />Î4 ‘Î'& Þ
As > p _ , the exponential term will decay, and the solution will oscillate about an
average
value of "# , with an amplitude of )ÎÈ'& Þ
________________________________________________________________________
page 26
—————————————————————————— CHAPTER 2. ——
29. The integrating factor is .a>b œ /$>Î# , and the differential equation can be written
w
as a/$>Î# Cb œ $> /$>Î# # />Î# Þ The general solution is C a>b œ #> %Î$ % />
- /$>Î# Þ Imposing the initial condition, C a>b œ #> %Î$ % /> aC! "'Î$b /$>Î# Þ
As > p _ , the term containing /$>Î# will dominate the solution. Its sign will determine
the divergence properties. Hence the critical value of the initial condition is
C! œ "'Î$Þ
The corresponding solution, Ca>b œ #> %Î$ % /> , will also decrease without
bound.
Note on Problems 31-34 :
Let 1a>b be given, and consider the function Ca>b œ C" a>b 1a>b , in which C" a>b p _
as > p _ . Differentiating, C w a>b œ C"w a>b 1 w a>b . Letting + be a constant, it follows
that C w a>b +C a>b œ C"w a>b +C" a>b 1 w a>b +1a>bÞ Note that the hypothesis on the
function C" a>b will be satisfied, if C"w a>b +C" a>b œ ! . That is, C" a>b œ - /+> Þ Hence
Ca>b œ - /+> 1a>b, which is a solution of the equation C w +C œ 1 w a>b +1a>bÞ
For convenience, choose + œ " .
31. Here 1a>b œ $ , and we consider the linear equation C w C œ $ Þ The integrating
w
factor is .a>b œ /> , and the differential equation can be written as a/> C b œ $/> Þ The
general solution is Ca>b œ $ - /> Þ
33. 1a>b œ $ > Þ Consider the linear equation C w C œ " $ > ÞThe integrating
w
factor is .a>b œ /> , and the differential equation can be written as a/> C b œ a# >b/> Þ
The general solution is Ca>b œ $ > - /> Þ
34. 1a>b œ % ># Þ Consider the linear equation C w C œ % #> ># ÞThe integrating
w
factor is .a>b œ /> , and the equation can be written as a/> C b œ a% #> ># b/> Þ
The general solution is Ca>b œ % ># - /> Þ
________________________________________________________________________
page 27
—————————————————————————— CHAPTER 2. ——
Section 2.2
2. For B Á " , the differential equation may be written as C .C œ cB# Îa" B$ bd.B Þ
Integrating both sides, with respect to the appropriate variables, we obtain the relation
C# Î# œ "$ 68k" B$ k - Þ That is, C aBb œ „É #$ 68k" B$ k - Þ
3. The differential equation may be written as C# .C œ =38 B .B Þ Integrating both
sides of the equation, with respect to the appropriate variables, we obtain the relation
C " œ -9= B - Þ That is, aG -9= BbC œ ", in which G is an arbitrary constant.
Solving for the dependent variable, explicitly, CaBb œ "ÎaG -9= Bb .
5. Write the differential equation as -9=# #C .C œ -9=# B .B, or =/- # #C .C œ -9=# B .BÞ
Integrating both sides of the equation, with respect to the appropriate variables, we obtain
the relation >+8 #C œ =38 B -9= B B - Þ
7. The differential equation may be written as aC /C b.C œ aB /B b.B Þ Integrating
both sides of the equation, with respect to the appropriate variables, we obtain the
relation
C # # / C œ B # # / B - Þ
8. Write the differential equation as a" C# b.C œ B# .B Þ Integrating both sides of the
equation, we obtain the relation C C $ Î$ œ B$ Î$ - , that is, $C C $ œ B$ GÞ
9a+b. The differential equation is separable, with C# .C œ a" #Bb.B Þ Integration
yields C" œ B B# - Þ Substituting B œ ! and C œ "Î' , we find that - œ ' Þ
Hence the specific solution is C" œ B# B ' . The explicit form is
CaBb œ "ÎaB# B 'bÞ
a, b
a- b. Note that B# B ' œ aB #baB $b . Hence the solution becomes singular at
B œ # and B œ $ Þ
10a+bÞ CaBb œ È#B #B# % Þ
________________________________________________________________________
page 28
—————————————————————————— CHAPTER 2. ——
10a,bÞ
11a+bÞ Rewrite the differential equation as B /B .B œ C .C Þ Integrating both sides
of the equation results in B /B /B œ C # Î# - Þ Invoking the initial condition, we
obtain - œ "Î# Þ Hence C # œ #/B #B /B "Þ The explicit form of the solution is
CaBb œ È#/B #B /B " Þ The positive sign is chosen, since C a!b œ "Þ
a, bÞ
a- bÞ The function under the radical becomes negative near B œ "Þ( and B œ !Þ(' Þ
11a+bÞ Write the differential equation as <# .< œ ) " . ) Þ Integrating both sides of the
equation results in the relation <" œ 68 ) - Þ Imposing the condition <a"b œ # , we
obtain - œ "Î# . The explicit form of the solution is <a) b œ #Îa" # 68 )bÞ
________________________________________________________________________
page 29
—————————————————————————— CHAPTER 2. ——
a, bÞ
a- bÞ Clearly, the solution makes sense only if ) ! Þ Furthermore, the solution becomes
singular when 68 ) œ "Î# , that is, ) œ È/ Þ
13a+bÞ CaBb œ È# 68a" B# b % Þ
a, bÞ
14a+b. Write the differential equation as C$ .C œ Ba" B# b
.B Þ Integrating both
sides of the equation, with respect to the appropriate variables, we obtain the relation
C# Î# œ È" B# - Þ Imposing the initial condition, we obtain - œ $Î# Þ
Hence the specific solution can be expressed as C# œ $ #È" B# Þ The explicit
form of the solution is CaBb œ "ÎÉ$ #È" B# Þ The positive sign is chosen to
"Î#
satisfy the initial condition.
________________________________________________________________________
page 30
—————————————————————————— CHAPTER 2. ——
a, bÞ
a- bÞ The solution becomes singular when #È" B# œ $ . That is, at B œ „È& Î# Þ
15a+bÞ CaBb œ "Î# ÈB# "&Î% Þ
a, bÞ
16a+b. Rewrite the differential equation as %C$ .C œ BaB# "b.B Þ Integrating both
sides
of the equation results in C% œ aB# "b# Î% - Þ Imposing the initial condition, we obtain
- œ ! Þ Hence the solution may be expressed as aB# "b# %C % œ ! Þ The explicit form
of the solution is CaBb œ ÈaB# "bÎ# Þ The sign is chosen based on C a!b œ "ÎÈ# Þ
________________________________________________________________________
page 31
—————————————————————————— CHAPTER 2. ——
a, bÞ
a- bÞ The solution is valid for all B − ‘ .
17a+bÞ CaBb œ &Î# ÈB$ /B "$Î% Þ
a, bÞ
a- b. The solution is valid for B "Þ%& Þ This value is found by estimating the root of
%B$ %/B "$ œ ! Þ
18a+b. Write the differential equation as a$ %Cb.C œ a/B /B b.B Þ Integrating both
sides of the equation, with respect to the appropriate variables, we obtain the relation
$C #C # œ a/B /B b - Þ Imposing the initial condition, C a!b œ " , we obtain
- œ (Þ Thus, the solution can be expressed as $C #C # œ a/B /B b (Þ Now by
completing the square on the left hand side, #aC $Î%b# œ a/B /B b '&Î) .
Hence the explicit form of the solution is CaBb œ $Î% È'&Î"' -9=2 B Þ
________________________________________________________________________
page 32
—————————————————————————— CHAPTER 2. ——
a, bÞ
a- b. Note the '& "' -9=2 B
the interval #Þ" B #Þ" .
! , as long as kBk #Þ" Þ Hence the solution is valid on
19a+bÞ C aBb œ 1Î$ "$ =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
________________________________________________________________________
page 33
—————————————————————————— CHAPTER 2. ——
a, b .
a- b. Evidently, the solution is defined for " Ÿ B Ÿ " Þ
22. The differential equation can be written as a$C# %b.C œ $B# .B Þ Integrating both
sides, we obtain C$ %C œ B$ - Þ Imposing the initial condition, the specific solution
is C$ %C œ B$ " Þ Referring back to the differential equation, we find that C w p _ as
C p „#ÎÈ$ Þ The respective values of the abscissas are B œ "Þ#(' , "Þ&*) Þ
Hence the solution is valid for "Þ#(' B "Þ&*) Þ
24. Write the differential equation as a$ #Cb.C œ a# /B b.B Þ Integrating both sides,
we obtain $C C # œ #B /B - Þ Based on the specified initial condition, the solution
can be written as $C C # œ #B /B " Þ Completing the square, it follows that
CaBb œ $Î# È#B /B "$Î% Þ The solution is defined if #B /B "$Î% ! ,
that is, "Þ& Ÿ B Ÿ # aapproximatelyb. In that interval, C w œ ! , for B œ 68 # Þ It can
be verified that C ww a68 #b ! . In fact, C ww aBb ! on the interval of definition. Hence
the solution attains a global maximum at B œ 68 # Þ
26. The differential equation can be written as a" C# b" .C œ #a" Bb.B Þ Integrating
both sides of the equation, we obtain +<->+8C œ #B B# - Þ Imposing the given
initial
condition, the specific solution is +<->+8C œ #B B# Þ Therefore, C aBb œ >+8a#B B# bÞ
Observe that the solution is defined as long as 1Î# #B B# 1Î# Þ It is easy to
see that #B B#
"Þ Furthermore, #B B# œ 1Î# for B œ #Þ' and !Þ' Þ Hence
the solution is valid on the interval #Þ' B !Þ' Þ Referring back to the differential
________________________________________________________________________
page 34
—————————————————————————— CHAPTER 2. ——
equation, the solution is stationary at B œ "Þ Since C ww aBb ! on the entire interval
of
definition, the solution attains a global minimum at B œ " Þ
28a+b. Write the differential equation as C" a% Cb" .C œ >a" >b" .> Þ Integrating
both sides of the equation, we obtain 68 kCk 68kC %k œ %> %68k" >k - Þ Taking
the exponential of both sides, it follows that kCÎaC %bk œ G /%> Îa" >b% Þ It follows
that as > p _ , kCÎaC %bk œ k" %ÎaC %bkp _ . That is, C a>b p % Þ
a,bÞ Setting Ca!b œ # , we obtain that G œ " . Based on the initial condition, the solution
may be expressed as CÎaC %b œ /%> Îa" >b% Þ Note that CÎaC %b ! , for all
> !Þ Hence C % for all > !Þ Referring back to the differential equation, it follows
that C w is always positive. This means that the solution is monotone increasing. We find
that the root of the equation /%> Îa" >b% œ $** is near > œ #Þ)%% Þ
a- bÞ Note the Ca>b œ % is an equilibrium solution. Examining the local direction field,
we see that if Ca!b ! , then the corresponding solutions converge to C œ % . Referring
back to part a+b, we have CÎaC %b œ cC! ÎaC! %bd/%> Îa" >b% , for C! Á % Þ Setting
%
> œ # , we obtain C! ÎaC! %b œ a$Î/# b C a#bÎaC a#b %bÞ Now since the function
0 aCb œ CÎaC %b is monotone for C % and C % , we need only solve the equations
C! ÎaC! %b œ $**a$Î/# b% and C! ÎaC! %b œ %!"a$Î/# b% Þ The respective solutions
are C! œ $Þ''## and C! œ %Þ%!%# Þ
30a0 bÞ
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—————————————————————————— CHAPTER 2. ——
31a- b
"
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
33a- bÞ
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—————————————————————————— 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
________________________________________________________________________
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—————————————————————————— CHAPTER 2. ——
a- bÞ
"
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 $# BC #" ˆ BC ‰ . Hence the
equation is homogeneous. The substitution C œ B @ results in B @ w œ a@# "bÎ#@, that
is, @##@" .@ œ B" .B Þ
a,bÞ Integrating both sides of the transformed equation yields 68k@# "k œ 68kBk - ,
that is, @# " œ G kBkÞ In terms of the original dependent variable, the general solution
is C# œ G B# kBk B# Þ
________________________________________________________________________
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—————————————————————————— CHAPTER 2. ——
a- bÞ
________________________________________________________________________
page 39
—————————————————————————— CHAPTER 2. ——
Section 2.3
5a+b. Let U be the amount of salt in the tank. Salt enters the tank of water at a rate of
# "% ˆ" "# =38 >‰ œ "# "% =38 > 9DÎ738 Þ It leaves the tank at a rate of # UÎ"!! 9DÎ738Þ
Hence the differential equation governing the amount of salt at any time is
.U
" "
œ =38 > UÎ&! Þ
.>
# %
The initial amount of salt is U! œ &! 9D Þ The governing ODE is linear, with integrating
w
factor .a>b œ />Î&! Þ Write the equation as ˆ/>Î&! U‰ œ />Î&! ˆ "# "% =38 >‰Þ The
specific solution is Ua>b œ #& "#Þ&=38 > '#&-9= > '$"&! />Î&! ‘Î#&!" 9D Þ
a, bÞ
a- bÞ The amount of salt approaches a steady state, which is an oscillation of amplitude
"Î% about a level of #& 9D Þ
6a+b. The equation governing the value of the investment is .WÎ.> œ < W . The value of
the investment, at any time, is given by W a>b œ W! /<> Þ Setting W aX b œ #W! , the required
time is X œ 68a#bÎ< Þ
a,bÞ 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
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page 40
—————————————————————————— CHAPTER 2. ——
a- b .
a. bÞ The two balances can never be equal.
11a+b. Let W be the value of the mortgage. The debt accumulates at a rate of <W , in
which < œ Þ!* is the annual interest rate. Monthly payments of $ )!! are equivalent to
$ *ß '!! per yearÞ The differential equation governing the value of the mortgage is
.WÎ.> œ Þ!* W *ß '!! Þ Given that W! is the original amount borrowed, the debt is
W a>b œ W! /Þ!*> "!'ß ''(a/Þ!*> "bÞ Setting W a$!b œ ! , it follows that
W! œ $ **ß &!! .
a,bÞ The total payment, over $! years, becomes $ #))ß !!! . The interest paid on this
purchase is $ "))ß &!! .
13a+b. The balance increases at a rate of < W $/yr, and decreases at a constant rate of 5
$ per year. Hence the balance is modeled by the differential equation .WÎ.> œ <W 5 .
The balance at any time is given by W a>b œ W! /<> 5< a/<> "bÞ
a,b. The solution may also be expressed as W a>b œ ÐW! 5< Ñ/<> 5< Þ Note that if the
withdrawal rate is 5! œ < W! , the balance will remain at a constant level W! Þ
5
a- b. Assuming that 5 5! , W aX! b œ ! for X! œ "< 68’ 55
“Þ
!
a. b. If < œ Þ!) and 5 œ #5! , then X! œ )Þ'' years .
a/b. Setting W a>b œ ! and solving for /<> in Parta,b, /<> œ
results in 5 œ <W! /<X Îa/<X "bÞ
5
5<W! Þ
Now setting > œ X
a0 bÞ In parta/b, let 5 œ "#ß !!! , < œ Þ!) , and X œ #! . The required investment
becomes W! œ $ ""*ß ("& .
14a+bÞ Let U w œ < U Þ The general solution is Ua>b œ U! /<> Þ Based on the
definition of half-life, consider the equation U! Î# œ U! /&($! < Þ It follows that
________________________________________________________________________
page 41
—————————————————————————— CHAPTER 2. ——
&($! < œ 68Ð"Î#Ñ, that is, < œ "Þ#!*( ‚ "!% per year.
a,b. Hence the amount of carbon-14 is given by Ua>b œ U! /"Þ#!*(‚"! > Þ
%
a- bÞ Given that UaX b œ U! Î& , we have the equation "Î& œ /"Þ#!*(‚"! X Þ Solving for
the decay time, the apparent age of the remains is approximately X œ "$ß $!%Þ'& years .
%
15. Let T a>b be the population of mosquitoes at any time > . The rate of increase of the
mosquito population is <T Þ The population decreases by #!ß !!! per day. Hence the
equation that models the population is given by .T Î.> œ <T #!ß !!! . Note that the
<>
variable > represents days . The solution is T a>b œ T! /<> #!ß!!!
< a/ "bÞ In the
absence of predators, the governing equation is .T" Î.> œ <T" , with solution
T" a>b œ T! /<> Þ Based on the data, set T" a(b œ #T! , that is, #T! œ T! /(< Þ The growth
rate is determined as < œ 68a#bÎ( œ Þ!**!# per dayÞ Therefore the population,
including the predation by birds, is T a>b œ # ‚ "!& /Þ!**> #!"ß **(a/Þ!**> "b œ
œ #!"ß **(Þ$ "*((Þ$ /Þ!**> Þ
16a+b. Ca>b œ /B:c#Î"! >Î"! #-9=Ð>ÑÎ"!dÞ The doubling-time is 7 ¸ #Þ*'$# Þ
a,bÞ The differential equation is .CÎ.> œ CÎ"! , with solution C a>b œ C a!b/>Î"! Þ The
doubling-time is given by 7 œ "!68a#b ¸ 'Þ*$"& Þ
a- b. Consider the differential equation .CÎ.> œ a!Þ& =38Ð#1>Ñb CÎ& Þ The equation is
separable, with C" .C œ ˆ!Þ" &" =38Ð#1>щ.> Þ Integrating both sides, with respect to the
appropriate variable, we obtain 68 C œ a1> -9=Ð#1>ÑbÎ"!1 - Þ Invoking the initial
condition, the solution is Ca>b œ /B:ca" 1> -9=Ð#1>ÑbÎ"!1dÞ The doubling-time is
7 ¸ 'Þ$)!% Þ The doubling-time approaches the value found in parta,b.
a. b .
17a+b. The differential equation .CÎ.> œ <a>b C 5 is linear, with integrating factor
.a>b œ /B: ' <a>b.>‘Þ Write the equation as a. C bw œ 5 .a>b Þ Integration of both
________________________________________________________________________
page 42
—————————————————————————— CHAPTER 2. ——
sides yields the general solution C œ 5 ' .a7 b. 7 C! .a!b‘Î.a>b . In this problem,
the
integrating factor is .a>b œ /B:ca-9= > >bÎ&dÞ
a,bÞ The population becomes extinct, if Ca>‡ b œ ! , for some > œ >‡ . Referring to
parta+b,
we find that Ca>‡ b œ ! Ê
(
>‡
!
/B:ca-9= 7 7 bÎ&d. 7 œ & /"Î& C- Þ
It can be shown that the integral on the left hand side increases monotonically, from zero
to a limiting value of approximately &Þ!)*$ . Hence extinction can happen only if
& /"Î& C- &Þ!)*$ , that is, C- !Þ)$$$ Þ
a- b. Repeating the argument in parta,b, it follows that Ca>‡ b œ ! Ê
(
!
>‡
/B:ca-9= 7 7 bÎ&d. 7 œ
" "Î&
/ C- Þ
5
Hence extinction can happen only if /"Î& C- Î5 &Þ!)*$ , that is, C- %Þ"''( 5 Þ
a. b. Evidently, C- is a linear function of the parameter 5 .
19a+b. Let Ua>b be the volume of carbon monoxide in the room. The rate of increase of
CO is aÞ!%ba!Þ"b œ !Þ!!% 0 >$ Î738 Þ The amount of CO leaves the room at a rate of
a!Þ"bUa>bÎ"#!! œ Ua>bÎ"#!!! 0 >$ Î738 Þ Hence the total rate of change is given by
the differential equation .UÎ.> œ !Þ!!% Ua>bÎ"#!!! Þ This equation is linear and
separable, with solution Ua>b œ %) %) /B:a >Î"#!!!b 0 >$ Þ Note that U! œ ! 0 >$ Þ
Hence the concentration at any time is given by Ba>b œ Ua>bÎ"#!! œ Ua>bÎ"# % .
a,b. The concentration of CO in the room is Ba>b œ % %/B:a >Î"#!!!b %Þ A level
of !Þ!!!"# corresponds to !Þ!"# % . Setting Ba7 b œ !Þ!"# , the solution of the equation
% %/B:a >Î"#!!!b œ !Þ!"# is 7 ¸ $' minutes .
________________________________________________________________________
page 43
—————————————————————————— CHAPTER 2. ——
20a+bÞ The concentration is - a>b œ 5 T Î< a-! 5 T Î<b/<>ÎZ Þ It is easy to see
that - a>p_b œ 5 T Î< Þ
a,bÞ - a>b œ -! /<>ÎZ . The reduction times are X&! œ 68Ð#ÑZ Î< and X"! œ 68Ð"!ÑZ Î<Þ
a- bÞ The reduction times, in years, are XW œ 68a"!ba'&Þ#bÎ"#ß #!! œ %$!Þ)&
XQ œ 68a"!ba"&)bÎ%ß *!! œ ("Þ% à XI œ 68a"!ba"(&bÎ%'! œ 'Þ!&
XS œ 68a"!ba#!*bÎ"'ß !!! œ "(Þ'$ Þ
21a- bÞ
22a+b. The differential equation for the motion is 7 .@Î.> œ @Î$! 71 Þ Given the
initial condition @a!b œ #! m/s , the solution is @a>b œ %%Þ" '%Þ" /B:a >Î%Þ&b .
Setting @a>" b œ ! , the ball reaches the maximum height at >" œ "Þ')$ sec . Integrating
@a>b , the position is given by Ba>b œ $")Þ%& %%Þ" > #))Þ%& /B:a >Î%Þ&bÞ Hence
the maximum height is Ba>" b œ %&Þ() m .
a,bÞ Setting Ba># b œ ! , the ball hits the ground at ># œ &Þ"#) sec .
a- bÞ
23a+bÞ The differential equation for the upward motion is 7 .@Î.> œ . @# 71 ,
in which . œ "Î"$#& . This equation is separable, with . @#7
71 .@ œ .> Þ Integrating
________________________________________________________________________
page 44
—————————————————————————— CHAPTER 2. ——
both sides and invoking the initial condition, @a>b œ %%Þ"$$ >+8aÞ%#& Þ### >bÞ Setting
@a>" b œ ! , the ball reaches the maximum height at >" œ "Þ*"' sec . Integrating @a>b , the
position is given by Ba>b œ "*)Þ(& 68c-9=a!Þ### > !Þ%#&bd %)Þ&( Þ Therefore the
maximum height is Ba>" b œ %)Þ&' m .
a,bÞ The differential equation for the downward motion is 7 .@Î.> œ .@# 71 Þ
This equation is also separable, with 717. @# .@ œ .> Þ For convenience, set > œ ! at
the top of the trajectory. The new initial condition becomes @a!b œ ! . Integrating both
sides and invoking the initial condition, we obtain 68ca%%Þ"$ @bÎa%%Þ"$ @bd œ >Î#Þ#&
Þ
Solving for the velocity, @a>b œ %%Þ"$a" />Î#Þ#& bÎa" />Î#Þ#& b Þ Integrating @a>b , the
#
position is given by Ba>b œ **Þ#* 68’/>Î#Þ#& Îa" />Î#Þ#& b “ ")'Þ# Þ To estimate the
duration of the downward motion, set Ba># b œ ! , resulting in ># œ $Þ#(' sec . Hence the
total time that the ball remains in the air is >1 ># œ &Þ"*# sec .
a- bÞ
24a+bÞ Measure the positive direction of motion downward . Based on Newton's #nd
law,
the equation of motion is given by
7
.@
!Þ(& @ 71 , ! > "!
œœ
"# @ 71
, > "!
.>
Þ
Note that gravity acts in the positive direction, and the drag force is resistive. During the
first ten seconds of fall, the initial value problem is .@Î.> œ @Î(Þ& $# , with initial
velocity @a!b œ ! fps Þ This differential equation is separable and linear, with solution
@a>b œ #%!a" />Î(Þ& b. Hence @a"!b œ "('Þ( fps Þ
a,b. Integrating the velocity, with Ba>b œ ! , the distance fallen is given by
Hence Ba"!b œ "!(%Þ& ft .
Ba>b œ #%! > ")!! />Î(Þ& ")!! .
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page 45
—————————————————————————— CHAPTER 2. ——
a- bÞ For computational purposes, reset time to > œ ! . For the remainder of the motion,
the initial value problem is .@Î.> œ $#@Î"& $# , with specified initial velocity
@a!b œ "('Þ( fps Þ The solution is given by @a>b œ "& "'"Þ( /$# >Î"& Þ As > p _ ,
@a>b p @P œ "& fps Þ Integrating the velocity, with Ba!b œ "!(%Þ& , the distance fallen
after the parachute is open is given by Ba>b œ "& > (&Þ) /$# >Î"& ""&!Þ$ Þ To find the
duration of the second part of the motion, estimate the root of the transcendental equation
"& X (&Þ) /$# X Î"& ""&!Þ$ œ &!!! Þ The result is X œ #&'Þ' sec Þ
a. bÞ
25a+b. Measure the positive direction of motion upward . The equation of motion is
given by 7.@Î.> œ 5 @ 71 . The initial value problem is .@Î.> œ 5@Î7 1 ,
with @a!b œ @! . The solution is @a>b œ 71Î5 a@! 71Î5 b/5>Î7 Þ Setting
@a>7 b œ !, the maximum height is reached at time >7 œ a7Î5 b68ca71 5 @! bÎ71dÞ
Integrating the velocity, the position of the body is
Ba>b œ 71 >Î5 ”Š
7 #
7 @!
5>Î7
ÑÞ
‹ 1
•Ð" /
5
5
Hence the maximum height reached is
B7 œ Ba>7 b œ
7 @!
7 #
71 5 @!
1Š ‹ 68”
•Þ
5
5
71
a,bÞ Recall that for $ ¥ " , 68a" $ b œ $ "# $ # "$ $ $ "% $ % á
26a,b.
a- bÞ
lim 71a5 @!571b/
5Ä !
lim
7Ä!
71
5
5>Î7
œ lim
5Ä !
>
7 a5 @ !
71b/5>Î7 œ 1> Þ
‰ 5>Î7 ‘ œ ! , since
ˆ 71
5 @! /
lim /5>Î7 œ ! Þ
7Ä!
28a+b. In terms of displacement, the differential equation is 7@ .@Î.B œ 5 @ 71 Þ
.@ .B
.@
This follows from the chain rule : .@
.> œ .B .> œ @ .> . The differential equation is
separable, with
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—————————————————————————— CHAPTER 2. ——
Ba@b œ
7@ 7# 1 71 5 @
# 68º
ºÞ
5
5
71
The inverse exists, since both B and @ are monotone increasing. In terms of the given
parameters, Ba@b œ "Þ#& @ "&Þ$" 68k!Þ!)"' @ "kÞ
a,bÞ Ba"!b œ "$Þ%& meters . The required value is 5 œ !Þ#% .
a- bÞ In parta+b, set @ œ "! m/s and B œ "! meters .
29a+bÞ Let B represent the height above the earth's surface. The equation of motion is
Q7
given by 7 .@
.> œ K aVBb# , in which K is the universal gravitational constant. The
symbols Q and V are the mass and radius of the earth, respectively. By the chain rule,
7@
.@
Q7
œ K
.
.B
aV Bb#
This equation is separable, with @ .@ œ KQ aV Bb# .B Þ Integrating both sides,
and
invoking the initial condition @a!b œ È#1V , the solution is @# œ #KQ aV Bb"
#1V #KQ ÎV Þ From elementary physics, it follows that 1 œ KQ ÎV # . Therefore
@aBb œ È#1 ’VÎÈV B “Þ a Note that 1 œ ()ß &%& mi/hr# .b
a,bÞ We now consider .BÎ.> œ È#1 ’VÎÈV B “. This equation is also separable,
with ÈV B .B œ È#1 V .> Þ By definition of the variable B, the initial condition is
#Î$
Ba!b œ !Þ Integrating both sides, we obtain Ba>b œ $# ˆÈ#1 V > #$ V $Î# ‰‘ V Þ
Setting the distance BaX b V œ #%!ß !!! , and solving for X , the duration of such a
flight would be X ¸ %* hours .
32a+bÞ Both equations are linear and separable. The initial conditions are @a!b œ ? -9=
E
and Aa!b œ ? =38 E . The two solutions are @a>b œ ? -9= E /<> and Aa>b œ 1Î<
a? =38 E 1Î<b/<> Þ
________________________________________________________________________
page 47
—————————————————————————— CHAPTER 2. ——
a,bÞ Integrating the solutions in parta+b, and invoking the initial conditions, the
coordinates are Ba>b œ ?< -9= Ea" /<> b and
a- bÞ
?
Ca>b œ 1>Î< ˆ1 ?< =38 E 2<# ‰Î<# Š =38 E 1Î<# ‹/<> Þ
<
a. bÞ Let X be the time that it takes the ball to go $&! ft horizontally. Then from above,
/X Î& œ a? -9= E (!bÎ? -9= E Þ At the same time, the height of the ball is given by
CaX b œ "'! X #'( "#&?=38 E Ð)!! &? =38 EÑca? -9= E (!bÎ? -9= E dÞ
Hence E and ? must satisfy the inequality
)!!68”
? -9= E (!
• #'( "#&?=38 E Ð)!! &? =38 EÑca? -9= E (!bÎ? -9= E d
? -9= E
33a+bÞ Solving equation a3b, C w aBb œ ca5 # C bÎCd
chosen, since C is an increasing function of B .
"Î#
"! Þ
. The positive answer is
a,b. Let C œ 5 # =38# > . Then .C œ #5 # =38 > -9= > .> Þ Substituting into the equation in
parta+b, we find that
#5 # =38 > -9= > .>
-9= >
œ
Þ
.B
=38 >
Hence #5 # =38# > .> œ .B Þ
a- bÞ Letting ) œ #> , we further obtain 5 # =38# )# . ) œ .B Þ Integrating both sides of the
equation and noting that > œ ) œ ! corresponds to the origin, we obtain the solutions
Ba) b œ 5 # a) =38 )bÎ# and cfrom parta,bd Ca)b œ 5 # a" -9= )bÎ# Þ
a. b. Note that CÎB œ a" -9= )bÎa) =38 )bÞ Setting B œ " , C œ # , the solution of
the equation a" -9= ) bÎa) =38 )b œ # is ) ¸ "Þ%!" . Substitution into either of the
expressions yields 5 ¸ #Þ"*$ Þ
________________________________________________________________________
page 48
—————————————————————————— CHAPTER 2. ——
Section 2.4
2. Considering the roots of the coefficient of the leading term, the ODE has unique
solutions on intervals not containing ! or % . Since # − a! ß %b , the initial value problem
has a unique solution on the interval a! ß %b Þ
3. The function >+8 > is discontinuous at odd multiples of 1# . Since
initial value problem has a unique solution on the interval ˆ 1# ß $#1 ‰Þ
1
#
1
$1
#
, the
5. :a>b œ #>Îa% ># b and 1a>b œ $># Îa% ># b. These functions are discontinuous at
B œ „# . The initial value problem has a unique solution on the interval a # ß #bÞ
6. The function 68 > is defined and continuous on the interval a! ß _b . Therefore the
initial value problem has a unique solution on the interval a! ß _b.
7. The function 0 a> ß C b is continuous everywhere on the plane, except along the straight
line C œ #>Î& Þ The partial derivative `0 Î`C œ (>Îa#> &C b# has the same
region of continuity.
9. The function 0 a> ß C b is discontinuous along the coordinate axes, and on the hyperbola
># C # œ " . Furthermore,
`0
„"
C 68k>C k
œ
#
#
#
`C
C a" > C b
a " ># C # b #
has the same points of discontinuity.
10. 0 a> ß C b is continuous everywhere on the plane. The partial derivative `0 Î`C is also
continuous everywhere.
12. The function 0 a> ß C b is discontinuous along the lines > œ „5 1 and C œ " . The
partial derivative `0 Î`C œ -9>a>bÎa" C b# has the same region of continuity.
14. The equation is separable, with .CÎC # œ #> .> Þ Integrating both sides, the solution
is given by Ca>b œ C! ÎÐ" C! ># ÑÞ For C! ! , solutions exist as long as ># "ÎC! .
For C! Ÿ ! , solutions are defined for all > .
15. The equation is separable, with .CÎC $ œ .> Þ Integrating both sides and invoking
the initial condition, Ca>b œ C! ÎÈ#C! > " Þ Solutions exist as long as #C! > " ! ,
that is, #C! > " . If C! ! , solutions exist for > "Î#C! . If C! œ ! , then the
solution Ca>b œ ! exists for all > . If C! ! , solutions exist for > "Î#C! .
16. The function 0 a> ß C b is discontinuous along the straight lines > œ " and C œ ! .
The partial derivative `0 Î`C is discontinuous along the same lines. The equation is
________________________________________________________________________
page 49
—————————————————————————— CHAPTER 2. ——
separable, with C .C œ ># .>Îa" >$ bÞ Integrating and invoking the initial condition, the
"Î#
solution is Ca>b œ #$ 68k" >$ k C!# ‘ Þ Solutions exist as long as
#
68¸" >$ ¸ C!#
$
!,
that is, C!#
#$ 68k" >$ kÞ For all C! Ðit can be verified that C! œ ! yields a valid
solution, even though Theorem #Þ%Þ# does not guarantee oneÑ , solutions exists as long as
k" >$ k /B:a $C!# Î#bÞ From above, we must have > " . Hence the inequality
may be written as >$ /B:a $C!# Î#b " Þ It follows that the solutions are valid for
"Î$
c/B:a $C!# Î#b "d > _ .
17.
18.
Based on the direction field, and the differential equation, for C! ! , the slopes
eventually become negative, and hence solutions tend to _ . For C! ! , solutions
increase without bound if >! ! Þ Otherwise, the slopes eventually become negative, and
solutions tend to zero . Furthermore, C! œ ! is an equilibrium solution. Note that slopes
are zero along the curves C œ ! and >C œ $ .
19.
________________________________________________________________________
page 50
—————————————————————————— CHAPTER 2. ——
For initial conditions a>! ß C! b satisfying >C $ , the respective solutions all tend to zero .
Solutions with initial conditions above or below the hyperbola >C œ $ eventually tend to
„_ . Also, C! œ ! is an equilibrium solution.
20.
Solutions with >! ! all tend to _ . Solutions with initial conditions a>! ß C! b to the
right of the parabola > œ " C # asymptotically approach the parabola as > p _ . Integral
curves with initial conditions above the parabola aand C! !b also approach the curve.
The slopes for solutions with initial conditions below the parabola aand C! !b are all
negative. These solutions tend to _ Þ
21. Define C- a>b œ #$ a> - b$Î# ?a> - b, in which ?a>b is the Heaviside step function.
Note that C- a- b œ C- a!b œ ! and C- ˆ- a$Î#b#Î$ ‰ œ "Þ
a+bÞ Let - œ " a$Î#b#Î$ Þ
a,bÞ Let - œ # a$Î#b#Î$ Þ
a- bÞ Observe that C! a#b œ #$ a#b$Î# , C- a>b #$ a#b$Î# for ! - #, and that C- a#b œ ! for
- #Þ So for any - !, „C- a#b − c #ß #dÞ
26a+bÞ Recalling Eq. a$&b in Section #Þ",
________________________________________________________________________
page 51
—————————————————————————— CHAPTER 2. ——
Cœ
It is evident that C" a>b œ
"
.a>b
"
Þ
( .a=b1a=b .=
.a>b
.a>b
and C# a>b œ
" '
.a>b
.a=b1a=b .=.
a,b. By definition, .a">b œ /B:ˆ ' :a>b.>‰. Hence C"w œ :a>b
That is, C"w :a>bC" œ !Þ
a- bÞ C#w œ Š :a>b
"
'>
.a>b ‹ ! .a=b1a=b .=
That is, C#w :a>bC# œ 1a>bÞ
"
.a>b
œ :a>bC" Þ
Š .a">b ‹.a>b1a>b œ :a>bC# 1a>bÞ
$
.C
C .@
$ .C
30. Since 8 œ $, set @ œ C # . It follows that .@
.> œ #C .> and .> œ # .> Þ
$
$
Substitution into the differential equation yields C# .@
.> &C œ 5 C , which further
results in @ w #&@ œ #5 Þ The latter differential equation is linear, and can be written as
w
a/#&> b œ #5 Þ The solution is given by @a>b œ #5> /#&> -/#&> Þ Converting back to
the original dependent variable, C œ „@"Î# Þ
$
.C
C .@
$ .C
31. Since 8 œ $, set @ œ C # . It follows that .@
.> œ #C .> and .> œ # .> Þ
$
$
The differential equation is written as C# .@
.> a>-9= > X bC œ 5 C , which upon
further substitution is @ w #a>-9= > X b@ œ #Þ This ODE is linear, with integrating
factor .a>b œ /B:ˆ#' a>-9= > X b.>‰ œ /B:a #>=38 > #X >bÞ The solution is
@a>b œ #/B:a#>=38 > #X >b( /B:a #>=38 7 #X 7 b. 7 - /B:a #>=38 > #X >bÞ
>
!
Converting back to the original dependent variable, C œ „@"Î# Þ
33. The solution of the initial value problem C"w #C" œ !, C" a!b œ " is C" a>b œ /#> Þ
Therefore ya" b œ C" a"b œ /# Þ On the interval a"ß _bß the differential equation is
C#w C# œ !, with C# a>b œ -/> Þ Therefore C a" b œ C# a"b œ -/" Þ Equating the limits
Ca" b œ Ca" b, we require that - œ /" Þ Hence the global solution of the initial value
problem is
Ca>b œ œ
/#> , ! Ÿ > Ÿ "
Þ
/"> , > "
Note the discontinuity of the derivative
C a >b œ œ
#/#> , ! > "
Þ
/"> , > "
________________________________________________________________________
page 52
—————————————————————————— CHAPTER 2. ——
Section 2.5
1.
For C! ! , the only equilibrium point is C ‡ œ ! . 0 w a!b œ + ! , hence the equilibrium
solution 9a>b œ ! is unstable.
2.
The equilibrium points are C‡ œ +Î, and C ‡ œ ! . 0 w a +Î, b ! , therefore the
equilibrium solution 9a>b œ +Î, is asymptotically stable.
3.
________________________________________________________________________
page 53
—————————————————————————— CHAPTER 2. ——
4.
The only equilibrium point is C‡ œ ! . 0 w a!b ! , hence the equilibrium solution
9a>b œ !
is unstable.
5.
The only equilibrium point is C‡ œ ! . 0 w a!b ! , hence the equilibrium solution
9a>b œ !
is +=C7:>9>3-+66C stable.
6.
________________________________________________________________________
page 54
—————————————————————————— CHAPTER 2. ——
7a,b.
8.
The only equilibrium point is C‡ œ " . Note that 0 w a"b œ ! , and that C w ! for C Á " .
As long as C! Á " , the corresponding solution is monotone decreasing. Hence the
equilibrium solution 9a>b œ " is semistable.
9.
________________________________________________________________________
page 55
—————————————————————————— CHAPTER 2. ——
10.
The equilibrium points are C‡ œ ! , „" . 0 w aC b œ " $C # . The equilibrium solution
9a>b œ ! is unstable, and the remaining two are asymptotically stable.
11.
12.
The equilibrium points are C‡ œ ! , „# . 0 w aC b œ )C %C $ . The equilibrium solutions
9a>b œ # and 9a>b œ # are unstable and asymptotically stable, respectively. The
equilibrium solution 9a>b œ ! is semistable.
________________________________________________________________________
page 56
—————————————————————————— CHAPTER 2. ——
13.
The equilibrium points are C‡ œ ! and " . 0 w aC b œ #C 'C # %C $ . Both equilibrium
solutions are semistable.
15a+b. Inverting the Solution a""b, Eq. a"$b shows > as a function of the population C
and
the carrying capacity O . With C! œ OÎ$,
" a"Î$bc" aCÎO bd
> œ 68º
ºÞ
<
aCÎO bc" a"Î$bd
Setting C œ #C! ,
" a"Î$bc" a#Î$bd
7 œ 68º
ºÞ
<
a#Î$bc" a"Î$bd
That is, 7 œ "< 68 % Þ If < œ !Þ!#& per year, 7 œ &&Þ%& years.
a,bÞ In Eq. a"$b, set C! ÎO œ ! and CÎO œ " . As a result, we obtain
" ! c" " d
X œ 68º
ºÞ
<
" c" ! d
Given ! œ !Þ", " œ !Þ* and < œ !Þ!#& per year, 7 œ "(&Þ() years.
16a+b.
________________________________________________________________________
page 57
—————————————————————————— CHAPTER 2. ——
17. Consider the change of variable ? œ 68aCÎO bÞ Differentiating both sides with
respect
to >, ? w œ C w ÎCÞ Substitution into the Gompertz equation yields ? w œ <?, with
solution ? œ ?! /<> Þ It follows that 68aCÎO b œ 68aC! ÎO b/<> Þ That is,
C
œ /B:68aC! ÎO b/<> ‘Þ
O
a+b. Given O œ )!Þ& ‚ "!' , C! ÎO œ !Þ#& and < œ !Þ(" per year, C a#b œ &(Þ&) ‚ "!' .
a,bÞ Solving for >,
"
68aCÎO b
> œ 68”
•Þ
<
68aC! ÎO b
Setting Ca7 b œ !Þ(&O , the corresponding time is 7 œ #Þ#" years.
19a+b. The rate of increase of the volume is given by rate of flow in rate of flow out.
That is, .Z Î.> œ 5 !+È#12 Þ Since the cross section is constant, .Z Î.> œ E.2Î.>Þ
Hence the governing equation is .2Î.> œ ˆ5 !+È#12 ‰ÎEÞ
" ˆ 5 ‰#
a,bÞ Setting .2Î.> œ !, the equilibrium height is 2/ œ #1
!+ Þ Furthermore, since
w
0 a2/ b !, it follows that the equilibrium height is asymptotically stable.
a- b. Based on the answer in parta,b, the water level will intrinsically tend to approach 2/ .
Therefore the height of the tank must be greater than 2/ ; that is, 2/ Z ÎE.
22a+b. The equilibrium points are at C‡ œ ! and C ‡ œ ". Since 0 w aC b œ ! #!C , the
equilibrium solution 9 œ ! is unstable and the equilibrium solution 9 œ " is
asymptotically stable.
a,b. The ODE is separable, with cCa" C bd" .C œ ! .> . Integrating both sides and
invoking the initial condition, the solution is
Ca>b œ
It is evident that aindependent of C! b
23a+b. Ca>b œ C! /"> Þ
C! /!>
Þ
" C! C! /!>
lim Ca>b œ ! and
> Ä _
lim Ca>b œ " .
>Ä_
a,b. From parta+b, .BÎ.> œ ! B C! /"> Þ Separating variables, .BÎB œ ! C! /"> .> Þ
Integrating both sides, the solution is Ba>b œ B! /B:! C! Î" ˆ" /"> ‰‘Þ
a- b. As > p _ , Ca>b p ! and Ba>b p B! /B:a! C! Î" bÞ Over a long period of time, the
________________________________________________________________________
page 58
—————————————————————————— CHAPTER 2. ——
proportion of carriers vanishes. Therefore the proportion of the population that escapes
the epidemic is the proportion of susceptibles left at that time, B! /B:a! C! Î" bÞ
25a+b. Note that 0 aBb œ BcaV V- b + B# d, and 0 w aBb œ aV V- b $+ B# . So if
aV V- b ! , the only equilibrium point is B‡ œ ! . 0 w a!b ! , and hence the solution
9a>b œ ! is asymptotically stable.
a,b. If aV V- b ! , there are three equilibrium points B‡ œ ! , „ÈaV V- bÎ+ Þ
Now 0 w a!b !, and 0 w ˆ„ÈaV V- bÎ+ ‰ !. Hence the solution 9 œ ! is unstable,
and the solutions 9 œ „ÈaV V- bÎ+ are asymptotically stable.
a- b .
________________________________________________________________________
page 59
—————————————————————————— CHAPTER 2. ——
Section 2.6
1. Q aBß Cb œ #B $ and R aBß C b œ #C # . Since QC œ RB œ ! , the equation is
exact. Integrating Q with respect to B , while holding C constant, yields <aBß Cb œ
œ B# $B 2aC b . Now <C œ 2 w aC b , and equating with R results in the possible
function 2aCb œ C # #C . Hence <aBß C b œ B# $B C # #C , and the solution is
defined implicitly as B# $B C # #C œ - .
2. Q aBß Cb œ #B %C and R aBß C b œ #B #C . Note that QC Á RB , and hence the
differential equation is not exact.
4. First divide both sides by a#BC #bÞ We now have Q aBß C b œ C and R aBß C b œ B .
Since QC œ RB œ ! , the resulting equation is exact. Integrating Q with respect to B ,
while holding C constant, results in <aBß Cb œ BC 2aC b . Differentiating with respect
to C , <C œ B 2 w aC b . Setting <C œ R , we find that 2 w aC b œ ! , and hence 2 aC b œ !
is acceptable. Therefore the solution is defined implicitly as BC œ - . Note that if
BC " œ ! , the equation is trivially satisfied.
6. Write the given equation as a+B ,C b.B a,B -C b.C . Now Q aBß C b œ +B ,C
and R aBß C b œ ,B -C . Since QC Á RB , the differential equation is not exact.
8. Q aBß Cb œ /B =38 C $C and R aBß C b œ $B /B =38 C . Note that QC Á RB , and
hence the differential equation is not exact.
10. Q aBß Cb œ CÎB 'B and R aBß C b œ 68 B #. Since QC œ RB œ "ÎB, the given
equation is exact. Integrating R with respect to C , while holding B constant, results in
<aBß Cb œ C 68 B #C 2 aBb Þ Differentiating with respect to B, <B œ CÎB 2 w aBb.
Setting <B œ Q , we find that 2w aBb œ 'B , and hence 2aBb œ $B# . Therefore the
solution
is defined implicitly as $B# C 68 B #C œ - .
11. Q aBß Cb œ B 68 C BC and R aBß C b œ C 68 B BC . Note that QC Á RB , and hence
the differential equation is not exact.
13. Q aBß Cb œ #B C and R aBß C b œ #C B. Since QC œ RB œ ", the equation is
exact. Integrating Q with respect to B , while holding C constant, yields <aBß Cb œ
œ B# BC 2aC b. Now <C œ B 2 w aC b. Equating <C with R results in 2 w aC b œ #C ,
and hence 2aCb œ C # . Thus <aBß C b œ B# BC C # , and the solution is given implicitly
as B# BC C # œ - . Invoking the initial condition C a"b œ $ , the specific solution is
B# BC C # œ (. The explicit form of the solution is CaBb œ "# ’B È#) $B# “Þ
Hence the solution is valid as long as $B# Ÿ #) Þ
16. Q aBß Cb œ C /#BC B and R aBß C b œ ,B /#BC . Note that QC œ /#BC #BC /#BC ,
and RB œ , /#BC #,BC /#BC Þ The given equation is exact, as long as , œ " . Integrating
________________________________________________________________________
page 60
—————————————————————————— CHAPTER 2. ——
R with respect to C , while holding B constant, results in <aBß Cb œ /#BC Î# 2aBb Þ Now
differentiating with respect to B, <B œ C /#BC 2w aBb. Setting <B œ Q , we find that
2w aBb œ B , and hence 2aBb œ B# Î# . Conclude that <aBß C b œ /#BC Î# B# Î# . Hence
the solution is given implicitly as /#BC B# œ - .
17. Integrating <C œ R , while holding B constant, yields
<aBß Cb œ ' R aBß C b.C 2 aBbÞ
`
Taking the partial derivative with respect to B , <B œ ' `B
R aBß C b.C 2 w aBbÞ Now
`
set <B œ Q aBß C b and therefore 2w aBb œ Q aBß C b ' `B
R aBß C b.C . Based on the fact
`
w
that QC œ RB , it follows that `C c2 aBbd œ ! . Hence the expression for 2w aBb can be
integrated to obtain
2aBb œ ( Q aBß C b.B ( ”(
18. Observe that
`
`C cQ aBbd
œ
`
`B cR aC bd
`
R aBß C b.C •.B .
`B
œ !Þ
20. QC œ C " -9= C C # =38 C and RB œ # /B a-9= B =38 BbÎC . Multiplying
both sides by the integrating factor .aBß Cb œ C /B , the given equation can be written as
a/B =38 C #C =38 Bb.B a/B -9= C #-9= Bb.C œ ! . Let Q œ .Q and R œ .R .
Observe that Q C œ R B , and hence the latter ODE is exact. Integrating R with respect
to C , while holding B constant, results in <aBß Cb œ /B =38 C #C -9= B 2 aBb Þ Now
differentiating with respect to B, <B œ /B =38 C #C =38 B 2 w aBb. Setting <B œ Q ,
we find that 2w aBb œ ! , and hence 2aBb œ ! is feasible. Hence the solution of the given
equation is defined implicitly by /B =38 C #C -9= B œ " .
21. QC œ " and RB œ # . Multiply both sides by the integrating factor .aBß C b œ C to
obtain C# .B a#BC C # /C b.C œ !. Let Q œ CQ and R œ CR . It is easy to see that
Q C œ R B , and hence the latter ODE is exact. Integrating Q with respect to B yields
<aBß Cb œ BC # 2aC b . Equating <C with R results in 2 w aC b œ C # /C , and hence
2aCb œ /C aC # #C #b. Thus <aBß C b œ BC # /C aC # #C #b, and the solution
is defined implicitly by BC# /C aC # #C #b œ - .
24. The equation .Q .R C w œ ! has an integrating factor if a.Q bC œ a.R bB , that is,
.C Q .B R œ .RB .QC . Suppose that RB QC œ V aBQ CR b, in which V is
some function depending only on the quantity D œ BC . It follows that the modified form
of the equation is exact, if .C Q .B R œ .V aBQ CR b œ V a. BQ . CR b. This
relation is satisfied if .C œ a. BbV and .B œ a. CbV . Now consider . œ .aBCbÞ Then
the partial derivatives are .B œ .w C and .C œ .w B . Note that .w œ . .Î.D . Thus . must
satisfy .w aD b œ V aD bÞ The latter equation is separable, with . . œ V aD b.D , and
.aD b œ ' V aD b.D Þ Therefore, given V œ V aBC b, it is possible to determine . œ .aBC b
which becomes an integrating factor of the differential equation.
________________________________________________________________________
page 61
—————————————————————————— CHAPTER 2. ——
28. The equation is not exact, since RB QC œ #C " . However, aRB QC bÎQ œ
œ a#C "bÎC is a function of C alone. Hence there exists . œ .aC b , which is a solution
of the differential equation .w œ a# "ÎCb. . The latter equation is separable, with
. .Î. œ # "ÎC . One solution is .aC b œ /B:a#C 68 C b œ /#C ÎC . Now rewrite the
given ODE as /#C .B a#B /#C "ÎC b.C œ ! . This equation is exact, and it is easy to
see that <aBß Cb œ B /#C 68 C . Therefore the solution of the given equation is defined
implicitly by B /#C 68 C œ - .
30. The given equation is not exact, since RB QC œ )B$ ÎC $ 'ÎC # . But note that
aRB QC bÎQ œ #ÎC is a function of C alone, and hence there is an integrating factor
. œ .aCb. Solving the equation .w œ a#ÎCb. , an integrating factor is .aCb œ C # Þ Now
rewrite the differential equation as a%B$ $Cb.B a$B %C $ b.C œ !. By inspection,
<aBß Cb œ B% $BC C % , and the solution of the given equation is defined implicitly by
B% $BC C % œ - .
32. Multiplying both sides of the ODE by . œ cBC a#B C bd" , the given equation is
equivalent to ca$B CbÎa#B# BC bd.B caB C bÎa#BC C # bd.C œ ! Þ Rewrite
the differential equation as
”
#
#
"
"
•.B ”
•.C œ ! .
B #B C
C #B C
It is easy to see that QC œ RB Þ Integrating Q with respect to B, while keeping C
constant, results in <aBß Cb œ #68kBk 68k#B C k 2aC b . Now taking the partial
derivative with respect to C , <C œ a#B C b" 2 w aC b . Setting <C œ R , we find that
2 w aCb œ "ÎC , and hence 2aC b œ 68 kC k . Therefore
<aBß Cb œ #68kBk 68k#B C k 68 kC k ,
and the solution of the given equation is defined implicitly by #B$ C B# C # œ - .
________________________________________________________________________
page 62
—————————————————————————— CHAPTER 2. ——
Section 2.7
2a+b. The Euler formula is C8" œ C8 2a#C8 "b œ a" #2 bC8 2 .
a. bÞ The differential equation is linear, with solution Ca>b œ a" /#> bÎ# Þ
4a+b. The Euler formula is C8" œ a" #2bC8 $2 -9= >8 .
a. b. The exact solution is Ca>b œ a'-9= > $=38 > ' /#> bÎ& .
5.
All solutions seem to converge to 9a>b œ #&Î* .
6.
Solutions with positive initial conditions seem to converge to a specific function. On the
other hand, solutions with negative coefficients decrease without bound. 9a>b œ ! is an
equilibrium solution.
7.
________________________________________________________________________
page 63
—————————————————————————— CHAPTER 2. ——
All solutions seem to converge to a specific function.
8.
Solutions with initial conditions to the 'left' of the curve > œ !Þ"C # seem to diverge. On
the other hand, solutions to the 'right' of the curve seem to converge to zero. Also, 9a>b
is an equilibrium solution.
9.
All solutions seem to diverge.
10.
________________________________________________________________________
page 64
—————————————————————————— CHAPTER 2. ——
Solutions with positive initial conditions increase without bound. Solutions with
negative
initial conditions decrease without bound. Note that 9a>b œ ! is an equilibrium solution.
11. The Euler formula is C8" œ C8 $2ÈC8 &2 . The initial value is C! œ # .
12. The iteration formula is C8" œ a" $2bC8 2 >8 C8# . a>! ß C! b œ a! ß !Þ&bÞ
14. The iteration formula is C8" œ a" 2 >8 bC8 2 C8$ Î"! Þ a>! ß C! b œ a! ß "bÞ
17. The Euler formula is
C8" œ C8
The initial point is a>! ß C! b œ a" ß #bÞ
2aC8# #>8 C8 b
.
$ >#8
18a+b. See Problem 8.
19a+bÞ
a,b. The iteration formula is C8" œ C8 2 C8# 2 >8# . The critical value of ! appears
to be near !! ¸ !Þ')"& . For C! !! , the iterations diverge.
________________________________________________________________________
page 65
—————————————————————————— CHAPTER 2. ——
20a+b. The ODE is linear, with general solution Ca>b œ > - /> . Invoking the specified
initial condition, Ca>! b œ C! , we have C! œ >! - />! Þ Hence - œ aC! >! b/>! Þ Thus
the solution is given by 9a>b œ aC! >! b/>>! > .
a,b. The Euler formula is C8" œ a" 2bC8 2 2 >8 . Now set 5 œ 8 " .
a- b. We have C" œ a" 2bC! 2 2 >! œ a" 2 bC! a>" >! b 2 >! . Rearranging
the terms, C" œ a" 2baC! >! b >" . Now suppose that C5 œ a" 2b5 aC! >! b >5 ,
for some 5 " . Then C5" œ a" 2bC5 2 2 >5 . Substituting for C5 , we find that
C5" œ a" 2b5" aC! >! b a" 2b >5 2 2 >5 œ a" 2 b5" aC! >! b >5 2 .
Noting that >5" œ >5 5 , the result is verified.
a. b. Substituting 2 œ a> >! bÎ8 , with >8 œ > ,
C8 œ Œ"
> >! 8
aC! >! b > .
8
Taking the limit of both sides, as 8 p _ , and using the fact that lim a" +Î8b8 œ /+ ,
8Ä_
pointwise convergence is proved.
21. The exact solution is 9a>b œ /> . The Euler formula is C8" œ a" 2bC8 . It is easy
to see that C8 œ a" 2b8 C! œ a" 2b8 Þ Given > ! , set 2 œ >Î8 . Taking the limit,
we find that lim C8 œ lim a" >Î8b8 œ /> Þ
8Ä_
8Ä_
23. The exact solution is 9a>b œ >Î# /#> . The Euler formula is C8" œ a" #2bC8
2Î# 2 >8 . Since C! œ " , C" œ a" #2 b 2Î# œ a" #2 b >" Î# Þ It is easy to
show by mathematical induction, that C8 œ a" #2b8 >8 Î# . For > ! , set 2 œ >Î8
and thus >8 œ > . Taking the limit, we find that lim C8 œ lim ca" #>Î8b8 >Î#d œ
8Ä_
œ /#> >Î# . Hence pointwise convergence is proved.
8Ä_
________________________________________________________________________
page 66
—————————————————————————— CHAPTER 2. ——
Section 2.8
2. Let D œ C $ and 7 œ > " . It follows that .DÎ. 7 œ a.DÎ.>ba.>Î. 7 b œ .DÎ.> .
Furthermore, .DÎ.> œ .CÎ.> œ " C $ . Hence .DÎ. 7 œ " aD $b$ . The new initial
condition is D a7 œ !b œ ! Þ
3. The approximating functions are defined recursively by 98" a>b œ '! #c98 a=b "d.= .
Setting 9! a>b œ ! , 9" a>b œ #> . Continuing, 9# a>b œ #># #> , 9$ a>b œ %$ >$ #># #> ,
9% a>b œ #$ >% %$ >$ #># #> , â . Given convergence, set
>
9a>b œ 9" a>b "c95" a>b 95 a>bd
_
5œ"
œ #> "
_
5œ#
+5 5
> .
5x
Comparing coefficients, +$ Î$x œ %Î$ , +% Î%x œ #Î$ , â . It follows that +$ œ ) ,
+% œ "',
and so on. We find that in general, that +5 œ #5 . Hence
9a>b œ "
_
5œ"
#>
#5 5
>
5x
œ / ".
5. The approximating functions are defined recursively by
98" a>b œ ( c 98 a=bÎ# =d.= .
>
!
Setting 9! a>b œ ! , 9" a>b œ ># Î# . Continuing, 9# a>b œ ># Î# >$ Î"# ,9$ a>b œ ># Î#
>$ Î"# >% Î*' , 9% a>b œ ># Î# >$ Î"# >% Î*' >& Î*'! , â . Given convergence, set
________________________________________________________________________
page 67
—————————————————————————— CHAPTER 2. ——
9a>b œ 9" a>b "c95" a>b 95 a>bd
_
5œ"
_
œ ># Î# "
5œ$
+5 5
> .
5x
Comparing coefficients, +$ Î$x œ "Î"# , +% Î%x œ "Î*' , +& Î&x œ "Î*'! , â . We
find that +$ œ "Î# , +% œ "Î%, +& œ "Î) , â Þ In general, +5 œ #5" . Hence
9a>b œ "
#5#
a >b 5
5
x
5œ#
_
œ % />Î# #> % .
6. The approximating functions are defined recursively by
98" a>b œ ( c98 a=b " =d.= .
>
!
Setting 9! a>b œ ! , 9" a>b œ > ># Î# , 92 a>b œ > >$ Î' , 9$ a>b œ > >% Î#% , 9% a>b œ
œ > >& Î"#! , â . Given convergence, set
9a>b œ 9" a>b "c95" a>b 95 a>bd
_
œ > > Î# ># Î# >$ Î'‘ >$ Î' >% Î#%‘ â
œ>!!â.
5œ"
#
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
98" a>b œ ( =# 98 a=b =‘.= .
>
!
Set 9! a>b œ !. The iterates are given by 9" a>b œ ># Î# , 92 a>b œ ># Î# >& Î"! ,
9$ a>b œ ># Î# >& Î"! >) Î)! , 9% a>b œ ># Î# >& Î"! >) Î)! >"" Î))! ,â .
Upon inspection, it becomes apparent that
8"
"
>$
>'
a >$ b
98 a>b œ > ”
â
•
# #†& #†&†)
# † & † )âc# $a8 "bd
#
œ ># "
8
a, b .
5œ"
a >$ b
5"
# † & † )âc# $a5 "bd
Þ
The iterates appear to be converging.
9a+b. The approximating functions are defined recursively by
98" a>b œ ( =# 98# a=b‘.= .
>
!
Set 9! a>b œ !. The first three iterates are given by 9" a>b œ >$ Î$ , 92 a>b œ >$ Î$ >( Î'$ ,
9$ a>b œ >$ Î$ >( Î'$ #>"" Î#!(* >"& Î&*&$& .
________________________________________________________________________
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—————————————————————————— CHAPTER 2. ——
a, b .
The iterates appear to be converging.
10a+b. The approximating functions are defined recursively by
98" a>b œ ( " 98$ a=b‘.= .
>
!
Set 9! a>b œ !. The first three iterates are given by 9" a>b œ > , 92 a>b œ > >% Î% ,
9$ a>b œ > >% Î% $>( Î#) $>"! Î"'! >"$ Î)$$ .
a, b .
The approximations appear to be diverging.
12a+b. The approximating functions are defined recursively by
98" a>b œ ( ”
>
!
Note that "Îa#C #b œ
above iteration formula by
"
#
$=# %= #
•.= .
#a98 a=b "b
! C 5 S ˆC( ‰. For computational purposes, replace the
'
5œ!
________________________________________________________________________
page 70
—————————————————————————— CHAPTER 2. ——
'
" >
98" a>b œ ( – ˆ$=# %= #‰" 985 a=b—.= .
# !
5œ!
Set 9! a>b œ !. The first four approximations are given by 9" a>b œ > ># >$ Î# ,
92 a>b œ > ># Î# >$ Î' >% Î% >& Î& >' Î#% â,
9$ a>b œ > ># Î# >% Î"# $>& Î#! %>' Î%& â,
9% a>b œ > ># Î# >% Î) (>& Î'! >' Î"& â
a, b .
The approximations appear to be converging to the exact solution,
9a>b œ " È" #> #># >$ Þ
13. Note that 98 a!b œ ! and 98 a"b œ " , a 8 " . Let + − a! ß "b . Then 98 a+b œ +8 .
Clearly, lim +8 œ ! . Hence the assertion is true.
8Ä_
14a+b. 98 a!b œ ! , a 8 " . Let + − Ð! ß "Ó. Then 98 a+b œ #8+ /8+ œ #8+Î/8+ Þ
#
#
Using l'Hospital's rule, lim #+DÎ/+D œ lim "ÎD/+D œ ! . Hence lim 98 a+b œ ! Þ
#
DÄ_
DÄ_
#
# "
"
a,b. '! #8B /8B .B œ /8B ¸! œ " /8 Þ Therefore,
#
8Ä_
lim ( 98 aBb.B Á ( lim 98 aBb.B .
"
8Ä_ !
"
! 8Ä_
15. Let > be fixed, such that a> ß C" bß a> ß C# b − H . Without loss of generality, assume that
C" C# . Since 0 is differentiable with respect to C , the mean value theorem asserts that
b 0 − aC" ß C# b such that 0 a> ß C" b 0 a> ß C# b œ 0C a> ß 0baC" C# bÞ Taking the absolute
value of both sides, k0 a> ß C" b 0 a> ß C# bk œ k0C a> ß 0bk kC" C# kÞ Since, by assumption,
`0 Î`C is continuous in H, 0C attains a maximum on any closed and bounded subset of H
.
________________________________________________________________________
page 71
—————————————————————————— CHAPTER 2. ——
Hence k0 a> ß C" b 0 a> ß C# bk Ÿ O kC" C# kÞ
16. For a sufficiently small interval of >, 98" a>b , 98 a>b − H . Since 0 satisfies a
Lipschitz condition, k0 a> ß 98 a>bb 0 a> ß 98" a>bbk Ÿ O k98 a>b 98" a>bkÞ Here
O œ 7+B k0C k .
k> k
k>k
>
17a+bÞ 9" a>b œ '! 0 a= ß !b.= . Hence k9" a>bk Ÿ '! k0 a= ß !bk.= Ÿ '! Q .= œ Q k>k , in
which Q is the maximum value of k0 a> ß C bk on H .
>
a,b. By definition, 9# a>b 9" a>b œ '! c0 a= ß 9" a=bb 0 a= ß !bd.= . Taking the absolute
k>k
value of both sides, k9# a>b 9" a>bk Ÿ '! kc0 a= ß 9" a=bb 0 a= ß !bdk.= . Based on the
k>k
k>k
results in Problems 16 and 17, k9# a>b 9" a>bk Ÿ '! O k9" a=b !k.= Ÿ OQ '! k=k.= .
Evaluating the last integral, we obtain k9# a>b 9" a>bk Ÿ Q O k>k# Î# .
a- b. Suppose that
k93 a>b 93" a>bk Ÿ
Q O 3" k>k3
3x
for some 3 " . By definition, 93" a>b 93 a>b œ '! c0 a> ß 93 a=bb 0 a= ß 93" a=bbd.= .
It follows that
>
k93" a>b 93 a>bk Ÿ (
Ÿ(
k> k
!
k> k
k0 a= ß 93 a=bb 0 a= ß 93" a=bbk.=
O k93 a=b 93" a=bk.=
Q O 3" k=k3
.=
3x
!
Q O 3 k>k3"
Q O 3 23"
œ
Ÿ
Þ
a 3 "b x
a 3 "b x
Ÿ(
!
k> k
O
Hence, by mathematical induction, the assertion is true.
18a+b. Use the triangle inequality, k+ ,k Ÿ k+k k,k .
a,b. For k>k Ÿ 2 , k9" a>bk Ÿ Q 2 , and k98 a>b 98" a>bk Ÿ Q O 8" 2 8 Îa8 xb . Hence
k98 a>bk Ÿ Q "
8
O 3" 23
3x
3œ"
œ
Q 8 aO2b3
"
Þ
O 3œ" 3 x
________________________________________________________________________
page 72
—————————————————————————— CHAPTER 2. ——
ˆ O2 "‰Þ By the comparison
a- b. The sequence of partial sums in a,b converges to Q
O /
test, the sums in a+b also converge. Furthermore, the sequence k98 a>bk is bounded, and
hence has a convergent subsequence. Finally, since individual terms of the series must
tend to zero, k98 a>b 98" a>bk p ! , and it follows that the sequence k98 a>bk is convergent.
>
>
19a+b. Let 9a>b œ '! 0 a= ß 9a=bb.= and <a>b œ '! 0 a= ß <a=bb.= Þ Then by linearity of
>
the integral, 9a>b <a>b œ '! c0 a= ß 9a=bb 0 a= ß <a=bbd.= Þ
a,b. It follows that k9a>b <a>bk Ÿ '! k0 a= ß 9a=bb 0 a= ß <a=bbk.= Þ
>
a- b. We know that 0 satisfies a Lipschitz condition,
k0 a> ß C" b 0 a> ß C# bk Ÿ O kC" C# k ,
based on k`0 Î`Ck Ÿ O in H. Therefore,
k9a>b <a>bk Ÿ ( k0 a= ß 9a=bb 0 a= ß <a=bbk.=
>
!
Ÿ ( O k9a=b <a=bk.= .
>
!
________________________________________________________________________
page 73
—————————————————————————— CHAPTER 2. ——
Section 2.9
1. Writing the equation for each 8 ! , C" œ !Þ* C! , C# œ !Þ* C" , C$ œ !Þ* C#
and so on, it is apparent that C8 œ a !Þ*b8 C! . The terms constitute an alternating
series, which converge to zero, regardless of C! .
3. Write the equation for each 8 ! , C" œ È$ C! , C# œ È%Î# C" , C$ œ È&Î$ C# ,â.
Upon substitution, we find that C# œ Èa% † $bÎ# C" , C$ œ Èa& † % † $bÎa$ † #b C! , â Þ
It can be proved by mathematical induction, that
"
a 8 #b x
Ê
C!
È#
8x
" È
œ
a8 "ba8 #b C! .
È#
C8 œ
This sequence is divergent, except for C! œ ! .
4. Writing the equation for each 8
and so on, it can be shown that
C!
C8 œ œ
C!
! , C" œ C! , C# œ C" , C$ œ C# , C% œ C$ ,
, for 8 œ %5 or 8 œ %5 "
, for 8 œ %5 # or 8 œ %5 $
The sequence is convergent only for C! œ ! .
6. Writing the equation for each 8
!,
C" œ !Þ& C! '
C# œ !Þ& C" ' œ !Þ&a!Þ& C! 'b ' œ a!Þ&b# C! ' a!Þ&b'
C$ œ !Þ& C# ' œ !Þ&a!Þ& C" 'b ' œ a!Þ&b$ C! 'c" a!Þ&b Ð!Þ&Ñ# d
ã
C8 œ a!Þ&b8 C! "#c" a!Þ&b8 d
which can be verified by mathematical induction. The sequence is convergent for all C! ,
and in fact C8 p "# .
7. Let C8 be the balance at the end of the 8-th day. Then C8" œ a" <Î$&'b C8 . The
solution of this difference equation is C8 œ a" <Î$'&b8 C! , in which C! is the initial
balance. At the end of one year, the balance is C$'& œ a" <Î$'&b$'& C! . Given that
< œ Þ!(, C$'& œ a" <Î$'&b$'& C! œ "Þ!(#& C! . Hence the effective annual yield is
a"Þ!(#& C! C! bÎC! œ (Þ#& % .
8. Let C8 be the balance at the end of the 8-th month. Then C8" œ a" <Î"#b C8 #& .
As in the previous solutions, we have
________________________________________________________________________
page 74
—————————————————————————— CHAPTER 2. ——
C8 œ 38 ”C!
#&
#&
,
•
"3
"3
in which 3 œ a" <Î"#b. Here < is the annual interest rate, given as ) % . Therefore
C$' œ a"Þ!!''b$' ’"!!!
a"#b#&
< “
a"#b#&
<
œ #ß #)$Þ'$ dollars.
9. Let C8 be the balance due at the end of the 8-th month. The appropriate difference
equation is C8" œ a" <Î"#b C8 T . Here < is the annual interest rate and T is the
monthly payment. The solution, in terms of the amount borrowed, is given by
C8 œ 38 ”C!
T
T
,
•
"3
"3
in which 3 œ a" <Î"#b and C! œ )ß !!! Þ To figure out the monthly payment, T , we
require that C$' œ !Þ That is,
336 ”C!
T
T
.
•œ
"3
"3
After the specified amounts are substituted, we find the T œ $ #&)Þ"% .
11. Let C8 be the balance due at the end of the 8-th month. The appropriate difference
equation is C8" œ a" <Î"#b C8 T , in which < œ Þ!* and T is the monthly payment.
The initial value of the mortgage is C! œ "!!ß !!! dollars. Then the balance due at the
end of the 8-th month is
C8 œ 38 ”C!
T
T
Þ
•
"3
"3
where 3 œ a" <Î"#b. In terms of the specified values,
C8 œ a!Þ!!(&b8 ”"!&
"#T
"#T
.
•
<
<
Setting 8 œ $!a"#b œ $'! , and C$'! œ ! , we find that T œ )!%Þ'# dollarsÞ For the
monthly payment corresponding to a #! year mortgage, set 8 œ #%! and C#%! œ ! .
12. Let C8 be the balance due at the end of the 8-th month, with C! the initial value of the
mortgage. The appropriate difference equation is C8" œ a" <Î"#b C8 T , in which
< œ !Þ" and T œ *!! dollars is the maximum monthly payment. Given that the life of
the mortgage is #! years, we require that C#%! œ !. The balance due at the end of the 8th month is
C8 œ 38 ”C!
T
T
Þ
•
"3
"3
In terms of the specified values for the parameters, the solution of
________________________________________________________________________
page 75
—————————————————————————— CHAPTER 2. ——
aÞ!!)$$b#%! ”C!
"#a"!!!b
"#a"!!!b
•œ
!Þ"
!Þ"
is C! œ "!$ß '#%Þ'# dollars.
15.
16. For example, take 3 œ $Þ& and ?! œ "Þ" :
________________________________________________________________________
page 76
—————————————————————————— CHAPTER 2. ——
19a+b. $# œ a3# 3" bÎa3$ 3# b œ a$Þ%%* $bÎa$Þ&%% $Þ%%*b œ %Þ(#'$ .
a,b. % diff œ
k$ $# k
$
‚ "!! œ
k%Þ''*#%Þ($'$k
%Þ''*#
‚ "!! ¸ "Þ## % Þ
a- b. Assuming a3$ 3# bÎa3% 3$ b œ $ , 3% ¸ $Þ&'%$
a. b. A period "' solutions appears near 3 ¸ $Þ&'& .
a/b. Note that a38" 38 b œ $8" a38 38" bÞ With the assumption that $8 œ $ , we have
a38" 38 b œ $ " a38 38" b, which is of the form C8" œ ! C8 , 8 $ . It follows that
a35 35" b œ $ $5 a3$ 3# b for 5 % . Then
38 œ 3" a3# 3" b a3$ 3# b a3% 3$ b â a38 38" b
œ 3" a3# 3" b a3$ 3# bc" $ " $ # â $ $8 d
" $ %8
œ 3" a3# 3" b a3$ 3# b”
•Þ
" $ "
$ ‘
Hence lim 38 œ 3# a3$ 3# b $ "
Þ Substitution of the appropriate values yields
8Ä_
lim 38 œ $Þ&'**
8Ä_
________________________________________________________________________
page 77
—————————————————————————— CHAPTER 2. ——
Miscellaneous Problems
1. Linear
Ò C œ -ÎB# B$ Î& ÓÞ
2. Homogeneous
Ò+<->+8aCÎBb 68ÈB# C # œ - Ó.
3. Exact
Ò B# BC $C C $ œ ! Ó.
4. Linear in BaCb
Ò B œ - / C C / C Ó.
5. Exact
Ò B# C BC # B œ - Ó.
6. Linear
Ò C œ B" a" /"B b Ó.
#
7. Let ? œ B#
Ò B# C # " œ - /C Ó.
8. Linear
Ò C œ a% -9= # -9= BbÎB# Ó.
9. Exact
Ò B# C B C # œ - Ó .
10. . œ .aBb
Ò C # ÎB$ CÎB# œ - Ó.
11. Exact
Ò B$ Î$ BC /C œ - Ó.
12. Linear
Ò C œ - /B /B 68a" /B b Ó.
13. Homogeneous Ò #ÈCÎB 68 kBk œ - Ó.
14. Exact/Homogeneous
Ò B# #BC #C # œ $% Ó.
15. Separable
Ò C œ -Î-9=2# aBÎ#b Ó.
16. Homogeneous Ò Š#ÎÈ$ ‹+<->+8’Ð#C BÑÎÈ$ B“ 68 kBk œ - Ó.
17.
18.
19.
20.
21.
22.
23.
Linear
Ò C œ - /$B /#B Ó.
Linear/Homogeneous
Ò C œ - B# B Ó.
. œ .aBb
Ò $C #BC $ "!B œ ! Ó.
Separable
Ò /B /C œ - Ó.
Homogeneous Ò /CÎB 68 kBk œ - Ó.
Separable
Ò C$ $C B$ $B œ # Ó.
Bernoulli
Ò "ÎC œ B' B# /#B .B -B Ó.
24. Separable
Ò =38# B =38 C œ - Ó.
25. Exact
Ò B# ÎC +<->+8aCÎBb œ - Ó.
26. . œ .aBb
Ò B# #B# C C # œ - Ó.
27. . œ .aBb
Ò =38 B -9= #C "# =38# B œ - Ó.
28. Exact
Ò #BC BC $ B$ œ - Ó.
29. Homogeneous Ò +<-=38aCÎBb 68 kBk œ - Ó.
30. Linear in BaCb Ò BC # 68 kC k œ ! Ó.
31. Separable
32. . œ .aCb
Ò B 68 kBk B" C # 68 kC k œ - Ó.
Ò B$ C # BC $ œ % Ó.
________________________________________________________________________
page 78
—————————————————————————— CHAPTER 3. ——
Chapter Three
Section 3.1
1. Let C œ /<> , so that C w œ < /<> and C ww œ < /<> . Direct substitution into the differential
equation yields a<# #< $b/<> œ ! . Canceling the exponential, the characteristic
equation is <# #< $ œ ! Þ The roots of the equation are < œ $ , " . Hence the
general solution is C œ -" /> -# /$> .
2. Let C œ /<> . Substitution of the assumed solution results in the characteristic equation
<# $< # œ ! Þ The roots of the equation are < œ # , " . Hence the general
solution is C œ -" /> -# /#> .
4. Substitution of the assumed solution C œ /<> results in the characteristic equation
#<# $< " œ ! Þ The roots of the equation are < œ "Î# , " . Hence the general
solution is C œ -" />Î# -# /> .
6. The characteristic equation is %<# * œ ! , with roots < œ „$Î# . Therefore the
general solution is C œ -" /$>Î# -# /$>Î# .
8. The characteristic equation is <# #< # œ ! , with roots < œ "„È$ . Hence the
general solution is C œ -" /B:Š" È$‹> -# /B:Š" È$‹>.
9. Substitution of the assumed solution C œ /<> results in the characteristic equation
<# < # œ ! Þ The roots of the equation are < œ # , " . Hence the general
solution is C œ -" /#> -# /> . Its derivative is C w œ #-" /#> -# /> . Based on the
first condition, Ca!b œ " , we require that -" -# œ " . In order to satisfy C w a!b œ " ,
we find that #-" -# œ " . Solving for the constants, -" œ ! and -# œ " . Hence the
specific solution is Ca>b œ /> .
11. Substitution of the assumed solution C œ /<> results in the characteristic equation
'<# &< " œ ! Þ The roots of the equation are < œ "Î$ , "Î# . Hence the general
solution is C œ -" />Î$ -# />Î# . Its derivative is C w œ -" />Î$ Î$ -# />Î# Î# . Based
on the first condition, Ca!b œ " , we require that -" -# œ % . In order to satisfy the
condition C w a!b œ " , we find that -" Î$ -# Î# œ ! . Solving for the constants, -" œ "#
and -# œ ) . Hence the specific solution is C a>b œ "# />Î$ ) />Î# .
12. The characteristic equation is <# $< œ ! , with roots < œ $ , ! . Therefore the
general solution is C œ -" -# /$> , with derivative C w œ $ -# /$> . In order to
satisfy the initial conditions, we find that -" -# œ # , and $ -# œ $ . Hence the
specific solution is Ca>b œ " /$> .
13. The characteristic equation is <# &< $ œ ! , with roots
________________________________________________________________________
page 83
—————————————————————————— CHAPTER 3. ——
& È"$
<"ß# œ „
.
#
#
The general solution is C œ -" /B:Š & È"$‹>Î# -# /B:Š & È"$‹>Î#, with
derivative
Cw œ
& È"$
& È"$
-" /B:Š & È"$‹>Î#
-# /B:Š & È"$‹>Î# .
#
#
In order to satisfy the initial conditions, we require that -" -# œ " , and
È
&È"$
-" &# "$ -# œ ! . Solving for the coefficients, -" œ Š" &ÎÈ"$ ‹Î# and
#
-# œ Š" &ÎÈ"$ ‹Î# Þ
14. The characteristic equation is #<# < % œ ! , with roots
" È$$
<"ß# œ „
.
%
%
The general solution is C œ -" /B:Š " È$$‹>Î% -# /B:Š " È$$‹>Î%, with
derivative
Cw œ
" È$$
" È$$
-" /B:Š " È$$‹>Î%
-# /B:Š " È$$‹>Î% .
%
%
In order to satisfy the initial conditions, we require that -" -# œ ! , and
È
"È$$
-" "% $$ -# œ " . Solving for the coefficients, -" œ #ÎÈ$$ and
%
-# œ #ÎÈ$$ Þ The specific solution is
Ca>b œ #’/B:Š " È$$‹>Î% /B:Š " È$$‹>Î%“ÎÈ$$ .
________________________________________________________________________
page 84
—————————————————————————— CHAPTER 3. ——
16. The characteristic equation is %<# " œ ! , with roots < œ „"Î# . Therefore the
general solution is C œ -" />Î# -# />Î# . Since the initial conditions are specified at
> œ # , is more convenient to write C œ ." /a>#bÎ# .# /a>#bÎ# Þ The derivative
is given by C w œ ." /a>#bÎ# ‘Î# .# /a>#bÎ# ‘Î# . In order to satisfy the initial
conditions, we find that ." .# œ " , and ." Î# .# Î# œ " . Solving for the
coefficients, ." œ $Î# , and .# œ "Î# . The specific solution is
Ca>b œ
$ a>#bÎ# " a>#bÎ#
/
/
#
#
$ >Î# / >Î#
œ /
/ Þ
#/
#
18. An algebraic equation with roots # and "Î# is #<# &< # œ ! . This is the
characteristic equation for the ODE #C ww &C w # C œ ! .
20. The characteristic equation is #<# $< " œ ! , with roots < œ "Î# , " . Therefore
the general solution is C œ -" />Î# -# /> , with derivative C w œ -" />Î# Î# -# /> . In
order to satisfy the initial conditions, we require -" -# œ # and -" Î# -# œ "Î# .
Solving for the coefficients, -" œ $ , and -# œ " . The specific solution is
Ca>b œ $/>Î# /> Þ To find the stationary point, set C w œ $/>Î# Î# /> œ ! . There is
a unique solution, with >" œ 68a*Î%b. The maximum value is then C a>" b œ *Î%Þ To find
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—————————————————————————— CHAPTER 3. ——
the x-intercept, solve the equation $/>Î# /> œ ! Þ The solution is readily found to be
># œ 68* ¸ #Þ"*(# .
22. The characteristic equation is %<# " œ ! , with roots < œ „"Î# . Hence the
general solution is C œ -" />Î# -# />Î# , with derivative C w œ -" />Î# Î# -# />Î# Î# .
Invoking the initial conditions, we require that -" -# œ # and -" -# œ " .
The specific solution is Ca>b œ a" " b/>Î# a" " b/>Î# Þ Based on the form of the
solution, it is evident that as > p _ , C a>b p ! as long as " œ " .
23. The characteristic equation is <# a#! "b< !a! "b œ ! . Examining the
coefficients, the roots are < œ ! , ! " . Hence the general solution of the differential
equation is Ca>b œ -" /!> -# /a!"b> Þ Assuming ! − ‘ , all solutions will tend to zero
as long as ! ! . On the other hand, all solutions will become unbounded as long as
! " ! , that is, ! " .
25. Ca>b œ # />Î# Î& $ /#> Î& .
The minimum occurs at a>! , C! b œ a!Þ("'( ß !Þ("&&b.
26a+b. The characteristic roots are < œ $ , # . The solution of the initial value
problem is Ca>b œ a' " b/#> a% " b/$> .
"b
a,b. The maximum point has coordinates >! œ 68’ $#aa%
'" b “, C! œ
a- b. C! œ
%a'" b$
#(a%" b#
a. b. lim >! œ 68 $# .
" Ä_
% , as long as "
' 'È $ .
%a'" b$
#(a%" b#
.
lim C! œ _ .
" Ä_
29. Set @ œ C w and @ w œ C ww . Substitution into the ODE results in the first order equation
> @ w @ œ " . The equation is linear, and can be written as a> @bw œ " . Hence the general
solution is @ œ " -" Î> . Hence C w œ " -" Î> , and C œ > -" 68 > -# .
31. Setting @ œ C w and @ w œ C ww , the transformed equation is #># @ w @$ œ #> @ . This
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is a Bernoulli equation, with 8 œ $ . Let A œ @# . Substitution of the new dependent
variable yields ># A w " œ #> A , or ># A w #> A œ œ " . Integrating, we find that
A œ a> -" bÎ># . Hence @ œ „>ÎÈ> -" , that is, C w œ „>ÎÈ> -" . Integrating one
more time results in Ca>b œ „ #$ a> #-" bÈ> -" -# Þ Ð Note that @ œ ! is also a
solution of the transformed equationÑ.
32. Setting @ œ C w and @ w œ C ww , the transformed equation is @ w @ œ /> . This ODE
is linear, with integrating factor .a>b œ /> . Hence @ œ C w œ a> -" b/> . Integrating,
we obtain Ca>b œ a> -" b/> -# Þ
33. Set @ œ C w and @ w œ C ww . The resulting equation is ># @ w œ @# . This equation is
separable, with solution @ œ C w œ >Îa" -" >bÞ Integrating, the general solution is
Ca>b œ >Î-" -"# 68k" -" >k -# ,
as long as -" Á ! Þ For -" œ ! , the solution is C a>b œ ># Î# -# . Note that @ œ ! is
also a solution of the transformed equation.
35. Let C w œ @ and C ww œ @ .@Î.C . Then @ .@Î.C C œ ! is the transformed equation
for @ œ @aC b . This equation is separable, with @ .@ œ C .CÞ The solution is given by
@# œ C # -" . Substituting for @ , we find that C w œ „È-" C # Þ This equation is
also
separable, with solution +<-=38ˆCÎÈ-" ‰ œ „ > -# , or C a>b œ ." =38a> .# b .
36. Let C w œ @ and C ww œ @ .@Î.C . It follows that @.@Î.C C@$ œ ! is the differential
equation for @ œ @aC b . This equation is separable, with @# .@ œ C .CÞ The solution
"
"
is given by @ œ cC # Î# -" d . Substituting for @ , we find that C w œ cC # Î# -" d . This
equation is also separable, with aC# Î# -" b.C œ .> . The solution is defined implicitly
by C$ Î' -" C -# œ > .
38. Setting C w œ @ and C ww œ @.@Î.C , the transformed equation is C @.@Î.C @$ œ !.
This equation is separable, with @# .@ œ .CÎC Þ The solution is @aC b œ c-" 68kC kd" Þ
Substituting for @ , we obtain a separable equation, a-" 68kC kb.C œ .B . The solution is
given implicitly by -# C C 68kC k -$ œ > .
39. Let C w œ @ and C ww œ @.@Î.C . It follows that @.@Î.C @# œ #/C is the equation
for @ œ @aC b . Inspection of the left hand side suggests a substitution A œ @# . The
resulting
equation is .AÎ.C #A œ %/C . This equation is linear, with integrating factor
. œ /#C Þ
We obtain . a/#C AbÎ.C œ % /C , which upon integration yields AaC b œ % /C -" /#C Þ
Converting back to the original dependent variable, C w œ „/C È% /C -" Þ Separating
variables, /C a% /C -" b"Î# .C œ „ .> . Integration yields È% /C -" œ „ #> -# .
41. Setting C w œ @ and C ww œ @.@Î.C , the transformed equation is @.@Î.C $C # œ !.
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—————————————————————————— CHAPTER 3. ——
This equation is separable, with @.@ œ $C # .C Þ The solution is C w œ @ œ È#C $ -" Þ
The positive root is chosen based on the initial conditions. Furthermore, when > œ ! ,
C œ # , and C w œ @ œ % . The initial conditions require that -" œ ! . It follows that
C w œ È#C $ Þ Separating variables and integrating, "ÎÈC œ >ÎÈ# -# . Hence
the solution is Ca>b œ #Îa" > b# Þ
42. Setting @ œ C w and @ w œ C ww , the transformed equation is a" ># b@ w #> @ œ
œ $># Þ Rewrite the equation as @ w #> @Îa" ># b œ $># Îa" ># bÞ This
equation is linear, with integrating factor . œ " ># . Hence we have
ˆ" ># ‰@‘w œ $># Þ
Integrating both sides, @ œ $>" Îa" ># b -" Îa" ># b. Invoking the initial condition
@a"b œ " , we require that -" œ & . Hence C w œ a$ &>bÎa> >$ b. Integrating,
we obtain Ca>b œ $# 68c># Îa" ># bd & +<->+8a>b -# . Based on the initial condition
Ca"b œ # , we find that -# œ $# 68 # &% 1 # .
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—————————————————————————— CHAPTER 3. ——
Section 3.2
1.
[ ˆ/#> ß /$>Î# ‰ œ º
/#>
#/#>
( >Î#
/$>Î#
$ $>Î# º œ / Þ
#
#/
3.
[ ˆ/#> ß > /#> ‰ œ º
/#>
#/#>
>/#>
%>
ºœ/ .
a" #>b/#>
5.
[ ˆ/> =38 > ß /> -9= >‰ œ º
/> =38 >
/> -9= >
œ /#> .
/> a=38 > -9= >b /> a-9= > =38 >b º
6.
[ ˆ-9=# ) ß " -9= #) ‰ œ º
-9=# )
#=38 ) -9= )
" -9= #)
œ !.
# =38 #) º
7. Write the equation as C ww a$Î>bC w œ " . :a>b œ $Î> is continuous for all > ! .
Since >! ! , the IVP has a unique solution for all > ! .
$
%
#
9. Write the equation as C ww >%
C w >a>%
b C œ >a>%b . The coefficients are not
continuous at > œ ! and > œ % . Since >! − a! ß %b , the largest interval is ! > % .
10. The coefficient $68k>k is discontinuous at > œ ! . Since >! ! , the largest interval
of existence is ! > _ .
kBk
B
11. Write the equation as C ww B$
C w 68
B$ C œ ! . The coefficients are discontinuous
at B œ ! and B œ $ . Since B! − a! ß $b , the largest interval is ! B $ .
13. C"ww œ # . We see that ># a#b #a># b œ ! . C#ww œ # >$ , with ># aC#ww b #aC# b œ ! .
Let C$ œ -" ># -# >" , then C$ww œ #-" #-# >$ . It is evident that C$ is also a solution.
16. No. Substituting C œ =38a># b into the differential equation,
%># =38ˆ># ‰ #-9=ˆ># ‰ #> -9=ˆ># ‰:a>b =38ˆ># ‰; a>b œ ! .
For the equation to be valid, we must have :a>b œ "Î> , which is not continuous, or
even defined, at > œ ! .
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—————————————————————————— CHAPTER 3. ——
17. [ a/#> ß 1a>bb œ /#> 1 w a>b #/#> 1a>b œ $/%> Þ Dividing both sides by /#> , we find
that 1 must satisfy the ODE 1 w #1 œ $/#> Þ Hence 1a>b œ $> /#> - /#> Þ
19. [ a0 ß 1b œ 0 1 w 0 w 1 . Also, [ a? ß @b œ [ a#0 1 ß 0 #1b . Upon evaluation,
[ a? ß @b œ &0 1 w &0 w 1 œ &[ a0 ß 1b.
20. [ a0 ß 1b œ 0 1 w 0 w 1 œ > -9= > =38 > , and [ a? ß @b œ %0 1 w %0 w 1 .
Hence [ a? ß @b œ %> -9= > %=38 > .
22. The general solution is C œ -" /$> -# /> . [ a/$> ß /> b œ #/%> , and hence
the exponentials form a fundamental set of solutions. On the other hand, the fundamental
solutions must also satisfy the conditions C" a"b œ " , C"w a"b œ ! ; C# a"b œ ! , C#w a"b œ " Þ
For C" , the initial conditions require -" -# œ / , $-" -# œ ! . The coefficients are
-" œ /$ Î# , -# œ $/Î# . For the solution, C# , the initial conditions require -" -# œ !
, $-" -# œ / . The coefficients are -" œ /$ Î# , -# œ /Î# . Hence the fundamental
solutions are ˜C" œ "# /$a>"b $# /a>"b ß C# œ "# /$a>"b "# /a>"b ™Þ
23. Yes. C"ww œ % -9= #> ; C#ww œ % =38 #> . [ a-9= #> ß =38 #>b œ # Þ
24. Clearly, C" œ /> is a solution. C#w œ a" >b/> , C#ww œ a# >b/> Þ Substitution into the
ODE results in a# >b/> #a" >b/> > /> œ ! . Furthermore, [ a/> ß >/> b œ /#> Þ
Hence the solutions form a fundamental set of solutions.
26. Clearly, C" œ B is a solution. C#w œ -9= B , C#ww œ =38 B Þ Substitution into the
ODE results in a" B -9> Bba =38 Bb Ba-9= Bb =38 B œ !. [ aC" , C# b œ B -9=
B =38 B,
which is nonzero for ! B 1 . Hence eB ß =38 Bf is a fundamental set of solutions.
28. T œ " , U œ B , V œ " . We have T ww U w V œ ! . The equation is exact. Note
that aC w bw aBCbw œ ! . Hence C w BC œ -" . This equation is linear, with integrating
#
factor . œ /B Î# Þ Therefore the general solution is
CaBb œ -" /B:ˆ B# Î#‰( /B:ˆ?# Î#‰.? -# /B:ˆ B# Î#‰Þ
B
B!
29. T œ " , U œ $B# , V œ B . Note that T ww U w V œ &B , and therefore the
differential equation is not exact.
31. T œ B# , U œ B , V œ " . We have T ww U w V œ ! . The equation is exact.
w
Write the equation as aB# C w b aBCbw œ ! . Integrating, we find that B# C w B C œ - .
Divide both sides of the ODE by B# . The resulting equation is linear, with integrating
factor . œ "ÎB . Hence aCÎBbw œ - B$ . The solution is C a>b œ -" B" -# B .
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—————————————————————————— CHAPTER 3. ——
33. T œ B# , U œ B , V œ B# / # . Hence the coefficients are #T w U œ $B and
T ww U w V œ B# " / # . The adjoint of the original differential equation is given
by B# . ww $B . w aB# " / # b. œ ! .
35. T œ " , U œ ! , V œ B . Hence the coefficients are given by #T w U œ ! and
T ww U w V œ B . Therefore the adjoint of the original equation is . ww B . œ ! .
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—————————————————————————— CHAPTER 3. ——
Section 3.3
1. Suppose that ! 0 a>b " 1a>b œ ! , that is, !a># &>b " a># &>b œ ! on some
interval M . Then a! " b># &a! " b> œ ! , a > − M . Since a quadratic .has at most
two
roots, we must have ! " œ ! and ! " œ ! . The only solution is ! œ " œ ! .
Hence the two functions are linearly independent.
3. Suppose that /-> -9= .> œ E /-> =38 .> , for some E Á ! , on an interval M . Since the
function =38 .> Á ! on some subinterval M! § M , we conclude that >+8 .> œ E on M! .
This is clearly a contradiction, hence the functions are linearly independent.
4. Obviously, 0 aBb œ / 1aBb for all real numbers B . Hence the functions are linearly
dependent.
5. Here 0 aBb œ $1aBb for all real numbers. Hence the functions are linearly dependent.
)Þ Note that 0 aBb œ 1aBb for B − Ò ! ß _Ñ, and 0 aBb œ 1aBb for B − Ð _ ß ! Ó. It
follows that the functions are linearly dependent on ‘ and ‘ . Nevertheless, they are
linearly independent on any open interval containing zero.
9. Since [ a>b œ > =38# > has only isolated zeros, [ a>b cannot identically vanish on any
open interval. Hence the functions are linearly independent.
10. Same argument as in Prob. 9.
11. By linearity of the differential operator, -" C" and -# C# are also solutions.
Calculating
the Wronskian, [ a-" C" ß -# C# b œ a-" C" ba-# C# bw a-" C" bw a-# C# b œ -" -# [ aC" ß C# b .
Since [ aC" ß C# b is not identically zero, neither is [ a-" C" ß -# C# b .
13. Direct calculation results in
[ a+" C" +# C# ß ," C" ,# C# b œ +" ,# [ aC" ß C# b ," +# [ aC" ß C# b
œ a+" ,# +# ," b[ aC" ß C# b Þ
Hence the combinations are also linearly independent as long as +" ,# +# ," Á ! .
14. Let !ai jb " ai jb œ ! i ! j . Then ! " œ ! and ! " œ ! . The only
solution is ! œ " œ ! . Hence the given vectors are linearly independent. Furthermore,
any vector +" i +# j œ ˆ +#" +## ‰ai jb ˆ +#" +## ‰ai jb .
16. Writing the equation in standard form, we find that T a>b œ =38 >Î-9= > . Hence the
=38 >
‰
Wronskian is [ a>b œ , /B:ˆ ' -9=
> .> œ , /B: a68k-9= >kb œ , -9= > , in which , is
some constant.
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—————————————————————————— CHAPTER 3. ——
17. After writing the equation in standard form, we have T aBb œ "ÎB . The Wronskian
is [ a>b œ - /B:ˆ ' B" .B‰ œ - /B:a 68kBkb œ -ÎkBk, in which - is some constant.
18. Writing the equation in standard form, we find that T aBb œ #BÎa" B# b. The
#B
#
# "
‰
Wronskian is [ a>b œ - /B:ˆ ' "B
,
# .B œ - /B: a 68k" B kb œ - k" B k
in which - is some constant.
19. Rewrite the equation as :a>bC ww : w a>bC w ; a>bC œ ! Þ After writing the equation
in standard form, we have T a>b œ : w a>bÎ:a>b . Hence the Wronskian is
[ a>b œ - /B:Œ (
: w a >b
.> œ - /B:a 68 :a>bb œ -Î:a>b .
: a >b
21. The Wronskian associated with the solutions of the differential equation is given by
‰
[ a>b œ - /B:ˆ ' #
># .> œ - /B: a #Î>b. Since [ a#b œ $ , it follows that for the
hypothesized set of solutions, - œ $ / . Hence [ a%b œ $È/ .
22. For the given differential equation, the Wronskian satisfies the first order differential
equation [ w :a>b[ œ ! . Given that [ is constant, it is necessary that :a>b ´ ! .
23. Direct calculation shows that
[ a0 1 ß 0 2b œ a0 1ba0 2bw a0 1bw a0 2 b
œ a0 1ba0 w 2 0 2 w b a0 w 1 0 1 w ba0 2 b
œ 0 # [ a1 ß 2 b Þ
25. Since C" and C# are solutions, they are differentiable. The hypothesis can thus be
restated as C"w a>! b œ C#w a>! b œ ! at some point >! in the interval of definition. This
implies that [ aC" ß C# ba>! b œ ! . But [ aC" ß C# ba>! b œ - /B:ˆ ' :a>b.>‰ , which
cannot be equal to zero, unless - œ ! . Hence [ aC" ß C# b ´ ! , which is ruled out for
a fundamental set of solutions.
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—————————————————————————— CHAPTER 3. ——
Section 3.4
2. /B:a# $3b œ /# /$3 œ /# a-9= $ 3 =38 $b.
3. /31 œ -9= 1 3 =38 1 œ " .
4. /B:ˆ# 1# 3‰ œ /# ˆ-9= 1# 3 =38 1# ‰ œ /# 3 .
6. 1"#3 œ /B:ca " #3b68 1d œ /B:a 68 1b/B:a# 68 1 3b œ
œ 1" c-9= a# 68 1b 3 =38 a# 68 1bdÞ
"
1
/B:a# 68 1 3b œ
8. The characteristic equation is <# #< ' œ ! , with roots < œ " „ 3È& . Hence the
general solution is C œ -" /> -9= È& > -# /> =38 È& >.
9. The characteristic equation is <# #< ) œ ! , with roots < œ % ß # . The roots
are real and different, hence the general solution is C œ -" /%> -# /#> .
10. The characteristic equation is <# #< # œ ! , with roots < œ " „ 3. Hence the
general solution is C œ -" /> -9= > -# /> =38 >.
12. The characteristic equation is %<# * œ ! , with roots < œ „ $# 3. Hence the
general solution is C œ -" -9= $# > -# =38 $# > .
13. The characteristic equation is <# #< "Þ#& œ !, with roots < œ " „ "# 3. Hence
the general solution is C œ -" /> -9= "# > -# /> =38 "# > .
15. The characteristic equation is <# < "Þ#& œ ! , with roots < œ "# „ 3. Hence
the general solution is C œ -" />Î# -9= > -# />Î# =38 > .
16. The characteristic equation is <# %< 'Þ#& œ ! , with roots < œ # „ $# 3. Hence
the general solution is C œ -" /#> -9= $# > -# /#> =38 $# > .
17. The characteristic equation is <# % œ ! , with roots < œ „ #3. Hence the general
solution is C œ -" -9= #> -# =38 #> . Its derivative is C w œ #-" =38 #> #-# -9= #> .
Based on the first condition, Ca!b œ ! , we require that -" œ ! . In order to satisfy the
condition C w a!b œ " , we find that #-# œ " . The constants are -" œ ! and -# œ "Î# .
Hence the specific solution is Ca>b œ "# =38 #> .
19. The characteristic equation is <# #< & œ ! , with roots < œ "„ #3. Hence the
general solution is C œ -" /> -9= #> -# /> =38 #> . Based on the condition, C a1Î#b œ ! ,
we require that -" œ ! . It follows that C œ -# /> =38 #> , and so the first derivative is
C w œ -# /> =38 #> #-# /> -9= #> . In order to satisfy the condition C w a1Î#b œ #, we find
that #/1Î# -# œ # . Hence we have -# œ /1Î# . Therefore the specific solution is
Ca>b œ />1Î# =38 #> .
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—————————————————————————— CHAPTER 3. ——
20. The characteristic equation is <# " œ ! , with roots < œ „ 3. Hence the general
solution is C œ -" -9= > -# =38 > . Its derivative is C w œ -" =38 > -# -9= > . Based
on the first condition, Ca1Î$b œ # , we require that -" È$ -# œ % . In order to satisfy
the condition C w a1Î$b œ % , we find that È$ -" -# œ ) . Solving these for
the constants, -" œ " #È$ and -# œ È$ # . Hence the specific solution is a steady
oscillation, given by Ca>b œ Š" #È$‹-9= > ŠÈ$ #‹=38 > .
21. From Prob. 15, the general solution is C œ -" />Î# -9= > -# />Î# =38 > . Invoking
the first initial condition, Ca!b œ $ , which implies that -" œ $ . Substituting, it follows
that C œ $/>Î# -9= > -# />Î# =38 > , and so the first derivative is
$
-#
C w œ />Î# -9= > $/>Î# =38 > -# />Î# -9= > />Î# =38 > .
#
#
Invoking the initial condition, C w a!b œ " , we find that $# -# œ " , and so -# œ
Hence the specific solution is Ca>b œ $/>Î# -9= > &$ />Î# =38 > Þ
24a+b. The characteristic equation is &<# #< ( œ ! , with roots < œ "& „ 3
È
È
&
#
.
È$%
& .
The solution is ? œ -" />Î& -9= &$% > -# />Î& =38 &$% > . Invoking the given initial
conditions, we obtain the equations for the coefficients : -" œ # , # È$% -# œ & .
That is, -" œ # , -# œ (ÎÈ$% . Hence the specific solution is
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—————————————————————————— CHAPTER 3. ——
?a>b œ #/>Î& -9=
È$%
&
>
(
/>Î& =38
È$%
È$%
&
>.
a,b. Based on the graph of ?a>b, X is in the interval "% > "' . A numerical solution
on that interval yields X ¸ "%Þ&""& .
26a+b. The characteristic equation is <# #+ < a+# "b œ ! , with roots < œ + „ 3 Þ
Hence the general solution is Ca>b œ -" /+> -9= > -# /+> =38 > . Based on the initial
conditions, we find that -" œ " and -# œ + Þ Therefore the specific solution is given by
in which 9 œ >+8" a+bÞ
Ca>b œ /+> -9= > + /+> =38 >
œ È" +# /+> -9= a> 9b,
a,b. For estimation, note that kCa>bk Ÿ È" +# /+> . Now consider the inequality
È" +# /+> Ÿ "Î"! . The inequality holds for > " 68’"!È" +# “Þ Therefore
X Ÿ +" 68’"!È" +# “. Setting + œ " , numerical analysis gives X ¸ "Þ)('$ .
+
a- b. Similarly, X"Î% ¸ (Þ%#)% , X"Î# ¸ %Þ$!!$ , X# ¸ "Þ&""' , X$ ¸ "Þ"%*' .
a. b .
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—————————————————————————— CHAPTER 3. ——
Note that the estimates X+ approach the graph of
"
È
+ 68’"! "
+# “ as + gets large.
27. Direct calculation gives the result. On the other hand, it was shown in Prob. 3.3.23
that [ a0 1 ß 0 2b œ 0 # [ a1 ß 2 bÞ Hence
[ ˆ/-> -9= .> ß /-> =38 .>‰ œ /#-> [ a-9= .> ß =38 .>b
œ /#-> -9= .>a=38 .>bw a-9= .>bw =38 .>‘
œ . /#-> Þ
28a+b. Clearly, C" and C# are solutions. Also, [ a-9= >ß =38 >b œ -9=# > =38# > œ "Þ
a,b. C w œ 3 /3> , C ww œ 3# /3> œ /3> Þ Evidently, C is a solution and so C œ -" C" -# C# .
a- b. Setting > œ ! , " œ -" -9= ! -# =38 ! , and -" œ ! . Differentiating, 3 /3> œ -# -9= > .
Setting > œ ! , 3 œ -# -9= ! and hence -# œ 3 . Therefore /3> œ -9= > 3 =38 > .
29. Euler's formula is /3> œ -9= > 3 =38 > . It follows that /3> œ -9= > 3 =38 > .
Adding these equation, /3> /3> œ # -9= > . Subtracting the two equations results in
/3> /3> œ #3 =38 > .
30. Let <" œ -" 3." , and <# œ -# 3.# . Then
/B:a<" <# b> œ /B:ca-" -# b> 3a." .# b>d
œ /a-" -# b> c-9=a." .# b> 3 =38a." .# b>d
œ /a-" -# b> ca-9= ." > 3=38 ." >ba-9= .# > 3=38 .# >bd
œ /-" > a-9= ." > 3=38 ." >b † /-# > a-9= ." > 3=38 ." >b
Hence /a<" <# b> œ /<" > /<# > Þ
32. If 9a>b œ ?a>b 3 @a>b is a solution, then
a? 3@bww :a>ba? 3@bw ; a>ba? 3@b œ ! ,
and a? ww 3@ ww b :a>ba? w 3@ w b ; a>ba? 3@b œ ! . After expanding the equation and
separating the real and imaginary parts,
?ww :a>b? w ; a>b? œ !
@ww :a>b@ w ; a>b@ œ !
Hence both ?a>b and @a>b are solutions.
.C w
.C
.D .B
34a+b. By the chain rule, CaBbw œ .B
B Þ In general, .D
.> œ .B .> . Setting D œ .> ,
#
. # C .B .B
.C . .B ‘ .B
.D .B
. .C .B ‘ .B
we have ..>C# œ .B
.> œ .B .B .> .> œ ’ .B# .> “ .> .B .B .> .> Þ However,
. .B ‘ .B
.B .> .>
.>
œ ’ ..>B# “ .B
†
#
.B
.>
œ
.#B
.># .
Hence
.# C
.>#
œ
. # C .B ‘#
.B# .>
.C . # B
.B .># .
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—————————————————————————— CHAPTER 3. ——
a,b. Substituting the results in Parta+b into the general ODE, C ww :a>bC w ; a>bC œ ! ,
`we find that
. # C .B # .C . # B
.C .B
:a>b
; a>bC œ ! .
” •
#
#
.B .>
.B .>
.B .>
Collecting the terms,
.B . # C
.#B
.B .C
:a>b •
; a>bC œ ! Þ
” •
”
#
#
.> .B
.>
.> .B
#
‘# œ 5 ; a>b , and ; a>b ! , we find that
a- b. Assuming .B
.>
be integrated. That is, B œ 0a>b œ ' È5 ; a>b .> .
a. b. Let 5 œ " . It follows that
.#B
.>#
:a>b .B
.> œ
.0
.>
.B
.>
œ È5 ; a>b , which can
:a>b0a>b œ
;w
#È ;
.# B
.B
.B #
; w a>b #:a>b; a>b
.
” # : a> b • Î ” • œ
.>
.>
.>
#c; a>bd$Î#
: È; Þ Hence
As long as .BÎ.> Á ! , the differential equation can be expressed as
.# C
; w a>b #:a>b; a>b .C
C œ !Þ
”
•
.B#
.B
#c; a>bd$Î#
‘# œ ; a>b .
* For the case ; a>b ! , write ; a>b œ c ; a>bd , and set .B
.>
36. :a>b œ $> and ; a>b œ ># . We have B œ ' > .> œ ># Î# . Furthermore,
; w a>b #:a>b; a>b
œ ˆ" $># ‰Î># .
#c; a>bd$Î#
The ratio is not constant, and therefore the equation cannot be transformed.
37. :a>b œ > "Î> and ; a>b œ ># . We have B œ ' > .> œ ># Î# . Furthermore,
; w a>b #:a>b; a>b
œ ".
#c; a>bd$Î#
The ratio is constant, and therefore the equation can be transformed. From Prob. 35,
the transformed equation is
.# C
.C
C œ !Þ
#
.B
.B
Based on the methods in this section, the characteristic equation is <# < " œ ! , with
È
roots < œ "# „ 3 #$ . The general solution is
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—————————————————————————— CHAPTER 3. ——
CaBb œ -" /BÎ# -9= È$ BÎ# -# /BÎ# =38 È$ BÎ# .
Since B œ ># Î# , the solution in the original variable > is
Ca>b œ /> Î% ’-" -9= ŠÈ$ ># Î%‹ -# =38 ŠÈ$ ># Î%‹“ .
#
40. :a>b œ %Î> and ; a>b œ #Î># . We have B œ È# ' >" .> œ È# 68 > . Furthermore,
; w a>b #:a>b; a>b
$
.
œ
$Î#
È#
#c; a>bd
The ratio is constant, and therefore the equation can be transformed. In fact, we obtain
.# C
$ .C
C œ !Þ
#
È# .B
.B
Based on the methods in this section, the characteristic equation is È# <# $< È# œ ! ,
with roots < œ È# , "ÎÈ# . The general solution is
CaBb œ -" /
È# B
-# /BÎ
È#
.
Since B œ È# 68 > , the solution in the original variable > is
Ca>b œ -" /# 68 > -# /68 >
œ -" ># -# >" Þ
41. :a>b œ $Î> and ; a>b œ "Þ#&Î># . We have B œ È"Þ#& ' >" .> œ È"Þ#& 68 > .
Checking the feasibility of the transformation,
; w a>b #:a>b; a>b
%
œ
.
$Î#
È&
#c; a>bd
The ratio is constant, and therefore the equation can be transformed. In fact, we obtain
.# C
% .C
C œ !Þ
È& .B
.B#
Based on the methods in this section, the characteristic equation is
È& <# %< È& œ ! , with roots < œ # „3 " . The general solution is
È&
È&
CaBb œ -" /#BÎ & -9= BÎÈ& -# /#BÎ & =38 BÎÈ& .
È
È
Since #BÎÈ& œ 68 > , the solution in the original variable > is
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—————————————————————————— CHAPTER 3. ——
Ca>b œ -" /68 > -9=Š68È> ‹ -# /68 > =38Š68È> ‹
œ >" ’-" -9=Š68È> ‹ -# =38Š68È> ‹“.
42. :a>b œ %Î> and ; a>b œ 'Î># . Set B œ È' ' >" .> œ È' 68 > .
Checking the feasibility of the transformation a‡ see Prob. 34 d , with ; !b,
; w a>b #:a>b; a>b
&
œ
.
$Î#
È'
#c ; a>bd
The ratio is constant, and therefore the equation can be transformed. In fact, we obtain
.# C
& .C
C œ !Þ
#
È
.B
' .B
Based on the methods in this section, the characteristic equation is È' <# &
< È' œ ! ,
with roots < œ È' , "ÎÈ' . The general solution is
CaBb œ -" /
È' B
-# /BÎ
È'
.
Since B œ È' 68 > , the solution in the original variable > is
Ca>b œ -" /'68 > -# /68 >
œ -" > ' -# >" Þ
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—————————————————————————— CHAPTER 3. ——
Section 3.5
2. The characteristic equation is *<# '< " œ ! , with the double root < œ "Î$ Þ
Based on the discussion in this section, the general solution is Ca>b œ -" />Î$ -# > />Î$ .
3. The characteristic equation is %<# %< $ œ ! , with roots < œ "Î# , $Î# Þ The
general solution is Ca>b œ -" />Î# -# /$>Î# .
4. The characteristic equation is %<# "#< * œ ! , with the double root < œ $Î# Þ
Based on the discussion in this section, the general solution is Ca>b œ a-" -# >b/$>Î# .
5. The characteristic equation is <# #< "! œ ! , with complex roots < œ " „ $3Þ
The general solution is Ca>b œ -" /> -9= $> -# /> =38 $> .
6. The characteristic equation is <# '< * œ ! , with the double root < œ $ Þ The
general solution is Ca>b œ -" /$> -# > /$> .
7. The characteristic equation is %<# "(< % œ ! , with roots < œ "Î% , % Þ
The general solution is Ca>b œ -" />Î% -# /%> .
8. The characteristic equation is "'<# #%< * œ ! , with the double root < œ $Î% Þ
The general solution is Ca>b œ -" /$>Î% -# > /$>Î% .
10. The characteristic equation is #<# #< " œ ! , with complex roots < œ "# „ "# 3Þ
The general solution is Ca>b œ -" />Î# -9= >Î# -# />Î# =38 >Î# .
11. The characteristic equation is *<# "#< % œ ! , with the double root < œ #Î$ Þ
The general solution is Ca>b œ -" /#>Î$ -# > /#>Î$ . Invoking the first initial condition, it
follows that -" œ # Þ Now C w a>b œ a%Î$ -# b/#>Î$ #-# > /#>Î$ Î$ . Invoking the second
initial condition, %Î$ -# œ " , or -# œ (Î$ . Hence C a>b œ #/#>Î$ ($ > /#>Î$ .
Since the second term dominates for large > , Ca>b p _ .
13. The characteristic equation is *<# '< )# œ !, with complex roots < œ "$ „$ 3Þ
The general solution is Ca>b œ -" />Î$ -9= $> -# />Î$ =38 $> . Based on the first initial
condition, -" œ " Þ Invoking the second initial condition, "Î$ $-# œ # , or -# œ &* .
Hence Ca>b œ />Î$ -9= $> &* />Î$ =38 $> .
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—————————————————————————— CHAPTER 3. ——
15a+b. The characteristic equation is %<# "#< * œ ! , with the double root < œ $# Þ
The general solution is Ca>b œ -" /$>Î# -# > /$>Î# . Invoking the first initial condition,
it follows that -" œ " Þ Now C w a>b œ a $Î# -# b/#>Î$ $# -# > /#>Î$ . The second
initial condition requires that $Î# -# œ % , or -# œ &Î# . Hence the specific
solution is Ca>b œ /$>Î# &# > /$>Î# .
a,b. The solution crosses the x-axis at > œ !Þ% .
a- b. The solution has a minimum at the point a"'Î"& ß &/)Î& Î$b.
a. b. Given that C w a!b œ , , we have $Î# -# œ , , or -# œ , $Î# . Hence the
solution is Ca>b œ /$>Î# ˆ, $# ‰> /$>Î# . Since the second term dominates, the longterm solution depends on the sign of the coefficient , $# . The critical value is , œ $# Þ
16. The characteristic roots are <" œ <# œ "Î# . Hence the general solution is given by
Ca>b œ -" />Î# -# > />Î# . Invoking the initial conditions, we require that -" œ # , and that
" -# œ ,. The specific solution is C a>b œ #/>Î# a, "b> />Î# . Since the second term
dominates, the long-term solution depends on the sign of the coefficient , ". The
critical value is , œ "Þ
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—————————————————————————— CHAPTER 3. ——
18a+b. The characteristic roots are <" œ <# œ #Î$ . Therefore the general solution is
given by Ca>b œ -" /#>Î$ -# > /#>Î$ . Invoking the initial conditions, we require that
-" œ + , and that #+Î$ -# œ " . After solving for the coefficients, the specific
‰ #>Î$ .
solution is Ca>b œ +/#>Î$ ˆ #+
$ " >/
a,b. Since the second term dominates, the long-term solution depends on the sign of the
coefficient #+
$ " . The critical value is + œ $Î# Þ
20a+b. The characteristic equation is <# #+< +# œ ! , with double root < œ +Þ
Hence one solution is C" a>b œ -" /+> .
a,b. Recall that the Wronskian satisfies the differential equation [ w #+[ œ ! . The
solution of this equation is [ a>b œ - /#+> .
a- b. By definition, [ œ C" C#w C"w C# . Hence -" /+> C#w +-" /+> C# œ - /#+> .
That is, C#w + C# œ -# /+> . This equation is first order linear, with general solution
C# a>b œ -# >/+> -$ /+> . Setting -# œ " and -$ œ ! , we obtain C# a>b œ >/+> Þ
22a+b. Write +<# ,< - œ +ˆ<# +, < +- ‰. It follows that +, œ #<" and +- œ <"# Þ
Hence +<# ,< - œ +<# #+<" < +<"# œ +a<# #<" < <"# b œ +a< <" b# Þ We
find that Pc/<> d œ a+<# ,< - b/<> œ +a< <" b# /<> Þ Setting < œ <" , Pc/<" > d œ ! .
a,b. Differentiating Eq.a3b with respect to <,
`
P/<> ‘ œ +>/<> a< <" b# #+/<> a< <" b.
`<
Now observe that
`
`
` # <>
`
ˆ/ ‰ , ˆ/<> ‰- ˆ/<> ‰•
P/<> ‘ œ
+
”
`<
`< `>#
`>
#
`
`
` `
`
œ ”+ # Œ /<> , Œ /<> - Œ /<> •
`> `<
`> `<
`<
#
`
`
œ + # ˆ>/<> ‰ , ˆ>/<> ‰- ˆ>/<> ‰.
`>
`>
Hence Pc>/<> d œ +>/<> a< <" b# #+/<> a< <" b. Setting < œ <" , Pc>/<" > d œ ! .
23. Set C# a>b œ ># @a>b . Substitution into the ODE results in
># ˆ># @ ww %>@ w #@‰ %>ˆ># @ w #>@‰ '># @ œ ! .
After collecting terms, we end up with >% @ ww œ ! . Hence @a>b œ -" -# > , and thus
C# a>b œ -" ># -# >$ . Setting -" œ ! and -# œ " , we obtain C# a>b œ >$ Þ
24. Set C# a>b œ > @a>b . Substitution into the ODE results in
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—————————————————————————— CHAPTER 3. ——
># a>@ ww #@ w b #>a>@ w @b #>@ œ ! .
After collecting terms, we end up with >$ @ ww %># @ w œ ! . This equation is linear in
the variable A œ @ w . It follows that @ w a>b œ - >% , and @a>b œ -" >$ -# . Thus
C# a>b œ -" ># -# > . Setting -" œ " and -# œ ! , we obtain C# a>b œ ># Þ
26. Set C# a>b œ > @a>b . Substitution into the ODE results in @ ww @ w œ ! . This ODE
is linear in the variable A œ @ w . It follows that @ w a>b œ -" /> , and @a>b œ -" /> -# .
Thus C# a>b œ -" >/> -# > . Setting -" œ " and -# œ ! , we obtain C# a>b œ >/> Þ
28. Set C# aBb œ /B @aBb . Substitution into the ODE results in
@ ww
B# w
@ œ !.
B"
This ODE is linear in the variable A œ @ w . An integrating factor is
. œ /B:Œ(
/B
œ
Þ
B"
B#
.B
B"
/ @ ‘
Rewrite the equation as B"
œ ! , from which it follows that @ w aBb œ - aB "b/B .
Hence @aBb œ -" B/B -# and C# aBb œ -" B -# /B . Setting -" œ " and -# œ ! , we
obtain C# aBb œ BÞ
B w
w
29. Set C# aBb œ C1 aBb @aBb , in which C1 aBb œ B"Î% /B:ˆ#ÈB‰. It can be verified that
C1 is a solution of the ODE, that is, B# C"ww aB !Þ")(&bC" œ ! . Substitution of the
given form of C# results in the differential equation
#B*Î% @ ww ˆ%B(Î% B&Î% ‰@ w œ ! .
This ODE is linear in the variable A œ @ w . An integrating factor is
. œ /B:Œ( ”#B"Î#
œ ÈB /B:ˆ%ÈB‰Þ
"
•.B
#B
Rewrite the equation as ÈB /B:ˆ%ÈB‰ @ w ‘ œ ! , from which it follows that
w
@ w aBb œ - /B:ˆ %ÈB‰ÎÈB .
Integrating, @aBb œ -" /B:ˆ %ÈB‰ -# and as a result,
C# aBb œ -" B"Î% /B:ˆ #ÈB‰ -# B"Î% /B:ˆ#ÈB‰.
Setting -" œ " and -# œ ! , we obtain C# aBb œ B"Î% /B:ˆ #ÈB‰Þ
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page 104
—————————————————————————— CHAPTER 3. ——
32. Direct substitution verifies that C" a>b œ /B:a $ B# Î#b is a solution of the ODE.
Now set C# aBb œ C1 aBb @aBb. Substitution of C# into the ODE results in
@ ww $ B@ w œ ! .
This ODE is linear in the variable A œ @ w . An integrating factor is . œ /B:a $ B# Î#b.
Rewrite the equation as c /B:a $ B# Î#b@ w dw œ ! , from which it follows that
@ w aBb œ -" /B:ˆ$ B# Î#‰ .
Integrating, we obtain
@aBb œ -" ( /B:ˆ$ ?# Î#‰.? @aB! b.
B
B!
Hence
C# aBb œ -" /B:ˆ $ B# Î#‰( /B:ˆ$ ?# Î#‰.? -# /B:ˆ $B# Î#‰.
B
B!
Setting -# œ ! , we obtain a second independent solution.
34. After writing the ODE in standard form, we have :a>b œ $Î> . Based on Abel's
identity, [ aC" ß C# b œ -" /B:ˆ ' $> .>‰ œ -" >$ Þ As shown in Prob. 33, two solutions
of a second order linear equation satisfy
aC# ÎC" bw œ [ aC" ß C# bÎC"# .
In the given problem, C" a>b œ >" . Hence a> C# bw œ -" >" Þ Integrating both sides of the
equation, C# a>b œ -" >" 68 > -# >" Þ
36. After writing the ODE in standard form, we have :aBb œ BÎaB "b. Based on
B
Abel's identity, [ aC" ß C# b œ - /B:ˆ' B"
.B‰ œ - /B aB "bÞ Two solutions of a
second order linear equation satisfy
aC# ÎC" bw œ [ aC" ß C# bÎC"# .
In the given problem, C" aBb œ /B . Hence a/B C# bw œ - /B aB "bÞ Integrating both
sides of the equation, C# aBb œ -" B -# /B Þ Setting -" œ " and -# œ ! , we obtain
C# aBb œ B Þ
37. Write the ODE in standard form to find :aBb œ "ÎB. Based on Abel's identity,
[ aC" ß C# b œ - /B:ˆ ' "B .B‰ œ - B" Þ Two solutions of a second order linear ODE
satisfy aC# ÎC" bw œ [ aC" ß C# bÎC"# . In the given problem, C" aBb œ B"Î# =38 B . Hence
ÈB
w
"
=38 B C# œ - =38# B Þ
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—————————————————————————— CHAPTER 3. ——
Integrating both sides of the equation, C# aBb œ -" B"Î# -9= B -# B"Î# =38 BÞ Setting
-" œ " and -# œ ! , we obtain C# aBb œ B"Î# -9= B Þ
39a+b. The characteristic equation is +<# - œ ! . If + , - ! , then the roots are
<"ß# œ „ 3È-Î+ . The general solution is
Ca>b œ -" -9=Ê > -# =38Ê > ,
+
+
which is bounded.
a,bÞ The characteristic equation is +<# ,< œ ! . The roots are <"ß# œ ! , ,Î+ ,
and hence the general solution is Ca>b œ -" -# /B:a ,>Î+bÞ Clearly, C a>b p -" .
40. Note that -9= > =38 > œ "# =38 #> . So that " 5 -9= > =38 > œ " 5# =38 #> . If
! 5 # , then 5# =38 #> k=38 #>k and 5# =38 #> k=38 #>k. Hence
5
=38 #>
#
" k=38 #>k
!Þ
" 5 -9= > =38 > œ "
41. :a>b œ $Î> and ; a>b œ %Î># . We have B œ #' >" .> œ # 68 > , and > œ /BÎ# Þ
Furthermore,
; w a>b #:a>b; a>b
œ #.
#c; a>bd$Î#
The ratio is constant, and therefore the equation can be transformed. In fact, we obtain
.# C
.C
#
C œ !Þ
#
.B
.B
The general solution of this ODE is CaBb œ -" /B -# B/B . In terms of the original
independent variable, Ca>b œ -" ># -# ># 68 > .
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page 106
—————————————————————————— CHAPTER 3. ——
Section 3.6
2. The characteristic equation for the homogeneous problem is <# #< & œ ! , with
complex roots < œ "„#3 . Hence C- a>b œ -" /> -9= #> -# /> =38 #> . Since the
function 1a>b œ $ =38 #> is not proportional to the solutions of the homogeneous equation,
set ] œ E -9= #> F =38 #> . Substitution into the given ODE, and comparing the
coefficients, results in the system of equations F %E œ $ and E %F œ ! . Hence
$
] œ "#
"( -9= #> "( =38 #> . The general solution is C a>b œ C- a>b ] Þ
3. The characteristic equation for the homogeneous problem is <# #< $ œ ! , with
roots < œ " , $ . Hence C- a>b œ -" /> -# /$> . Note that the assignment ] œ E>/>
is not sufficient to match the coefficients. Try ] œ E>/> F># /> . Substitution into
the differential equation, and comparing the coefficients, results in the system of
$
equations %E #F œ ! and )F œ $ . Hence ] œ "'
>/> )$ ># /> . The
general
solution is Ca>b œ C- a>b ] Þ
5. The characteristic equation for the homogeneous problem is <# * œ ! , with
complex roots < œ „$3. Hence C- a>b œ -" -9= $> -# =38 $> . To simplify the analysis,
set 1" a>b œ ' and 1# a>b œ ># /$> . By inspection, we have ]" œ #Î$ . Based on the form
of 1# , set ]# œ E/$> F>/$> G># /$> . Substitution into the differential equation, and
comparing the coefficients, results in the system of equations ")E 'F #G œ ! ,
")F "#G œ ! , and ")G œ " . Hence
]# œ
" $>
"
"
/ >/$> ># /$> .
"'#
#(
")
The general solution is Ca>b œ C- a>b ]" ]# Þ
7. The characteristic equation for the homogeneous problem is #<# $< " œ ! , with
roots < œ " , "Î# . Hence C- a>b œ -" /> -# />Î# . To simplify the analysis,
set 1" a>b œ ># and 1# a>b œ $ =38 > . Based on the form of 1" , set ]" œ E F> G># Þ
Substitution into the differential equation, and comparing the coefficients, results in the
system of equations E $F %G œ ! , F 'G œ ! , and G œ " . Hence we obtain
]" œ "% '> ># Þ On the other hand, set ]# œ H -9= > I =38 > . After substitution
into the ODE, we find that H œ *Î"! and I œ $Î"! . The general solution is
C a >b œ C - a > b ] " ] # Þ
9. The characteristic equation for the homogeneous problem is <# =#! œ ! , with
complex roots < œ „ =! 3. Hence C- a>b œ -" -9= =! > -# =38 =! > . Since = Á =! ,
set ] œ E -9= => F =38 => . Substitution into the ODE and comparing the coefficients
results in the system of equations a=#! =# bE œ " and a=#! =# bF œ ! . Hence
] œ
=#
!
The general solution is Ca>b œ C- a>b ] .
"
-9= => Þ
=#
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page 107
—————————————————————————— CHAPTER 3. ——
10. From Prob. 9, C- a>b œ - . Since -9= =! > is a solution of the homogeneous problem,
set ] œ E> -9= =! > F> =38 =! > . Substitution into the given ODE and comparing the
coefficients results in E œ ! and F œ #=" ! Þ Hence the general solution is
Ca>b œ -" -9= =! > -# =38 =! > #=> ! =38 =! > .
12. The characteristic equation for the homogeneous problem is <# < # œ ! , with
roots < œ " , # . Hence C- a>b œ -" /> -# /#> . Based on the form of the right hand
side, that is, -9=2Ð#>Ñ œ a/#> /#> bÎ# , set ] œ E> /#> F/#> . Substitution into the
given ODE and comparing the coefficients results in E œ "Î' and F œ "Î) Þ Hence the
general solution is C a>b œ -" /> -# /#> > /#> Î' /#> Î) .
14. The characteristic equation for the homogeneous problem is <# % œ ! , with roots
< œ „ #3. Hence C- a>b œ -" -9= #> -# =38 #> . Set ]" œ E F> G># Þ Comparing
the coefficients of the respective terms, we find that E œ "Î) , F œ ! , G œ "Î% .
Now set ]# œ H /> , and obtain H œ $Î& . Hence the general solution is
Ca>b œ -" -9= #> -# =38 #> "Î) ># Î% $ /> Î& .
Invoking the initial conditions, we require that "*Î%! -" œ ! and $Î& #-# œ # .
Hence -" œ "*Î%! and -# œ (Î"! .
15. The characteristic equation for the homogeneous problem is <# #< " œ ! , with
a double root < œ " . Hence C- a>b œ -" /> -# > /> . Consider 1" a>b œ > /> . Note that
1" is a solution of the homogeneous problem. Set ]" œ E># /> F>$ /> Ðthe first term is
not sufficient for a matchÑ. Upon substitution, we obtain ]" œ >$ /> Î' . By inspection,
]# œ % . Hence the general solution is C a>b œ -" /> -# > /> >$ /> Î' % . Invoking the
initial conditions, we require that -" % œ " and -" -# œ " . Hence -" œ $ and
-# œ % .
17. The characteristic equation for the homogeneous problem is <# % œ ! , with roots
< œ „#3. Hence C- a>b œ -" -9= #> -# =38 #> . Since the function =38 #> is a solution of
the homogeneous problem, set ] œ E> -9= #> F> =38 #> . Upon substitution, we obtain
] œ $% > -9= #> . Hence the general solution is C a>b œ -" -9= #> -# =38 #> "% > -9= #> .
Invoking the initial conditions, we require that -" œ # and #-# $% œ " . Hence
-" œ # and -# œ "Î) .
18. The characteristic equation for the homogeneous problem is <# #< & œ ! , with
complex roots < œ "„ #3. Hence C- a>b œ -" /> -9= #> -# /> =38 #> . Based on the
form of 1a>b, set ] œ E> /> -9= #> F> /> =38 #> . After comparing coefficients, we
obtain ] œ > /> =38 #> . Hence the general solution is
Ca>b œ -" /> -9= #> -# /> =38 #> > /> =38 #> .
Invoking the initial conditions, we require that -" œ " and -" #-# œ ! . Hence
-" œ " and -# œ "Î# .
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—————————————————————————— CHAPTER 3. ——
20. The characteristic equation for the homogeneous problem is <# " œ ! , with
complex roots < œ „ 3. Hence C- a>b œ -" -9= > -# =38 >. Let 1" a>b œ > =38 > and
1# a>b œ > . By inspection, it is easy to see that ]# a>b œ " . Based on the form of 1" a>b,
set ]" a>b œ E> -9= > F> =38 > G># -9= > H># =38 > Þ Substitution into the equation
and comparing the coefficients results in E œ ! , F œ "Î% , G œ "Î% , and H œ ! Þ
Hence ] a>b œ " "% > =38 > "% ># -9= > .
21. The characteristic equation for the homogeneous problem is <# &< ' œ ! , with
roots < œ # , $. Hence C- a>b œ -" /#> -# /$> . Consider 1" a>b œ /#> a$> %b=38 > , and
1# a>b œ /> -9= #>. Based on the form of these functions on the right hand side of the
ODE,
set ]# a>b œ /> aE" -9= #> E# =38 #>b, ]" a>b œ aF" F# > b/#> =38 > aG" G#>b/#> -9= >.
Substitution into the equation and comparing the coefficients results in
] a>b œ
" >
$
ˆ/ -9= #> $/> =38 #>‰ >/#> a-9= > =38 >b /#> Š " -9= > &=38 >‹.
#
#!
#
23. The characteristic roots are < œ # , #. Hence C- a>b œ -" /#> -# >/#> . Consider the
functions 1" a>b œ #># , 1# a>b œ %>/#> , and 1$ a>b œ > =38 #> . The corresponding forms of
the respective parts of the particular solution are ]" a>b œ E! E" > E# ># , ]# a>b œ
œ /#> aF# ># F$ >$ b, and ]$ a>b œ >aG" -9= #> G# =38 #>b aH" -9=#> H# =38#>b.
Substitution into the equation and comparing the coefficients results in
] a>b œ
"
#
"
"
ˆ$ %> #># ‰ >$ /#> > -9= #> a-9= #> =38 #>b .
%
$
)
"'
24. The homogeneous solution is C- a>b œ -" -9= #> -# =38 #>. Since -9= #> and =38 #>
are both solutions of the homogeneous equation, set
] a>b œ >ˆE! E" > E# ># ‰-9= #> >ˆF! F" > F# ># ‰=38 #> .
Substitution into the equation and comparing the coefficients results in
] a>b œ Œ
"$
"
"
> >$ -9= #> ˆ#)> "$># ‰=38 #> .
$#
"#
"'
25. The homogeneous solution is C- a>b œ -" /> -# >/#> . None of the functions on the
right hand side are solutions of the homogenous equation. In order to include all possible
combinations of the derivatives, consider ] a>b œ /> aE! E" > E# ># b-9= #>
/> aF! F" > F# ># b=38 #> /> aG" -9= > G# =38 >b H/> . Substitution into the
differential equation and comparing the coefficients results in
] a>b œ /> ˆE! E" > E# ># ‰-9= #> /> ˆF! F" > F# ># ‰=38 #>
/> Š -9= > =38 >‹ #/> Î$ ,
#
$
#
$
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—————————————————————————— CHAPTER 3. ——
in which E! œ %"!&Î$&"&# ß E" œ ($Î'(' ß E# œ &Î&# ß F! œ "#$$Î$&"&# ß
F" œ "!Î"'* ß F# œ "Î&# Þ
26. The homogeneous solution is C- a>b œ -" /> -9= #> -# /> =38 #>. None of the terms
on the right hand side are solutions of the homogenous equation. In order to include the
appropriate combinations of derivatives, consider ] a>b œ /> aE" > E# ># b-9= #>
/> aF" > F# ># b=38 #> /#> aG! G" >b-9= #> /#> aH! H" >b=38 #>.
Substitution into the differential equation and comparing the coefficients results in
] a>b œ
$ >
$
"
>/ -9= #> ># /> =38 #> /#> a( "!>b-9= #>
"'
)
#&
" #>
/ a" &>b=38 #> .
#&
27. The homogeneous solution is C- a>b œ -" -9= -> -# =38 ->. Since the differential
operator does not contain a first derivative aand - Á 71b, we can set
] a>b œ "G7 =38 71> .
R
7œ"
Substitution into the ODE yields
" 7# 1# G7 =38 71> -# "G7 =38 71> œ "+7 =38 71> .
R
R
R
7œ"
7œ"
7œ"
Equating coefficients of the individual terms, we obtain
+7
G7 œ #
, 7 œ "ß # â R .
- 7# 1#
29. The homogeneous solution is C- a>b œ -" /> -9= #> -# /> =38 #>. The input function
is independent of the homogeneous solutions, on any interval. Since the right hand side
is
piecewise constant, it follows by inspection that
] a>b œ œ
"Î& ,
! ,
! Ÿ > Ÿ 1Î#
Þ
> 1Î#
For ! Ÿ > Ÿ 1Î# , the general solution is C a>b œ -" /> -9= #> -# /> =38 #> "Î& .
Invoking the initial conditions Ca!b œ C w a!b œ ! , we require that -" œ "Î& , and that
-# œ "Î"! . Hence
C a >b œ
"
"
ˆ#/> -9= #> /> =38 #>‰
& "!
on the interval ! Ÿ > Ÿ 1Î# . We now have the values C a1Î#b œ ˆ" /1Î# ‰Î& , and
C w a1Î#b œ ! . For > 1Î# , the general solution is C a>b œ ." /> -9= #> .# /> =38 #> .
It follows that Ca1Î#b œ /1Î# ." and C w a1Î#b œ /1Î# ." #/1Î# .# . Since the
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—————————————————————————— CHAPTER 3. ——
solution is continuously differentiable, we require that
/1Î# ." œ ˆ" /1Î# ‰Î&
/1Î# ." #/1Î# .# œ ! .
Solving for the coefficients, ." œ #.# œ ˆ/1Î# "‰Î& .
31. Since + ß ,ß - ! , the roots of the characteristic equation has negative real parts.
That is, < œ !„" 3 , where ! ! . Hence the homogeneous solution is
C- a>b œ -" /!> -9= " > -# /!> =38 " > .
If 1a>b œ . , then the general solution is
Ca>b œ .Î- -" /!> -9= " > -# /!> =38 " > .
Since ! ! , Ca>b p .Î- as > p _ . If - œ !, then that characteristic roots are < œ ! and
< œ ,Î+ . The ODE becomes +C ww ,C w œ . . Integrating both sides, we find that
+C w ,C œ . > -" . The general solution can be expressed as
Ca>b œ . >Î, -" -# /,>Î+ Þ
In this case, the solution grows without bound. If , œ !, also, then the differential
equation
can be written as C ww œ .Î+ , which has general solution C a>b œ . ># Î#+ -" -# .
Hence the assertion is true only if the coefficients are positive.
32a+b. Since H is a linear operator,
H# C ,HC -C œ H# C a<" <# bHC <" <# C
œ H# C <# HC <" HC <" <# C
œ HaHC <# C b <" aHC <# C b
œ aH <" baH <# bC .
a,b. Let ? œ aH <# bC . Then the ODE a3b can be written as aH <" b? œ 1a>b, that is,
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—————————————————————————— CHAPTER 3. ——
? w <" ? œ 1a>b. The latter is a linear first order equation in ? . Its general solution is
?a>b œ /<" > ( /<" 7 1a7 b. 7 -" /<" > .
>
>!
From above, we have C w <# C œ ?a>b. This equation is also a first order ODE. Hence
the general solution of the original second order equation is
Ca>b œ / ( /<# 7 ?a7 b. 7 -# /<# > .
>
<# >
>!
Note that the solution Ca>b contains two arbitrary constants.
34. Note that a#H# $H "bC œ a#H "baH "bC . Let ? œ aH "bC , and solve
the ODE #? w ? œ ># $=38 > Þ This equation is a linear first order ODE, with solution
?a>b œ />Î# ( /7 Î# ’7 # Î# =38 7 “. 7 - />Î#
>
$
#
>!
'
$
œ ># %> ) -9= > =38 > - />Î# Þ
&
&
Now consider the ODE C w C œ ?a>b. The general solution of this first order ODE is
Ca>b œ /> ( /7 ?a7 b. 7 -# /> ,
>
>!
in which ?a>b is given above. Substituting for ?a>b and performing the integration,
Ca>b œ ># '> "%
*
$
-9= > =38 > -" />Î# -# /> Þ
"!
"!
35. We have aH# #H "bC œ aH "baH "bC . Let ? œ aH "bC , and consider
the ODE ? w ? œ #/> Þ The general solution is ?a>b œ #> /> - /> Þ We therefore
have the first order equation ? w ? œ #> /> -" /> Þ The general solution of the latter
differential equation is
Ca>b œ / ( c#7 -" d. 7 -# />
>
œ/
>
>!
> ˆ #
> -" > -# ‰Þ
36. We have aH# #HbC œ HaH #bC . Let ? œ aH #bC , and consider the equation
? w œ $ %=38 #> Þ Direct integration results in ?a>b œ $> #-9= #> - Þ The problem
is reduced to solving the ODE C w #C œ $> #-9= #> - Þ The general solution of this
first order differential equation is
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—————————————————————————— CHAPTER 3. ——
Ca>b œ /#> ( /#7 c$7 #-9= #7 - d. 7 -# /#>
>
>!
$
"
œ > a-9= #> =38 #>b -" -# /#> Þ
#
#
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—————————————————————————— CHAPTER 3. ——
Section 3.7
1. The solution of the homogeneous equation is C- a>b œ -" /#> -# /$> . The functions
C" a>b œ /#> and C# a>b œ /$> form a fundamental set of solutions. The Wronskian of
these functions is [ aC" ß C# b œ /&> . Using the method of variation of parameters, the
particular solution is given by ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which
?" a>b œ (
œ #/>
/$> a#/> b
.>
[ a>b
/#> a#/> b
.>
[ a>b
œ /#>
?# a>b œ (
Hence the particular solution is ] a>b œ #/> /> œ /> .
3. The solution of the homogeneous equation is C- a>b œ -" /> -# >/> . The functions
C" a>b œ /> and C# a>b œ >/> form a fundamental set of solutions. The Wronskian of
these functions is [ aC" ß C# b œ /#> . Using the method of variation of parameters, the
particular solution is given by ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which
>/> a$/> b
.>
[ a>b
œ $># Î#
?" a>b œ (
?# a>b œ (
œ $>
/> a$/> b
.>
[ a>b
Hence the particular solution is ] a>b œ $># /> Î# $># /> œ $># /> Î# .
4. The functions C" a>b œ />Î# and C# a>b œ >/>Î# form a fundamental set of solutions.
The Wronskian of these functions is [ aC" ß C# b œ /> . First write the equation in standard
form, so that 1a>b œ %/>Î# . Using the method of variation of parameters, the particular
solution is given by ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which
?" a>b œ (
>/>Î# ˆ%/>Î# ‰
.>
[ a>b
œ #>#
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—————————————————————————— CHAPTER 3. ——
/>Î# ˆ%/>Î# ‰
?# a>b œ (
.>
[ a>b
œ %>
Hence the particular solution is ] a>b œ #># />Î# %># />Î# œ #># />Î# .
6. The solution of the homogeneous equation is C- a>b œ -" -9= $> -# =38 $>. The two
functions C" a>b œ -9= $> and C# a>b œ =38 $> form a fundamental set of solutions, with
[ aC" ß C# b œ $ . The particular solution is given by ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in
which
=38 $>a* =/- # $>b
.>
[ a>b
œ -=- $>
?" a>b œ (
-9= $>a* =/- # $>b
.>
[ a>b
œ 68k=/- $> >+8 $>k
?# a>b œ (
Hence the particular solution is ] a>b œ " a=38 $>b68k=/- $> >+8 $>k. The general
solution is given by Ca>b œ -" -9= $> -# =38 $> a=38 $>b68k=/- $> >+8 $>k " .
7. The functions C" a>b œ /#> and C# a>b œ >/#> form a fundamental set of solutions.
The Wronskian of these functions is [ aC" ß C# b œ /%> . The particular solution is given
by ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which
>/#> a># /#> b
?" a>b œ (
.>
[ a>b
œ 68 >
/#> a># /#> b
.>
[ a>b
œ "Î>
?# a>b œ (
Hence the particular solution is ] a>b œ /#> 68 > /#> . Since the second term is a
solution of the homogeneous equation, the general solution is given by Ca>b œ -" /#>
-# >/#> /#> 68 > .
8. The solution of the homogeneous equation is C- a>b œ -" -9= #> -# =38 #>. The two
functions C" a>b œ -9= #> and C# a>b œ =38 #> form a fundamental set of solutions, with
[ aC" ß C# b œ # . The particular solution is given by ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in
which
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—————————————————————————— CHAPTER 3. ——
=38 #>a$ -=- #>b
.>
[ a>b
œ $>Î#
?" a>b œ (
?# a>b œ (
-9= #>a$ -=- #>b
.>
[ a>b
œ 68k=38 #>k
$
%
Hence the particular solution is ] a>b œ $# >-9= #> $% a=38 $>b68k=38 #>k. The general
solution is given by Ca>b œ -" -9= #> -# =38 #> $# >-9= #> $% a=38 $>b68k=38 #>k.
9. The functions C" a>b œ -9= a>Î#b and C# a>b œ =38a>Î#b form a fundamental set of
solutions. The Wronskian of these functions is [ aC" ß C# b œ "Î# . First write the ODE
in standard form, so that 1a>b œ =/- a>Î#bÎ#. The particular solution is given by
] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which
-9= a>Î#bc=/- a>Î#bd
.>
#[ a>b
œ # 68c-9= a>Î#bd
?" a>b œ (
?# a>b œ (
œ>
=38a>Î#bc=/- a>Î#bd
.>
#[ a>b
The particular solution is ] a>b œ #-9=a>Î#b68c-9= a>Î#bd > =38a>Î#b. The general
solution is given by
Ca>b œ -" -9= a>Î#b -# =38a>Î#b # -9=a>Î#b 68c-9= a>Î#bd > =38a>Î#b.
10. The solution of the homogeneous equation is C- a>b œ -" /> -# >/> . The functions
C" a>b œ /> and C# a>b œ >/> form a fundamental set of solutions, with [ aC" ß C# b œ /#> .
The particular solution is given by ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which
?" a>b œ (
>/> a/> b
.>
[ a>ba" ># b
"
œ 68ˆ" ># ‰
#
/ > a/ > b
.>
[ a>ba" ># b
œ +<->+8 >
?# a>b œ (
The particular solution is ] a>b œ "# /> 68a" ># b >/> +<->+8a>b. Hence the general
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—————————————————————————— CHAPTER 3. ——
solution is given by Ca>b œ -" /> -# >/> "# /> 68a" ># b >/> +<->+8a>b.
12. The functions C" a>b œ -9= #> and C# a>b œ =38 #> form a fundamental set of
solutions, with [ aC" ß C# b œ # . The particular solution is given by ] a>b œ ?" a>b
C" a>b ?# a>b C# a>b,
in which
" >
?" a>b œ ( 1a=b =38 #= .=
#
?# a>b œ
" >
( 1a=b -9= #= .=
#
Hence the particular solution is
>
>
"
"
] a>b œ -9= #>( 1a=b =38 #= .= =38 #>( 1a=b -9= #= .= .
#
#
Note that =38 #> -9= #= -9= #> =38 #= œ =38a#> #=b. It follows that
] a>b œ
" >
( 1a=b=38a#> #=b.= .
#
The general solution of the differential equation is given by
" >
Ca>b œ -" -9= #> -# =38 #> ( 1a=b=38a#> #=b.= .
#
13. Note first that :a>b œ ! , ; a>b œ #Î># and 1a>b œ a$># "bÎ># . The functions
C" a>b and C# a>b are solutions of the homogeneous equation, verified by substitution. The
Wronskian of these two functions is [ aC" ,C# b œ $ . Using the method of variation of
parameters, the particular solution is ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which
>" a$># "b
?" a>b œ (
.>
># [ a > b
œ ># Î' 68 >
># a$># "b
.>
># [ a>b
œ >$ Î$ >Î$
?# a>b œ (
Therefore ] a>b œ "Î' ># 68 > ># Î$ "Î$ Þ Hence the general solution is
Ca>b œ -" ># -# >" ># 68 > "Î# .
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—————————————————————————— CHAPTER 3. ——
15. Observe that 1a>b œ > /#> . The functions C" a>b and C# a>b are a fundamental set of
solutions. The Wronskian of these two functions is [ aC" ,C# b œ > /> . Using the method
of variation of parameters, the particular solution is ] a>b œ ?" a>b C" a>b ?# a>b C# a>b,
in which
/> a> /#> b
?" a>b œ (
.>
[ a>b
œ /#> Î#
?# a>b œ (
œ > />
a" >ba> /#> b
.>
[ a>b
Therefore ] a>b œ a" >b/#> Î# > /#> œ /#> Î# > /#> Î# Þ
16. Observe that 1a>b œ #a" >b /> . Direct substitution of C" a>b œ /> and C# a>b œ >
verifies that they are solutions of the homogeneous equation. The Wronskian of the two
solutions is [ aC" ,C# b œ a" >b /> . Using the method of variation of parameters, the
particular solution is ] a>b œ ?" a>b C" a>b ?# a>b C# a>b, in which
#>a" >b/>
?" a>b œ (
.>
[ a>b
œ >/#> /#> Î#
#a" >b
.>
[ a>b
œ # />
?# a>b œ (
Therefore ] a>b œ >/> /> Î# #> /> œ >/> /> Î# Þ
17. Note that 1aBb œ 68 B . The functions C" aBb œ B# and C# aBb œ B# 68 B are solutions
of the homogeneous equation, as verified by substitution. The Wronskian of the solutions
is [ aC" ,C# b œ B$ . Using the method of variation of parameters, the particular solution is
] aBb œ ?" aBb C" aBb ?# aBb C# aBb,
in which
B# 68 Ba68 Bb
.B
[ aBb
œ a68 Bb$ Î$
?" aBb œ (
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—————————————————————————— CHAPTER 3. ——
B# a68 Bb
.B
[ aBb
œ a68 Bb# Î#
?# aBb œ (
Therefore ] aBb œ B# a68 Bb$ Î$ B# a68 Bb$ Î# œ B# a68 Bb$ Î' Þ
19. First write the equation in standard form. Note that the forcing function becomes
1aBbÎa" Bb . The functions C" aBb œ /B and C# aBb œ B are a fundamental set of
solutions,
as verified by substitution. The Wronskian of the solutions is [ aC" ,C# b œ a" Bb/B .
Using the method of variation of parameters, the particular solution is
] aBb œ ?" aBb C" aBb ?# aBb C# aBb,
in which
?" aBb œ (
?# aBb œ (
Therefore
B
B
7 a1 a7 bb
.7
a" 7 b[ a 7 b
/ 7 a1 a 7 b b
.7
a" 7 b[ a 7 b
7 a1 a7 bb
. 7 B(
a" 7 b[ a 7 b
aB/7 /B 7 b1a7 b
.7 Þ
a" 7 b # / 7
] aBb œ /B (
œ(
B
B
B
/ 7 a1 a 7 bb
.7
a" 7 b[ a 7 b
20. First write the equation in standard form. The forcing function becomes 1aBbÎB# .
The functions C" aBb œ B"Î# =38 B and C# aBb œ B"Î# -9= B are a fundamental set of
solutions. The Wronskian of the solutions is [ aC" ,C# b œ "ÎB . Using the method
of variation of parameters, the particular solution is
] aBb œ ?" aBb C" aBb ?# aBb C# aBb,
in which
?" aBb œ (
B
?# aBb œ (
-9= 7 a1a7 bb
.7
7 È7
B
=38 7 a1a7 bb
.7
7 È7
Therefore
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—————————————————————————— CHAPTER 3. ——
=38 B B -9= 7 a1a7 bb
-9= B
.>
(
ÈB
ÈB (
7 È7
B
"
=38aB 7 b 1a7 b
œ
.7 Þ
ÈB (
7 È7
] aBb œ
B
=38 7 a1a7 bb
.7
7 È7
21. Let C" a>b and C# a>b be a fundamental set of solutions, and [ a>b œ [ aC" ß C# b be the
corresponding Wronskian. Any solution, ?a>b, of the homogeneous equation is a linear
combination ?a>b œ !" C" a>b !# C# a>b. Invoking the initial conditions, we require that
C! œ !" C" a>! b !# C# a>! b
C!w œ !" C"w a>! b !# C#w a>! b
Note that this system of equations has a unique solution, since [ a>! b Á !. Now consider
the nonhomogeneous problem, Pc@d œ 1a>b ,with homogeneous initial conditions. Using
the method of variation of parameters, the particular solution is given by
] a>b œ C" a>b(
>
>!
>
C# a=b 1a=b
C" a=b 1a=b
.= C# a>b(
.= Þ
[ a =b
[ a=b
>!
The general solution of the IVP a333b is
@a>b œ "" C" a>b "# C# a>b ] a>b
œ "" C" a>b "# C# a>b C" a>b?" a>b C# a>b?# a>b
in which ?" and ?# are defined above. Invoking the initial conditions, we require that
! œ "" C" a>! b "# C# a>! b ] a>! b
! œ "" C"w a>! b "# C#w a>! b ] w a>! b
Based on the definition of ?" and ?# , ] a>! b œ ! . Furthermore, since C" ?"w C# ?#w œ ! ,
it follows that ] w a>! b œ ! . Hence the only solution of the above system of equations is
the trivial solution. Therefore @a>b œ ] a>b . Now consider the function C œ ?@ . Then
PcCd œ Pc? @d œ Pc?d Pc@d œ 1a>b. That is, C a>b is a solution of the
nonhomogeneous
problem. Further, Ca>! b œ ?a>! b @a>! b œ C! , and similarly, C w a>! b œ C!w . By the
uniqueness theorems, Ca>b is the unique solution of the initial value problem.
23. A fundamental set of solutions is C" a>b œ -9= > and C# a>b œ =38 > . The Wronskian
[ a>b œ C" C#w C"w C# œ " . By the result in Prob. ##,
] a>b œ (
>
>!
>
-9=a=b =38a>b -9=a>b =38a=b
1a=b.=
[ a =b
œ ( c-9=a=b =38a>b -9=a>b =38a=bd1a=b.= Þ
>!
Finally, we have -9=a=b =38a>b -9=a>b =38a=b œ =38a> =b.
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—————————————————————————— CHAPTER 3. ——
24. A fundamental set of solutions is C" a>b œ /+> and C# a>b œ /,> . The Wronskian
[ a>b œ C" C#w C"w C# œ a, +b/B:ca+ , b>d . By the result in Prob. ##,
] a>b œ (
/ / /+> /,=
1a=b.=
[ a =b
>!
> += ,>
"
/ / /+> /,=
œ
1a=b.= Þ
(
, + >! /B:ca+ , b=d
> += ,>
Hence the particular solution is
>
"
,a>=b
] a>b œ
/+a>=b ‘1a=b.= Þ
( /
, + >!
26. A fundamental set of solutions is C" a>b œ /+> and C# a>b œ >/+> . The Wronskian
[ a>b œ C" C#w C"w C# œ /#+> . By the result in Prob. ##,
/ / /+> /,=
] a>b œ (
1a=b.=
[ a =b
>!
> += ,>
"
/ / /+> /,=
œ
1a=b.= Þ
(
, + >! /B:ca+ , b=d
> += ,>
Hence the particular solution is
] a>b œ
>
"
,a>=b
/+a>=b ‘1a=b.= Þ
( /
, + >!
26. A fundamental set of solutions is C" a>b œ /+> and C# a>b œ >/+> . The Wronskian
[ a>b œ C" C#w C"w C# œ /#+> . By the result in Prob. ##,
] a>b œ (
>/+=+> = /+>+=
1a=b.=
[ a =b
>!
>
a> =b/+=+>
œ(
1a=b.= Þ
/#+=
>!
>
Hence the particular solution is
] a>b œ ( a> =b/+a>=b 1a=b.= Þ
>
>!
27. Depending on the values of +, , and - , the operator +H# ,H - can have three
types of fundamental solutions.
a 3b
The characteristic roots <"ß# œ ! , " ; ! Á " . C" a>b œ /!> and C# a>b œ /"> .
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—————————————————————————— CHAPTER 3. ——
O a>b œ
"
/"> /!> ‘Þ
"!
a33b The characteristic roots <"ß# œ ! , " ; ! œ " . C" a>b œ /!> and C# a>b œ >/!> .
O a>b œ > /!> Þ
a333b The characteristic roots <"ß# œ -„ 3 . . C" a>b œ /-> -9= .> and C# a>b œ /-> =38 .>.
O a>b œ
" ->
/ =38 .> Þ
.
28. Let Ca>b œ @a>bC" a>b, in which C" a>b is a solution of the homogeneous equation.
Substitution into the given ODE results in
@ ww C" #@ w C"w @C"ww :a>bc@ w C" @C"w d ; a>b@C" œ 1a>b .
By assumption, C"ww :a>bC" ; a>bC" œ ! , hence @a>b must be a solution of the ODE
@ ww C" c#C"w :a>bC" d@ w œ 1a>b .
Setting A œ @ w , we also have A w C" c#C"w :a>bC" dA œ 1a>b .
30. First write the equation as C ww (>" C &># C œ >" . As shown in Prob. #), the
function Ca>b œ >" @a>b is a solution of the given ODE as long as @ is a solution of
>" @ ww c #># (># d@ w œ >" ,
that is, @ ww &>" @ w œ " . This ODE is linear and first order in @ w . The integrating
factor is . œ >& . The solution is @ w œ >Î' - >& . Direct integration now results in
@a>b œ ># Î"# -" >% -# . Hence C a>b œ >Î"# -" >& -# >" .
31. Write the equation as C ww >" a" >bC >" C œ > /#> . As shown in Prob. #), the
function Ca>b œ a" >b@a>b is a solution of the given ODE as long as @ is a solution of
a" >b @ ww # >" a" >b# ‘@ w œ > /#> ,
that is, @ ww
">#
>a>"b
> #>
>" / .
# >
@w œ
This equation is first order linear in @ w , with integrating
factor . œ >" a" >b / . The solution is @ w œ a># /#> -" >/> bÎa" >b# . Integrating,
we obtain @a>b œ /#> Î# /#> Îa> "b -" /> Îa> "b -# Þ Hence the solution of the
original ODE is Ca>b œ a> "b/#> Î# -" /> -# a> "b .
32. Write the equation as C ww >a" >b" C a" >b C œ #a" >b /> . The function
Ca>b œ /> @a>b is a solution to the given ODE as long as @ is a solution of
"
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—————————————————————————— CHAPTER 3. ——
/> @ ww #/> >a" >b" /> ‘@ w œ #a" >b /> ,
that is, @ ww ca# >bÎa" >bd@ w œ #a" >b /#> . This equation is first order linear in
@ w , with integrating factor . œ /> Îa> "b. The solution is
@ w œ a> "bˆ#/#> -" /> ‰.
Integrating, we obtain @a>b œ a"Î# >b/#> -" >/> -# Þ Hence the solution of the
original ODE is Ca>b œ a"Î# >b/> -" > -# /> .
Section 3.8
1. V-9= $ œ $ and V=38 $ œ % Ê V œ È#& œ & and $ œ +<->+8Ð%Î$ÑÞ Hence
? œ & -9=a#> !Þ*#($bÞ
3. V-9= $ œ % and V=38 $ œ # Ê V œ È#! œ #È& and $ œ +<->+8Ð"Î#ÑÞ
Hence
? œ #È& -9=a$> !Þ%'$'bÞ
4. V-9= $ œ # and V=38 $ œ $ Ê V œ È"$ and $ œ 1 +<->+8Ð$Î#Ñ.
Hence
? œ È"$ -9=a1> %Þ"#%%bÞ
5. The spring constant is 5 œ #Îa"Î#b œ % lb/ft . Mass 7 œ #Î$# œ "Î"' lb-s# /ft .
Since there is no damping, the equation of motion is
" ww
? %? œ ! ,
"'
that is, ? ww '%? œ ! . The initial conditions are ?a!b œ "Î% ft , ?w a!b œ ! fps . The
general solution is ?a>b œ E -9= )> F =38 )> . Invoking the initial conditions, we have
?a>b œ "% -9= )> . V œ $ inches, $ œ ! rad , =! œ ) rad/s , and X œ 1Î% sec .
7. The spring constant is 5 œ $Îa"Î%b œ "# lb/ft . Mass 7 œ $Î$# lb-s# /ft . Since
there is no damping, the equation of motion is
$ ww
? "#? œ ! ,
$#
that is, ? ww "#)? œ ! . The initial conditions are ?a!b œ "Î"# ft , ?w a!b œ # fps .
The general solution is ?a>b œ E -9= )È# > F =38 )È# > . Invoking the initial
conditions, we have
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—————————————————————————— CHAPTER 3. ——
?a>b œ
"
"
-9= )È# >
=38 )È# > .
È
"#
% #
V œ È""Î"# ft , $ œ 1 +>+8Š$ÎÈ# ‹ rad , =! œ )È# rad/s, and X œ 1Ί%È# ‹ sec.
10. The spring constant is 5 œ "'Îa"Î%b œ '% lb/ft . Mass 7 œ "Î# lb-s# /ft . The
damping coefficient is # œ # lb-sec/ft . Hence the equation of motion is
" ww
? #? w '%? œ ! ,
#
that is, ? ww %? w "#)? œ ! . The initial conditions are ?a!b œ ! ft, ?w a!b œ "Î% fps.
The general solution is ?a>b œ E -9= #È$" > F =38 #È$" > . Invoking the initial
conditions, we have
?a>b œ
"
/#> =38 #È$" > .
È
) $"
Solving ?a>b œ ! , on the interval c!Þ# , !Þ%d, we obtain > œ 1Î#È$" œ !Þ#)#" sec .
Based on the graph, and the solution of ?a>b œ !Þ!" , we have k?a>bk Ÿ !Þ!" for
> 7 œ !Þ#"%& .
11. The spring constant is 5 œ $ÎaÞ"b œ $! N/m . The damping coefficient is given as
# œ $Î& N-sec/m . Hence the equation of motion is
$
#? ww ? w $!? œ ! ,
&
that is, ? ww !Þ$? w "&? œ ! . The initial conditions are ?a!b œ !Þ!& 7 and
?w a!b œ !Þ!" 7Î= . The general solution is ?a>b œ E -9= .> F =38 .> , in which
. œ $Þ)(!!) rad/s . Invoking the initial conditions, we have
?a>b œ /0.15> a!Þ!&-9= .> !Þ!!%&#=38 .>b .
Also, .Î=! œ $Þ)(!!)ÎÈ"& ¸ !Þ***#& .
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—————————————————————————— CHAPTER 3. ——
13. The frequency of the undamped motion is =! œ " . The quasi frequency of the
damped
motion is . œ "# È% # # Þ Setting . œ #$ =! , we obtain # œ #$ È& Þ
14. The spring constant is 5 œ 71ÎP . The equation of motion for an undamped system
is
71
7? ww
? œ !.
P
Hence the natural frequency of the system is =! œ É P1 . The period is X œ #1Î=! .
15. The general solution of the system is ?a>b œ E -9= # a> >! b F =38 # a> >! b .
Invoking the initial conditions, we have ?a>b œ ?! -9=# a> >! b a?!w Î# b=38# a> >! b.
Clearly, the functions @ œ ?! -9=# a> >! b and A œ a?!w Î# b=38# a> >! b satisfy the given
criteria.
16. Note that < =38a =! > ) b œ < =38=! > -9= ) < -9==! > =38) . Comparing the given
expressions, we have E œ < =38 ) and F œ < -9= ) . That is, < œ V œ ÈE# F # ,
and >+8 ) œ EÎF œ "Î>+8 $ . The latter relation is also >+8 ) -9> $ œ " .
18. The system is critically damped, when V œ #ÈPÎG Þ Here V œ "!!! ohms .
21a+b. Let ? œ V/# >Î#7 -9=a.> $ b . Then attains a maximum when .>5 $ œ #5 1.
Hence X. œ >5" >5 œ #1Î. .
a,b. ?a>5 bÎ?a>5" b œ /B:a # >5 Î#7bÎ/B:a # >5" Î#7b œ /B:ca# >5" # >5 bÎ#7dÞ
Hence ?a>5 bÎ?a>5" b œ /B:c# a#1Î.bÎ#7d œ /B:a# X. Î#7bÞ
a- b. ? œ 68c?a>5 bÎ?a>5" bd œ # a#1Î.bÎ#7 œ 1# Î.7 .
22. The spring constant is 5 œ "'Îa"Î%b œ '% lb/ft . Mass 7 œ "Î# lb-s# /ft . The
damping coefficient is # œ # lb-sec/ft . The quasi frequency is . œ #È$" rad/s .
1
Hence ? œ È#$"
¸ "Þ"#)& Þ
25a+b. The solution of the IVP is ?a>b œ />Î) Š# -9= $) È( > !Þ#&#=38 $) È( >‹.
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—————————————————————————— CHAPTER 3. ——
Using the plot, and numerical analysis, 7 ¸ %"Þ("& .
a,b. For # œ !Þ& , 7 ¸ #!Þ%!# à for # œ "Þ! , 7 ¸ *Þ"') à for # œ "Þ& , 7 ¸ (Þ")% .
a- bÞ
a. b. For # œ "Þ' , 7 ¸ (Þ#") à for # œ "Þ( , 7 ¸ 'Þ('( à for # œ "Þ) , 7 ¸ &Þ%($ à
for # œ "Þ* , 7 ¸ 'Þ%'! . 7 steadily decreases to about 7738 ¸ %Þ)($ , corresponding to
the critical value #! ¸ "Þ($ .
a/b. We have ?a>b œ
%/# >Î#
È%# #
-9=a.> $ b , in which . œ "# È% # # , and
#
$ œ >+8" È%
. Hence k?a>bk Ÿ
##
%/# >Î#
È%# #
.
26a+b. The characteristic equation is 7<# # < 5 œ ! . Since # # %57 , the roots
È%75# #
#
are <"ß# œ #7
„3 #7 . The general solution is
?a>b œ /# >Î#7 –E -9=
È%75 # #
#7
> F =38
È%75 # #
#7
> —Þ
Invoking the initial conditions, E œ ?! and
Fœ
a#7@! # ?! b
.
È%75 # #
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—————————————————————————— CHAPTER 3. ——
a,b. We can write ?a>b œ V /# >Î#7 -9=a.> $ b , in which
V œ Ë?#!
a#7@! # ?! b#
,
%75 # #
$ œ +<->+8–
a#7@! # ?! b
Þ
?! È%75 # # —
and
a- b. V œ É?#!
a#7@! # ?! b#
%75# #
# ?! @! 7@! b
+,#
œ #É 7a5?!%75
œ É %75
##
## .
#
#
It is evident that V increases amonotonicallyb without bound as # p Š#È75 ‹ .
28a+b. The general solution is ?a>b œ E-9= È# > F=38 È# > . Invoking the initial
conditions, we have ?a>b œ È# =38 È# > .
a, b .
a- bÞ
The condition ? w a!b œ # implies that ?a>b initially increases. Hence the phase point
travels clockwise.
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—————————————————————————— CHAPTER 3. ——
29. ?a>b œ
"'
È"#(
/>Î) =38
È"#(
) >.
31. Based on Newton's second law, with the positive direction to the right,
"J œ 7? ww
where
"J œ 5? # ? w .
Hence the equation of motion is 7? ww # ? w 5? œ ! . The only difference in this
problem is that the equilibrium position is located at the unstretched configuration of
the spring.
32a+b. The restoring force exerted by the spring is J= œ a5? &?$ bÞ The opposing
viscous force is J. œ # ? w . Based on Newton's second law, with the positive direction
to the right,
J= J. œ 7? ww .
Hence the equation of motion is 7? ww # ? w 5? &?$ œ ! .
a,b. With the specified parameter values, the equation of motion is ? ww ? œ ! . The
general solution of this ODE is ?a>b œ E -9= > F =38 > . Invoking the initial
conditions,
the specific solution is ?a>b œ =38 > . Clearly, the amplitude is V œ ", and the period of
the motion is X œ #1 .
a- b. Given & œ !Þ" , the equation of motion is ? ww ? !Þ" ?$ œ ! . A solution of the
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—————————————————————————— CHAPTER 3. ——
IVP can be generated numerically:
a. b .
a/bÞ The amplitude and period both seem to decrease.
a0 b .
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—————————————————————————— CHAPTER 3. ——
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—————————————————————————— CHAPTER 3. ——
Section 3.9
2. We have =38a!„" b œ =38 ! -9= " „ -9= ! =38 " . Subtracting the two identities, we
obtain =38a! " b =38a! " b œ # -9= ! =38 " . Setting ! " œ (> and ! " œ '>,
! œ 'Þ&> and " œ !Þ&> . Hence =38 (> =38 '> œ # =38 #> -9= "$>
# .
3. Consider the trigonometric identity -9=a!„" b œ -9= ! -9= " … =38 ! =38 " . Adding
the two identities, we obtain -9=a! " b -9=a! " b œ # -9= ! -9= " . Comparing the
expressions, set ! " œ #1> and ! " œ 1>. Hence ! œ $1>Î# and " œ 1>Î# . Upon
substitution, we have -9=a1>b -9=a#1>b œ # -9=a$1>Î#b -9=a1>Î#b .
4. Adding the two identities =38a!„" b œ =38 ! -9= " „ -9= ! =38 " , it follows that
=38a! " b =38a! " b œ #=38 ! -9= " . Setting ! " œ %> and ! " œ $>, we
have ! œ (>Î# and " œ >Î# . Hence =38 $> =38 %> œ # =38a(>Î#b -9=a>Î#b .
6. Using mks units, the spring constant is 5 œ &a*Þ)bÎ!Þ" œ %*! N/m , and the damping
coefficient is # œ #Î!Þ!% œ &! N-sec/m . The equation of motion is
&? ww &!? w %*!? œ "! =38a>Î#b .
The initial conditions are ?a!b œ ! m and ? w a!b œ !Þ!$ m/s .
8a+b. The homogeneous solution is ?- a>b œ E/&> -9=È($ > F/&> =38È($ > Þ Based
on the method of undetermined coefficients, the particular solution is
Y a>b œ
"
c "'! -9=a>Î#b $"#) =38a>Î#bdÞ
"&$#)"
Hence the general solution of the ODE is ?a>b œ ?- a>b Y a>b. Invoking the initial
conditions, we find that E œ "'!Î"&$#)" and F œ $)$%%$È($ Î""")*&"$!! Þ Hence
the response is
?a>b œ
È
"
&>
È($ > $)$%%$ ($ /&> =38 È($ > Y a>b.
"'!
/
-9=
—
"&$#)" –
($!!
a,b. ?- a>b is the transient part and Y a>b is the steady state part of the response.
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—————————————————————————— CHAPTER 3. ——
a- b .
a. b. Based on Eqs. a*b and a"!b, the amplitude of the forced response is given by
V œ #Î?, in which
? œ É#&a*) =# b# #&!! =# Þ
The maximum amplitude is attained when ? is a minimum. Hence the amplitude is
maximum at = œ %È$ rad/s .
9. The spring constant is 5 œ "# lb/ft and hence the equation of motion is
' ww
? "#? œ % -9= (> ,
$#
w
that is, ? ww '%? œ '%
$ -9= (> . The initial conditions are ?a!b œ ! ft, ? a!b œ ! fps.
The general solution is ?a>b œ E-9= )> F=38 )> '%
%& -9= (> Þ Invoking the initial
'%
'%
conditions, we have ?a>b œ %& -9= )> %& -9= (> œ "#)
%& =38a>Î#b=38a"&>Î#b Þ
12. The equation of motion is
#? ww ? w $? œ $-9= $> #=38 $> .
Since the system is damped, the steady state response is equal to the particular solution.
Using the method of undetermined coefficients, we obtain
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—————————————————————————— CHAPTER 3. ——
?== a>b œ
"
a=38 $> -9= $>b.
'
Further, we find that V œ È# Î' and $ œ +<->+8a "b œ $1Î% . Hence we can write
È
?== a>b œ '# -9=a$> $1Î%b.
13. The amplitude of the steady-state response is given by
Vœ
J!
É7# a=!# =# b# # # =#
.
Since J! is constant, the amplitude is maximum when the denominator of V is minimum .
#
Let D œ =# , and consider the function 0 aD b œ 7# a=!# D b # # D . Note that 0 aD b is
a quadratic, with minimum at D œ =#! # # Î#7# . Hence the amplitude V attains a
maximum at =#7+B œ =!# # # Î#7# . Furthermore, since =!# œ 5Î7 , and therefore
=#7+B œ =!# ”"
##
•.
#57
Substituting =# œ =#7+B into the expression for the amplitude,
J!
È# % Î%7# # # a=!# # # Î#7# b
J!
œ
È=#! # # # % Î%7#
J!
œ
Þ
#=! È" # # Î%75
V œ
14a+b. The forced response is ?== a>b œ E-9= => F=38 => Þ The constants are obtain by
the method of undetermined coefficients. That is, comparing the coefficients of -9= =>
and =38 =>, we find that
7=# E #=F 5E œ J! , and 7=# F #=E 5F œ ! .
Solving this system results in
E œ 7ˆ=#! =# ‰Î? and F œ #=Î? ,
in which ? œ É7# a=!# =# b # # =# Þ It follows that
#
>+8 $ œ FÎE œ
#=
.
7a=#! =# b
a,b. Here 7 œ " , # œ !Þ"#& , =! œ " . Hence >+8 $ œ !Þ"#&=Îa" =# bÞ
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—————————————————————————— CHAPTER 3. ——
17a+b. Here 7 œ " , # œ !Þ#& , =#! œ # , J! œ # . Hence ?== a>b œ ?# -9=a=> $ b ,
where ? œ Éa# =# b# =# Î"' œ "% È'% '$=# "' =% , and >+8 $ œ %a#= =# b Þ
a,bÞ The amplitude is
Vœ
a- b .
È'% '$=# "' =%
)
.
a. b. See Prob. 13. The amplitude is maximum when the denominator of V is minimum.
That is, when = œ =7+B œ $È"% Î) ¸ "Þ%!$" . Hence V a= œ =7+B b œ '%ÎÈ"#( .
18a+b. The homogeneous solution is ?- a>b œ E-9= > F=38 > Þ Based on the method of
undetermined coefficients, the particular solution is
Y a>b œ
$
-9= => Þ
" =#
Hence the general solution of the ODE is ?a>b œ ?- a>b Y a>b. Invoking the initial
conditions, we find that E œ $Îa=# "b and F œ ! Þ Hence the response is
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—————————————————————————— CHAPTER 3. ——
a, b .
?a>b œ
$
c -9= => -9= > d.
" =#
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—————————————————————————— CHAPTER 3. ——
Note that
?a>b œ
'
a" =b>
a = "b >
=38”
•=38”
•.
#
"=
#
#
19a+b. The homogeneous solution is ?- a>b œ E-9= > F=38 > Þ Based on the method of
undetermined coefficients, the particular solution is
Y a>b œ
$
-9= => Þ
" =#
Hence the general solution is ?a>b œ ?- a>b Y a>b. Invoking the initial conditions, we
find that E œ a=# #bÎa=# "b and F œ " Þ Hence the response is
a,.b
?a>b œ
"
$ -9= => ˆ=# #‰-9= > ‘ =38 > .
#
"=
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—————————————————————————— CHAPTER 3. ——
________________________________________________________________________
page 137
—————————————————————————— CHAPTER 3. ——
Note that
?a>b œ
'
a" =b>
a = "b >
=38”
•=38”
• -9= > =38 > .
#
"=
#
#
20.
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—————————————————————————— CHAPTER 3. ——
21. The general solution is ?a>b œ ?- a>b Y a>b, in which
?- a>b œ />Î"' –
È#&&
È#&&
"("$&)
#&((&)
-9=
>
=38
>
"$#(#"
"'
"' —
"$#(#"È#&&
and
Y a>b œ
"
c%$')!! -9=aÞ$>b ")!!! =38aÞ$>bd Þ
"$#(#"
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—————————————————————————— CHAPTER 3. ——
a+ b .
a, b .
23. The general solution is ?a>b œ ?- a>b Y a>b, in which
?- a>b œ />Î"' –
È#&&
È#&&
*(%'
"#&)
-9=
>
=38
>
%"!&
"'
"' —
)#"È#&&
and
Y a>b œ
"
c "&$' -9=a$>b (# =38a$>bd Þ
%"!&
________________________________________________________________________
page 140
—————————————————————————— CHAPTER 3. ——
a+ b .
a, b .
24.
25a+b.
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—————————————————————————— CHAPTER 3. ——
________________________________________________________________________
page 142
—————————————————————————— CHAPTER 3. ——
________________________________________________________________________
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—————————————————————————— CHAPTER 3. ——
a, b .
a- b. The amplitude for a similar system with a linear spring is given by
Vœ
È#& %*=# #&=%
&
.
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—————————————————————————— CHAPTER 3. ——
________________________________________________________________________
page 145
—————————————————————————— CHAPTER 4. ——
Chapter Four
Section 4.1
1. The differential equation is in standard form. Its coefficients, as well as the function
1a>b œ > , are continuous everywhere. Hence solutions are valid on the entire real line.
3. Writing the equation in standard form, the coefficients are rational functions with
singularities at > œ ! and > œ " . Hence the solutions are valid on the intervals a _,!b,
a! , "b , and a" , _b .
4. The coefficients are continuous everywhere, but the function 1a>b œ 68 > is defined
and
continuous only on the interval a! , _b. Hence solutions are defined for positive reals.
5. Writing the equation in standard form, the coefficients are rational functions with a
singularity at B! œ " . Furthermore, :% aBb œ >+8 BÎaB "b is undefined, and hence not
continuous, at B5 œ „a#5 "b1Î# , 5 œ !ß "ß #ß â Þ Hence solutions are defined on any
interval that does not contain B! or B5 .
6. Writing the equation in standard form, the coefficients are rational functions with
singularities at B œ „ # . Hence the solutions are valid on the intervals a _, #b,
a # , #b , and a# , _b .
7. Evaluating the Wronskian of the three functions, [ a0" ß 0# ß 0$ b œ "% . Hence the
functions are linearly independent.
9. Evaluating the Wronskian of the four functions, [ a0" ß 0# ß 0$ ß 0% b œ ! . Hence the
functions are linearly dependent. To find a linear relation among the functions, we need
to find constants -" ß -# ß -$ ß -% , not all zero, such that
-" 0" a>b -# 0# a>b -$ 0$ a>b -% 0% a>b œ ! .
Collecting the common terms, we obtain
a-# #-$ -% b># a#-" -$ -% b> a $-" -# -% b œ ! ,
which results in three equations in four unknowns. Arbitrarily setting -% œ " , we can
solve the equations -# #-$ œ " , #-" -$ œ " , $-" -# œ " , to find that -" œ #Î(,
-# œ "$Î(, -$ œ $Î( . Hence
#0" a>b "$0# a>b $0$ a>b (0% a>b œ ! .
10. Evaluating the Wronskian of the three functions, [ a0" ß 0# ß 0$ b œ "&' . Hence the
functions are linearly independent.
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—————————————————————————— CHAPTER 4. ——
11. Substitution verifies that the functions are solutions of the ODE. Furthermore, we
have
[ a"ß -9= >ß =38 >b œ " .
12. Substitution verifies that the functions are solutions of the ODE. Furthermore, we
have [ a"ß >ß -9= >ß =38 >b œ " .
14. Substitution verifies that the functions are solutions of the ODE. Furthermore, we
have [ a"ß >ß /> ß > /> b œ /#> .
15. Substitution verifies that the functions are solutions of the ODE. Furthermore, we
have [ a"ß Bß B$ b œ 'B .
16. Substitution verifies that the functions are solutions of the ODE. Furthermore, we
have [ aBß B# ß "ÎBb œ 'ÎB .
18. The operation of taking a derivative is linear, and hence
a-" C" -# C# ba5 b œ -" C" -# C# .
a5 b
a5 b
It follows that
Pc-" C" -# C# d œ -" C"a8b -# C#a8b :" -" C"a8"b -# C#a8"b ‘ â :8 c-" C" -# C# dÞ
Rearranging the terms, we obtain Pc-" C" -# C# d œ -" PcC" d -# PcC# dÞ Since C" and C#
are solutions, Pc-" C" -# C# d œ ! . The rest follows by induction.
19a+b. Note that . 5 a>8 bÎ.>5 œ 8a8 "bâa8 5 "b>85 , for 5 œ "ß #ß âß 8 .
Hence
Pc>8 d œ +! 8x +" c8a8 "bâ#d> â +8" 8 >8" +8 >8 Þ
a,b. We have . 5 a/<> bÎ.>5 œ <5 /<> , for 5 œ !ß "ß #ß â . Hence
P/<> ‘ œ +! <8 /<> +" <8" /<> â +8" < /<> +8 /<>
œ +! <8 +" <8" â +8" < +8 ‘/<> Þ
a- b. Set C œ /<> , and substitute into the ODE. It follows that <% &<# % œ ! , with
< œ „ "ß „ # . Furthermore, [ a/> ß /> ß /#> ß /#> b œ (# .
20a+b. Let 0 a>b and 1a>b be arbitrary functions. Then [ a0 ß 1b œ 0 1 w 0 w 1 . Hence
[ w a0 ß 1b œ 0 w 1 w 0 1 ww 0 ww 1 0 w 1 w œ 0 1 ww 0 ww 1 . That is,
[ w a0 ß 1 b œ º
0
0 ww
1
Þ
1 ww º
Now expand the $-by-$ determinant as
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—————————————————————————— CHAPTER 4. ——
[ aC" ß C# ß C$ b œ C" º
C#w
C#ww
C$w
C"w
C
# º ww
C$ww º
C"
C$w
C"w
C
$ º ww
C$ww º
C"
C#w
Þ
C#ww º
Differentiating, we obtain
[ w aC" ß C# ß C$ b œ C"w º
C#w
C#ww
C#w
C" º www
C#
w
w
C$w
C$w
C#w
w C"
w C"
ww º C# º ww
ww º C$ º ww
C$
C" C $
C" C#ww º
C$w
C"w C$w
C"w C#w
Þ
www º C# º www
www º C$ º www
C$
C"
C$
C"
C#www º
The second line follows from the observation above. Now we find that
â w
â â
â
C#
C$ â
â C" C#w C$w â â C"
â
â â
â
[ w aC" ß C# ß C$ b œ â C"w C#w C$w â â C"w C#w C$w âÞ
â ww
â
â
â
â C" C#ww C3ww â â C"www C#www C3www â
Hence the assertion is true, since the first determinant is equal to zero.
a,b. Based on the properties of determinants,
â
â :$ C"
â
w
:# a>b:$ a>b[ œ â :# C"w
â www
â C"
: $ C#
:# C#w
C#www
â
: $ C$ â
â
:# C$w â
â
C3www â
Adding the first two rows to the third row does not change the value of the determinant.
Since the functions are assumed to be solutions of the given ODE, addition of the rows
results in
â
â
: $ C#
: $ C$ â
â :$ C"
â
â
:# C#w
:# C$w âÞ
:# a>b:$ a>b[ w œ â :# C"w
â
â
â :" C"ww :" C#ww :" C3ww â
It follows that :# a>b:$ a>b[ w œ :" a>b:# a>b:$ a>b[ . As long as the coefficients are not
zero, we obtain [ w œ :" a>b[ .
a- b. The first order equation [ w œ :" a>b[ is linear, with integrating factor .a>b œ
œ /B:ˆ' :" a>b.>‰ . Hence [ a>b œ - /B:ˆ ' :" a>b.>‰ . Furthermore, [ a>b is zero
only if - œ ! .
a. b. It can be shown, by mathematical induction, that
â C
C#
â
â "
â Cw
w
C
â
â "
#
â ã
w
[ aC" ß C# ß âß C8 b œ â
â a82b
C#a82b â
â C"
â a8b
â C"
C#a8b â
C8"
w
C8"
a82b
C8"
a8b
C8"
â
â
â
â
â
âÞ
â
C8a82b â
â
C8a8b â
C8
C8w
ã
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page 148
—————————————————————————— CHAPTER 4. ——
Based on the reasoning in Parta,b, it follows that
:# a>b:$ a>bâ:8 a>b[ w œ :" a>b:# a>b:$ a>bâ:8 a>b[ ,
and hence [ w œ :" a>b[ .
22. Inspection of the coefficients reveals that :" a>b œ ! . Based on Prob. #!, we find
that [ w œ ! , and hence [ œ - .
23. After writing the equation in standard form, observe that :" a>b œ #Î> . Based on the
results in Prob. #!, we find that [ w œ a #Î>b[ , and hence [ œ -Î># .
24. Writing the equation in standard form, we find that :" a>b œ "Î>. Using Abel's
formula, the Wronskian has the form [ a>b œ - /B:ˆ ' "> .>‰ œ -Î> .
25a+b. Assuming that -" C" a>b -# C# a>b â -8 C8 a>b œ ! , then taking the first 8 "
derivatives of this equation results in
-" C" a>b -# C# a>b â -8 C8a5 b a>b œ !
a5 b
a5 b
for 5 œ !ß "ß âß 8 " . Setting > œ >! , we obtain a system of 8 algebraic equations with
unknowns -" ß -# ß âß -8 . The Wronskian, [ aC" ß C# ß âß C8 ba>! b, is the determinant of the
coefficient matrix. Since system of equations is homogeneous, [ aC" ß C# ß âß C8 ba>! b Á !
implies that the only solution is the trivial solution, -" œ -# œ â œ -8 œ ! .
a,b. Suppose that [ aC" ß C# ß âß C8 ba>! b œ ! for some >! . Consider the system of
algebraic
equations
-" C" a>! b -# C# a>! b â -8 C8a5 b a>! b œ ! ,
a5 b
a5 b
5 œ !ß "ß âß 8 " , with unknowns -" ß -# ß âß -8 . Vanishing of the Wronskian, which is
the determinant of the coefficient matrix, implies that there is a nontrivial solution of the
system of homogeneous equations. That is, there exist constants -" ß -# ß âß -8 , not all
zero, which satisfy the above equations. Now let
Ca>b œ -" C" a>b -# C# a>b â -8 C8 a>b.
Since the ODE is linear, Ca>b is also a nonzero solution. Based on the system of algebraic
equations above, Ca>! b œ C w a>! b œ â œ C a8"b a>! b œ ! . This contradicts the uniqueness
of the identically zero solution.
26. Let Ca>b œ C" a>b@a>b. Then C w œ C"w @ C" @ w , C ww œ C"ww @ #C"w @ w C" @ ww , and
C www œ C"www @ $C"ww @ w $C"w @ ww C" @ www Þ Substitution into the ODE results in
C"www @ $C"ww @ w $C"w @ ww C" @ www :" cC"ww @ #C"w @ w C" @ ww d :# cC"w @ C"@ w d :$C"@ œ !.
Since C" is assumed to be a solution, all terms containing the factor @a>b vanish. Hence
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—————————————————————————— CHAPTER 4. ——
C1 @ www c:" C1 $C1w d@ ww c$C1ww #:1 C1w :2 C1 d@ w œ ! ,
which is a second order ODE in the variable ? œ @ w .
28. First write the equation in standard form:
C www $
> # ww
>"
'
C ' #
Cw #
C œ !.
>a > $ b
> a> $ b
> a> $b
Let Ca>b œ ># @a>bÞ Substitution into the given ODE results in
># @ www $
>a> %b ww
@ œ !.
>$
Set A œ @ ww . Then A is a solution of the first order differential equation
Aw $
>%
A œ !Þ
>a> $b
This equation is linear, with integrating factor .a>b œ >% Îa> $b. The general solution
is A œ - a> $bÎ>% . Integrating twice, it follows that @a>b œ -" >" -" ># -# > -$ Þ
Hence Ca>b œ -" > -" -# >$ -$ ># Þ Finally, since C" a>b œ ># and C# a>b œ >$ are given
solutions, the third independent solution is C$ a>b œ -" > -" Þ
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page 150
—————————————————————————— CHAPTER 4. ——
Section 4.2
1. The magnitude of " 3 is V œ È# and the polar angle is 1Î% . Hence the polar
form is given by " 3 œ È# /31Î% Þ
3. The magnitude of $ is V œ $ and the polar angle is 1 Þ Hence $ œ $ /31 Þ
4. The magnitude of 3 is V œ " and the polar angle is $1Î# Þ Hence 3 œ /$13Î# Þ
5. The magnitude of È$ 3 is V œ # and the polar angle is 1Î' œ ""1Î' . Hence
the polar form is given by È$ 3 œ # /""13Î' Þ
6. The magnitude of " 3 is V œ È# and the polar angle is &1Î% . Hence the polar
form is given by " 3 œ È# /&13Î% Þ
7. Writing the complex number in polar form, " œ /#713 , where 7 may be any integer.
Thus ""Î$ œ /#713Î$ Þ Setting 7 œ !ß "ß # successively, we obtain the three roots as
""Î$ œ " , ""Î$ œ /#13Î$ , ""Î$ œ /%13Î$ . Equivalently, the roots can also be written as
" , -9=a#1Î$b 3 =38a#1Î$b œ "# Š " È$ ‹, -9=a%1Î$b 3=38a%1Î$b œ "# Š " È$ ‹.
9. Writing the complex number in polar form, " œ /#713 , where 7 may be any integer.
Thus ""Î% œ /#713Î% Þ Setting 7 œ !ß "ß #ß $ successively, we obtain the three roots as
""Î% œ " , ""Î% œ /13Î# , ""Î% œ /13 , ""Î% œ /$13Î# . Equivalently, the roots can also be
written as " , -9=a1Î#b 3 =38a1Î#b œ 3, -9=a1b 3=38a1b œ " , -9=a$1Î#b
3=38a$1Î#b œ 3.
10. In polar form, #a-9= 1Î$ 3 =38 1Î$b œ # /31Î$#71 , in which 7 is any integer.
Thus c#a-9= 1Î$ 3 =38 1Î$bd"Î# œ #"Î# /31Î'71 . With 7 œ ! , one square root is
given by #"Î# /31Î' œ ŠÈ$ 3‹ÎÈ# . With 7 œ " , the other root is given by
#"Î# /3(1Î' œ Š È$ 3‹ÎÈ# .
11. The characteristic equation is <$ <# < " œ !Þ The roots are < œ "ß "ß " .
One root is repeated, hence the general solution is C œ -" /> -# /> -$ >/> Þ
13. The characteristic equation is <$ #<# < # œ ! , with roots < œ "ß "ß # Þ The
roots are real and distinct, hence the general solution is C œ -" /> -# /> -$ /#> Þ
14. The characteristic equation can be written as <# a<# %< %b œ ! Þ The roots are
< œ !ß !ß #ß # . There are two repeated roots, and hence the general solution is given by
C œ -" -# > -$ /#> -% >/#> Þ
15. The characteristic equation is <' " œ ! . The roots are given by < œ a "b"Î' ,
that is, the six sixth roots of " . They are /13Î'713Î$ , 7 œ !ß "ß âß & . Explicitly,
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—————————————————————————— CHAPTER 4. ——
< œ ŠÈ$ 3‹Î# , ŠÈ$ 3‹Î# , 3 , 3 , Š È$ 3‹Î# , Š È$ 3‹Î# . Hence
the general solution is given by C œ / $ >Î# c-" -9= a>Î#b -# =38 a>Î#bd -$ -9= >
È
-% =38 > / $ >Î# c-& -9= a>Î#b -' =38 a>Î#bdÞ
È
16. The characteristic equation can be written as a<# "ba<# %b œ ! . The roots
are given by < œ „ "ß „ # . The roots are real and distinct, hence the general solution is
C œ -" /> -# /> -$ /#> -% /#> Þ
17. The characteristic equation can be written as a<# "b œ ! . The roots are given by
< œ „ " , each with multiplicity three. Hence the general solution is
$
C œ -" /> -# >/> -$ ># /> -% /> -& >/> -' ># /> Þ
18. The characteristic equation can be written as <# ˆ<% "‰ œ ! . The roots are given
by < œ !ß !ß „ "ß „ 3 . The general solution is C œ -" -# > -$ /> -% /> -& -9= >
-' =38 > .
19. The characteristic equation can be written as <ˆ<% $<$ $<# $< #‰ œ ! .
Examining the coefficients, it follows that <% $<$ $<# $< # œ a< "ba< #b ‚
a<# "bÞ Hence the roots are < œ !ß "ß #ß „ 3 . The general solution of the ODE is given
by C œ -" -# /> -$ /#> -% -9= > -& =38 > .
20. The characteristic equation can be written as <a<$ )b œ ! , with roots < œ ! ,
# /#713Î$ , 7 œ !ß "ß # . That is, < œ !ß #ß " „ 3È$ . Hence the general solution is
C œ -" -# /#> /> ’-$ -9=È$ > -% =38È$ > “Þ
#
21. The characteristic equation can be written as ˆ<% %‰ œ ! . The roots of the
equation <% % œ ! are < œ " „ 3 , " „ 3 . Each of these roots has 7?6>3:63-3>C >A9Þ
The general solution is C œ /> c-" -9= > -# =38 > d >/> c-$ -9= > -% =38 > d
/> c-& -9= > -' =38 > d >/> c-( -9= > -) =38 > d.
22. The characteristic equation can be written as a<# "b œ ! . The roots are given
by < œ „ 3 , each with 7?6>3:63-3>C >A9Þ The general solution is C œ -" -9= > -# =38
>
>c-$ -9= > -% =38 > dÞ
#
24. The characteristic equation is <$ &<# '< # œ !Þ Examining the coefficients,
we find that <$ &<# '< # œ a< "ba<# %< #b. Hence the roots are deduced as
È
È
< œ " , # „È# . The general solution is C œ -" /> -# /Š# #‹> -$ /Š# #‹> Þ
25. The characteristic equation is ")<$ #"<# "%< % œ !Þ By examining the first
and last coefficients, we find that ")<$ #"<# "%< % œ a#< "ba*<# '< %b.
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—————————————————————————— CHAPTER 4. ——
Hence the roots are < œ "Î# , Š "„È$ ‹Î$ . The general solution of the ODE is
given by C œ -" />Î# />Î$ ’-# -9=Š>ÎÈ$ ‹ -$ =38Š>ÎÈ$ ‹ “Þ
26. The characteristic equation is <% (<$ '<# $!< $' œ !Þ By examining the
first and last coefficients, we find that
<% (<$ '<# $!< $' œ a< $ba< #bˆ<# '< '‰.
The roots are < œ #ß $ß $ „È$ . The general solution is
C œ -" /#> -# /$> -$ /Š$
È$‹>
-% /Š$
È$‹>
Þ
28. The characteristic equation is <% '<$ "(<# ##< "% œ !Þ It can be shown
that <% '<$ "(<# ##< "% œ a<# #< #ba<# %< (b. Hence the roots are
< œ " „ 3 , # „ 3È$ . The general solution is
C œ /> c-" -9= > -# =38 > d /#> ’-$ -9=È$ > -% =38È$ > “Þ
30. Ca>b œ "# />Î
È#
=38Š>ÎÈ#‹ "# />Î
È#
=38Š>ÎÈ#‹.
32. The characteristic equation is <$ <# < " œ ! , with roots < œ " , „ 3 . Hence
the general solution is Ca>b œ -" /> -# -9= > -$ =38 > . Invoking the initial conditions,
we obtain the system of equations
-" - # œ #
-" - $ œ "
-" - # œ #
with solution -" œ ! , -# œ # , -$ œ " . Therefore the solution of the initial value
problem is Ca>b œ #-9= > =38 > .
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—————————————————————————— CHAPTER 4. ——
33. The characteristic equation is #<% <$ *<# %< % œ ! , with roots < œ "Î# ,
" , „ # . Hence the general solution is Ca>b œ -" />Î# -# /> -$ /#> -% /#> . Applying
the initial conditions, we obtain the system of equations
-" -# - $ - % œ #
"
-" -# #-$ #-% œ !
#
"
-" -# %-$ %-% œ #
%
"
-" -# )-$ )-% œ !
)
with solution -" œ "'Î"& , -# œ #Î$ , -$ œ "Î' , -% œ "Î"! . Therefore the
" #>
>Î#
solution of the initial value problem is Ca>b œ "'
#$ /> "' /#> "!
/ .
"& /
The solution decreases without bound.
34. Ca>b œ
# >
"$ /
#%
/>Î# "$
-9= >
$
‘
"$ =38 > .
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page 154
—————————————————————————— CHAPTER 4. ——
The solution is an oscillation with increasing amplitude.
35. The characteristic equation is ' <$ &<# < œ ! , with roots < œ ! , "Î$ , "Î# .
The general solution is Ca>b œ -" -# />Î$ -$ />Î# Þ Invoking the initial conditions,
we require that
-" - # - $ œ #
"
"
-# - $ œ #
$
#
"
"
-# - $ œ !
*
%
with solution -" œ ) , -# œ ") , -$ œ ) . Therefore the solution of the initial value
problem is Ca>b œ ) ")/>Î$ )/>Î# .
36. The general solution is derived in Prob.a#)b as
Ca>b œ /> c-" -9= > -# =38 > d /#> ’-$ -9=È$ > -% =38È$ > “Þ
Invoking the initial conditions, we obtain the system of equations
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—————————————————————————— CHAPTER 4. ——
-" - $ œ "
-" -# #-$ È$ -% œ #
#-# -$ %È$ -% œ !
#-" #-# "!-$ *È$ -% œ $
with solution -" œ #"Î"$ , -# œ $)Î"$ , -$ œ )Î"$ , -% œ "(È$ Î$* .
The solution is a rapidly-decaying oscillation.
38.
[ ˆ/> ß /> ß -9= >ß =38 >‰ œ )
[ a-9=2 >ß =382 >ß -9= >ß =38 >b œ %
40. Suppose that -" /<" > -# /<# > â -8 /<8 > œ ! , and each of the <5 are real and
different. Multiplying this equation by /<" > , -" -# /a<# <" b> â -8 /a<8 <" b> œ ! .
Differentiation results in
-# a<# <" b/a<# <" b> â -8 a<8 <" b/a<8 <" b> œ ! .
Now multiplying the latter equation by /a<# <" b> , and differentiating, we obtain
-$ a<$ <# ba<$ <" b/a<$ <# b> â -8 a<8 <# ba<8 <" b/a<8 <# b> œ ! .
Following the above steps in a similar manner, it follows that
-8 a<8 <8" bâa<8 <" b/a<8 <8" b> œ ! .
Since these equations hold for all >, and all the <5 are different, we have -8 œ ! . Hence
-" /<" > -# /<# > â -8" /<8" > œ ! , _ > _ .
The same procedure can now be repeated, successively, to show that
-" œ - # œ â œ - 8 œ ! .
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—————————————————————————— CHAPTER 4. ——
Section 4.3
2. The general solution of the homogeneous equation is C- œ -" /> -# /> -$ -9= >
-% =38 > . Let 1" a>b œ $> and 1# a>b œ -9= >. By inspection, we find that ]" a>b œ $>.
Since 1# a>b is a solution of the homogeneous equation, set ]# a>b œ >aE-9= > F=38 >bÞ
Substitution into the given ODE and comparing the coefficients of similar term results in
E œ ! and F œ "Î% . Hence the general solution of the nonhomogeneous problem is
>
Ca>b œ C- a>b $> =38 > .
%
3. The characteristic equation corresponding to the homogeneous problem can be written
as a< "ba<# "b œ ! . The solution of the homogeneous equation is C- œ -" />
-# -9= > -$ =38 > . Let 1" a>b œ /> and 1# a>b œ %>. Since 11 a>b is a solution of the
homogeneous equation, set ]1 a>b œ E>/> Þ Substitution into the ODE results in E œ "Î#.
Now let ]# a>b œ F> G . We find that F œ G œ % . Hence the general solution of
the nonhomogeneous problem is Ca>b œ C- a>b >/> Î# %a> "b .
4. The characteristic equation corresponding to the homogeneous problem can be written
as <a< "ba< "b œ ! . The solution of the homogeneous equation is C- œ -" -# />
-$ /> . Since 1a>b œ # =38 > is not a solution of the homogeneous problem, we can set
] a>b œ E -9= > F =38 > . Substitution into the ODE results in E œ " and F œ ! .
Thus
the general solution is Ca>b œ -" -# /> -$ /> -9= > .
6. The characteristic equation corresponding to the homogeneous problem can be written
as a<# "b# œ ! . It follows that C- œ -" -9= > -# =38 > >a-$ -9= > -% =38 >bÞ Since
1a>b is not a solution of the homogeneous problem, set ] a>b œ E F-9= #> G=38 #> .
Substitution into the ODE results in E œ $, F œ "Î*, G œ ! . Thus the general solution
is Ca>b œ C- a>b $ "* -9= #> .
7. The characteristic equation corresponding to the homogeneous problem can be written
as <$ a<$ "b œ ! . Thus the homogeneous solution is
C- œ -" -# > -$ ># -% /> />Î# ’-& -9=ŠÈ$ >Î#‹ -& =38ŠÈ$ >Î#‹“Þ
Note the 1a>b œ > is a solution of the homogenous problem. Consider a particular
solution
of the form ] a>b œ >$ aE> F bÞ Substitution into the ODE results in E œ "Î#% and
F œ !. Thus the general solution is Ca>b œ C- a>b >% Î#% .
8. The characteristic equation corresponding to the homogeneous problem can be written
as <$ a< "b œ !. Hence the homogeneous solution is C- œ -" -# > -$ ># -% /> .
Since 1a>b is not a solution of the homogeneous problem, set ] a>b œ E-9= #> F=38 #> .
Substitution into the ODE results in E œ "Î%! and F œ "Î#! . Thus the general solution
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—————————————————————————— CHAPTER 4. ——
is Ca>b œ C- a>b a-9= #> #=38 #>bÎ%! .
10. From Prob. ## in Section %Þ# , the homogeneous solution is
C- œ -" -9= > -# =38 > >c-$ -9= > -% =38 > dÞ
Since 1a>b is not a solution of the homogeneous problem, substitute ] a>b œ E> F into
the ODE to obtain E œ $ and F œ %. Thus the general solution is C a>b œ C- a>b $> %.
Invoking the initial conditions, we find that -" œ % , -# œ % , -$ œ " , -% œ $Î# .
Therefore the solution of the initial value problem is
Ca>b œ a> %b-9= > a$>Î# %b=38 > $> % .
11. The characteristic equation can be written as < a<# $< #b œ !. Hence the
homogeneous solution is C- œ -" -# /> -$ /#> . Let 1" a>b œ /> and 1# a>b œ >. Note
that 1" is a solution of the homogeneous problem. Set ]" a>b œ E>/> . Substitution into
the ODE results in E œ ". Now let ]# a>b œ F># G>. Substitution into the ODE
results in F œ "Î% and G œ $Î% . Therefore the general solution is
Ca>b œ -" -# /> -$ /#> >/> ˆ># $>‰Î% .
Invoking the initial conditions, we find that -" œ " , -# œ -$ œ ! . The solution of the
initial value problem is Ca>b œ " >/> a># $>bÎ% .
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—————————————————————————— CHAPTER 4. ——
12. The characteristic equation can be written as a< "ba< $ba<# %b œ !. Hence
the homogeneous solution is C- œ -" /> -# /$> -$ -9= #> -% =38 #>. None of the
terms in 1a>b is a solution of the homogeneous problem. Therefore we can assume a form
] a>b œ E/> F-9= > G=38 > Þ Substitution into the ODE results in E œ "Î#! ,
F œ #Î& , G œ %Î& . Hence the general solution is
Ca>b œ -" /> -# /$> -$ -9= #> -% =38 #> /> Î#! a#-9= > %=38 >bÎ& .
Invoking the initial conditions, we find that -" œ )"Î%! , -# œ ($Î&#! , -$ œ ((Î'& ,
-% œ %*Î"$! .
14. From Prob. %, the homogeneous solution is C- œ -" -# /> -$ /> . Consider the
terms 1" a>b œ >/> and 1# a>b œ #-9= >. Note that since < œ " is a simple root of the
characteristic equation, Table %Þ$Þ" suggests that we set ]" a>b œ >aE> F b/> Þ The
function #-9= > is not a solution of the homogeneous equation. We can simply choose
]# a>b œ G-9= > H=38 > . Hence the particular solution has the form
] a>b œ >aE> F b/> G-9= > H=38 > .
15. The characteristic equation can be written as a<# "b œ !. The roots are given
#
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—————————————————————————— CHAPTER 4. ——
as < œ „ ", each with multiplicity two. Hence the solution of the homogeneous problem
is C- œ -" /> -# >/> -$ /> -% >/> . Let 1" a>b œ /> and 1# a>b œ =38 >. The function
/> is a solution of the homogeneous problem. Since < œ " has multiplicity >A9, we set
]" a>b œ E># /> . The function =38 > is not a solution of the homogeneous equation. We
can set ]# a>b œ F-9= > G=38 > . Hence the particular solution has the form
] a>b œ E># /> F-9= > G=38 > .
16. The characteristic equation can be written as <# a<# %b œ ! , with roots < œ !, „#3Þ
The root < œ ! has multiplicity >A9, hence the homogeneous solution is C- œ -" -# >
-$ -9= #> -% =38 #> . The functions 1" a>b œ =38 #> and 1# a>b œ % are solutions of the
homogenous equation. The complex roots have multiplicity one, therefore we need to set
]" a>b œ E> -9= #> F> =38 #> . Now 1# a>b œ % is associated with the double root < œ !Þ
Based on Table %Þ$Þ", set ]# a>b œ G># . Finally, 1$ a>b œ >/> aand its derivativesb is
independent of the homogeneous solution. Therefore set ]$ a>b œ aH> I b/> . Conclude
that the particular solution has the form
] a>b œ E> -9= #> F> =38 #> G># aH> I b/> Þ
18. The characteristic equation can be written as <# a<# #< #b œ ! , with roots < œ !,
with multiplicity two, and < œ " „ 3 . The homogeneous solution is C- œ -" -# >
-$ /> -9= > -% /> =38 > . The function 1" a>b œ $/> #>/> , and all of its derivatives,
is independent of the homogeneous solution. Therefore set ]" a>b œ E/> aF> G b/> Þ
Now 1# a>b œ /> =38 > is a solution of the homogeneous equation, associated with the
complex roots. We need to set ]# a>b œ >aH /> -9= > I /> =38 >bÞ It follows that the
particular solution has the form
] a>b œ E/> aF> G b/> >ˆH /> -9= > I /> =38 >‰Þ
19. Differentiating C œ ?a>b@a>b, successively, we have
C w œ ? w @ ?@ w
C ww œ ? ww @ #? w @ w ?@ ww
ã
8
8 a84b a4b
a8b
"
C œ
@
Π?
4
4œ !
Setting @a>b œ /!> , @a4b œ !4 /!> . So for any : œ "ß #ß âß 8 ,
C
a:b
:
œ / "Œ !4 ?a: 4b .
4
4œ !
:
!>
It follows that
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—————————————————————————— CHAPTER 4. ——
8
:
P/!> ?‘ œ /!> " –+ 8: "Œ !4 ?a: 4b —
4
:œ!
4œ !
:
a‡bÞ
It is evident that the right hand side of Eq. a‡b is of the form
/!> 5! ?a8b 5" ?a8"b â 5 8" ? w 58 ?‘ Þ
Hence operator equation Pc/!> ?d œ /!> a,! >7 ," >7" â , 7" > ,7 b can be
written as
5! ?a8b 5" ?a8"b â 5 8" ? w 58 ? œ
œ ,! >7 ," >7" â , 7" > ,7 Þ
The coefficients 5 3 , 3 œ !ß "ß âß 8 can be determined by collecting the like terms in
the double summation in Eq. a‡b. For example, 5! is the coefficient of ?a8b . The only
term that contains ?a8b is when : œ 8 and 4 œ ! . Hence 5! œ +! . On the other hand,
58 is the coefficient of ?a>b. The inner summation in a‡b contains terms with ?, given by
!: ? awhen 4 œ :b, for each : œ !ß "ß âß 8 . Hence
58 œ "+ 8: !: .
8
:œ!
21a+b. Clearly, /#> is a solution of C w #C œ ! , and >/> is a solution of the differential
equation C ww #C w C œ ! . The latter ODE has characteristic equation a< "b# œ ! .
Hence aH #bc$/#> d œ $aH #bc/#> d œ ! and aH "b# c>/> d œ ! . Furthermore,
we have aH #baH "b# c>/> d œ aH #bc!d œ !, and aH #baH "b# c$/#> d œ
œ aH "b# aH #bc$/#> d œ aH "b# c!d œ ! .
a,b. Based on Part a+b,
aH #baH "b# aH #b$ aH "b] ‘ œ aH #baH "b# $/#> >/> ‘
œ !,
since the operators are linear. The implied operations are associative and commutative.
Hence
a H #b % a H "b $ ] œ ! Þ
The operator equation corresponds to the solution of a linear homogeneous ODE with
characteristic equation a< #b% a< "b$ œ ! . The roots are < œ # , with multiplicity %
and < œ " , with multiplicity $ . It follows that the given homogeneous solution is
] a>b œ -" /#> -# >/#> -$ ># /#> -% >$ /#> -& /> -' >/> -( ># /> ,
which is a linear combination of seven independent solutions.
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—————————————————————————— CHAPTER 4. ——
22a"&b. Observe that aH "bc/> d œ ! and aH# "bc=38 >d œ ! . Hence the operator
L aHb œ aH "baH# "b is an annihilator of /> =38 > . The operator corresponding
#
to the left hand side of the given ODE is aH# "b Þ It follows that
aH "b# aH "b$ ˆH# "‰] œ ! .
The resulting ODE is homogeneous, with solution
] a>b œ -" /> -# >/> -$ /> -% >/> -& >$ /> -' -9= > -( =38 > .
After examining the homogeneous solution of Prob. "& , and eliminating duplicate terms,
we have
] a>b œ -& >$ /> -' -9= > -( =38 > .
22a"'b. We find that Hc%d œ ! , aH "b# c>/> d œ ! , and aH# %bc=38 #>d œ ! .
The operator L aHb œ HaH "b# aH# %bis an annihilator of ># >/> =38 #>. The
operator corresponding to the left hand side of the ODE is H# aH# %bÞ It follows that
H$ aH "b# ˆH# %‰ ] œ ! .
#
The resulting ODE is homogeneous, with solution
] a>b œ -" -# > -$ ># -% /> -& >/> -' -9= #> -( =38 #> -) >-9= #> -* >=38 #> .
After examining the homogeneous solution of Prob. "' , and eliminating duplicate terms,
we have
] a>b œ -$ ># -% /> -& >/> -) >-9= #> -* >=38 #> .
22a")b. Observe that aH "bc/> d œ ! , aH "b# c>/> d œ ! Þ The function /> =38 > is
a solution of a second order ODE with characteristic roots < œ " „ 3 . It follows that
aH# #H #bc/> =38 >d œ ! . Therefore the operator
L aHb œ aH "baH "b# ˆH# #H #‰
is an annihilator of $/> #>/> /> =38 > . The operator corresponding to the left hand
side of the given ODE is H# aH# #H #bÞ It follows that
H# aH "baH "b# ˆH# #H #‰ ] œ ! .
#
The resulting ODE is homogeneous, with solution
] a>b œ -" -# > -$ /> -% /> -& >/>
/> a-' -9= > -( =38 >b >/> a-) -9= > -* =38 > b Þ
After examining the homogeneous solution of Prob. ") , and eliminating duplicate terms,
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—————————————————————————— CHAPTER 4. ——
we have
] a>b œ -$ /> -% /> -& >/> >/> a-) -9= > -* =38 > b.
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—————————————————————————— CHAPTER 4. ——
Section 4.4
2. The characteristic equation is <a<# "b œ ! . Hence the homogeneous solution is
C- a>b œ -" -# /> -$ /> . The Wronskian is evaluated as [ a"ß /> ß /> b œ # . Now
compute the three determinants
â
â
/> â
â ! />
â
â
[" a>b œ â ! /> /> â œ #
â
â
/> â
â " />
â
â"
â
[# a>b œ â !
â
â!
â
/> â
â
/> ⠜ />
â
/> â
!
!
"
â
â"
â
[$ a>b œ â !
â
â!
/>
/>
/>
The solution of the system of equations a"!b is
?"w a>b œ
?#w a>b œ
?$w a>b œ
â
!â
â
! ⠜ />
â
"â
> [ " a >b
œ >
[ a>b
> [ # a >b
œ >/> Î#
[ a>b
> [ $ a >b
œ >/> Î#
[ a>b
Hence ?" a>b œ ># Î# , ?2 a>b œ /> a> "bÎ# , ?$ a>b œ /> a> "bÎ# Þ The particular
solution becomes ] a>b œ ># Î# a> "bÎ# a> "bÎ# œ ># Î# " . The constant
is a solution of the homogeneous equation, therefore the general solution is
Ca>b œ -" -# /> -$ /> ># Î# .
3. From Prob. "$ in Section %Þ#, C- a>b œ -" /> -# /> -$ /#> Þ The Wronskian is
evaluated as [ a/> ß /> ß /#> b œ ' /#> . Now compute the three determinants
â
â
â ! /> /#> â
â
â
[" a>b œ â ! /> #/#> â œ /$>
â
â
â " /> %/#> â
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—————————————————————————— CHAPTER 4. ——
â >
â /
â
[# a>b œ â />
â >
â /
â
! /#> â
â
! #/#> ⠜ $/>
â
" %/#> â
â >
â /
â
[$ a>b œ â />
â >
â /
/>
/>
/>
â
!â
â
!⠜ #
â
"â
Hence ?"w a>b œ /&> Î' , ?#w a>b œ /$> Î# , ?$w a>b œ /#> Î$ . Therefore the particular solution
can be expressed as
] a>b œ /> /&> Î$!‘ /> /$> Î'‘ /#> /#> Î'‘
œ /%> Î$! Þ
6. From Prob. ## in Section %Þ#, C- a>b œ -" -9= > -# =38 > >c-$ -9= > -% =38 > dÞ The
Wronskian is evaluated as [ a-9= >ß =38 >ß > -9= >ß > =38 >b œ % . Now compute the four
auxiliary determinants
â
â
â!
=38 >
> -9= >
> =38 >
â
â
â
-9= >
-9= > > =38 >
=38 > > -9= > â
â!
[" a>b œ â
⠜ #=38 > #> -9= >
#-9= > > =38 > â
â ! =38 > #=38 > > -9= >
â
â
â " -9= > $-9= > > =38 > $=38 > > -9= > â
â
â
â
â
[# a>b œ â
â
â
â
-9= >
!
=38 > !
-9= > !
=38 >
"
â
â
â
â
[$ a>b œ â
â
â
â
â
â
â
â
[% a>b œ â
â
â
â
> -9= >
-9= > > =38 >
#=38 > > -9= >
$-9= > > =38 >
-9= >
=38 >
-9= >
=38 >
=38 >
!
-9= >
!
=38 > !
-9= > "
-9= >
=38 >
-9= >
=38 >
=38 >
-9= >
=38 >
-9= >
â
> =38 >
â
â
=38 > > -9= > â
⠜ #> =38 > #-9= >
#-9= > > =38 > â
â
$=38 > > -9= > â
â
â
> =38 >
â
=38 > > -9= > â
⠜ #-9= >
#-9= > > =38 > â
â
$=38 > > -9= > â
â
> -9= >
!â
â
-9= > > =38 >
!â
⠜ #=38 >
#=38 > > -9= > ! â
â
$-9= > > =38 > " â
It follows that ?"w a>b œ c =38# > > =38 > -9= >dÎ# , ?#w a>b œ c>=38# > =38 > -9= >dÎ# ,
?$w a>b œ =38 > -9= >Î# , ?%w a>b œ =38# >Î# . Hence
?" a>b œ $=38 > -9= > #> -9=# > >‘Î)
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—————————————————————————— CHAPTER 4. ——
?# a>b œ =38# > #-9=# > #> =38 > -9= > ># ‘Î)
?$ a>b œ =38# >Î%
?% a>b œ c-9= > =38 > >dÎ%
Therefore the particular solution can be expressed as
] a>b œ -9= > c?" a>bd =38 > c?# a>bd > -9= > c?$ a>bd > =38 > c?% a>bd
œ =38 > $> -9= > ># =38 >‘Î) Þ
Note that only the last term is not a solution of the homogeneous equation. Hence the
general solution is
Ca>b œ -" -9= > -# =38 > >c-$ -9= > -% =38 > d ># =38 > Î) Þ
8. Based on the results in Prob. # , C- a>b œ -" -# /> -$ /> . It was also shown that
[ a"ß /> ß /> b œ # , with [" a>b œ # , [# a>b œ /> , [$ a>b œ /> . Therefore we have
?"w a>b œ -=- > , ?#w a>b œ /> -=- > Î# , ?$w a>b œ /> -=- > Î# . The particular solution can
be expressed as ] a>b œ c?" a>bd /> c?# a>bd /> c?$ a>bdÞ More specifically,
/> > =
/> > =
] a>b œ 68k-=- a>b -9>a>bk ( / -=- a=b.=
( / -=- a=b.=
# >!
# >!
œ 68k-=- a>b -9>a>bk ( -9=2 a> =b-=- a=b.= .
>
>!
9. Based on Prob. % , ?"w a>b œ =/- > , ?#w a>b œ " , ?$w a>b œ >+8 > . The particular
solution can be expressed as ] a>b œ c?" a>bd -9= > c?# a>bd =38 > c?$ a>bdÞ That is,
] a>b œ 68k=/- a>b >+8a>bk > -9= > =38 > 68k-9=a>bk.
Hence the general solution of the initial value problem is
Ca>b œ -" -# -9= > -$ =38 > 68k=/- a>b >+8a>bk > -9= > =38 > 68k-9=a>bkÞ
Invoking the initial conditions, we require that -" -# œ # , -$ œ " , -# œ # .
Therefore
Ca>b œ #-9= > =38 > 68k=/- a>b >+8a>bk > -9= > =38 > 68k-9=a>bk
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—————————————————————————— CHAPTER 4. ——
10. From Prob. ' , Ca>b œ -" -9= > -# =38 > -$ > -9= > -% > =38 > ># =38 > Î) Þ In
order to satisfy the initial conditions, we require that -" œ # , -# -$ œ ! ,
-" #-% œ " , $Î% -# $-$ œ " Þ Therefore
Ca>b œ #-9= > (=38 > (> -9= > %> =38 > ># =38 > ‘Î) Þ
12. From Prob. ) , the general solution of the initial value problem is
Ca>b œ -" -# /> -$ /> 68k-=- a>b -9>a>bk
/> > =
/> > =
( / -=- a=b.=
( / -=- a=b.=.
# >!
# >!
In this case, >! œ 1Î# . Observe that C a1Î#b œ C- a1Î#b , C w a1Î#b œ C-w a1Î#b , and
C ww a1Î#b œ C-ww a1Î#b . Therefore we obtain the system of equations
-" -# /1Î# -$ /1Î# œ #
-# /1Î# -$ /1Î# œ "
-# /1Î# -$ /1Î# œ "
Hence the solution of the initial value problem is
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—————————————————————————— CHAPTER 4. ——
Ca>b œ $ />1Î# 68k-=- a>b -9>a>bk ( -9=2 a> =b-=- a=b.= Þ
>
>!
13. First write the equation as C www B" C ww #B# C w #B$ C œ #B Þ The Wronskian
is evaluated as [ aBß B# ß "ÎBb œ 'ÎB . Now compute the three determinants
â
â
â ! B#
"ÎB â
â
â
[" aBb œ â ! #B "ÎB# â œ $
â
â
#ÎB$ â
â" #
â
âB !
â
[# aBb œ â " !
â
â! "
â
"ÎB â
â
"ÎB# â œ #ÎB
â
#ÎB$ â
â
â
â B B# ! â
â
â
[$ aBb œ â " #B ! â œ B#
â
â
â! # "â
Hence ?"w aBb œ B# , ?#w aBb œ #BÎ$ , ?$w aBb œ B% Î$ . Therefore the particular solution
can be expressed as
] aBb œ B B$ Î$‘ B# B# Î$‘
" &
B Î"&‘
B
œ B% Î"& Þ
15. The homogeneous solution is C- a>b œ -" -9= > -# =38 > -$ -9=2 > -% =382 > Þ The
Wronskian is evaluated as [ a-9= >ß =38 >ß -9=2 >ß =382 >b œ % . Now the four additional
determinants are given by [" a>b œ # =38 > , [# a>b œ # -9= > , [$ a>b œ # =382 > ,
[% a>b œ # -9=2 > . If follows that ?"w a>b œ 1a>b =38a>bÎ# , ?#w a>b œ 1a>b -9=a>bÎ# ,
?$w a>b œ 1a>b =382a>bÎ# , ?%w a>b œ 1a>b -9=2a>bÎ# . Therefore the particular solution
________________________________________________________________________
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—————————————————————————— CHAPTER 4. ——
can be expressed as
-9=a>b >
=38a>b >
( 1a=b =38a=b .=
( 1a=b -9=a=b .=
#
#
>!
>!
-9=2a>b >
=382a>b >
( 1a=b =382a=b .=
( 1a=b -9=2a=b .= Þ
#
#
>!
>!
] a>b œ
Using the appropriate identities, the integrals can be combined to obtain
] a>b œ
" >
" >
1
a
=
b
=382
a
>
=
b
.=
(
( 1a=b =38a> =b .= Þ
# >!
# >!
17. First write the equation as C www $ B" C ww ' B# C w ' B$ C œ 1aBb Î B$ Þ It can
be shown that C- aBb œ -" B -# B# -$ B$ is a solution of the homogeneous equation.
The Wronskian of this fundamental set of solutions is [ aBß B# ß B$ b œ #B$ Þ The three
additional determinants are given by [" aBb œ B% , [# aBb œ #B$ , [$ aBb œ B# Þ
Hence ?"w aBb œ 1aBbÎ#B# , ?#w aBb œ 1aBb ÎB$ , ?$w aBb œ 1aBbÎ#B% . Therefore the
particular solution can be expressed as
B
B
1a>b
1a>b
1a>b
#
$
.>
B
.>
B
.>
(
(
#
$
%
B! #>
B! >
B! #>
" B B
# B#
B$
œ ( ” # $ % •1a>b.> Þ
# B! >
>
>
] aBb œ B(
B
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page 169
—————————————————————————— CHAPTER 5. ——
Chapter Five
Section 5.1
1. Apply the ratio test :
lim
8Ä_
¸aB $b8" ¸
k a B $b 8 k
œ lim kB $k œ kB $kÞ
8Ä_
Hence the series converges absolutely for kB $k " . The radius of convergence is
3 œ " . The series diverges for B œ # and B œ % , since the n-th term does not approach
zero.
3. Applying the ratio test,
k8x B#8# k
B#
œ
lim
œ !Þ
8 Ä _ ka8 "bx B#8 k
8Ä_ 8 "
lim
The series converges absolutely for all values of B . Thus the radius of convergence is
3 œ _.
4. Apply the ratio test :
k#8" B8" k
œ lim #kBk œ #kBkÞ
8 Ä _ k#8 B8 k
8Ä_
lim
Hence the series converges absolutely for #kBk, or kBk "Î# . The radius of convergence
is 3 œ "Î# . The series diverges for B œ „"Î# , since the n-th term does not approach
zero.
6. Applying the ratio test,
¸8aB B! b8" ¸
8 Ä _ k a8
lim
"baB B! b k
8
8
kaB B! bk œ kaB B! bkÞ
8Ä_ 8 "
œ lim
Hence the series converges absolutely for kaB B! bk " . The radius of convergence is
3 œ " . At B œ B! " , we obtain the harmonic series, which is divergent. At the other
endpoint, B œ B! " , we obtain
a "b 8
,
8
8œ"
"
_
which is conditionally convergent.
7. Apply the ratio test :
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—————————————————————————— CHAPTER 5. ——
lim
8Ä_
¸$ 8 a8 "b# aB #b8" ¸
k$ 8" 8# aB #b8 k
a 8 "b #
"
œ lim
k a B # b k œ k aB # b k Þ
#
8Ä_
$8
$
Hence the series converges absolutely for "$ kB #k ", or kB #k $ . The radius of
convergence is 3 œ $ . At B œ & and B œ " , the series diverges, since the n-th
term does not approach zero.
8. Applying the ratio test,
k88 a8 "bx B8" k
88
"
œ
lim
kBk œ kBk,
8
8"
8 Ä _ ¸ a 8 "b
8 Ä _ a 8 "b
/
8x B8 ¸
lim
since
88
8 œ
8 Ä _ a 8 "b
lim
lim Œ"
8Ä_
" 8
"
œ/ .
8
Hence the series converges absolutely for kBk /. The radius of convergence is 3 œ / .
At B œ „ /, the series diverges, since the n-th term does not approach zero. This follows
from the fact that
lim
8Ä_
8x /8
œ ".
88 È # 1 8
10. We have 0 aBb œ /B , with 0 a8b aBb œ /B , for 8 œ "ß #ß â . Therefore 0 a8b a!b œ ".
Hence the Taylor expansion about B! œ ! is
/B œ "
_
B8
.
8x
8œ!
Applying the ratio test,
k8xB8" k
"
lim
œ lim
kBk œ !.
8 Ä _ ka8 "bx B8 k
8Ä_ 8 "
The radius of convergence is 3 œ _ .
11. We have 0 aBb œ B , with 0 w aBb œ " and 0 a8b aBb œ ! , for 8 œ #ß â . Clearly,
0 a"b œ " and 0 w a"b œ " , with all other derivatives equal to zero. Hence the Taylor
expansion about B! œ " is
B œ " aB "bÞ
Since the series has only a finite number of terms, the converges absolutely for all B .
14. We have 0 aBb œ "Îa" Bb, 0 w aBb œ "Îa" Bb# , 0 ww aBb œ #Îa" Bb$ , â
with 0 a8b aBb œ a "b8 8xÎa" Bb8" , for 8 " . It follows that 0 a8b a!b œ a "b8 8x
________________________________________________________________________
page 170
—————————————————————————— CHAPTER 5. ——
for 8
! . Hence the Taylor expansion about B! œ ! is
_
"
œ " a "b8 B8 .
" B 8œ!
Applying the ratio test,
kB8" k
œ lim kBk œ kBk.
lim
8 Ä _ k B8 k
8Ä_
The series converges absolutely for kBk " , but diverges at B œ „ " .
15. We have 0 aBb œ "Îa" Bb, 0 w aBb œ "Îa" Bb# , 0 ww aBb œ #Îa" Bb$ , â
with 0 a8b aBb œ 8xÎa" Bb8" , for 8 " . It follows that 0 a8b a!b œ 8x, for 8 ! .
Hence the Taylor expansion about B! œ ! is
_
"
œ " B8 .
" B 8œ!
Applying the ratio test,
kB8" k
œ lim kBk œ kBk.
8 Ä _ k B8 k
8Ä_
lim
The series converges absolutely for kBk " , but diverges at B œ „ " .
16. We have 0 aBb œ "Îa" Bb, 0 w aBb œ "Îa" Bb# , 0 ww aBb œ #Îa" Bb$ , â
with 0 a8b aBb œ 8xÎa" Bb8" , for 8 " . It follows that 0 a8b a#b œ a "b8" 8x for
8 ! . Hence the Taylor expansion about B! œ # is
_
"
œ " a " b 8 aB # b 8 .
"B
8œ!
Applying the ratio test,
lim
8Ä_
¸aB #b8" ¸
k a B #b 8 k
œ lim kB #k œ kB #k.
8Ä_
The series converges absolutely for kB #k " , but diverges at B œ " and B œ $ .
17. Applying the ratio test,
ka8 "bB8" k
8"
œ lim
kBk œ kBk.
8
8Ä_
8
Ä
_
k8B k
8
lim
The series converges absolutely for kBk " . Term-by-term differentiation results in
________________________________________________________________________
page 171
—————————————————————————— CHAPTER 5. ——
C œ " 8# B8" œ " %B *B# "' B$ â
_
w
8œ"
C œ " 8# a8 "b B8# œ % ")B %)B# "!!B$ â
_
ww
8œ#
Shifting the indices, we can also write
C w œ " a8 "b# B8 and C ww œ " a8 #b# a8 "b B8 Þ
_
_
8œ!
8œ!
20. Shifting the index in the second series, that is, setting 8 œ 5 " ,
" +5 B5" œ " +8" B8 .
_
_
5œ!
8œ"
Hence
" +5" B " +5 B
_
_
5
5œ!
5œ!
œ " +5" B " +5" B5
_
5"
_
5
5œ!
5œ"
œ +" " a+5" +5" bB5" Þ
_
5œ"
21. Shifting the index by # , that is, setting 7 œ 8 # ,
" 8a8 "b+8 B8# œ " a7 #ba7 "b+7# B7
_
_
8œ#
7œ!
_
œ " a8 #ba8 "b+8# B8 Þ
8œ!
22. Shift the index down by # , that is, set 7 œ 8 # . It follows that
" +8 B8# œ " +7# B7
_
_
8œ!
7œ#
_
œ " +8# B8 Þ
8œ#
24. Clearly,
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—————————————————————————— CHAPTER 5. ——
ˆ" B# ‰" 8a8 "b+8 B8# œ " 8a8 "b+8 B8# " 8a8 "b+8 B8 Þ
_
_
_
8œ#
8œ#
8œ#
Shifting the index in the first series, that is, setting 5 œ 8 # ,
" 8a8 "b+8 B8# œ " a5 #ba5 "b+5# B5
_
_
8œ#
5œ!
_
œ " a8 #ba8 "b+8# B8 Þ
8œ!
Hence
ˆ" B
#‰
" 8a8 "b+8 B
œ " a8 #ba8 "b+8# B " 8a8 "b+8 B8 .
_
_
8#
_
8
8œ#
8œ!
8œ#
Note that when 8 œ ! and 8 œ " , the coefficients in the second series are zero. So that
ˆ" B# ‰" 8a8 "b+8 B8#
_
8œ#
œ " ca8 #ba8 "b+8# 8a8 "b+8 dB8 Þ
_
8œ!
26. Clearly,
" 8+8 B8" B " +8 B8 œ " 8+8 B8" " +8 B8" Þ
_
_
_
_
8œ"
8œ!
8œ"
8œ!
Shifting the index in the first series, that is, setting 5 œ 8 " ,
" 8+8 B8" œ " a5 "b+5" B5 Þ
_
_
8œ"
5œ!
Shifting the index in the second series, that is, setting 5 œ 8 " ,
" +8 B
_
8"
8œ!
œ " +5" B5 Þ
_
5œ"
Combining the series, and starting the summation at 8 œ " ,
" 8+8 B8" B " +8 B8 œ +" " ca8 "b+8" +8" dB8 Þ
_
_
_
8œ"
8œ!
8œ"
27. We note that
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—————————————————————————— CHAPTER 5. ——
B " 8a8 "b+8 B8# " +8 B8 œ " 8a8 "b+8 B8" " +8 B8 Þ
_
_
_
_
8œ#
8œ!
8œ#
8œ!
Shifting the index in the first series, that is, setting 5 œ 8 " ,
" 8a8 "b+8 B8" œ " 5 a5 "b+5" B5
_
_
8œ#
5œ"
_
œ " 5 a5 "b+5" B5 ,
5œ!
since the coefficient of the term associated with 5 œ ! is zero. Combining the series,
B " 8a8 "b+8 B
_
8#
8œ#
" +8 B œ " c8a8 "b+8" +8 dB8 Þ
_
_
8
8œ!
8œ!
________________________________________________________________________
page 174
—————————————————————————— CHAPTER 5. ——
Section 5.2
1. Let C œ +! +" B +# B# â +8 B8 â . Then
C ww œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ
_
_
8œ#
8œ!
Substitution into the ODE results in
" a8 #ba8 "b+8# B8 " +8 B8 œ !
_
_
8œ!
8œ!
or
" ca8 #ba8 "b+8# +8 dB8 œ ! Þ
_
8œ!
Equating all the coefficients to zero,
a8 #ba8 "b+8# +8 œ ! ,
8 œ !ß "ß #ß â Þ
We obtain the recurrence relation
+8# œ
+8
,
a8 "ba8 #b
8 œ !ß "ß #ß â Þ
The subscripts differ by two, so for 5 œ "ß #ß â
+#5#
+#5%
+!
+#5 œ
œ
œâœ
a#5 "b#5
a#5 $ba#5 #ba#5 "b#5
a#5 bx
and
+#5" œ
+#5"
+#5$
+"
œ
œâœ
Þ
#5 a#5 "b
a#5 #ba#5 "b#5 a#5 "b
a#5 "bx
Hence
C œ +! "
_
B#5
B#5"
+" "
Þ
a
#5
b
x
a
#5
"
b
x
5œ!
5œ!
_
The linearly independent solutions are
C" œ +! Œ"
B#
B%
B'
⠜ +! -9=2 B
#x
%x
'x
C# œ +" ŒB
B$
B&
B(
⠜ +" =382 B .
$x
&x
(x
________________________________________________________________________
page 175
—————————————————————————— CHAPTER 5. ——
4. Let C œ +! +" B +# B# â +8 B8 â . Then
C ww œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ
_
_
8œ#
8œ!
Substitution into the ODE results in
" a8 #ba8 "b+8# B8 5 # B# " +8 B8 œ ! .
_
_
8œ!
8œ!
Rewriting the second summation,
" a8 #ba8 "b+8# B8 " 5 # +8# B8 œ ! ,
_
_
8œ!
8œ#
that is,
#+# $ † # +$ B " a8 #ba8 "b+8# 5 # +8# ‘B8 œ ! Þ
_
8œ#
Setting the coefficients equal to zero, we have +# œ ! , +$ œ ! , and
a8 #ba8 "b+8# 5 # +8# œ ! ,
for 8 œ #ß $ß %ß â Þ
The recurrence relation can be written as
+8# œ
5 # +8#
, 8 œ #ß $ß %ß â .
a8 #ba8 "b
The indices differ by four, so +% , +) , +"# , â are defined by
5 # +!
5 # +%
5 # +)
+% œ
, +) œ
, +"# œ
, â.
%†$
)†(
"# † ""
Similarly, +& , +* , +"$ , â are defined by
5 # +"
5 # +&
5 # +*
+& œ
, +* œ
, +"$ œ
, â.
&†%
*†)
"$ † "#
The remaining coefficients are zero. Therefore the general solution is
5# %
5%
5'
)
C œ + ! ”"
B
B
B"# â•
%†$
)†(†%†$
"# † "" † ) † ( † % † $
#
%
5
5
5'
+ " ”B
B&
B*
B"$ â•.
&†%
*†)†&†%
"$ † "# † * † ) † % † %
Note that for the even coefficients,
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page 176
—————————————————————————— CHAPTER 5. ——
5 # +%7%
, 7 œ "ß #ß $ß â
a%7 "b%7
+%7 œ
and for the odd coefficients,
5 # +%7$
, 7 œ "ß #ß $ß â .
%7a%7 "b
+%7" œ
Hence the linearly independent solutions are
7"
a "b7" ˆ5 # B% ‰
$ † % † ( † ) â a%7 $ba%7 %b
7œ!
C" aBb œ " "
_
7"
a "b7" ˆ5 # B% ‰
C# aBb œ B–" "
Þ
% † & † ) † * â a%7 %ba%7 &b —
7œ!
_
6. Let C œ +! +" B +# B# â +8 B8 â . Then
C w œ " 8+8 B8" œ " a8 "b+8" B8
_
_
8œ"
8œ!
and
C ww œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ
_
_
8œ#
8œ!
Substitution into the ODE results in
ˆ# B# ‰" a8 #ba8 "b+8# B8 B" a8 "b+8" B8 %" +8 B8 œ ! .
_
_
_
8œ!
8œ!
8œ!
Before proceeding, write
B# " a8 #ba8 "b+8# B8 œ " 8a8 "b+8 B8
_
_
8œ!
8œ#
and
B" a8 "b+8" B8 œ " 8 +8 B8 Þ
_
_
8œ!
8œ"
It follows that
%+! %+# a$+" "#+$ bB " c#a8 #ba8 "b+8# 8a8 "b+8 8 +8 %+8 dB
_
8
œ !.
8œ#
________________________________________________________________________
page 177
—————————————————————————— CHAPTER 5. ——
Equating the coefficients to zero, we find that +# œ +! , +$ œ +" Î% , and
8# #8 %
+8 , 8 œ !ß "ß #ß â .
#a8 #ba8 "b
+8# œ
The indices differ by two, so for 5 œ !ß "ß #ß â
+#5# œ
and
a#5 b# %5 %
+#5
#a#5 #ba#5 "b
a#5 "b# %5 #
œ
+#5" Þ
#a#5 $ba#5 #b
+#5$
Hence the linearly independent solutions are
B%
B'
C" aBb œ " B
â
'
$!
#
C# aBb œ B
B$
(B&
"*B(
âÞ
%
"'! "*#!
7. Let C œ +! +" B +# B# â +8 B8 â . Then
C œ " 8+8 B8" œ " a8 "b+8" B8
_
_
8œ"
8œ!
w
and
C ww œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ
_
_
8œ#
8œ!
Substitution into the ODE results in
" a8 #ba8 "b+8# B B" a8 "b+8" B #" +8 B8 œ ! .
_
_
_
8
8œ!
8
8œ!
8œ!
First write
B" a8 "b+8" B8 œ " 8 +8 B8 Þ
_
_
8œ!
8œ"
We then obtain
#+# #+! " ca8 #ba8 "b+8# 8 +8 #+8 dB8 œ ! Þ
_
8œ"
________________________________________________________________________
page 178
—————————————————————————— CHAPTER 5. ——
It follows that +# œ +! and +8# œ +8 Îa8 "b , 8 œ !ß "ß #ß â . Note that the
indices differ by two, so for 5 œ "ß #ß â
+#5
+#5#
+#5%
a "b 5 + !
œ
œ
œâœ
#5 "
a#5 $ba#5 "b
" † $ † & âa#5 "b
and
+#5"
+#5"
+#5$
a "b 5 + "
œ
œ
œâœ
Þ
#5
a#5 #b#5
# † % † ' âa#5 b
Hence the linearly independent solutions are
_
B#
B%
B'
a "b8 B#8
"
C" aBb œ "
âœ"
"
"†$ "†$†&
" † $ † & âa#8 "b
8œ"
_
B$
B&
B(
a "b8 B#8"
"
C# aBb œ B
âœB
Þ
#
#†% #†%†'
# † % † ' âa#8b
8œ"
9. Let C œ +! +" B +# B# â +8 B8 â . Then
C œ " 8+8 B
_
w
8"
8œ"
œ " a8 "b+8" B8
_
8œ!
and
C ww œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ
_
_
8œ#
8œ!
Substitution into the ODE results in
ˆ" B# ‰" a8 #ba8 "b+8# B8 %B" a8 "b+8" B8 '" +8 B8 œ ! .
_
_
_
8œ!
8œ!
8œ!
Before proceeding, write
B# " a8 #ba8 "b+8# B8 œ " 8a8 "b+8 B8
_
_
8œ!
8œ#
and
B" a8 "b+8" B8 œ " 8 +8 B8 Þ
_
_
8œ!
8œ"
It follows that
________________________________________________________________________
page 179
—————————————————————————— CHAPTER 5. ——
'+! #+# a#+" '+$ bB " ca8 #ba8 "b+8# 8a8 "b+8 %8 +8 '+8 dB
_
8
œ !.
8œ#
Setting the coefficients equal to zero, we obtain +# œ $+! , +$ œ +" Î$ , and
+8# œ
a8 #ba8 $b
+8 , 8 œ !ß "ß #ß â .
a8 "ba8 #b
Observe that for 8 œ # and 8 œ $ , we obtain +% œ +& œ ! . Since the indices differ by
two, we also have +8 œ ! for 8 % . Therefore the general solution is a polynomial
C œ +! +" B $+! B# +" B$ Î$ .
Hence the linearly independent solutions are
C" aBb œ " $B#
and C# aBb œ B B$ Î$ .
10. Let C œ +! +" B +# B# â +8 B8 â . Then
C ww œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ
_
_
8œ#
8œ!
Substitution into the ODE results in
ˆ% B# ‰" a8 #ba8 "b+8# B8 #" +8 B8 œ ! Þ
_
_
8œ!
8œ!
First write
B# " a8 #ba8 "b+8# B8 œ " 8a8 "b+8 B8 .
_
_
8œ!
8œ#
It follows that
#+! )+# a#+" #%+$ bB " c%a8 #ba8 "b+8# 8a8 "b+8 #+8 dB8 œ ! .
_
8œ#
We obtain +# œ +! Î% , +$ œ +" Î"# and
%a8 #b+8# œ a8 #b+8 , 8 œ !ß "ß #ß â .
Note that for 8 œ # , +% œ ! . Since the indices differ by two, we also have +#5 œ ! for
5 œ #ß $ß â . On the other hand, for 5 œ "ß #ß â ,
+#5" œ
+"
a#5 $b+#5"
a#5 &ba#5 $b+#5$
œ
œâœ 5
Þ
#
%a#5 "b
% a#5 "ba#5 "b
% a#5 "ba#5 "b
Therefore the general solution is
________________________________________________________________________
page 180
—————————————————————————— CHAPTER 5. ——
_
B#
B#8"
"
C œ +! + " B + ! + "
.
%
%8 a#8 "ba#8 "b
8œ"
Hence the linearly independent solutions are C" aBb œ " B# Î% and
C# aBb œ B
_
B#8"
B$
B&
B(
.
âœB" 8
% a#8 "ba#8 "b
"# #%! ##%!
8œ"
11. Let C œ +! +" B +# B# â +8 B8 â . Then
C w œ " 8+8 B8" œ " a8 "b+8" B8
_
_
8œ"
8œ!
and
C œ " 8a8 "b+8 B
_
ww
œ " a8 #ba8 "b+8# B8 Þ
_
8#
8œ#
8œ!
Substitution into the ODE results in
ˆ$ B# ‰" a8 #ba8 "b+8# B8 $B" a8 "b+8" B8 " +8 B8 œ ! .
_
_
_
8œ!
8œ!
8œ!
Before proceeding, write
_
#"
B
8œ!
a8 #ba8 "b+8# B œ " 8a8 "b+8 B8
_
8
8œ#
and
B" a8 "b+8" B8 œ " 8 +8 B8 Þ
_
_
8œ!
8œ"
It follows that
8
'+# +! a %+" ")+$ bB " c$a8 #ba8 "b+8# 8a8 "b+8 $8 +8 +8 dB œ !.
_
8œ#
We obtain +# œ +! Î' , #+$ œ +" Î* , and
$a8 #b+8# œ a8 "b+8 , 8 œ !ß "ß #ß â .
The indices differ by two, so for 5 œ "ß #ß â
+#5 œ
$ † & âa#5 "b +!
a#5 "b+#5#
a#5 $ba#5 "b+#5%
œ
œâœ 5
#
$a#5 b
$ a#5 #ba#5 b
$ † # † % âa#5 b
and
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—————————————————————————— CHAPTER 5. ——
+#5" œ
a#5 b+#5"
a#5 #ba#5 b+#5$
# † % † ' âa#5 b +"
œ #
œâœ 5
Þ
$a#5 "b
$ a#5 "ba#5 "b
$ † $ † & âa#5 "b
Hence the linearly independent solutions are
_
B#
B%
&B'
$ † & âa#8 "b B#8
C" aBb œ "
âœ"" 8
'
#% %$#
$ † # † % âa#8b
8œ"
_
#B$
)B&
"'B(
# † % † ' âa#8b B#8"
âœB" 8
Þ
*
"$&
*%&
$ † $ † & âa#8 "b
8œ"
C# aBb œ B
12. Let C œ +! +" B +# B# â +8 B8 â . Then
C w œ " 8+8 B8" œ " a8 "b+8" B8
_
_
8œ"
8œ!
and
C œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ
_
_
8œ#
8œ!
ww
Substitution into the ODE results in
a" Bb" a8 #ba8 "b+8# B8 B" a8 "b+8" B8 " +8 B8 œ ! .
_
_
_
8œ!
8œ!
8œ!
Before proceeding, write
B " a8 #ba8 "b+8# B8 œ " a8 "b8 +8" B8
_
_
8œ!
8œ"
and
B" a8 "b+8" B8 œ " 8 +8 B8 Þ
_
_
8œ!
8œ"
It follows that
#+# +! " ca8 #ba8 "b+8# a8 "b8 +8" 8 +8 +8 dB8 œ !.
_
8œ"
We obtain +# œ +! Î# and
a8 #ba8 "b+8# a8 "b8 +8" a8 "b+8 œ !
for 8 œ !ß "ß #ß â . Writing out the individual equations,
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—————————————————————————— CHAPTER 5. ——
$ † # +$ # † " + #
% † $ +% $ † # + $ + #
& † % +& % † $ + % # + $
' † & +' & † % + & $ + %
ã
œ!
œ!
œ!
œ!
The coefficients can be calculated successively as +$ œ +! Îa# † $b, +% œ +$ Î# +# Î"#
œ +! Î#% , +& œ $+% Î& +$ Î"! œ +! Î"#! , â . We can now see that for 8 # , +8 is
proportional to +! . In fact, for 8 # , +8 œ +! Îa8xb . Therefore the general solution is
C œ +! + " B
+! B#
+! B$
+ ! B%
â.
#x
$x
%x
Hence the linearly independent solutions are C# aBb œ B and
C" aBb œ " "
_
B8
Þ
8x
8œ#
13. Let C œ +! +" B +# B# â +8 B8 â . Then
C œ " 8+8 B
_
w
8œ"
œ " a8 "b+8" B8
_
8"
8œ!
and
C ww œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ
_
_
8œ#
8œ!
Substitution into the ODE results in
# " a8 #ba8 "b+8# B8 B" a8 "b+8" B8 $" +8 B8 œ ! .
_
_
_
8œ!
8œ!
8œ!
First write
B" a8 "b+8" B8 œ " 8 +8 B8 Þ
_
_
8œ!
8œ"
We then obtain
%+# $+! " c#a8 #ba8 "b+8# 8 +8 $+8 dB8 œ ! Þ
_
8œ"
It follows that +# œ $+! Î% and
#a8 #ba8 "b+8# a8 $b+8 œ !
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—————————————————————————— CHAPTER 5. ——
for 8 œ !ß "ß #ß â . The indices differ by two, so for 5 œ "ß #ß â
a#5 "b+#5#
a#5 "ba#5 "b+#5%
œ #
ω
#a#5 "ba#5 b
# a#5 $ba#5 #ba#5 "ba#5 b
a "b5 $ † & âa#5 "b
œ
+! Þ
#5 a#5 bx
+#5 œ
and
a#5 #b+#5"
a#5 ba#5 #b+#5$
œ #
ω
#a#5 ba#5 "b
# a#5 #ba#5 "ba#5 ba#5 "b
a "b5 % † ' âa#5 ba#5 #b
œ
+" Þ
#5 a#5 "bx
+#5" œ
Hence the linearly independent solutions are
_
$
&
( '
a "b8 $ † & âa#8 "b #8
C" aBb œ " B# B%
B ✠"
B
%
$#
$)%
#8 a#8bx
8œ!
_
"
"
" (
a "b8 % † ' âa#8 #b #8"
C# aBb œ B B$ B&
B âœB"
B
Þ
$
#!
#"!
#8 a#8 "bx
8œ"
15a+b. From Prob. # , we have
C" aBb œ "
_
B#8
#8 8x
8œ!
C# aBb œ "
_
and
#8 8x B#8"
.
a#8 "bx
8œ!
Since +! œ C a!b and +" œ C w a!b , we have C aBb œ # C" aBb C# aBb . That is,
"
"
"
"
CaBb œ # B B# B$ B% B& B' â .
$
%
"&
#%
The four- and five-term polynomial approximations are
:% œ # B B# B$ Î$
:& œ # B B# B$ Î$ B% Î% Þ
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—————————————————————————— CHAPTER 5. ——
a, b .
a- b. The four-term approximation :% appears to be reasonably accurate awithin "!%b
on the interval kBk !Þ( .
17a+b. From Prob. (, the linearly independent solutions are
a "b8 B#8
" † $ † & âa#8 "b
8œ"
C" aBb œ " "
_
a "b8 B#8"
Þ
#
†
%
†
'
â
a
#8
b
8œ"
C# aBb œ B "
_
Since +! œ C a!b and +" œ C w a!b , we have C aBb œ % C" aBb C# aBb . That is,
"
%
"
%
CaBb œ % B %B# B$ B% B& B' â .
#
$
)
"&
The four- and five-term polynomial approximations are
"
:% œ % B %B# B$
#
"
%
:& œ % B %B# B$ B% Þ
#
$
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—————————————————————————— CHAPTER 5. ——
a, b .
a- b. The four-term approximation :% appears to be reasonably accurate awithin "!%b
on the interval kBk !Þ& .
18a+b. From Prob. "# , we have
C" aBb œ " "
_
B8
8x
8œ#
and
C# aBb œ B .
Since +! œ C a!b and +" œ C w a!b , we have C aBb œ $ C" aBb # C# aBb . That is,
$
"
"
"
" '
CaBb œ $ #B B# B$ B% B&
B â.
#
#
)
%!
#%!
The four- and five-term polynomial approximations are
$
:% œ $ #B B#
#
$
:& œ $ #B B#
#
" $
B
#
" $ " %
B B Þ
#
)
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—————————————————————————— CHAPTER 5. ——
a, b .
a- b. The four-term approximation :% appears to be reasonably accurate awithin "!%b
on the interval kBk !Þ* .
20. Two linearly independent solutions of Airy's equation aabout B! œ !b are
C" aBb œ " "
_
B$8
# † $ âa$8 "ba$8b
8œ"
C# aBb œ B "
_
B$8"
Þ
$ † % âa$8ba$8 "b
8œ"
Applying the ratio test to the terms of C" aBb ,
k# † $ âa$8 "ba$8b B$8$ k
"
œ lim
kBk$ œ ! Þ
$8
8 Ä _ k# † $ âa$8 #ba$8 $b B k
8 Ä _ a$8 "ba$8 #ba$8 $b
lim
Similarly, applying the ratio test to the terms of C# aBb ,
8 Ä _ k$
lim
¸$ † % âa$8ba$8 "b B$8% ¸
† % âa$8 $ba$8
%b B$8" k
"
kBk$ œ ! Þ
8 Ä _ a$8 #ba$8 $ba$8 %b
œ lim
Hence both series converge absolutely for all B .
21. Let C œ +! +" B +# B# â +8 B8 â . Then
C œ " 8+8 B
_
w
8"
8œ"
œ " a8 "b+8" B8
_
8œ!
and
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—————————————————————————— CHAPTER 5. ——
C œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ
_
_
8œ#
8œ!
ww
Substitution into the ODE results in
" a8 #ba8 "b+8# B8 #B" a8 "b+8" B8 - " +8 B8 œ ! .
_
_
_
8œ!
8œ!
8œ!
First write
B" a8 "b+8" B8 œ " 8 +8 B8 Þ
_
_
8œ!
8œ"
We then obtain
#+# - +! " ca8 #ba8 "b+8# #8 +8 - +8 dB8 œ ! Þ
_
8œ"
Setting the coefficients equal to zero, it follows that
+8# œ
a#8 -b
+8
a8 "ba8 #b
for 8 œ !ß "ß #ß â . Note that the indices differ by two, so for 5 œ "ß #ß â
a%5 % -b+#5#
a%5 ) -ba%5 % -b+#5%
œ
ω
a#5 "b#5
a#5 $ba#5 #ba#5 "b#5
- âa- %5 )ba- %5 %b
œ a "b 5
+! Þ
a#5 bx
+#5 œ
and
a%5 # -b+#5"
a%5 ' -ba%5 # -b+#5$
œ
ω
#5 a#5 "b
a#5 #ba#5 "b#5 a#5 "b
a- #b âa- %5 'ba- %5 #b
œ a "b 5
+" Þ
a#5 "bx
+#5" œ
Hence the linearly independent solutions of the Hermite equation aabout B! œ !b are
C" aBb œ "
C# aBb œ B
- # -a- %b % -a- %ba- )b '
B
B
B â
#x
%x
'x
- # $ a- #ba- 'b & a- #ba- 'ba- "!b (
B
B
B âÞ
$x
&x
(x
a,b. Based on the recurrence relation
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—————————————————————————— CHAPTER 5. ——
+8# œ
a#8 -b
+8 ,
a8 "ba8 #b
the series solution will terminate as long as - is a nonnegative even integer. If - œ #7,
then one or the other of the solutions in Part a,b will contain at most 7Î# " terms. In
particular, we obtain the polynomial solutions corresponding to - œ !ß #ß %ß 'ß )ß "! À
C" aBb œ "
C# aBb œ B
C" aBb œ " #B#
C# aBb œ B #B$ Î$
C" aBb œ " %B# %B% Î$
C# aBb œ B %B$ Î$ %B& Î"&
-œ!
-œ#
-œ%
-œ'
-œ)
- œ "!
a- b. Observe that if - œ #8 , and +! œ +" œ " , then
+#5 œ a "b5
and
+#5" œ a "b5
#8 âa#8 %5 )ba#8 %5 %b
a#5 bx
a#8 #b âa#8 %5 'ba#8 %5 #b
Þ
a#5 "bx
for 5 œ "ß #ß â c8Î#d. It follows that the coefficient of B8 , in C" and C# , is
Ú
a "b5 a%#55xbx for 8 œ #5
+8 œ Û
5 %5 5x
Ü a "b a#5"bx for 8 œ #5 "
Then by definition,
L8 aBb œ
5
a#5 bx
a#5 bx
a "b5 #8 %5 5x C" aBb œ a "b5 5x C" aBb for 8 œ #5
a#5"bx
# a#5"bx
a "b5 #8 %5 5x C# aBb œ a "b5
C# aBb for 8 œ
5x
#5 "
Therefore the first six Hermite polynomials are
L! aBb œ "
L" aBb œ #B
L# aBb œ %B# #
L$ aBb œ )B) "#B
L% aBb œ "'B% %)B# "#
L& aBb œ $#B& "'!B$ "#!B
23. The series solution is given by
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—————————————————————————— CHAPTER 5. ——
"
"
"
"
CaBb œ " B# # B% $ B' % B) â Þ
#
# #x
# $x
# %x
24. The series solution is given by
C aBb œ " B#
B%
B'
B)
â.
'
$! "#!
25. The series solution is given by
C aBb œ B
B$
B&
B(
B*
â.
#
#†% #†%†' #†%†'†)
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—————————————————————————— CHAPTER 5. ——
26. The series solution is given by
C aBb œ B
B$
B&
B(
B*
âÞ
"# #%! ##%! "'"#)
27. The series solution is given by
CaBb œ "
B%
B)
B"#
â.
"# '(# ))(!%
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—————————————————————————— CHAPTER 5. ——
28. Let C œ +! +" B +# B# â +8 B8 â . Then
C œ " 8+8 B8" œ " a8 "b+8" B8
_
_
8œ"
8œ!
w
and
C ww œ " 8a8 "b+8 B8# œ " a8 #ba8 "b+8# B8 Þ
_
_
8œ#
8œ!
Substitution into the ODE results in
a" Bb" a8 #ba8 "b+8# B8 B" a8 "b+8" B8 # " +8 B8 œ ! .
_
_
_
8œ!
8œ!
8œ!
After appropriately shifting the indices, it follows that
#+# #+! " ca8 #ba8 "b+8# a8 "b8 +8" 8 +8 # +8 dB8 œ !.
_
8œ"
We find that +# œ +! and
a8 #ba8 "b+8# a8 "b8 +8" a8 #b+8 œ !
for 8 œ "ß #ß â . Writing out the individual equations,
$ † # +$ # † " + # + "
% † $ +% $ † # + $
& † % +& % † $ + % + $
' † & +' & † % + & # + %
ã
œ!
œ!
œ!
œ!
Since +! œ ! and +" œ " , the remaining coefficients satisfy the equations
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—————————————————————————— CHAPTER 5. ——
$ † # +$ " œ !
% † $ +% $ † # + $ œ !
& † % +& % † $ + % + $ œ !
' † & +' & † % + & # + % œ !
ã
That is, +$ œ "Î' , +% œ "Î"# , +& œ "Î#% , +' œ "Î%& , â . Hence the series solution
of the initial value problem is
"
"
"
"
"$ (
CaBb œ B B$ B% B& B'
B âÞ
'
"#
#%
%&
"!!)
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—————————————————————————— CHAPTER 5. ——
Section 5.3
2. Let C œ 9aBb be a solution of the initial value problem. First note that
C ww œ a=38 BbC w a-9= BbC Þ
Differentiating twice,
C www œ a=38 BbC ww #a-9= BbC w a=38 BbC
C 3@ œ a=38 BbC www $a-9= BbC ww $a=38 BbC w a-9= BbC Þ
Given that 9a!b œ ! and 9 w a!b œ " , the first equation gives 9 ww a!b œ ! and the last
two equations give 9 www a!b œ # and 9 3@ a!b œ ! .
3. Let C œ 9aBb be a solution of the initial value problem. First write
C ww œ
" B w $ 68 B
C
CÞ
B#
B#
Differentiating twice,
C www œ
"
ˆB B# ‰C ww a$B 68 B B #bC w a$ ' 68 BbC ‘Þ
B$
"
’ˆB# B$ ‰C www ˆ$B# 68 B #B# %B‰C ww
B%
a' )B "#B 68 BbC w a") 68 B "&bC “Þ
C 3@ œ
Given that 9a"b œ # and 9 w a"b œ ! , the first equation gives 9 ww a"b œ ! and the last
two equations give 9 www a!b œ ' and 9 3@ a!b œ %# .
4. Let C œ 9aBb be a solution of the initial value problem. First note that
C ww œ B# C w a=38 BbC Þ
Differentiating twice,
C www œ B# C ww a#B =38 BbC w a-9= BbC
C 3@ œ B# C www a%B =38 BbC ww a# #-9= BbC w a=38 BbC Þ
Given that 9a!b œ +! and 9 w a!b œ +" , the first equation gives 9 ww a!b œ ! and the last
two equations give 9 www a!b œ +! and 9 3@ a!b œ %+" .
5. Clearly, :aBb œ % and ; aBb œ 'B are analytic for all B . Hence the series solutions
converge everywhere.
7. The zeroes of T aBb œ " B$ are the three cube roots of " . They all lie on the
unit circle in the complex plane. So for B! œ ! , 3738 œ " . For B! œ # , the nearest
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—————————————————————————— CHAPTER 5. ——
root is /31Î$ œ Š" 3È$ ‹Î# , hence 3738 œ È$ .
8. The only root of T aBb œ B is zero. Hence 3738 œ " .
9a,b. :aBb œ B and ; aBb œ " are analytic for all B .
a- b. :aBb œ B and ; aBb œ " are analytic for all B .
a. b. :aBb œ ! and ; aBb œ 5B# are analytic for all B .
a/b. The only root of T aBb œ " B is " . Hence 3738 œ " .
a1b. :aBb œ B and ; aBb œ # are analytic for all B .
a3bÞ The zeroes of T aBb œ " B# are „ 3 . Hence 3738 œ " .
a4bÞ The zeroes of T aBb œ % B# are „# . Hence 3738 œ # .
a5 bÞ The zeroes of T aBb œ $ B# are „È$ . Hence 3738 œ È$ .
a6bÞ The only root of T aBb œ " B is " . Hence 3738 œ " .
a7b. :aBb œ BÎ# and ; aBb œ $Î# are analytic for all B .
a8b. :aBb œ a" BbÎ# and ; aBb œ $Î# are analytic for all B .
12. The Taylor series expansion of /B , about B! œ ! , is
/ œ"
_
B
B8
Þ
8x
8œ!
Let C œ +! +" B +# B# â +8 B8 â . Substituting into the ODE,
–"
_
_
B8
8
"
"
a8 #ba8 "b+8# B — B
+ 8 B8 œ ! Þ
—–
8 x 8œ!
8œ!
8œ!
_
First note that
B " +8 B8 œ " +8" B8 œ +! B +" B# +# B$ â +8" B8 â .
_
_
8œ!
8œ"
The coefficient of B8 in the product of the two series is
-8 œ #+#
"
"
"
'+$
"#+%
â a8 "b8 +8" a8 #ba8 "b+8# Þ
a 8 "b x
a8 #b x
8x
Expanding the individual series, it follows that
#+# a#+# '+$ bB a+# '+$ "#+% bB# a+# '+$ "#+% #!+& bB$ â
+ ! B + " B# + # B $ â œ ! Þ
Setting the coefficients equal to zero, we obtain the system #+# œ !, #+# '+$ +! œ !,
+# '+$ "#+% +" œ ! , +# '+$ "#+% #!+& +# œ ! , â . Hence the
general solution is
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—————————————————————————— CHAPTER 5. ——
CaBb œ +! +" B +!
B$
B%
B&
B'
%
a+! +" b a#+" +! b Π+! #+"
â.
$
'
"#
%!
"#!
We find that two linearly independent solutions are
C" aBb œ "
B$
B%
B&
â
'
"# %!
C# aBb œ B
B%
B&
B'
â
"# #! '!
Since :aBb œ ! and ; aBb œ B/B converge everywhere, 3 œ _ .
13. The Taylor series expansion of -9= B , about B! œ ! , is
-9= B œ "
_
8œ!
a "b8 B#8
a#8b x
Þ
Let C œ +! +" B +# B# â +8 B8 â . Substituting into the ODE,
a "b8 B#8
"
" a8 #ba8 "b+8# B8 — " 8+8 B8 #" +8 B8 œ ! Þ
–
a#8b x —–
_
_
_
_
8œ!
8œ!
8œ"
8œ!
The coefficient of B8 in the product of the two series is
-8 œ #+# ,8 '+$ ,8" "#+% ,8# â a8 "b8 +8" ," a8 #ba8 "b+8# ,! ,
in which -9= B œ ,! ," B ,# B# â ,8 B8 â . It follows that
#+# #+! " -8 B8 " a8 #b+8 B8 œ ! Þ
_
_
8œ"
8œ"
Expanding the product of the series, it follows that
#+# #+! '+$ B a +# "#+% bB# a $+$ #!+& bB$ â
+1 B +$ B$ #+% B% â œ ! Þ
Setting the coefficients equal to zero, +# +! œ ! , '+$ +" œ ! , +# "#+% œ ! ,
$+$ #!+& +$ œ ! , â Þ Hence the general solution is
CaBb œ +! +" B +! B# +"
B$
B%
B&
B'
B(
+!
+"
+!
+"
â.
'
"#
'!
"#!
&'!
We find that two linearly independent solutions are
C" aBb œ " B#
B%
B'
â
"# "#!
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—————————————————————————— CHAPTER 5. ——
C# aBb œ B
B$
B&
B(
â
'
'! &'!
The nearest zero of T aBb œ -9= B is at B œ „1Î# Þ Hence 3738 œ 1Î# Þ
14. The Taylor series expansion of 68a" Bb , about B! œ ! , is
68a" Bb œ "
_
a "b8" B8
8œ"
8
Þ
Let C œ +! +" B +# B# â +8 B8 â . Substituting into the ODE,
–"
_
a "b8 B8
8x
8œ!
–"
_
8œ"
a "b
8
— " a8 #ba8 "b+8# B
_
8œ!
8"
B8
8
8
8
— " a8 "b+8" B B " +8 B œ ! Þ
_
_
8œ!
8œ!
The first product is the series
#+# a #+# '+$ bB a+# '+$ "#+% bB# a +# '+$ "#+% #!+& bB$ â .
The second product is the series
+" B a#+# +" Î#bB# a$+$ +# +" Î$bB$ a%+% $+$ Î# #+# Î$ +" Î%bB$ â Þ
Combining the series and equating the coefficients to zero, we obtain
#+# œ !
#+# '+$ +" +! œ !
"#+% '+$ $+# $+" Î# œ !
#!+& "#+% *+$ $+# +" Î$ œ !
ã
Hence the general solution is
B$
B%
(B&
B'
&
CaBb œ +! +" B a+! +" b a#+! +" b +"
Π+" + !
â.
$
'
#%
"#!
"#!
We find that two linearly independent solutions are
C" aBb œ "
B$
B%
B'
â
'
"# "#!
C# aBb œ B
B$
B%
(B&
â
'
#% "#!
The coefficient :aBb œ /B 68a" Bb is analytic at B! œ ! , but its power series has a
radius of convergence 3 œ " .
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—————————————————————————— CHAPTER 5. ——
15. If C" œ B and C# œ B# are solutions, then substituting C# into the ODE results in
# T aBb #B UaBb B# V aBb œ ! .
Setting B œ ! , we find that T a!b œ ! . Similarly, substituting C" into the ODE results
in Ua!b œ ! . Therefore T aBbÎUaBb and V aBbÎT aBb may not be analytic. If they were,
Theorem $Þ#Þ" would guarantee that C" and C# were the only two solutions. But note
that an arbitrary value of Ca!b cannot be a linear combination of C" a!b and C# a!bÞ Hence
B! œ ! must be a singular point.
16. Let C œ +! +" B +# B# â +8 B8 â . Substituting into the ODE,
" a8 "b+8" B8 " +8 B8 œ ! Þ
_
_
8œ!
8œ!
That is,
" ca8 "b+8" +8 dB8 œ ! Þ
_
8œ!
Setting the coefficients equal to zero, we obtain
+8" œ
+8
8"
for 8 œ !ß "ß #ß â . It is easy to see that +8 œ +! Îa8 xb . Therefore the general solution
is
CaBb œ +! ”" B
œ +! / B Þ
B#
B$
â•
#x $x
The coefficient +! œ C a!b, which can be arbitrary.
17. Let C œ +! +" B +# B# â +8 B8 â . Substituting into the ODE,
" a8 "b+8" B8 B " +8 B8 œ ! Þ
_
_
8œ!
8œ!
That is,
" a8 "b+8" B8 " +8" B8 œ ! Þ
_
_
8œ!
8œ"
Combining the series, we have
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—————————————————————————— CHAPTER 5. ——
+" " ca8 "b+8" +8" d B8 œ ! Þ
_
8œ"
Setting the coefficient equal to zero, +" œ ! and +8" œ +8" Îa8 "b for 8 œ "ß #ß â.
Note that the indices differ by two, so for 5 œ "ß #ß â
+#5#
+#5%
+!
+#5 œ
œ
œâœ
a#5 b
a#5 #ba#5 b
# † % âa#5 b
and
+#5" œ ! Þ
Hence the general solution is
CaBb œ +! ”"
B#
B%
B'
B#8
# $ â 8
â•
#
# #x # $x
# 8x
œ +! /B:ˆB# Î#‰.
The coefficient +! œ C a!b, which can be arbitrary.
19. Let C œ +! +" B +# B# â +8 B8 â . Substituting into the ODE,
a" Bb" a8 "b+8" B8 " +8 B8 œ ! Þ
_
_
8œ!
8œ!
That is,
" a8 "b+8" B8 " 8 +8 B8 " +8 B8 œ ! Þ
_
_
_
8œ!
8œ"
8œ!
Combining the series, we have
+" +! " ca8 "b+8" 8 +8 +8 d B8 œ ! Þ
_
8œ"
Setting the coefficients equal to zero, +" œ +! and +8" œ +8 for 8 œ !ß "ß #ß â .
Hence the general solution is
CaBb œ +! " B B# B$ â B8 â‘
"
œ +!
Þ
"B
The coefficient +! œ C a!b, which can be arbitrary.
21. Let C œ +! +" B +# B# â +8 B8 â . Substituting into the ODE,
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—————————————————————————— CHAPTER 5. ——
" a8 "b+8" B8 B " +8 B8 œ " B Þ
_
_
8œ!
8œ!
That is,
" a8 "b+8" B8 " +8" B8 œ " B Þ
_
_
8œ!
8œ"
Combining the series, and the nonhomogeneous terms, we have
a+" "b a#+# +! "bB " ca8 "b+8" +8" d B8 œ ! Þ
_
8œ#
Setting the coefficients equal to zero, we obtain +" œ " , #+# +! " œ ! , and
+8#
+8 œ
, 8 œ $ß %ß â .
8
The indices differ by two, so for 5 œ #ß $ß â
+#5 œ
+#5#
+#5%
a "b5" +#
a " b 5 a+ ! " b
œ
œâœ
œ
,
% † ' âa#5 b
# † % † ' âa#5 b
a#5 b
a#5 #ba#5 b
and for 5 œ "ß #ß â
+#5" œ œ
+#5"
+#5$
a "b 5
œ
œâœ
Þ
$ † & âa#5 "b
a#5 "b
a#5 "ba#5 "b
Hence the general solution is
CaBb œ +! B
" +! # B$
B%
B&
B'
B
+! #
+! $ â
#
$
# #x $ † &
# $x
Collecting the terms containing +! ,
CaBb œ +! ”"
B#
B%
B'
# $ â•
#
# #x # $x
#
$
B
B
B%
B&
B'
B(
”B
#
$
â •Þ
#
$
# #x $ † & # $x $ † & † (
Upon inspection, we find that
CaBb œ +! /B:ˆ B# Î#‰ ”B
B#
B$
B%
B&
B'
B(
#
$
â •Þ
#
$
# #x $ † & # $x $ † & † (
Note that the given ODE is first order linear, with integrating factor .a>b œ /B Î# . The
general solution is given by
#
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—————————————————————————— CHAPTER 5. ——
CaBb œ /B Î# ( /? Î# .? aC a!b "b/B Î# " Þ
B
#
#
#
!
23. If ! œ ! , then C" aBb œ " . If ! œ #8 , then +#7 œ ! for 7
C" aBb œ " " a "b7
8
7œ"
#7 8a8 "bâa8 7 "ba#8 "ba#8 $bâa#8 #7 "b #7
B Þ
a#7bx
!œ!
!œ#
!œ%
"
" $B#
" "!B#
If ! œ #8 " , then +#7" œ ! for 7
C# aBb œ B " a "b7
8
7œ"
8 " . As a result,
$& %
$ B
8 " . As a result,
#7 8a8 "bâa8 7 "ba#8 $ba#8 &bâa#8 #7 "b #7"
Þ
B
a#7 "bx
!œ"
!œ$
!œ&
B
B &$ B$
$
B "%
$ B
24a+b. Based on Prob. #$,
!œ!
!œ#
!œ%
"
" $B#
" "!B#
$& %
$ B
#" &
& B
C" a"b œ "
C" a"b œ #
C" a"b œ )$
Normalizing the polynomials, we obtain
T! aBb œ "
T# aBb œ
" $ #
B
# #
$ "&
$&
T% aBb œ B# B%
)
%
)
!œ"
!œ$
!œ&
B
B &$ B$
$
B "%
$ B
#" &
& B
C# a " b œ "
C# a"b œ
)
C# a"b œ "&
#
$
Similarly,
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—————————————————————————— CHAPTER 5. ——
T" aBb œ B
$
&
T$ aBb œ B B$
#
#
"&
$& $ '$ &
T& aBb œ
B B B
)
%
)
a, b .
a- b. T! aBb has no roots. T" aBb has one root at B œ ! . The zeros of T# aBb are at
B œ „ "ÎÈ$ Þ The zeros of T$ aBb are B œ !, „È$Î& . The roots of T% aBb are given
by B# œ Š"& #È$!‹Î$& , Š"& #È$!‹Î$& . The roots of T& aBb are given by
B œ ! and B# œ Š$& #È(!‹Î'$ , Š$& #È(!‹Î'$ .
25. Observe that
T8 a "b œ
a "b 8
a "b5 a#8 #5 bx
"
#8
5 xa8 5 bxa8 #5 bx
5œ!
Ò8Î#Ó
œ a " b 8 T 8 a "b .
But T8 a"b œ " for all nonnegative integers 8.
27. We have
a "b85 8 x #5
B ,
5
x
a
8
5
b
x
5œ!
ˆB# "‰8 œ "
8
which is a polynomial of degree #8. Differentiating n times,
8
.8 #
a "b85 8 x
ˆB "‰8 œ "
a#5 ba#5 "bâa#5 8 "bB#58 ,
.B8
5
x
a
8
5
b
x
5œ.
in which the lower index is . œ c8Î#d " . Note that if 8 œ #7 ", then . œ 7 " .
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—————————————————————————— CHAPTER 5. ——
Now shift the index, by setting
5 œ 8 4.
Hence
a "b 4 8 x
.8 #
ˆB "‰8 œ "
a#8 #4ba#8 #4 "bâa8 #4 "bB8#4
a
8
4
b
x4
x
.B8
4œ!
c8Î#d
a "b4 a#8 #4bx 8#4
B
Þ
a
8
4
b
x4
x
a
8
#4
b
x
4œ!
c8Î#d
œ 8x "
Based on Prob. #&,
.8 #
ˆB "‰8 œ 8x #8 T8 aBbÞ
8
.B
29. Since the 8 " polynomials T! , T" , â, T8 are linearly independent, and the degree
of T5 is 5 , any polynomial, 0 , of degree 8 can be expressed as a linear combination
0 aBb œ " +5 T5 aBb .
8
5œ!
Multiplying both sides by T7 and integrating,
( 0 aBbT7 aBb.B œ " +5 ( T5 aBbT7 aBb.B .
"
8
"
"
5œ!
"
Based on Prob. #),
( T5 aBbT7 aBb.B œ
#
$57 .
#7 "
( 0 aBbT7 aBb.B œ
#
+7 Þ
#7 "
"
"
Hence
"
"
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—————————————————————————— CHAPTER 5. ——
Section 5.4
2. We see that T aBb œ ! when B œ ! and " . Since the three coefficients have no
factors
in common, both of these points are singular points. Near B œ !,
lim B :aBb œ lim B
BÄ!
BÄ!
lim B# ; aBb œ lim B#
BÄ!
#B
B# a"
BÄ!
Bb#
B# a"
%
Bb#
œ #Þ
œ %Þ
The singular point B œ ! is regular. Considering B œ ",
lim aB "b:aBb œ lim aB "b
BÄ"
BÄ"
#B
B# a" Bb#
Þ
The latter limit does not exist. Hence B œ " is an irregular singular point.
3. T aBb œ ! when B œ ! and " . Since the three coefficients have no common factors,
both of these points are singular points. Near B œ !,
lim B :aBb œ lim B
BÄ!
BÄ!
B#
Þ
B# a" Bb
The limit does not exist, and so B œ ! is an irregular singular point. Considering B œ ",
lim aB "b:aBb œ lim aB "b
BÄ"
BÄ"
lim aB "b# ; aBb œ lim aB "b#
BÄ"
B#
œ "Þ
Bb
B# a"
BÄ"
$B
œ !Þ
B# a" Bb
Hence B œ " is a regular singular point.
4. T aBb œ ! when B œ ! and „ " . Since the three coefficients have no common factors,
both of these points are singular points. Near B œ !,
lim B :aBb œ lim B
BÄ!
BÄ!
B$ a"
#
Þ
B# b
The limit does not exist, and so B œ ! is an irregular singular point. Near B œ ",
lim aB "b:aBb œ lim aB "b
BÄ"
BÄ"
#
œ "Þ
B$ a" B# b
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page 204
—————————————————————————— CHAPTER 5. ——
lim aB "b# ; aBb œ lim aB "b#
BÄ"
BÄ"
B$ a"
#
œ !Þ
B# b
Hence B œ " is a regular singular point. At B œ " ,
lim aB "b:aBb œ lim aB "b
BÄ"
BÄ"
B$ a"
lim aB "b# ; aBb œ lim aB "b#
BÄ"
BÄ"
#
œ "Þ
B# b
B$ a"
#
œ !Þ
B# b
Hence B œ " is a regular singular point.
6. The only singular point is at B œ ! . We find that
B
lim B :aBb œ lim B # œ " Þ
BÄ!
BÄ! B
lim B# ; aBb œ lim B#
BÄ!
BÄ!
B# / #
œ / #Þ
B#
Hence B œ ! is a regular singular point.
7. The only singular point is at B œ $ . We find that
lim aB $b:aBb œ lim aB $b
BÄ$
BÄ$
#B
œ 'Þ
B$
lim aB $b# ; aBb œ lim aB $b#
BÄ$
BÄ$
" B#
œ !Þ
B$
Hence B œ $ is a regular singular point.
8. Dividing the ODE by Ba" B# b , we find that
$
:aBb œ
"
#
and
;
a
B
b
œ
.
Ba" B# b
Ba" Bb# a" Bb$
The singular points are at B œ ! and „ " . For B œ !,
lim B :aBb œ lim B
BÄ!
BÄ!
lim B# ; aBb œ lim B#
BÄ!
BÄ!
"
œ "Þ
Ba" B# b
#
Ba" Bb# a" Bb$
œ !Þ
Hence B œ ! is a regular singular point. For B œ ",
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—————————————————————————— CHAPTER 5. ——
lim aB "b:aBb œ lim aB "b
BÄ"
BÄ"
lim aB "b# ; aBb œ lim aB "b#
BÄ"
"
"
œ Þ
#
Ba" B b
#
#
Ba" Bb a" Bb
#
BÄ"
$
œ
"
Þ
%
Hence B œ " is a regular singular point. For B œ ",
lim aB "b:aBb œ lim aB "b
BÄ"
BÄ"
"
"
œ Þ
#
Ba" B b
#
lim aB "b# ; aBb œ lim aB "b#
BÄ"
BÄ"
#
Ba" Bb# a" Bb$
Þ
The latter limit does not exist. Hence B œ " is an irregular singular point.
9. Dividing the ODE by aB #b# aB "b, we find that
:aBb œ
$
a B #b
#
and ; aBb œ
#
.
aB #baB "b
The singular points are at B œ # and " . For B œ #,
lim aB #b:aBb œ lim aB #b
BÄ#
BÄ#
$
a B #b #
Þ
The limit does not exist. Hence B œ # is an irregular singular point. For B œ ",
lim aB "b:aBb œ lim aB "b
BÄ"
BÄ"
lim aB "b# ; aBb œ lim aB "b#
BÄ"
BÄ"
$
a B #b #
œ !Þ
#
œ !Þ
aB #baB "b
Hence B œ " is a regular singular point.
10. T aBb œ ! when B œ ! and $ . Since the three coefficients have no common factors,
both of these points are singular points. Near B œ !,
lim B :aBb œ lim B
BÄ!
BÄ!
B"
"
œ Þ
Ba$ Bb
$
lim B# ; aBb œ lim B#
BÄ!
BÄ!
#
œ !Þ
Ba$ Bb
Hence B œ ! is a regular singular point. For B œ $,
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page 206
—————————————————————————— CHAPTER 5. ——
lim aB $b:aBb œ lim aB $b
BÄ$
BÄ$
B"
%
œ Þ
Ba$ Bb
$
lim aB $b# ; aBb œ lim aB $b#
BÄ$
BÄ$
#
œ !Þ
Ba$ Bb
Hence B œ $ is a regular singular point.
11. Dividing the ODE by aB# B #b, we find that
:aBb œ
B"
#
and ; aBb œ
.
aB #baB "b
aB #baB "b
The singular points are at B œ # and " . For B œ #,
lim aB #b:aBb œ lim
BÄ#
B"
"
œ Þ
BÄ# B "
$
#aB #b
œ !Þ
BÄ# B "
lim aB #b# ; aBb œ lim
BÄ#
Hence B œ # is a regular singular point. For B œ ",
lim aB "b:aBb œ lim
BÄ"
B"
#
œ Þ
BÄ" B #
$
lim aB "b# ; aBb œ lim
BÄ"
BÄ"
#aB "b
œ !Þ
a B #b
Hence B œ " is a regular singular point.
13. Note that :aBb œ 68kBk and ; aBb œ $B . Evidently, :aBb is not analytic at B! œ ! .
Furthermore, the function B :aBb œ B 68kBk does not have a Taylor series about B! œ !.
Hence B œ ! is an irregular singular point.
14. T aBb œ ! when B œ ! . Since the three coefficients have no common factors, B œ !
is a singular point. The Taylor series of /B ", about B œ !, is
/B " œ B B# Î# B$ Î' â .
Hence the function B :aBb œ #a/B "bÎB is analytic at B œ ! . Similarly, the Taylor
series of /B -9= B , about B œ !, is
/B -9= B œ " B B$ Î$ B% Î' â .
The function B# ; aBb œ /B -9= B is also analytic at B œ ! . Hence B œ ! is a regular
singular point.
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page 207
—————————————————————————— CHAPTER 5. ——
15. T aBb œ ! when B œ ! . Since the three coefficients have no common factors, B œ !
is a singular point. The Taylor series of =38 B , about B œ !, is
=38 B œ B B$ Î$x B& Î&x â .
Hence the function B :aBb œ $=38 BÎB is analytic at B œ ! . On the other hand, ; aBb
is a rational function, with
lim B# ; aBb œ lim B#
BÄ!
BÄ!
" B#
œ "Þ
B#
Hence B œ ! is a regular singular point.
16. T aBb œ ! when B œ ! . Since the three coefficients have no common factors, B œ !
is a singular point. We find that
lim B :aBb œ lim B
BÄ!
BÄ!
"
œ "Þ
B
Although the function V aBb œ -9> B does not have a Taylor series about B œ ! , note that
B# ; aBb œ B -9> B œ " B# Î$ B% Î%& #B' Î*%& â . Hence B œ ! is a regular
singular point. Furthermore, ;aBb œ -9> BÎB# is undefined at B œ „ 81 . Therefore the
points B œ „ 81 are also singular points. First note that
lim aB…81b:aBb œ lim aB…81b
BÄ„81
BÄ„81
"
œ !Þ
B
Furthermore, since -9> B has period 1 ,
; aBb œ -9> BÎB œ -9>aB … 81bÎB
"
œ -9>aB … 81b
Þ
a B … 81 b „ 8 1
Therefore
aB … 81b# ; aBb œ aB … 81b-9>aB … 81b”
a B … 81 b
•Þ
a B … 81 b „ 8 1
From above,
aB … 81b-9>aB … 81b œ " aB … 81b# Î$ aB … 81b% Î%& â .
Note that the function in brackets is analytic near B œ „ 81. It follows that the function
aB … 81b# ; aBb is also analytic near B œ „ 81 . Hence all the singular points are regular.
18. The singular points are located at B œ „ 81 , 8 œ !ß "ß â . Dividing the ODE by
B =38 B , we find that B :aBb œ $ -=- B and B# ; aBb œ B# -=- B . Evidently, B :aBb is
not even defined at B œ !. Hence B œ ! is an irregular singular point. On the other
hand, the Taylor series of B -=- B, about B œ !, is
________________________________________________________________________
page 208
—————————————————————————— CHAPTER 5. ——
B -=- B œ " B# Î' (B% $'! â .
Noting that -=- aB … 81b œ a "b8 -=- B ,
aB … 81b:aBb œ $a "b8 aB … 81b-=- aB … 81bÎB
œ $a "b8 aB … 81b-=- aB … 81b”
"
•Þ
a B … 81 b „ 8 1
It is apparent that aB … 81b:aBb is analytic at B œ „ 81 . Similarly,
aB … 81b# ; aBb œ aB … 81b# -=- B
œ a "b8 aB … 81b# -=- aB … 81b,
which is also analytic at B œ „ 81 . Hence all other singular points are regular.
20. B œ ! is the only singular point. Dividing the ODE by #B# , we have :aBb œ $Îa#Bb
and ; aBb œ B# a" BbÎ#Þ It follows that
lim B :aBb œ lim B
BÄ!
BÄ!
lim B# ; aBb œ lim B#
BÄ!
BÄ!
$
$
œ ,
#B
#
a" Bb
"
œ Þ
#
#B
#
Hence B œ ! is a regular singular point. Let C œ +! +" B +# B# â +8 B8 â .
Substitution into the ODE results in
#B# " a8 #ba8 "b+8# B8 $B" a8 "b+8" B8 a" Bb " +8 B8 œ ! .
_
_
_
8œ!
8œ!
8œ!
That is,
#" 8a8 "b+8 B $" 8 +8 B " +8 B " +8" B8 œ ! .
_
_
8
_
8
8œ#
8œ"
_
8
8œ!
8œ"
It follows that
+! a#+" +! bB " c#8a8 "b+8 $8 +8 +8 +8" dB8 œ ! Þ
_
8œ#
Equating the coefficients to zero, we find that +! œ ! , #+" +! œ ! , and
a#8 "ba8 "b+8 œ +8" , 8 œ #ß $ß â .
We conclude that all the +8 are equal to zero. Hence CaBb œ ! is the only solution that
can be obtained.
22. Based on Prob. #", the change of variable, B œ "Î0 , transforms the ODE into the
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page 209
—————————————————————————— CHAPTER 5. ——
form
0%
.# C
.C
#0$
C œ !.
#
.0
.0
Evidently, 0 œ ! is a singular point. Now :a0b œ #Î0 and ; a0b œ "Î0% . Since the value
of lim 0# ; a0b does not exist, 0 œ ! , that is, B œ _ , is an irregular singular point.
0 Ä!
24. Under the transformation B œ "Î0 , the ODE becomes
0 % Œ"
" .# C
"
$
# " .C
!a! "bC œ ! ,
# ”#0 Œ" # #0 •
#
0 .0
0
0 .0
that is,
ˆ0% 0# ‰
.# C
.C
#0$
!a! "bC œ ! .
#
.0
.0
Therefore 0 œ ! is a singular point. Note that
: a0 b œ
#0
! a ! "b
and ; a0b œ # #
.
0 a0 "b
"
0#
It follows that
lim0 :a0b œ lim 0
0Ä!
0Ä!
lim0# ; a0b œ lim 0#
0Ä!
0Ä!
#0
œ !,
"
0#
! a ! "b
œ ! a ! "b Þ
0 # a 0 # "b
Hence 0 œ ! aB œ _b is a regular singular point.
26. Under the transformation B œ "Î0 , the ODE becomes
0%
.# C
" .C
”#0$ #0# •
-C œ !,
#
.0
0 .0
that is,
0%
.# C
.C
ˆ0$ 0‰
#
-C œ !.
. 0#
.0
Therefore 0 œ ! is a singular point. Note that
: a0 b œ
#a0# "b
and ; a0b œ % .
$
0
0
It immediately follows that the limit lim0 :a0b does not exist. Hence 0 œ ! aB œ _b
0Ä!
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—————————————————————————— CHAPTER 5. ——
is an irregular singular point.
27. Under the transformation B œ "Î0 , the ODE becomes
0%
.# C
.C "
#0$
C œ !.
#
.0
.0
0
Therefore 0 œ ! is a singular point. Note that
: a0 b œ
#
"
and ; a0b œ & .
0
0
We find that
lim0 :a0b œ lim 0
0Ä!
0Ä!
but
lim0# ; a0b œ lim 0#
0Ä!
0Ä!
#
œ #,
0
a "b
.
0&
The latter limit does not exist. Hence 0 œ ! aB œ _b is an irregular singular point.
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—————————————————————————— CHAPTER 5. ——
Section 5.5
1. Substitution of C œ B< results in the quadratic equation J a<b œ ! , where
J a<b œ <a< "b %< #
œ <# $< # Þ
The roots are < œ # , " . Hence the general solution, for B Á ! , is
C œ -" B# -# B" Þ
3. Substitution of C œ B< results in the quadratic equation J a<b œ ! , where
J a<b œ <a< "b $< %
œ <# %< % Þ
The root is < œ # , with multiplicity two . Hence the general solution, for B Á ! , is
C œ a-" -# 68kBkb B# Þ
5. Substitution of C œ B< results in the quadratic equation J a<b œ ! , where
J a < b œ < a < "b < "
œ <# #< " Þ
The root is < œ " , with multiplicity two . Hence the general solution, for B Á ! , is
C œ a-" -# 68kBkb B Þ
6. Substitution of C œ aB "b< results in the quadratic equation J a<b œ ! , where
J a<b œ <# (< "# Þ
The roots are < œ $ , % . Hence the general solution, for B Á " , is
C œ -" aB "b$ -# aB "b% Þ
7. Substitution of C œ B< results in the quadratic equation J a<b œ ! , where
J a<b œ <# &< " Þ
The roots are < œ Š& „ È#*‹Î# . Hence the general solution, for B Á ! , is
C œ -" kBk
Š& È#*‹Î#
-# kBk
Š& È#*‹Î#
Þ
8. Substitution of C œ B< results in the quadratic equation J a<b œ ! , where
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—————————————————————————— CHAPTER 5. ——
J a<b œ <# $< $ Þ
The roots are complex, with < œ Š$ „ 3È$ ‹Î# . Hence the general solution, for B Á !,
is
C œ -" kBk$Î# -9=Œ
È$
#
68kBk -# kBk$Î# =38Œ
È$
68kBkÞ
#
10. Substitution of C œ aB #b< results in the quadratic equation J a<b œ ! , where
J a<b œ <# %< ) Þ
The roots are complex, with < œ # „ #3 . Hence the general solution, for B Á #, is
C œ -" aB #b# -9=a# 68kB #kb -# aB #b# =38a# 68kB #kbÞ
11. Substitution of C œ B< results in the quadratic equation J a<b œ ! , where
J a< b œ < # < % Þ
The roots are complex, with < œ Š" „ 3È"& ‹Î# . Hence the general solution, for
B Á !, is
C œ -" kBk"Î# -9=Œ
È"&
#
68kBk -# kBk"Î# =38Œ
È"&
#
68kBkÞ
12. Substitution of C œ B< results in the quadratic equation J a<b œ ! , where
J a<b œ <# &< % Þ
The roots are < œ " , % . Hence the general solution, for B Á ! , is
C œ - " B - # B% Þ
14. Substitution of C œ B< results in the quadratic equation J a<b œ ! , where
J a<b œ %<# %< "( Þ
The roots are complex, with < œ "Î# „ #3 . Hence the general solution, for
B !, is
C œ -" B"Î# -9=a# 68 Bb -# B"Î# =38a# 68 BbÞ
Invoking the initial conditions, we obtain the system of equations
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—————————————————————————— CHAPTER 5. ——
-" œ #
"
-" #-# œ $
#
Hence the solution of the initial value problem is
CaBb œ # B"Î# -9=a# 68 Bb B"Î# =38a# 68 BbÞ
As B p ! , the solution decreases without bound.
15. Substitution of C œ B< results in the quadratic equation J a<b œ ! , where
J a<b œ <# %< % Þ
The root is < œ # , with multiplicity two . Hence the general solution, for B ! , is
C œ a-" -# 68 kBkb B# Þ
Invoking the initial conditions, we obtain the system of equations
-" œ #
#-" -# œ $
Hence the solution of the initial value problem is
CaBb œ a# ( 68 kBkb B# Þ
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—————————————————————————— CHAPTER 5. ——
We find that CaBb p ! as B p ! .
18. Substitution of C œ B< results in the quadratic equation <# < " œ ! . The roots
are
<œ
" „ È " %"
.
#
If " "Î% , the roots are complex, with <"ß# œ ˆ" „ 3È%" " ‰Î# . Hence the general
solution, for B Á ! , is
"
"
C œ -" kBk"Î# -9=Œ È%" " 68kBk -# kBk"Î# =38Œ È%" " 68kBkÞ
#
#
Since the trigonometric factors are bounded, CaBb p ! as B p ! . If " œ "Î% , the roots
are equal, and
C œ -" kBk"Î# -# kBk"Î# 68 kBk Þ
Since lim ÈkBk 68 kBk œ ! , C aBb p ! as B p ! . If " "Î% , the roots are real, with
<"ß# œ ˆ" „ È" %" ‰Î# . Hence the general solution, for B Á ! , is
BÄ!
C œ -" kBk"Î#
È"%" Î#
-# kBk"Î#
È"%" Î#
Þ
Evidently, solutions approach zero as long as "Î# È" %" Î# ! Þ That is,
! " "Î% .
Hence all solutions approach zero, for " ! .
19. Substitution of C œ B< results in the quadratic equation <# < # œ ! . The roots
are < œ " , # . Hence the general solution, for B Á ! , is
C œ -" B" -# B# Þ
Invoking the initial conditions, we obtain the system of equations
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—————————————————————————— CHAPTER 5. ——
-" - # œ "
-" #-# œ #
Hence the solution of the initial value problem is
CaBb œ
# # " " # #
B
BÞ
$
$
The solution is bounded, as B p ! , if # œ # .
20. Substitution of C œ B< results in the quadratic equation <# a! "b< &Î# œ ! .
Formally, the roots are given by
< œ
œ
" ! „ È ! # #! *
#
" ! „ ÊŠ! " È"!‹Š! " È"!‹
#
Þ
a3b The roots <"ß# will be complex, if k" !k È"! . For solutions to approach zero,
as B p _ , we need È"! " ! ! Þ
a33b The roots will be equal, if k" !k œ È"! Þ In this case, all solutions approach
zero as long as " ! œ È"! .
a333b The roots will be real and distinct, if k" !k È"! . It follows that
<7+B œ
" ! È ! # #! *
.
#
For solutions to approach zero, we need " ! È!# #! * ! Þ That is,
" ! È"! .
Hence all solutions approach zero, as B p _ , as long as ! " Þ
23a+b. Given that B œ /D , C aBb œ C a/D b œ AaD b. By the chain rule,
.C
.
.A .D
" .A
œ
A aD b œ
œ
.
.B
.B
.D .B
B .D
Similarly,
.# C
. " .A
" .A " . # A .D
œ
œ
”
•
.B#
.B B .D
B# .D
B .D # .B
" .A
" .#A
œ #
#
Þ
B .D
B .D #
a,b. Direct substitution results in
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—————————————————————————— CHAPTER 5. ——
B# ”
" .# A
" .A
" .A
B
!
•
”
• "A œ !,
B# .D #
B# .D
B .D
that is,
.# A
.A
a ! "b
"A œ !.
#
.D
.D
The associated characteristic equation is <# a! "b< " œ !. Since D œ 68 B ,
it follows that CaBb œ Aa68 BbÞ
a- b. If the roots <"ß# are real and distinct, then
C œ - " / <" D - # / < # D
œ -" B<" -# B<# Þ
a. b. If the roots <"ß# are real and equal, then
C œ - " / <" D - # D / < " D
œ -" B<" -# B<" 68 B Þ
a/b. If the roots are complex conjugates, then < œ - „ 3. , and
C œ /-D a-" -9= .D -# =38 .D b
œ B- c-" -9=a. 68 Bb -# =38a. 68 BbdÞ
24. Based on Prob. #$, the change of variable B œ /D transforms the ODE into
. # A .A
#A œ ! .
.D #
.D
The associated characteristic equation is <# < # œ ! , with roots < œ " , # .
Hence AaD b œ -" /D -# /#D , and C aBb œ -" B" -# B# Þ
26. The change of variable B œ /D transforms the ODE into
.# A
.A
'
&A œ /D .
#
.D
.D
The associated characteristic equation is <# ' < & œ ! , with roots < œ & , " .
Hence A- aD b œ -" /D -# /&D Þ Since the right hand side is not a solution of the
homogeneous equation, we can use the method of undetermined coefficients to show
that a particular solution is [ œ /D Î"# . Therefore the general solution is given by
AaD b œ -" /D -# /&D /D Î"# , that is, C aBb œ -" B" -# B& BÎ"# Þ
27. The change of variable B œ /D transforms the given ODE into
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—————————————————————————— CHAPTER 5. ——
.# A
.A
$
#A œ $/#D #D .
.D #
.D
The associated characteristic equation is <# $ < # œ ! , with roots < œ " , # .
Hence A- aD b œ -" /D -# /#D Þ Using the method of undetermined coefficients, let
[ œ E/#D FD/#D GD H . It follows that the general solution is given by
AaD b œ -" /D -# /#D $D/#D D $Î# , that is,
CaBb œ -" B -# B# $ B# 68 B 68 B $Î# Þ
28. The change of variable B œ /D transforms the given ODE into
.# A
%A œ =38 D .
.D #
The solution of the homogeneous equation is A- aD b œ -" -9= #D -# =38 #D Þ The right
hand side is not a solution of the homogeneous equation. We can use the method of
undetermined coefficients to show that a particular solution is [ œ "$ =38 D . Hence
the general solution is given by AaD b œ -" -9= #D -# =38 #D "$ =38 D , that is,
CaBb œ -" -9=a# 68 Bb -# =38a# 68 Bb "$ =38a68 Bb Þ
29. After dividing the equation by $ , the change of variable B œ /D transforms the ODE
into
.# A
.A
$
$A œ ! .
#
.D
.D
The associated characteristic equation is <# $ < $ œ ! , with complex roots
< œ Š$ „ 3È$ ‹Î# . Hence the general solution is
AaD b œ /$DÎ# ’-" -9=ŠÈ$ DÎ#‹ -# =38ŠÈ$ DÎ#‹“,
and therefore
CaBb œ B$Î# ”-" -9=Œ
È$
#
68 B -# =38Œ
È$
#
68 B•.
30. Let B !. Setting C œ a Bb< , successive differentiation gives C w œ <a Bb<"
and C ww œ <a< "ba Bb<# . It follows that
Pca Bb< d œ <a< "bB# a Bb<# ! < Ba Bb<" " a Bb< Þ
Since B# œ a Bb# , we find that
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—————————————————————————— CHAPTER 5. ——
Pca Bb< d œ <a< "ba Bb< ! <a Bb< " a Bb<
œ a Bb< c<a< "b ! < " dÞ
Given that <" and <# are roots of J a<b œ <a< "b ! < " , we have Pca Bb<3 d œ !.
Therefore C" œ a Bb<" and C# œ a Bb<# are linearly independent solutions of the
differential equation, PcCd œ ! , for B ! , as long as <" Á <# .
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—————————————————————————— CHAPTER 5. ——
Section 5.6
1. T aBb œ ! when B œ ! . Since the three coefficients have no common factors, B œ !
is a singular point. Near B œ !,
lim B :aBb œ lim B
BÄ!
BÄ!
"
"
œ Þ
#B
#
lim B# ; aBb œ lim B#
BÄ!
BÄ!
"
œ !Þ
#
Hence B œ ! is a regular singular point. Let
C œ B< ˆ+! +" B +# B# â +8 B8 ≠œ " +8 B<8 .
_
8œ!
Then
C w œ " a< 8b+8 B<8"
_
8œ!
and
C œ " a< 8ba< 8 "b+8 B<8# Þ
_
ww
8œ!
Substitution into the ODE results in
# " a< 8ba< 8 "b+8 B<8" " a< 8b+8 B<8"
_
_
8œ!
8œ!
_
" +8 B<8" œ ! .
8œ!
That is,
# " a< 8ba< 8 "b+8 B<8 " a< 8b+8 B<8 " +8# B<8 œ ! .
_
_
_
8œ!
8œ!
8œ#
It follows that
+! c#<a< "b <dB< +" c#a< "b< < "dB<"
" c#a< 8ba< 8 "b+8 a< 8b+8 +8# dB<8 œ ! Þ
_
8œ#
Assuming that +! Á ! , we obtain the indicial equation #<# < œ !, with roots <" œ "Î#
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—————————————————————————— CHAPTER 5. ——
and <# œ ! . It immediately follows that +" œ ! . Setting the remaining coefficients
equal
to zero, we have
+8#
, 8 œ #ß $ß â .
+8 œ
a< 8bc#a< 8b "d
For < œ "Î# , the recurrence relation becomes
+8#
+8 œ
, 8 œ #ß $ß â .
8a" #8b
Since +" œ ! , the odd coefficients are zero. Furthermore, for 5 œ "ß #ß â ,
+#5 œ
+#5#
+#5%
a "b5 + !
œ
œ 5
Þ
#5 a" %5 b
a#5 #ba#5 ba%5 $ba%5 "b
# 5x & † * † "$ âa%5 "b
For < œ ! , the recurrence relation becomes
+8#
+8 œ
, 8 œ #ß $ß â .
8a#8 "b
Since +" œ ! , the odd coefficients are zero, and for 5 œ "ß #ß â ,
+#5 œ
+#5#
+#5%
a "b5 + !
œ
œ 5
Þ
#5 a%5 "b
# 5x $ † ( † ""âa%5 "b
a#5 #ba#5 ba%5 &ba%5 "b
The two linearly independent solutions are
_
a "b5 B#5
C" aBb œ ÈB –" " 5
# 5x & † * † "$ âa%5 "b —
5œ"
a "b5 B#5
Þ
#5 5x $ † ( † ""âa%5 "b
5œ"
C# aBb œ " "
_
3. Note that B :aBb œ ! and B# ; aBb œ B , which are both analytic at B œ ! . Set
C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in
" a< 8ba< 8 "b+8 B<8" " +8 B<8 œ ! ,
_
_
8œ!
8œ!
and after multiplying both sides of the equation by B ,
" a< 8ba< 8 "b+8 B<8 " +8" B<8 œ ! .
_
_
8œ!
8œ"
It follows that
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—————————————————————————— CHAPTER 5. ——
+! c<a< "bdB " ca< 8ba< 8 "b+8 +8" dB<8 œ ! Þ
_
<
8œ"
Setting the coefficients equal to zero, the indicial equation is <a< "b œ ! . The roots
are <" œ " and <# œ ! . Here <" <# œ " . The recurrence relation is
+8"
, 8 œ "ß #ß â .
+8 œ
a< 8ba< 8 "b
For < œ " ,
+8 œ
Hence for 8
",
+8 œ
+8"
, 8 œ "ß #ß â .
8a8 "b
+8"
+8#
a "b8 + !
œ
œ
â
œ
.
a8 "b8# a8 "b
8a8 "b
8xa8 "bx
Therefore one solution is
a "b8 B8
.
8xa8 "bx
8œ!
C" aBb œ B "
_
5. Here B :aBb œ #Î$ and B# ; aBb œ B# Î$ , which are both analytic at B œ ! . Set
C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in
$" a< 8ba< 8 "b+8 B
_
8œ!
#" a< 8b+8 B
_
<8
8œ!
" +8 B<8# œ ! Þ
_
<8
8œ!
It follows that
+! c$<a< "b #<dB< +" c$a< "b< #a< "bdB<"
" c$a< 8ba< 8 "b+8 #a< 8b+8 +8# dB<8 œ ! .
_
8œ#
Assuming +! Á ! , the indicial equation is $<# < œ ! , with roots <" œ "Î$ , <# œ ! Þ
Setting the remaining coefficients equal to zero, we have +" œ ! , and
+8#
+8 œ
, 8 œ #ß $ß â .
a< 8bc$a< 8b "d
It immediately follows that the odd coefficients are equal to zero. For < œ "Î$ ,
+8#
, 8 œ #ß $ß â .
+8 œ
8a" $8b
So for 5 œ "ß #ß â ,
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—————————————————————————— CHAPTER 5. ——
+#5
+#5#
+#5%
a "b5 + !
œ
œ
œ 5
.
#5 a'5 "b
# 5x ( † "$ âa'5 "b
a#5 #ba#5 ba'5 &ba'5"b
For < œ ! ,
+8 œ
+8#
, 8 œ #ß $ß â .
8a$8 "b
So for 5 œ "ß #ß â ,
+#5 œ
a "b5 + !
+#5#
+#5%
.
œ
œ 5
a#5 #ba#5 ba'5 (ba'5 "b
#5 a'5 "b
# 5x & † "" âa'5 "b
The two linearly independent solutions are
C" aBb œ B
5
a "b 5
B#
Œ —
5x ( † "$ âa'5 "b #
5œ"
–" "
_
"Î$
5
a "b 5
B#
Œ Þ
5x & † ""âa'5 "b #
5œ"
C# aBb œ " "
_
6. Note that B :aBb œ " and B# ; aBb œ B #, which are both analytic at B œ ! . Set
C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in
" a< 8ba< 8 "b+8 B
_
8œ!
" a< 8b+8 B<8
_
<8
8œ!
" +8 B<8" #" +8 B<8 œ ! Þ
_
_
8œ!
8œ!
After adjusting the indices in the second-to-last series, we obtain
+! c<a< "b < #dB " ca< 8ba< 8 "b+8 a< 8b+8 # +8 +8" dB<8 œ !Þ
_
<
8œ"
Assuming +! Á ! , the indicial equation is <# # œ ! , with roots < œ „ È# . Setting
the remaining coefficients equal to zero, the recurrence relation is
+8"
, 8 œ "ß #ß â .
+8 œ
a < 8b # #
First note that a< 8b# # œ Š< 8 È# ‹Š< 8 È# ‹. So for < œ È# ,
+8 œ
+8"
, 8 œ "ß #ß â .
8Š8 #È# ‹
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—————————————————————————— CHAPTER 5. ——
It follows that
a "b 8 + !
, 8 œ "ß #ß â .
+8 œ
8xŠ" #È# ‹Š# #È# ‹âŠ8 #È# ‹
For < œ È# ,
+8 œ
+8"
, 8 œ "ß #ß â ,
8Š8 #È# ‹
and therefore
a "b 8 + !
, 8 œ "ß #ß â .
+8 œ
8xŠ" #È# ‹Š# #È# ‹âŠ8 #È# ‹
The two linearly independent solutions are
C" aBb œ B
È#
C# aBb œ B
Ô
×
_
a "b8 B8
Ö" "
Ù
È
È
È
8 œ " 8xŠ" # # ‹Š# # # ‹âŠ8 # # ‹ Ø
Õ
Ô
×
_
a "b8 B8
ÙÞ
""
È
È
È
8x
"
#
#
#
#
#
â
8
#
#
Š
‹Š
‹ Š
‹Ø
8œ"
Õ
È# Ö
7. Here B :aBb œ " B and B# ; aBb œ B , which are both analytic at B œ ! . Set
C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in
" a< 8ba< 8 "b+8 B<8" " a< 8b+8 B<8"
_
_
8œ!
8œ!
" a< 8b+8 B<8 " +8 B<8 œ ! Þ
_
_
8œ!
8œ!
After multiplying both sides by B ,
" a< 8ba< 8 "b+8 B<8 " a< 8b+8 B<8
_
_
8œ!
8œ!
" a < 8 b+ 8 B
_
<8+1
8œ!
" +8 B<8+1 œ !Þ
_
8œ!
After adjusting the indices in the last two series, we obtain
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—————————————————————————— CHAPTER 5. ——
+! c<a< "b <dB " ca< 8ba< 8 "b+8 a< 8b+8 a< 8b+8" dB<8 œ ! Þ
_
<
8œ"
Assuming +! Á ! , the indicial equation is <# œ ! , with roots <" œ <# œ ! . Setting
the remaining coefficients equal to zero, the recurrence relation is
+8"
+8 œ
, 8 œ "ß #ß â .
<8
With < œ ! ,
+8 œ
+8"
, 8 œ "ß #ß â .
8
Hence one solution is
C" aBb œ "
B
B#
B8
â
⠜ /B .
"x
#x
8x
8. Note that B :aBb œ $Î# and B# ; aBb œ B# "Î#, which are both analytic at B œ ! .
Set C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in
#" a< 8ba< 8 "b+8 B<8 $" a< 8b+8 B<8
_
_
8œ!
8œ!
#" +8 B<8# " +8 B<8 œ ! Þ
_
_
8œ!
8œ!
After adjusting the indices in the second-to-last series, we obtain
+! c#<a< "b $< "dB< +" c#a< "b< $a< "b "d
" c#a< 8ba< 8 "b+8 $a< 8b+8 +8 # +8# dB<8 œ !Þ
_
8œ#
Assuming +! Á ! , the indicial equation is #<# < " œ ! , with roots <" œ "Î# and
<# œ " . Setting the remaining coefficients equal to zero, the recurrence relation is
+8 œ
#+8#
, 8 œ #ß $ß â .
a< 8 "bc#a< 8b "d
Setting the remaining coefficients equal to zero, we have +" œ ! , which implies that all
of the odd coefficients are zero. With < œ "Î# ,
+8 œ
#+8#
, 8 œ #ß $ß â .
8a#8 $b
So for 5 œ "ß #ß â ,
________________________________________________________________________
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—————————————————————————— CHAPTER 5. ——
+#5
+#5#
+#5%
a "b5 + !
œ
œ
œ
.
5 a%5 $b
5x ( † "" â a%5 $b
a5 "b5 a%5 &ba%5 $b
With < œ " ,
+8 œ
#+8#
, 8 œ #ß $ß â .
8a#8 $b
So for 5 œ "ß #ß â ,
+#5 œ
+#5#
+#5%
a "b5 + !
œ
œ
.
5 a%5 $b
a5 "b5 a%5 ""ba%5 $b
5x & † * â a%5 $b
The two linearly independent solutions are
a "b8 B#8
8x ( † "" â a%8 $b —
8œ"
C" aBb œ B"Î# –" "
_
a "b8 B#8
Þ
8x & † * â a%8 $b —
8œ"
C# aBb œ B" –" "
_
9. Note that B :aBb œ B $ and B# ; aBb œ B $, which are both analytic at B œ ! .
Set C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in
" a< 8ba< 8 "b+8 B<8 " a< 8b+8 B<8" $" a< 8b+8 B<8
_
_
_
8œ!
8œ!
8œ!
" +8 B<8" $" +8 B<8 œ ! Þ
_
_
8œ!
8œ!
After adjusting the indices in the second-to-last series, we obtain
+! c<a< "b $< $dB<
" ca< 8ba< 8 "b+8 a< 8 #b+8" $a< 8 "b+8 dB<8 œ !Þ
_
8œ"
Assuming +! Á ! , the indicial equation is <# %< $ œ ! , with roots <" œ $ and
<# œ " . Setting the remaining coefficients equal to zero, the recurrence relation is
+8 œ
a< 8 #b+8"
, 8 œ "ß #ß â .
a< 8 "ba< 8 $b
With < œ $ ,
________________________________________________________________________
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—————————————————————————— CHAPTER 5. ——
+8 œ
It follows that for 8
+8 œ
a8 "b+8"
, 8 œ "ß #ß â .
8a8 #b
",
+8#
# +!
a8 "b+8"
œ
œâœ
.
8a8 #b
8x a8 #b
a8 "ba8 #b
Therefore one solution is
# B8
C" aBb œ B –" "
.
8x a8 #b —
_
$
8œ"
10. Here B :aBb œ ! and B# ; aBb œ B# "Î% , which are both analytic at B œ ! .
Set C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in
" a< 8ba< 8 "b+8 B<8 " +8 B<8#
_
_
8œ!
8œ!
" _
" +8 B<8 œ ! Þ
% 8œ!
After adjusting the indices in the second series, we obtain
"
"
+! ”<a< "b •B< +" ”a< "b< •B<"
%
%
_
"
" ”a< 8ba< 8 "b+8 +8 +8# •B<8 œ ! Þ
%
8œ#
Assuming +! Á ! , the indicial equation is <# < "% œ !, with roots <" œ <# œ "Î#Þ
Setting the remaining coefficients equal to zero, we find that +" œ ! . The recurrence
relation is
+8 œ
%+8#
a#< #8 "b#
, 8 œ #ß $ß â .
With < œ "Î# ,
+8 œ
+8#
, 8 œ #ß $ß â .
8#
Since +" œ ! , the odd coefficients are zero. So for 5
+#5 œ
",
+#5#
+#5%
a "b5 + !
œ
œ
â
œ
.
%5 #
%# a5 "b# 5 #
%5 a5xb#
Therefore one solution is
________________________________________________________________________
page 227
—————————————————————————— CHAPTER 5. ——
C" aBb œ ÈB –" "
_
a "b8 B#8
8œ"
##8 a8xb#
—.
12a+b. Dividing through by the leading coefficient, the ODE can be written as
C ww
B
!#
w
C
C œ !.
" B#
" B#
For B œ " ,
:! œ lim aB "b:aBb œ lim
B
"
œ Þ
BÄ" B "
#
BÄ"
!# a" Bb
œ !Þ
BÄ"
B"
;! œ lim aB "b# ; aBb œ lim
BÄ"
For B œ " ,
:! œ lim aB "b:aBb œ lim
B
"
œ Þ
BÄ" B "
#
BÄ"
! # a B "b
;! œ lim aB "b ; aBb œ lim
œ !Þ
BÄ"
BÄ" a" Bb
#
Hence both B œ " and B œ " are regular singular points. As shown in Example " ,
the indicial equation is given by
< a < "b : ! < ; ! œ ! .
In this case, both sets of roots are <" œ "Î# and <# œ ! .
a,b. Let > œ B " , and ?a>b œ C a> "bÞ Under this change of variable, the differential
equation becomes
ˆ># #>‰? ww a> "b? w !# ? œ ! Þ
Based on Part a+b, > œ ! is a regular singular point. Set ? œ ! +8 ><8 Þ Substitution
_
8œ!
into the ODE results in
" a< 8ba< 8 "b+8 ><8 #" a< 8ba< 8 "b+8 ><8"
_
_
8œ!
8œ!
" a< 8b+8 ><8 " a< 8b+8 ><8" !# " +8 ><8 œ ! Þ
_
_
_
8œ!
8œ!
8œ!
Upon inspection, we can also write
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page 228
—————————————————————————— CHAPTER 5. ——
" a< 8b# +8 ><8 #" a< 8bŒ< 8 " +8 ><8" !# " +8 ><8 œ !Þ
_
_
_
8œ!
8œ!
#
8œ!
After adjusting the indices in the second series, it follows that
+! ”#<Œ< •><" " ”a< 8b# +8 #a< 8 "bŒ< 8 +8" !# +8 •><8 œ !Þ
#
#
_
"
"
8œ!
Assuming that +! Á ! , the indicial equation is #<# < œ ! , with roots < œ ! , "Î# .
The recurrence relation is
"
a< 8b# +8 #a< 8 "bŒ< 8 +8" !# +8 œ ! , 8 œ !ß "ß #ß â .
#
With <" œ "Î# , we find that for 8
",
%!# a#8 "b#
+8 œ
+8"
%8a#8 "b
#
#
#
#‘
8 c" %! dc* %! dâ a#8 "b %!
œ a "b
+! Þ
#8 a#8 "bx
With <# œ ! , we find that for 8
",
! # a8 "b #
+8 œ
+8"
8a#8 "b
!a !bc" !# dc% !# dâa8 "b# !# ‘
œ a "b 8
+! Þ
8x † $ † & âa#8 "b
The two linearly independent solutions of the Chebyshev equation are
C" aBb œ kB "k"Î# –" " a "b8
_
8œ"
C# aBb œ " " a "b8
_
8œ"
c" %!# dc* %!# dâa#8 "b# %!# ‘
aB "b 8 —
#8 a#8 "bx
!a !bc" !# dc% !# dâa8 "b# !# ‘
aB " b 8 Þ
8x † $ † & âa#8 "b
13. Here B :aBb œ " B and B# ; aBb œ - B , which are both analytic at B œ ! .
In fact,
:! œ lim B :aBb œ " and ;! œ lim B# ; aBb œ ! .
BÄ!
BÄ!
Hence the indicial equation is <a< "b < œ ! , with roots <"ß# œ ! . Set
________________________________________________________________________
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—————————————————————————— CHAPTER 5. ——
C œ +! + " B + # B# â + 8 B 8 â .
Substitution into the ODE results in
" 8a8 "b+8 B8" " 8+8 B8"
_
_
8œ#
8œ"
" 8+8 B8 -" +8 B8 œ ! Þ
_
_
8œ!
8œ!
That is,
" 8a8 "b+8" B " a8 "b+8" B8
_
_
8
8œ"
8œ!
" 8+8 B8 -" +8 B8 œ ! Þ
_
_
8œ"
8œ!
It follows that
+" - +! " a8 "b# +8" a8 -b+8 ‘B8 œ ! .
_
8œ"
Setting the coefficients equal to zero, we find that +" œ -+! , and
+8 œ
That is, for 8
a8 " - b
+8" , 8 œ #ß $ß â .
8#
#,
+8 œ
a8 " - b
a -ba" -bâa8 " -b
+
œ
â
œ
+! .
8"
8#
a8xb#
Therefore one solution of the Laguerre equation is
C" aBb œ " "
a -ba" -bâa8 " -b
_
8œ"
a8xb
#
Note that if - œ 7 , a positive integer, then +8 œ ! for 8
solution is a polynomial
C" aBb œ " "
7
8œ"
7 " . In that case, the
a -ba" -bâa8 " -b
a8xb
#
B8 Þ
B8 Þ
________________________________________________________________________
page 230
—————————————————————————— CHAPTER 5. ——
Section 5.7
2. T aBb œ ! only for B œ ! . Furthermore, B :aBb œ # B and B# ; aBb œ # B# Þ
It follows that
:! œ lim a # Bb œ #
;! œ lim ˆ# B# ‰ œ #
BÄ!
BÄ!
and therefore B œ ! is a regular singular point. The indicial equation is given by
<a< "b #< # œ ! ,
that is, <# $< # œ ! , with roots <" œ # and <# œ " .
4. The coefficients T aBb , UaBb , and V aBb are analytic for all B − ‘ . Hence there are
no singular points.
5. T aBb œ ! only for B œ ! . Furthermore, B :aBb œ $ =38B B and B# ; aBb œ # Þ It
follows that
=38 B
œ$
BÄ!
B
;! œ lim # œ #
:! œ lim $
BÄ!
and therefore B œ ! is a regular singular point. The indicial equation is given by
<a< "b $< # œ ! ,
that is, <# #< # œ ! , with roots <" œ " È$ and <# œ " È$ .
6. T aBb œ ! for B œ ! and B œ # . We note that :aBb œ B" aB #b" Î# , and
; aBb œ aB #b" Î# Þ For the singularity at B œ ! ,
"
"
œ
BÄ! #aB #b
%
#
B
;! œ lim
œ!
BÄ! #aB #b
:! œ lim
and therefore B œ ! is a regular singular point. The indicial equation is given by
"
< a < "b < œ ! ,
%
that is, <# $% < œ ! , with roots <" œ
$
%
and <# œ ! . For the singularity at B œ # ,
________________________________________________________________________
page 231
—————————————————————————— CHAPTER 5. ——
:! œ lim aB #b:aBb œ lim
"
"
œ
BÄ#
BÄ# #B
%
a
B
#
b
;! œ lim aB #b# ; aBb œ lim
œ!
BÄ#
BÄ#
#
and therefore B œ # is a regular singular point. The indicial equation is given by
"
< a < "b < œ ! ,
%
that is, <# &% < œ ! , with roots <" œ
&
%
and <# œ ! .
7. T aBb œ ! only for B œ ! . Furthermore, B :aBb œ
follows that
"
#
=38 B
#B
and B# ; aBb œ " Þ It
:! œ lim B:aBb œ "
;! œ lim B# ; aBb œ "
BÄ!
BÄ!
and therefore B œ ! is a regular singular point. The indicial equation is given by
< a < "b < " œ ! ,
that is, <# " œ ! , with complex conjugate roots < œ „ 3 .
8. Note that T aBb œ ! only for B œ " . We find that :aBb œ $aB "bÎaB "b , and
; aBb œ $ÎaB "b# . It follows that
:! œ lim aB "b:aBb œ lim $aB "b œ '
;! œ lim aB "b ; aBb œ lim $ œ $
BÄ"
BÄ"
#
BÄ"
BÄ"
and therefore B œ " is a regular singular point. The indicial equation is given by
<a< "b '< $ œ ! ,
that is, <# (< $ œ ! , with roots <" œ Š( È$( ‹Î# and <# œ Š( È$( ‹Î# .
10. T aBb œ ! for B œ # and B œ # . We note that :aBb œ #BaB #b# aB #b" ,
and ; aBb œ $aB #b" aB #b" Þ For the singularity at B œ # ,
lim aB #b:aBb œ lim
BÄ#
#B
,
%
BÄ# B#
which is undefined. Therefore B œ ! is an irregular singular point. For the singularity
at B œ # ,
________________________________________________________________________
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—————————————————————————— CHAPTER 5. ——
:! œ lim aB #b:aBb œ lim
#B
a B #b
$aB #b
;! œ lim aB #b# ; aBb œ lim
œ!
BÄ#
BÄ#
B#
BÄ#
BÄ#
#
œ
"
%
and therefore B œ # is a regular singular point. The indicial equation is given by
"
< a < "b < œ ! ,
%
that is, <# &% < œ ! , with roots <" œ
&
%
and <# œ ! .
11. T aBb œ ! for B œ # and B œ # . We note that :aBb œ #BÎa% B# b , and
; aBb œ $Îa% B# bÞ For the singularity at B œ # ,
:! œ lim aB #b:aBb œ lim
#B
œ "
BÄ#
BÄ# B #
$a# Bb
;! œ lim aB #b# ; aBb œ lim
œ!
BÄ#
BÄ#
B#
and therefore B œ # is a regular singular point. The indicial equation is given by
< a < "b < œ ! ,
that is, <# #< œ ! , with roots <" œ # and <# œ ! . For the singularity at B œ # ,
:! œ lim aB #b:aBb œ lim
#B
œ "
BÄ#
BÄ# # B
$aB #b
;! œ lim aB #b# ; aBb œ lim
œ!
BÄ#
BÄ#
#B
and therefore B œ # is a regular singular point. The indicial equation is given by
< a < "b < œ ! ,
that is, <# #< œ ! , with roots <" œ # and <# œ ! .
12. T aBb œ ! for B œ ! and B œ $ . We note that :aBb œ #B" aB $b" , and
; aBb œ "ÎaB $b# Þ For the singularity at B œ ! ,
:! œ lim B :aBb œ lim
#
#
œ
BÄ!
BÄ! B $
$
#
B
;! œ lim B# ; aBb œ lim
œ!
BÄ!
BÄ! aB $b#
and therefore B œ ! is a regular singular point. The indicial equation is given by
________________________________________________________________________
page 233
—————————————————————————— CHAPTER 5. ——
#
< a < "b < œ ! ,
$
that is, <# &$ < œ ! , with roots <" œ
&
$
and <# œ ! . For the singularity at B œ $ ,
:! œ lim aB $b:aBb œ lim
#
#
œ
BÄ$
BÄ$ B
$
#
;! œ lim aB $b ; aBb œ lim a "b œ "
BÄ$
BÄ$
and therefore B œ $ is a regular singular point. The indicial equation is given by
#
< a < "b < " œ ! ,
$
that is, <# "$ < " œ ! , with roots <" œ Š" È$( ‹Î' and <# œ Š" È$( ‹Î' .
13a+b. Note the :aBb œ "ÎB and ; aBb œ "ÎB . Furthermore, B :aBb œ " and
B# ; aBb œ B Þ It follows that
:! œ lim a"b œ "
;! œ lim a Bb œ !
BÄ!
BÄ!
and therefore B œ ! is a regular singular point.
a,bÞ The indicial equation is given by
< a < "b < œ ! ,
that is, <# œ ! , with roots <" œ <# œ ! .
a- b. Let C œ +! +" B +# B# â +8 B8 â . Substitution into the ODE results in
" a8 #ba8 "b+8# B8" " a8 "b+8" B8 " +8 B8 œ ! .
_
_
_
8œ!
8œ!
8œ!
After adjusting the indices in the first series, we obtain
+" +! " c8a8 "b+8" a8 "b+8" +8 dB8 œ ! .
_
8œ"
Setting the coefficients equal to zero, it follows that for 8
+8
+8" œ
.
a 8 "b #
So for 8
!,
",
________________________________________________________________________
page 234
—————————————————————————— CHAPTER 5. ——
+8 œ
+8"
+8#
"
œ
œâœ
+! Þ
#
#
8
8# a 8 " b
a8xb#
With +! œ " , one solution is
"
"
"
C" aBb œ " B B# B$ â
B8 â .
#
%
$'
a8xb
For a second solution, set C# aBb œ C" aBb 68 B ," B ,# B# â ,8 B8 â .
Substituting into the ODE, we obtain
PcC" aBbd † 68 B # C" aBb P– " ,8 B8 — œ ! .
_
w
8œ"
Since PcC" aBbd œ ! , it follows that
P– " ,8 B8 — œ # C"w aBb .
_
8œ"
More specifically,
," " c8a8 "b,8" a8 "b,8" ,8 dB8 œ
_
8œ"
"
"
" %
œ # B B# B $
B â.
'
(#
"%%!
Equating the coefficients, we obtain the system of equations
,"
%,# ,"
*,$ ,#
"',% ,$
œ
œ
œ
œ
#
"
"Î'
"Î(#
ã
Solving these equations for the coefficients, ," œ #, ,# œ $Î%, ,$ œ ""Î"!),
,% œ #&Î$%&' , â . Therefore a second solution is
$
"" $
#& %
C# aBb œ C" aBb 68 B ” #B B#
B
B â•Þ
%
"!)
$%&'
14a+b. Here B :aBb œ #B and B# ; aBb œ ' B/B Þ Both of these functions are analytic at
B œ ! , therefore B œ ! is a regular singular point. Note that :! œ ;! œ ! Þ
a,bÞ The indicial equation is given by
________________________________________________________________________
page 235
—————————————————————————— CHAPTER 5. ——
< a < "b œ ! ,
that is, <# < œ ! , with roots <" œ " and <# œ ! .
a- b. In order to find the solution corresponding to <" œ " , set C œ B ! +8 B8 Þ Upon
_
8œ!
substitution into the ODE, we have
" a8 #ba8 "b+8" B8" #" a8 "b+8 B8" ' /B " +8 B8" œ ! .
_
_
_
8œ!
8œ!
8œ!
After adjusting the indices in the first two series, and expanding the exponential function,
" 8a8 "b+8 B #" 8 +8" B8 ' +! B a'+! '+" bB#
_
_
8
a'+# '+" $+! bB a'+$ '+# $+" +! bB% ⠜ ! .
8œ"
8œ"
$
Equating the coefficients, we obtain the system of equations
#+" #+! '+!
'+# %+" '+! '+"
"#+$ '+# '+# '+" $+!
#!+% )+$ '+$ '+# $+" +!
œ!
œ!
œ!
œ!
ã
Setting +! œ " , solution of the system results in +" œ %, +# œ "(Î$, +$ œ %(Î"#,
+% œ "*"Î"#! , â . Therefore one solution is
C" aBb œ B %B#
"( $ %( %
B B â.
$
"#
The exponents differ by an integer. So for a second solution, set
C# aBb œ + C" aBb 68 B " -" B -# B# â -8 B8 â .
Substituting into the ODE, we obtain
+ PcC" aBbd † 68 B #+ C"w aBb #+ C" aBb +
Since PcC" aBbd œ ! , it follows that
_
C" aBb
P–" " -8 B8 — œ ! .
B
8œ"
P–" " -8 B8 — œ #+ C"w aBb #+ C" aBb +
_
8œ"
C" aBb
.
B
More specifically,
________________________________________________________________________
page 236
—————————————————————————— CHAPTER 5. ——
" 8a8 "b-8" B8 #" 8 -8 B8 ' a' '-" bB
_
_
8œ"
8œ"
a'-# '-" $bB# ⠜ + "!+B
'" # "*$ $
+B
+B â .
$
"#
Equating the coefficients, we obtain the system of equations
'œ +
#-# )-" ' œ "!+
'"
'-$ "!-# '-" $ œ +
$
"*$
"#-% "#-$ '-# $-" " œ
+
"#
ã
Solving these equations for the coefficients, + œ ' . In order to solve the remaining
equations, set -" œ ! . Then -# œ $$, -$ œ %%*Î' , -% œ "&*&Î#% , â Þ
Therefore a second solution is
C# aBb œ ' C" aBb 68 B ”" $$B#
%%* $ "&*& %
B
B â•Þ
'
#%
15a+b. Note the :aBb œ 'BÎaB "b and ; aBb œ $B" aB "b" . Furthermore,
B :aBb œ 'B# ÎaB "b and B# ; aBb œ $BÎaB "b Þ It follows that
'B#
œ!
BÄ! B "
$B
;! œ lim
œ!
BÄ! B "
:! œ lim
and therefore B œ ! is a regular singular point.
a,bÞ The indicial equation is given by
< a < "b œ ! ,
that is, <# < œ ! , with roots <" œ " and <# œ ! .
a- b. In order to find the solution corresponding to <" œ " , set C œ B ! +8 B8 Þ Upon
_
substitution into the ODE, we have
8œ!
________________________________________________________________________
page 237
—————————————————————————— CHAPTER 5. ——
" 8a8 "b+8 B8" " 8a8 "b+8 B8
_
_
8œ"
8œ"
_
'" a8 "b+8 B8# $" +8 B8" œ ! .
_
8œ!
8œ!
After adjusting the indices, it follows that
" 8a8 "b+8" B8 " 8a8 "b+8 B8
_
_
8œ#
8œ"
_
'" a8 "b+8# B8 $" +8" B8 œ ! .
_
8œ#
8œ"
That is,
#+" $+! " c 8a8 "b+8 ˆ8# 8 $‰+8" 'a8 "b+8# dB8 œ !Þ
_
8œ#
Setting the coefficients equal to zero, we have +" œ $+! Î# , and for 8
8a8 "b+8 œ ˆ8# 8 $‰+8" 'a8 "b+8# .
#,
If we assign +! œ " , then we obtain +" œ $Î# , +# œ *Î% , +$ œ &"Î"' , â .
Hence one solution is
$
*
&"
""" &
C" aBb œ B B# B$ B%
B â.
#
%
"'
%!
The exponents differ by an integer. So for a second solution, set
C# aBb œ + C" aBb 68 B " -" B -# B# â -8 B8 â .
Substituting into the ODE, we obtain
#+B C"w aBb #+ C"w aBb '+B C" aBb + C" aBb +
since PcC" aBbd œ ! . It follows that
_
C" aBb
P –" " - 8 B8 — œ ! ,
B
8œ"
P–" " -8 B8 — œ #+ C"w aBb #+B C"w aBb + C" aBb '+B C" aBb +
_
8œ"
C" aBb
.
B
Now
P–" " -8 B8 — œ $ a #-# $-" bB a '-$ &-# '-" bB#
8œ"
a "#-% *-$ "#-# bB$ a #!-& "&-% ")-$ bB% â Þ
_
Substituting for C" aBb , the right hand side of the ODE is
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—————————————————————————— CHAPTER 5. ——
(
$
$$
)'( % %%" &
+ +B +B# +B$
+B
+B â .
#
%
"'
)!
"!
Equating the coefficients, we obtain the system of equations
$œ+
(
#-# $-" œ +
#
$
'-$ &-# '-" œ +
%
$$
"#-% *-$ "#-# œ
+
"'
ã
We find that + œ $. In order to solve the second equation, set -" œ !. Solution of the
remaining equations results in -# œ #"Î% , -$ œ "*Î% , -% œ &*(Î'% , â.
Hence a second solution is
C# aBb œ $ C" aBb 68 B ”"
#" # "* $ &*( %
B B
B â•Þ
%
%
'%
16a+b. After multiplying both sides of the ODE by B , we find that B :aBb œ ! and
B# ; aBb œ B . Both of these functions are analytic at B œ ! , hence B œ ! is a regular
singular point.
a,b. Furthermore, :! œ ;! œ ! . So the indicial equation is <a< "b œ ! , with roots
<" œ " and <# œ ! .
a- b. In order to find the solution corresponding to <" œ " , set C œ B ! +8 B8 Þ Upon
_
8œ!
substitution into the ODE, we have
" 8a8 "b+8 B8 " +8 B8" œ ! .
_
_
8œ"
8œ!
That is,
" c8a8 "b+8 +8" d B8 œ ! .
_
8œ"
Setting the coefficients equal to zero, we find that for 8
+8"
+8 œ
.
8a8 "b
",
It follows that
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—————————————————————————— CHAPTER 5. ——
+8 œ
a "b8 + !
+8"
+8#
.
œ
œ
â
œ
a8 "b8# a8 "b
8a8 "b
a8xb# a8 "b
Hence one solution is
"
"
" %
" &
C" aBb œ B B# B$
B
B â.
#
"#
"%%
#))!
The exponents differ by an integer. So for a second solution, set
C# aBb œ + C" aBb 68 B " -" B -# B# â -8 B8 â .
Substituting into the ODE, we obtain
+ PcC" aBbd † 68 B #+ C"w aBb +
Since PcC" aBbd œ ! , it follows that
_
C" aBb
P–" " -8 B8 — œ ! .
B
8œ"
P–" " -8 B8 — œ #+ C"w aBb +
_
8œ"
C" aBb
.
B
Now
P–" " -8 B8 — œ " a#-# -" bB a'-$ -# bB# a"#-% -$ bB$
8œ"
a#!-& -% bB% a$!-' -& bB& â Þ
_
Substituting for C" aBb , the right hand side of the ODE is
$
&
(
"
+ +B +B#
+B$
+B% â .
#
"#
"%%
$#!
Equating the coefficients, we obtain the system of equations
"œ +
$
#-# -" œ +
#
&
'-$ -# œ +
"#
(
"#-% -$ œ
+
"%%
ã
Evidently, + œ " . In order to solve the second equation, set -" œ ! . We then find
that -# œ $Î% , -$ œ (Î$' , -% œ $&Î"(#) , â . Therefore a second solution is
$
(
$& %
C# aBb œ C" aBb 68 B ”" B# B$
B â•Þ
%
$'
"(#)
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—————————————————————————— CHAPTER 5. ——
19a+b. After dividing by the leading coefficient, we find that
# a " ! " bB
œ #Þ
BÄ!
"B
:! œ lim B :aBb œ lim
BÄ!
;! œ lim B# ; aBb œ lim
BÄ!
BÄ!
!" B
œ !Þ
"B
Hence B œ ! is a regular singular point. The indicial equation is <a< "b # < œ ! ,
with roots <" œ " # and <# œ ! .
a,bÞ For B œ ",
:! œ lim aB "b:aBb œ lim
BÄ"
BÄ"
# a" ! " bB
œ "#!"Þ
B
!"aB "b
œ !Þ
BÄ"
B
;! œ lim aB "b# ; aBb œ lim
BÄ"
Hence B œ " is a regular singular point. The indicial equation is
< # a# ! " b < œ ! ,
with roots <" œ # ! " and <# œ ! .
a- b. Given that <" <# is not a positive integer, we can set C œ ! +8 B8 Þ Substitution
_
8œ!
into the ODE results in
Ba" Bb" 8a8 "b+8 B8# c# a" ! " bBd" 8 +8 B8" !" " +8 B8 œ !Þ
_
_
_
8œ#
8œ"
8œ!
That is,
" 8a8 "b+8" B8 " 8a8 "b+8 B8 # " a8 "b+8" B8
_
_
8œ"
8œ#
_
8œ!
a" ! " b" 8 +8 B8 !" " +8 B8 œ !Þ
_
_
8œ"
8œ!
Combining the series, we obtain
# +" !" +! ca# ## b+# a" ! " !" b+" dB " E8 B8 œ ! ,
_
8œ#
in which
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—————————————————————————— CHAPTER 5. ——
E8 œ a8 "ba8 # b+8" c8a8 "b a" ! " b8 !" d+8 .
Note that 8a8 "b a" ! " b8 !" œ a8 !ba8 " b . Setting the coefficients
equal to zero, we have # +" !" +! œ ! , and
+8" œ
for 8
a8 !ba8 " b
+8
a8 "ba8 # b
" . Hence one solution is
!"
!a! "b" a" "b #
B
B
# † "x
# a# "b † #x
!a! "ba! #b" a" "ba" #b $
B âÞ
# a# "ba# #b † $x
C" aBb œ "
Since the nearest other singularity is at B œ " , the radius of convergence of C" aBb will
be at least 3 œ " .
a. b. Given that <" <# is not a positive integer, we can set C œ B"# ! ,8 B8 Þ Then
_
8œ!
Substitution into the ODE results in
Ba" Bb" a8 " # ba8 # b+8 B8# "
_
8œ!
c# a" ! " bBd" a8 " # b+8 B8# !" " +8 B8"# œ !Þ
_
_
8œ!
8œ!
That is,
" a8 " # ba8 # b+8 B8# " a8 " # ba8 # b+8 B8"#
_
_
8œ!
8œ!
# " a8 " # b+8 B8# a" ! " b" a8 " # b+8 B8"# !" " +8 B8"# œ !Þ
_
_
_
8œ!
8œ!
8œ!
After adjusting the indices,
" a8 " # ba8 # b+8 B8# " a8 # ba8 " # b+8" B8#
_
_
8œ!
8œ"
# " a8 " # b+8 B8# a" ! " b" a8 # b+8" B8# !" " +8" B8# œ !Þ
_
_
_
8œ!
8œ"
8œ"
Combining the series, we obtain
" F8 B8# œ ! ,
_
8œ"
in which
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—————————————————————————— CHAPTER 5. ——
F8 œ 8a8 " # b,8 ca8 # ba8 # ! " b !" d,8" .
Note that a8 # ba8 # ! " b !" œ a8 ! # ba8 " # b . Setting F8 œ !,
it follows that for 8 " ,
,8 œ
a8 ! # ba8 " # b
,8" Þ
8a8 " # b
Therefore a second solution is
a" ! # ba" " # b
B
a# # b"x
a" ! # ba# ! # ba" " # ba# " # b #
B â •Þ
a# # ba$ # b# x
C# aBb œ B"# ”"
a/b. Under the transformation B œ "Î0 , the ODE becomes
"
" .# C
"
"
"
.C
0% Œ" # œ#0$ Œ" 0# ”# a" ! " b •
!" C œ ! .
.0
0
0 .0
0
0
0
That is,
ˆ0 $ 0 # ‰
.# C
.C
#0# # 0# a " ! " b0‘
!" C œ ! .
. 0#
.0
Therefore 0 œ ! is a singular point. Note that
: a0 b œ
It follows that
a# # b 0 a " ! " b
!"
and ; a0b œ $
.
#
0 0
0 0#
a# # b 0 a " ! " b
œ "!",
0Ä!
0"
:! œ lim0 :a0b œ lim
0Ä!
;! œ lim0# ; a0b œ lim
0Ä!
0Ä!
!"
œ !" Þ
0"
Hence 0 œ ! aB œ _b is a regular singular point. The indicial equation is
<a< "b a" ! " b< !" œ ! ,
or <# a! " b< !" œ ! Þ Evidently, the roots are < œ ! and < œ " .
21a+b. Note that
:aBb œ
!
"
and ; a0b œ > .
=
B
B
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—————————————————————————— CHAPTER 5. ——
It follows that
lim B :aBb œ lim ! B"= ,
BÄ!
BÄ!
lim0# ; a0b œ lim " B#= Þ
0Ä!
0Ä!
Hence if = " or > # , one or both of the limits does not exist. Therefore B œ ! is an
irregular singular point.
a- b. Let C œ +! B< +" B<" â +8 B<8 â Þ Write the ODE as
B$ C ww ! B# C w " C œ ! .
Substitution of the assumed solution results in
" a8 <ba8 < "b+8 B8<" !" a8 <b+8 B8<" " " +8 B8< œ !Þ
_
_
_
8œ!
8œ!
8œ!
Adjusting the indices, we obtain
" a8 " <ba8 < #b+8" B8< !" a8 " <b+8" B8< " " +8 B8< œ !Þ
_
_
_
8œ"
8œ"
8œ!
Combining the series,
" +! " E8 B8< œ ! ,
_
8œ"
in which E8 œ " +8 a8 " <ba8 < ! #b+8" . Setting the coefficients equal
to zero, we have +! œ ! . But for 8 " ,
+8 œ
a8 " <ba8 < ! #b
+8" Þ
"
Therefore, regardless of the value of <, it follows that +8 œ ! , for 8 œ "ß #ß â .
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—————————————————————————— CHAPTER 5. ——
Section 5.8
3. Here B :aBb œ " and B# ; aBb œ #B , which are both analytic everywhere. We set
C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in
" a< 8ba< 8 "b+8 B<8 " a< 8b+8 B<8 #" +8 B<8" œ ! Þ
_
_
_
8œ!
8œ!
8œ!
After adjusting the indices in the last series, we obtain
+! c<a< "b <dB< " ca< 8ba< 8 "b+8 a< 8b+8 # +8" dB<8 œ !Þ
_
8œ"
Assuming +! Á ! , the indicial equation is <# œ ! , with double root < œ ! . Setting the
remaining coefficients equal to zero, we have for 8 " ,
+ 8 a< b œ
It follows that
+ 8 a< b œ
#
a8 < b #
+8" a<b .
a "b 8 # 8
ca8 <ba8 < "bâa" <bd#
Since < œ ! , one solution is given by
C" aBb œ "
_
8œ!
a "b 8 # 8
a8xb#
+! , 8
".
B8 .
For a second linearly independent solution, we follow the discussion in Section &Þ( .
First
note that
+8w a<b
"
"
"
œ #”
â
•.
+ 8 a< b
8< 8<"
"<
Setting < œ ! ,
+8w a!b
œ # L8 +8 a!b œ # L8
Therefore,
C# aBb œ C" aBb 68 B #"
_
8œ!
a "b 8 # 8
a8xb#
a "b 8 # 8 L 8
a8xb#
Þ
B8 .
4. Here B :aBb œ % and B# ; aBb œ # B , which are both analytic everywhere. We set
C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in
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—————————————————————————— CHAPTER 5. ——
" a< 8ba< 8 "b+8 B<8 %" a< 8b+8 B<8
_
_
8œ!
8œ!
_
" +8 B<8" #" +8 B<8 œ ! Þ
_
8œ!
8œ!
After adjusting the indices in the second-to-last series, we obtain
+! c<a< "b %< #dB< " ca< 8ba< 8 "b+8 %a< 8b+8 # +8 +8" dB<8 œ !Þ
_
8œ"
Assuming +! Á ! , the indicial equation is <# $< # œ ! , with roots <" œ " and
<# œ # . Setting the remaining coefficients equal to zero, we have for 8 " ,
+ 8 a< b œ
"
+8" a<b .
a8 < "ba8 < #b
It follows that
+ 8 a< b œ
a "b 8
+! , 8
ca8 < "ba8 <bâa# <bdca8 < #ba8 <bâa$ <bd
".
Since <" œ " , one solution is given by
a "b 8
B8 .
a
8
b
x
a
8
"
b
x
8œ!
C" aBb œ B" "
_
For a second linearly independent solution, we follow the discussion in Section &Þ( .
Since <" <# œ R œ " , we find that
+ " a< b œ
"
,
a< #ba< $b
with +! œ " . Hence the leading coefficient in the solution is
+ œ lim a< #b +" a<b œ " .
<Ä#
Further,
a < #b + 8 a < b œ
a "b 8
a8 < #b ca8 < "ba8 <bâa$ <bd#
Þ
Let E8 a<b œ a< #b +8 a<b . It follows that
E8w a<b
"
"
"
"
œ
#”
â
•.
E8 a<b
8<#
8<" 8<
$<
Setting < œ <# œ # ,
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—————————————————————————— CHAPTER 5. ——
E8w a #b
"
"
"
œ #”
â "•
E8 a #b
8
8" 8#
œ L8 L8" Þ
Hence
-8 a #b œ aL8 L8" b E8 a #b
a "b 8
œ aL8 L8" b
.
8xa8 "bx
Therefore,
a "b8 aL8 L8" b 8
B —.
8xa8 "bx
8œ"
C# aBb œ C" aBb 68 B B –" "
_
#
6. Let CaBb œ @aBbÎÈB . Then C w œ B"Î# @ w B$Î# @Î# and C ww œ B"Î# @ ww
B$Î# @ w $ B&Î# @Î% . Substitution into the ODE results in
"
B$Î# @ ww B"Î# @ w $ B"Î# @Î%‘ B"Î# @ w B"Î# @Î#‘ ŒB# B"Î# @ œ ! .
%
Simplifying, we find that
@ ww @ œ ! ,
with general solution @aBb œ -" -9= B -# =38 B . Hence
CaBb œ -" B"Î# -9= B -# B"Î# =38 B .
8. The absolute value of the ratio of consecutive terms is
+#7# B#7#
kBk#7# ##7 a7 "bx 7x
kBk#
œ
.
º
ºœ
+#7 B#7
%a7 #ba7 "b
kBk#7 ##7# a7 #bxa7 "bx
Applying the ratio test,
lim º
7Ä_
+#7# B#7#
kBk#
lim
œ
œ !.
º 7Ä_
+#7 B#7
%a7 #ba7 "b
Hence the series for N" aBb converges absolutely for all values of B . Furthermore,
since the series for N! aBb also converges absolutely for all B, term-by-term differentiation
results in
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—————————————————————————— CHAPTER 5. ——
a "b7 B#7"
##7" 7xa7 "bx
7œ"
N! aBb œ "
_
w
a "b7" B#7"
œ " #7"
#
a7 "bx 7x
7œ!
_
œ
Therefore, N!w aBb œ N" aBb .
B _
a "b7 B#7
" #7
.
# 7 œ ! # a7 "bx 7x
9a+b. Note that B :aBb œ " and B# ; aBb œ B# / # , which are both analytic at B œ ! .
Thus B œ ! is a regular singular point. Furthermore, :! œ " and ;! œ / # . Hence
the indicial equation is <# / # œ ! , with roots <" œ / and <# œ / .
a,b. Set C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE
results in
" a< 8ba< 8 "b+8 B<8 " a< 8b+8 B<8
_
_
8œ!
8œ!
" +8 B<8# / # " +8 B<8 œ ! Þ
_
_
8œ!
8œ!
After adjusting the indices in the second-to-last series, we obtain
+! <a< "b < / # ‘B< +" a< "b< a< "b / # ‘
" a< 8ba< 8 "b+8 a< 8b+8 / # +8 +8# ‘B<8 œ !Þ
_
8œ#
Setting the coefficients equal to zero, we find that +" œ ! , and
+8 œ
"
a < 8b # / #
+8# ,
for 8 # . It follows that +$ œ +& œ â œ +#7" œ â œ ! . Furthermore, with
<œ/,
+8 œ
"
+8# .
8a8 #/ b
So for 7 œ "ß #ß â ,
"
+#7#
#7a#7 #/ b
a "b 7
œ #7
+! .
# 7xa" / ba# / bâa7 " / ba7 / b
+#7 œ
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—————————————————————————— CHAPTER 5. ——
Hence one solution is
a "b 7
B #7
Š ‹ —.
7xa" / ba# / bâa7 " / ba7 / b #
7œ"
C" aBb œ B –" "
_
/
a- b. Assuming that <" <# œ #/ is not an integer, simply setting < œ / in the above
results in a second linearly independent solution
a "b 7
B #7
Š ‹ —Þ
7xa" / ba# / bâa7 " / ba7 / b #
7œ"
C# aBb œ B/ –" "
_
a. b. The absolute value of the ratio of consecutive terms in C" aBb is
+#7# B#7#
kBk#7# ##7 7xa" @bâa7 / b
œ
º
º
+#7 B#7
kBk#7 ##7# a7 "bxa" @bâa7 " / b
kBk#
œ
.
%a7 "ba7 " / b
Applying the ratio test,
lim º
7Ä_
+#7# B#7#
kBk#
œ
œ !.
lim
º 7Ä_
+#7 B#7
%a7 "ba7 " / b
Hence the series for C" aBb converges absolutely for all values of B . The same can be
shown for C# aBb . Note also, that if / is a positive integer, then the coefficients in the
series for C# aBb are undefined.
10a+b. It suffices to calculate PcN! aBb 68 Bd. Indeed,
cN! aBb 68 Bd w œ N!w aBb 68 B
and
N!w aBb N! aBb
cN! aBb 68 Bd œ N! aBb 68 B #
.
B
B#
ww
Hence
N! aBb
B
ww
PcN! aBb 68 Bd œ B# N!ww aBb 68 B #B N!w aBb N! aBb
B N!w aBb 68 B N! aBb B# N! aBb 68 B Þ
Since B# N!ww aBb B N!w aBb B# N! aBb œ ! ,
PcN! aBb 68 Bd œ #B N!w aBb .
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—————————————————————————— CHAPTER 5. ——
a,b. Given that PcC# aBbd œ ! , after adjusting the indices in Part a+b, we have
," B ## ,# B# " ˆ8# ,8 ,8# ‰B8 œ #B N!w aBb .
_
8œ$
Using the series representation of N!w aBb in Problem ) ,
," B # ,# B " ˆ8# ,8 ,8# ‰B8 œ # "
#
_
_
8œ$
8œ"
#
a "b8 a#8bB#8
##8 a8xb#
.
a- b. Equating the coefficients on both sides of the equation, we find that
," œ ,$ œ â œ ,#7" œ â œ ! Þ
Also, with 8 œ " , ## ,# œ "Îa"xb# , that is, ,# œ "Î## a"xb# ‘. Furthermore, for 7
a#7b# ,#7 ,#7# œ #
a "b7 a#7b
##7 a7xb#
#,
.
More explicitly,
"
"
Œ"
#
%
#
"
" "
,' œ # # # Œ"
# % '
# $
ã
,% œ
##
It can be shown, in general, that
,#7 œ a "b7"
L7
##7 a7xb#
.
11. Bessel's equation of order one is
B# C ww B C w ˆB# "‰C œ ! .
Based on Problem *, the roots of the indicial equation are <" œ " and <# œ " . Set
C œ B< a+! +" B +# B# â +8 B8 âb. Substitution into the ODE results in
" a< 8ba< 8 "b+8 B<8 " a< 8b+8 B<8
_
_
8œ!
8œ!
" +8 B<8# " +8 B<8 œ ! Þ
_
_
8œ!
8œ!
After adjusting the indices in the second-to-last series, we obtain
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—————————————————————————— CHAPTER 5. ——
+! c<a< "b < "dB< +" ca< "b< a< "b "d
" ca< 8ba< 8 "b+8 a< 8b+8 +8 +8# dB<8 œ !Þ
_
8œ#
Setting the coefficients equal to zero, we find that +" œ ! , and
+ 8 a< b œ
+8# a<b
a < 8b # "
"
œ
+8# a<b ,
a8 < "ba8 < "b
"
for 8 # . It follows that +$ œ +& œ â œ +#7" œ â œ !. Solving the recurrence
relation,
+#7 a<b œ
With < œ <" œ ",
a "b 7
a#7 < "ba#7 < "b# âa< $b# a< "b
+! .
a "b 7
+#7 a"b œ #7
+! Þ
# a7 "bx 7x
For a second linearly independent solution, we follow the discussion in Section &Þ( .
Since <" <# œ R œ # , we find that
+ # a< b œ
"
,
a< $ba< "b
with +! œ " . Hence the leading coefficient in the solution is
+ œ lim a< "b +# a<b œ
<Ä"
Further,
a< "b +#7 a<b œ
a "b 7
"
.
#
a#7 < "b ca#7 < "bâa$ <bd#
Þ
Let E8 a<b œ a< "b +8 a<b . It follows that
w
E#7
a< b
"
"
"
œ
#”
â
•.
E#7 a<b
#7 < "
#7 < "
$<
Setting < œ <# œ " , we calculate
________________________________________________________________________
page 251
—————————————————————————— CHAPTER 5. ——
"
-#7 a "b œ aL7 L7" bE#7 a "b
#
"
a "b 7
œ aL7 L7" b
#
#7 ca#7 #bâ#d#
"
a "b 7
œ aL7 L7" b #7"
Þ
#
#
7xa7 "bx
Note that +#7" a<b œ ! implies that E#7" a<b œ ! , so
-#7" a "b œ ”
.
E#7" a<b•
œ !.
.<
<œ<#
Therefore,
_
_
"
"
B #7
a "b 7
a "b7 aL7 L7" b B #7
C# aBb œ –B "
Š ‹ —68 B –" "
Š ‹ —Þ
# 7 œ ! a7 "bx 7x #
7xa7 "bx
#
B
7œ"
Based on the definition of N" aBb,
_
"
a "b7 aL7 L7" b B #7
C# aBb œ N" aBb 68 B –" "
Š ‹ —Þ
B
7xa7 "bx
#
7œ"
12. Consider a solution of the form
CaBb œ ÈB 0 ˆ!B" ‰.
Then
Cw œ
.0 !" B"
0 a0 b
†
. 0 ÈB
#È B
in which 0 œ !B" . Hence
C ww œ
. # 0 !# " # B#"
.0 !" # B"
0 a0b
†
†
,
. 0#
. 0 BÈB
BÈB
%BÈB
and
B# C ww œ !# " # B#" ÈB
.# 0
"È
# " È .0
!"
B
B
B 0 a0b Þ
. 0#
.0
%
Substitution into the ODE results in
!# " # B#"
.# 0
.0
"
"
!" # B"
0 a0b Œ!# " # B#" / # " # 0 a0b œ ! Þ
#
.0
.0
%
%
Simplifying, and setting 0 œ !B" , we find that
________________________________________________________________________
page 252
—————————————————————————— CHAPTER 5. ——
0#
.# 0
.0
0
ˆ0# / # ‰0 a0b œ ! , a‡b
. 0#
.0
which is a Bessel equation of order / . Therefore, the general solution of the given ODE
is
CaBb œ ÈB -" 0" ˆ!B" ‰ -# 0# ˆ!B" ‰‘,
in which 0" a0b and 0# a0b are the linearly independent solutions of a‡b.
________________________________________________________________________
page 253
—————————————————————————— CHAPTER 6. ——
Chapter Six
Section 6.1
3.
The function 0 a>b is continuous.
4.
The function 0 a>b has a jump discontinuity at > œ " .
7. Integration is a linear operation. It follows that
(
E
(
E
" E ,> =>
" E ,> =>
( / † / .> ( / † / .>
# !
# !
E
"
" E
œ ( /a,=b> .> ( /a,=b> .> Þ
# !
# !
-9=2 ,> † /=> .> œ
!
Hence
-9=2 ,> † /
!
=>
" " /a,=bE
" " /a,=bE
.> œ –
—Þ
#
= , — #–
=,
Taking a limit, as E p _ ,
________________________________________________________________________
page 254
—————————————————————————— CHAPTER 6. ——
(
_
"
"
"
"
”
• ”
•
# =,
# =,
=
œ #
Þ
= ,#
-9=2 ,> † /=> .> œ
!
Note that the above is valid for = k,k .
8. Proceeding as in Prob. ( ,
(
E
=382 ,> † /=> .> œ
!
" " /a,=bE
" " /a,=bE
—Þ
#– = , — #–
=,
Taking a limit, as E p _ ,
(
_
"
"
"
"
”
• ”
•
# =,
# =,
,
œ #
Þ
= ,#
=382 ,> † /=> .> œ
!
The limit exists as long as = k,k .
10. Observe that /+> =382 ,> œ ˆ/Ð+,Ñ> /Ð+,Ñ> ‰Î# Þ It follows that
(
E
/+> =382 ,> † /=> .> œ
!
" " /a+,=bE
" " /a,+=bE
Þ
#– = + , — #– = , + —
Taking a limit, as E p _ ,
(
_
"
"
"
"
”
• ”
•
# =+,
# =,+
,
œ
Þ
a= + b # , #
/+> =382 ,> † /=> .> œ
!
The limit exists as long as = + k,k .
11. Using the linearity of the Laplace transform,
_c =38 ,>d œ
"
"
_/3,> ‘ _/3,> ‘Þ
#3
#3
Since
(
_
!
/a+3,b> /=> .> œ
"
,
= + 3,
we have
________________________________________________________________________
page 255
—————————————————————————— CHAPTER 6. ——
(
_
/„ 3,> /=> .> œ
!
"
.
= … 3,
Therefore
_c=38 ,>d œ
"
"
"
”
•
#3 = 3, = 3,
,
.
œ #
= ,#
12. Using the linearity of the Laplace transform,
_c -9= ,>d œ
"
"
_/3,> ‘ _/3,> ‘Þ
#
#
From Prob. "" , we have
(
_
/„ 3,> /=> .> œ
!
"
.
= … 3,
Therefore
_c-9= ,>d œ
"
"
"
”
•
# = 3, = 3,
=
œ #
.
= ,#
14. Using the linearity of the Laplace transform,
_ /+> -9= ,>‘ œ
"
"
_/a+3,b> ‘ _/a+3,b> ‘Þ
#
#
Based on the integration in Prob. "" ,
(
_
/a+ „ 3,b> /=> .> œ
!
"
Þ
= + … 3,
Therefore
_/+> -9= ,>‘ œ
"
"
"
”
•
# = + 3, = + 3,
=+
œ
.
a= + b # , #
The above is valid for = + .
15. Integrating by parts,
________________________________________________________________________
page 256
—————————————————————————— CHAPTER 6. ——
(
E
+>
>/ † /
=>
!
E
> /a+=b> E
" a+=b>
.> œ
/
.>
º (
=+ !
! =+
" /Ea+=b Ea+ =b/Ea+=b
œ
Þ
a= + b #
Taking a limit, as E p _ ,
(
_
>/+> † /=> .> œ
!
"
a= + b #
Þ
Note that the limit exists as long as = + .
17. Observe that > -9=2 +> œ a> /+> > /+> bÎ# Þ For any value of - ,
(
E
->
>/ † /
!
=>
E
> /a-=b> E
" a-=b>
.> œ
/
.>
º (
=- !
! =" /Ea-=b Ea- =b/Ea-=b
œ
Þ
a= - b #
Taking a limit, as E p _ ,
(
_
>/-> † /=> .> œ
!
"
a= - b #
Þ
Note that the limit exists as long as = k- k . Therefore,
(
_
> -9=2 +> † /=> .> œ
!
œ
18. Integrating by parts,
(
E
>8 /+> † /=> .> œ
!
œ
"
"
"
–
—
#
# a= + b
a= + b #
=# + #
a= + b # a= + b #
.
E
>8 /a+=b> E
8 8" a+=b>
> /
.>
(
º
=+ !
! =+
E
E8 /a=+bE
8 8" a+=b>
(
> /
.> Þ
=+
! =+
Continuing to integrate by parts, it follows that
________________________________________________________________________
page 257
—————————————————————————— CHAPTER 6. ——
(
E
E8 /a+=bE
8E8" /a+=bE
=+
a= + b #
8xˆ/a+=bE "‰
8xE/a+=bE
â
Þ
a8 #bxa= +b$
a= +b8"
>8 /+> † /=> .> œ
!
That is,
(
E
!
>8 /+> † /=> .> œ :8 aEb † /a+=bE
8x
a= +b8"
,
in which :8 a0b is a polynomial of degree 8 . For any given polynomial,
lim :8 aEb † /a=+bE œ ! ,
EÄ_
as long as = + . Therefore,
(
_
>8 /+> † /=> .> œ
!
8x
a= +b8"
.
20. Observe that ># =382 +> œ a># /+> ># /+> bÎ# Þ Using the result in Prob. ") ,
(
_
># =382 +> † /=> .> œ
!
œ
"
#x
#x
–
—
# a= + b $
a= + b $
#+a$=# +# b
The above is valid for = k+k.
a =# + # b $
.
22. Integrating by parts,
(
E
!
> /> .> œ > /> º (
E
!
œ"/
E
E/
E
/> .>
!
E
Þ
Taking a limit, as E p _ ,
(
_
> /> .> œ " /E Þ
!
Hence the integral converges .
23. Based on a series expansion, note that for > ! ,
/> " > ># Î# ># Î# .
________________________________________________________________________
page 258
—————————————————————————— CHAPTER 6. ——
It follows that for > ! ,
"
.
#
># />
Hence for any finite E " ,
(
E
># /> .>
"
E"
.
#
It is evident that the limit as E p _ does not exist.
24. Using the fact that k-9= >k Ÿ " , and the fact that
(
_
/> .> œ " ,
!
it follows that the given integral converges.
25a+b. Let : ! . Integrating by parts,
(
E
!
/B B: .B œ /B B: º :(
E
!
œ E: /E :(
!
E
E
/B B:" .B
/B B:" .BÞ
!
Taking a limit, as E p _ ,
(
That is, >a: "b œ : >a:b .
_
!
/B B: .B œ :(
a,b. Setting : œ ! ,
>a"b œ (
_
_
/B B:" .B .
!
/B .B œ " .
!
a- b. Let : œ 8 . Using the result in Part a,b,
>a8 "b œ 8 >a8b
œ 8a8 "b>a8 "b
ã
œ 8a8 "ba8 #bâ# † " † >a"b .
Since >a"b œ " , >a8 "b œ 8x .
a. b. Using the result in Part a,b,
________________________________________________________________________
page 259
—————————————————————————— CHAPTER 6. ——
>a: 8b œ a: 8 "b >a: 8 "b
œ a: 8 "ba: 8 #b>a: 8 #b
ã
œ a: 8 "ba: 8 #bâa: "b: >a:b .
Hence
>a: 8b
œ :a: "ba: "bâa: 8 "b Þ
>a : b
Given that >a"Î#b œ È1 , it follows that
È1
$
"
"
>Œ œ >Œ œ
#
#
#
#
and
>Œ
*%&È1
""
* ( & $
$
.
œ † † † >Œ œ
#
# # # #
#
$#
________________________________________________________________________
page 260
—————————————————————————— CHAPTER 6. ——
Section 6.2
1. Write the function as
Hence _" c] a=bd œ
$
$ #
œ
.
=# %
# =# %
$
#
=38 #> .
3. Using partial fractions,
=#
#
#
"
"
œ ”
•.
$= %
& =" =%
Hence _" c] a=bd œ #& ˆ/> /%> ‰.
5. Note that the denominator =# #= & is irreducible over the reals. Completing the
square, =# #= & œ a= "b# % . Now convert the function to a rational function
of the variable 0 œ = " . That is,
=#
#= #
#a= "b
œ
.
#= &
a = "b # %
We know that
_" ”
#0
• œ # -9= #> .
%
0#
Using the fact that _c/+> 0 a>bd œ _c0 a>bd=p=+ ,
_" ”
=#
#= #
>
• œ #/ -9= #> .
#= &
6. Using partial fractions,
#= $
"
"
(
œ ”
•.
#
= %
% =# =#
Hence _" c] a=bd œ "% a/#> (/#> b. Note that we can also write
#= $
=
$ #
œ# #
.
#
= %
= % # =# %
8. Using partial fractions,
)=# %= "#
"
=
#
œ$ & #
# #
.
#
= a = %b
=
= %
= %
________________________________________________________________________
page 261
—————————————————————————— CHAPTER 6. ——
Hence _" c] a=bd œ $ & -9= #> # =38 #> .
9. The denominator =# %= & is irreducible over the reals. Completing the square,
=# %= & œ a= #b# " . Now convert the function to a rational function of the
variable 0 œ = # . That is,
=#
" #=
& #a = # b
œ
.
%= &
a = #b # "
We find that
_" ”
0#
&
#0
#
• œ & =38 > # -9= > .
" 0 "
Using the fact that _c/+> 0 a>bd œ _c0 a>bd=p=+ ,
_" ”
=#
" #=
#>
• œ / a& =38 > # -9= >b .
%= &
10. Note that the denominator =# #= "! is irreducible over the reals. Completing
the square, =# #= "! œ a= "b# * . Now convert the function to a rational
function of the variable 0 œ = " . That is,
=#
#= $
#a= "b &
œ
.
#= "!
a = "b # *
We find that
_" ”
#0
&
&
#
• œ # -9= $> =38 $> .
* 0 *
$
0#
Using the fact that _c/+> 0 a>bd œ _c0 a>bd=p=+ ,
_" ”
=#
&
#= $
>
• œ / Œ# -9= $> =38 $> .
$
#= "!
12. Taking the Laplace transform of the ODE, we obtain
=# ] a=b = Ca!b C w a!b $c= ] a=b C a!bd # ] a=b œ ! .
Applying the initial conditions,
=# ] a=b $= ] a=b # ] a=b = $ œ ! .
Solving for ] a=b, the transform of the solution is
] a =b œ
=#
=$
.
$= #
________________________________________________________________________
page 262
—————————————————————————— CHAPTER 6. ——
Using partial fractions,
#
"
=$
œ
.
=" =#
=# $= #
Hence Ca>b œ _" c] a=bd œ # /> /#> .
13. Taking the Laplace transform of the ODE, we obtain
=# ] a=b = Ca!b C w a!b #c= ] a=b C a!bd # ] a=b œ ! .
Applying the initial conditions,
=# ] a=b #= ] a=b # ] a=b " œ ! .
Solving for ] a=b, the transform of the solution is
] a =b œ
"
.
=# #= #
Since the denominator is irreducible, write the transform as a function of 0 œ = " .
That is,
=#
"
"
œ
.
#= #
a = "b # "
First note that
_" ”
0#
"
• œ =38 > .
"
Using the fact that _c/+> 0 a>bd œ _c0 a>bd=p=+ ,
Hence Ca>b œ /> =38 > .
_" ”
=#
"
>
• œ / =38 > .
#= #
15. Taking the Laplace transform of the ODE, we obtain
=# ] a=b = Ca!b C w a!b #c= ] a=b C a!bd # ] a=b œ ! .
Applying the initial conditions,
=# ] a=b #= ] a=b # ] a=b #= % œ ! .
Solving for ] a=b, the transform of the solution is
] a =b œ
=#
#= %
.
#= #
Since the denominator is irreducible, write the transform as a function of 0 œ = " .
Completing the square,
________________________________________________________________________
page 263
—————————————————————————— CHAPTER 6. ——
=#
First note that
_" ”
#= %
#a= "b #
œ
.
#= #
a = "b # $
#0
#
#
#
• œ # -9=2 È$ > È =382 È$ > .
$ 0 $
$
0#
Using the fact that _c/+> 0 a>bd œ _c0 a>bd=p=+ , the solution of the IVP is
Ca>b œ _" ”
=#
#= %
#
>
• œ / # -9=2 È$ > È =382 È$ > .
#= #
$
16. Taking the Laplace transform of the ODE, we obtain
=# ] a=b = Ca!b C w a!b #c= ] a=b C a!bd & ] a=b œ ! .
Applying the initial conditions,
=# ] a=b #= ] a=b & ] a=b #= $ œ ! .
Solving for ] a=b, the transform of the solution is
] a =b œ
=#
#= $
.
#= &
Since the denominator is irreducible, write the transform as a function of 0 œ = " .
That is,
#= $
#a= "b "
œ
.
=# #= &
a = "b # %
We know that
_" ”
#0
"
"
#
• œ # -9= #> =38 #> .
% 0 %
#
0#
Using the fact that _c/+> 0 a>bd œ _c0 a>bd=p=+ , the solution of the IVP is
Ca>b œ _" ”
=#
"
#= $
>
• œ / Œ# -9= #> =38 #> .
#= &
#
17. Taking the Laplace transform of the ODE, we obtain
=% ] a=b =$ Ca!b =# C w a!b = C ww a!b C www a!b %=$ ] a=b =# Ca!b = C w a!b C ww a!b‘
'=# ] a=b = Ca!b C w a!b‘ %c= ] a=b C a!bd ] a=b œ !
Applying the initial conditions,
________________________________________________________________________
page 264
—————————————————————————— CHAPTER 6. ——
=% ] a=b %=$ ] a=b '=# ] a=b %= ] a=b ] a=b =# %= ( œ ! .
Solving for the transform of the solution,
] a =b œ
=# %= (
=# %= (
œ
.
=% %=$ '=# %= "
a = "b %
Using partial fractions,
=# %= (
a= "b
%
œ
%
a= "b
%
#
a= "b
$
"
a= "b#
.
Note that _c >8 d œ a8xbÎ=8" and _c/+> 0 a>bd œ _c0 a>bd=p=+ . Hence the solution
of the IVP is
Ca>b œ _" –
=# %= (
a = "b
%
# $ >
# >
>
— œ $ > / > / >/ .
18. Taking the Laplace transform of the ODE, we obtain
=% ] a=b =$ C a!b =# C w a!b = C ww a!b C www a!b ] a=b œ ! Þ
Applying the initial conditions,
=% ] a=b ] a=b = $ = œ ! .
Solving for the transform of the solution,
] a =b œ
=#
=
.
"
By inspection, it follows that Ca>b œ _" =#=" ‘ œ -9=2 > Þ
19. Taking the Laplace transform of the ODE, we obtain
=% ] a=b =$ C a!b =# C w a!b = C ww a!b C www a!b % ] a=b œ ! Þ
Applying the initial conditions,
=% ] a=b %] a=b =$ #= œ ! .
Solving for the transform of the solution,
] a =b œ
=#
It follows that Ca>b œ _" =#=# ‘ œ -9= È# > Þ
=
.
#
20. Taking the Laplace transform of both sides of the ODE, we obtain
________________________________________________________________________
page 265
—————————————————————————— CHAPTER 6. ——
=# ] a=b = Ca!b C w a!b =# ] a=b œ
=#
=
.
%
Applying the initial conditions,
=# ] a=b =# ] a=b = œ
=#
=
.
%
Solving for ] a=b, the transform of the solution is
=
=
] a =b œ #
#
.
#
#
a= = ba= %b
= =#
Using partial fractions on the first term,
a =#
=
=# ba=#
%b
œ
"
=
=
#
” #
•.
#
#
%= = =
= %
First note that
_" ”
=#
=
• œ -9= =>
=#
and _" ”
=#
=
• œ -9= #> .
%
Hence the solution of the IVP is
Ca>b œ
"
"
-9= =>
-9= #> -9= =>
#
%=
% =#
& =#
"
œ
-9= =>
-9= #> .
#
%=
% =#
21. Taking the Laplace transform of both sides of the ODE, we obtain
=# ] a=b = Ca!b C w a!b #c= ] a=b C a!bd # ] a=b œ
=#
=
.
"
Applying the initial conditions,
=# ] a=b #= ] a=b # ] a=b = # œ
Solving for ] a=b, the transform of the solution is
] a =b œ
a =#
=#
=
.
"
=
=#
#
.
#
#= #ba= "b
= #= #
Using partial fractions on the first term,
a =#
=
" =#
=%
œ ” #
#
•.
#
#= #ba= "b
& = " = #= #
Thus we can write
________________________________________________________________________
page 266
—————————————————————————— CHAPTER 6. ——
] a =b œ
" =
# "
# #= $
.
#
#
& = " & = " & =# #= #
For the last term, we note that =# #= # œ a= "b# " . So that
=#
#= $
#a= "b "
œ
.
#= #
a = "b # "
We know that
_" ”
#0
"
#
• œ # -9= > =38 > .
" 0 "
0#
Based on the translation property of the Laplace transform,
_" ”
#= $
>
• œ / a# -9= > =38 >b.
#
= #= #
Combining the above, the solution of the IVP is
C a >b œ
"
#
#
-9= > =38 > /> a# -9= > =38 >b Þ
&
&
&
23. Taking the Laplace transform of both sides of the ODE, we obtain
=# ] a=b = Ca!b C w a!b #c= ] a=b C a!bd ] a=b œ
%
.
="
Applying the initial conditions,
=# ] a=b #= ] a=b ] a=b #= $ œ
Solving for ] a=b, the transform of the solution is
] a =b œ
First write
#= $
a= "b
#
œ
%
a= "b
$
#a= " b "
a= "b
#
#= $
%
.
="
a= "b#
œ
#
"
Þ
= " a= "b#
.
We note that
_" ”
%
#
"
# • œ # ># # > .
$
0
0
0
So based on the translation property of the Laplace transform, the solution of the IVP is
________________________________________________________________________
page 267
—————————————————————————— CHAPTER 6. ——
Ca>b œ # ># /> > /> # /> Þ
25. Let 0 a>b be the forcing function on the right-hand-side. Taking the Laplace
transform
of both sides of the ODE, we obtain
=# ] a=b = Ca!b C w a!b ] a=b œ _c0 a>bd.
Applying the initial conditions,
=# ] a=b ] a=b œ _c0 a>bd.
Based on the definition of the Laplace transform,
_c0 a>bd œ (
_
!
0 a>b /=> .>
œ ( > /=> .>
"
!
"
/=
/=
œ #
# .
=
=
=
Solving for the transform,
] a =b œ
"
="
/= # #
Þ
"b
= a = "b
=# a =#
Using partial fractions,
"
"
"
œ # #
"b
=
= "
=# a =#
and
=
"
=
œ #
Þ
"b
= = "
=# a =#
We find, by inspection, that
_" ”
"
• œ > =38 > .
"b
=# a =#
Referring to Line "$ , in Table 'Þ#Þ" ,
_c ?- a>b0 a> - bd œ /-= _c0 a>bd.
Let
_c 1a>b d œ
="
"
"
=
"
œ # #
#
.
"b
= =
= " = "
=# a =#
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page 268
—————————————————————————— CHAPTER 6. ——
Then 1a>b œ " > -9= > =38 > . It follows, therefore, that
_" ”/= †
="
• œ ?" a>bc " a> "b -9=a> "b =38a> "b d .
"b
=# a =#
Combining the above, the solution of the IVP is
Ca>b œ > =38 > ?" a>bc " a> "b -9=a> "b =38a> "b d .
26. Let 0 a>b be the forcing function on the right-hand-side. Taking the Laplace
transform
of both sides of the ODE, we obtain
=# ] a=b = Ca!b C w a!b % ] a=b œ _c0 a>bd.
Applying the initial conditions,
=# ] a=b % ] a=b œ _c0 a>bd.
Based on the definition of the Laplace transform,
_c0 a>bd œ (
_
!
0 a>b /=> .>
œ ( >/
"
!
=>
.> (
"
/=
œ # # .
=
=
_
/=> .>
"
Solving for the transform,
] a =b œ
"
"
/= # #
Þ
%b
= a = %b
=# a =#
Using partial fractions,
"
" "
"
œ ” # #
•Þ
%b
% =
= %
=# a =#
We find that
_" ”
"
"
"
• œ > =38 > .
%b
%
)
=# a =#
Referring to Line "$ , in Table 'Þ#Þ" ,
_c ?- a>b0 a> - bd œ /-= _c0 a>bd.
It follows that
________________________________________________________________________
page 269
—————————————————————————— CHAPTER 6. ——
_" ”/= †
"
"
"
• œ ?" a>b” a> "b =38a> "b•.
%b
%
)
=# a =#
Combining the above, the solution of the IVP is
C a >b œ
"
"
"
"
> =38 > ?" a>b” a> "b =38a> "b•.
%
)
%
)
28a+b. Assuming that the conditions of Theorem 'Þ#Þ" are satisfied,
. _ =>
J a =b œ
( / 0 a>b.>
.= !
_
` =>
/ 0 a>b‘.>
œ(
! `=
w
œ(
œ(
_
!
_
!
> /=> 0 a>b‘.>
/=> c >0 a>bd.> Þ
a,b. Using mathematical induction, suppose that for some 5
J a5 b a=b œ (
_
!
",
/=> ’a >b5 0 a>b“.> .
Differentiating both sides,
J a5"b a=b œ
. _ =>
5
( / ’a >b 0 a>b“.>
.= !
_
` =>
œ(
’/ a >b5 0 a>b“.>
! `=
œ(
œ(
29. We know that
_
!
!
_
’ > /=> a >b5 0 a>b“.>
/=> ’a >b5" 0 a>b“.> .
_/+> ‘ œ
"
.
=+
Based on Prob. #) ,
_ > /+> ‘ œ
.
"
”
•.
.= = +
________________________________________________________________________
page 270
—————————————————————————— CHAPTER 6. ——
Therefore,
_> /+> ‘ œ
"
a= + b #
.
31. Based on Prob. #) ,
.8
_ c a >b d œ 8 _ c " d
.=
.8 "
œ 8 ” •Þ
.= =
8
Therefore,
_c >8 d œ a "b8
œ
8x
=8"
a "b8 8x
=8"
Þ
33. Using the translation property of the Laplace transform,
_/+> =38 ,>‘ œ
,
a= + b # , #
Þ
Therefore,
_> /+> =38 ,>‘ œ
œ
.
,
–
.= a= +b# ,# —
#,a= +b
a=# #+= +# ,# b#
.
34. Using the translation property of the Laplace transform,
=+
_/+> -9= ,>‘ œ
Þ
a= + b # , #
Therefore,
_> /+> -9= ,>‘ œ
œ
.
=+
–
.= a= +b# ,# —
a= + b # , #
a=# #+= +# ,# b#
.
35a+b. Taking the Laplace transform of the given Bessel equation,
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—————————————————————————— CHAPTER 6. ——
_c > C ww d _c C w d _c > C d œ ! .
Using the differentiation property of the transform,
That is,
.
.
_c C ww d _c C w d _c C d œ ! .
.=
.=
. #
.
= ] a=b = C a!b C w a!b‘ = ] a=b C a!b ] a=b œ ! .
.=
.=
It follows that
ˆ" =# ‰] w a=b = ] a=b œ ! .
a,b. We obtain a first-order linear ODE in ] a=b:
=
] w a =b #
] a=b œ ! ,
= "
with integrating factor
.a=b œ /B:Œ(
=#
=
.= œ È=# " .
"
The first-order ODE can be written as
. È #
’ = " † ] a=b“ œ ! ,
.=
with solution
] a =b œ
È =# "
-
.
a- b. In order to obtain negative powers of = , first write
"
" "Î#
œ ”" # •
Þ
È =# "
=
=
"
Expanding Š"
" "Î#
=# ‹
in a binomial series,
"
" # " † $ % " † $ † & '
œ
"
=
=
= â,
È" a"Î=# b
#
#†%
#†%†'
valid for =# " . Hence, we can formally express ] a=b as
" " "
"†$ "
"†$†& "
] a =b œ - ”
â•Þ
= # =$
# † % =&
# † % † ' =(
________________________________________________________________________
page 272
—————————————————————————— CHAPTER 6. ——
Assuming that term-by-term inversion is valid,
C a >b œ - ” "
" >#
" † $ >%
" † $ † & >'
â•
# #x # † % %x # † % † ' 'x
#x >#
%x >%
'x
>'
œ - ”" #
# #
# # #
â•Þ
# #x # † % %x # † % † ' 'x
It follows that
Ca>b œ - ”"
" #
"
"
> # # >% # # # > ' â •
#
#
# †%
# †% †'
8
_
a "b #8
œ -"
> Þ
#
#8
8 œ ! # a8xb
The series is evidently the expansion, about B œ ! , of N! a>bÞ
36a,b. Taking the Laplace transform of the given Legendre equation,
_c C ww d _ ># C ww ‘ # _c > C w d !a! "b_c C d œ ! .
Using the differentiation property of the transform,
.#
.
_c C d # _c C ww d # _c C w d !a! "b_c C d œ ! .
.=
.=
ww
That is,
=# ] a=b = Ca!b C w a!b‘
.# #
= ] a=b = C a!b C w a!b‘
#
.=
.
# c= ] a=b C a!bd !a! "b] a=b œ ! Þ
.=
Invoking the initial conditions, we have
.# #
.
= ] a=b " # = ] a=b "‘ # c= ] a=bd !a! "b] a=b œ ! Þ
.=
.=
#
After carrying out the differentiation, the equation simplifies to
.# #
.
= ] a=b‘ # c= ] a=bd =# !a! "b‘] a=b œ " Þ
.=#
.=
That is,
=#
.#
.
]
a
=
b
#=
] a=b =# !a! "b‘] a=b œ " .
.=#
.=
37. By definition of the Laplace transform, given the appropriate conditions,
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page 273
—————————————————————————— CHAPTER 6. ——
_c 1a>bd œ (
_
!
œ(
/=> ”( 0 a7 b. 7 •.>
>
!
_
!
=>
( / 0 a7 b. 7 .> Þ
>
!
Assuming that the order of integration can be exchanged,
_c 1a>bd œ (
_
!
œ(
_
!
0 a7 b ”(
_
7
/=> .>•. 7
/=7
0 a7 b ”
•. 7 Þ
=
cNote the region of integration is the area between the lines 7 a>b œ > and 7 a>b œ ! Þd
Hence
_c 1a>bd œ
" _
=7
( 0 a7 b / . 7
= !
"
œ _ c 0 a> b d Þ
=
________________________________________________________________________
page 274
—————————————————————————— CHAPTER 6. ——
Section 6.3
1.
3.
5.
________________________________________________________________________
page 275
—————————————————————————— CHAPTER 6. ——
6.
7. Using the Heaviside function, we can write
0 a >b œ a > # b # ? # a > b .
The Laplace transform has the property that
_c ?- a>b0 a> - bd œ /-= _c0 a>bd.
Hence
_ a> #b# ?# a>b‘ œ
# /2=
.
=#
9. The function can be expressed as
0 a>b œ a> 1bc?1 a>b ?#1 a>bdÞ
Before invoking the ><+8=6+>398 :<9:/<>C of the transform, write the function as
0 a>b œ a> 1b ?1 a>b a> #1b ?#1 a>b 1 ?#1 a>b Þ
It follows that
_ c 0 a >b d œ
/ 1 =
/#1=
1/#1=
.
=#
=#
=
10. It follows directly from the ><+8=6+>398 :<9:/<>C of the transform that
_c 0 a>bd œ
/=
/$=
/%=
#
'
.
=
=
=
11. Before invoking the ><+8=6+>398 :<9:/<>C of the transform, write the function as
0 a>b œ a> #b ?# a>b ?# a>b a> $b ?$ a>b ?$ a>b Þ
________________________________________________________________________
page 276
—————————————————————————— CHAPTER 6. ——
It follows that
_c 0 a>bd œ
/#=
/#=
/$=
/$=
.
=#
=
=#
=
12. It follows directly from the ><+8=6+>398 :<9:/<>C of the transform that
_c 0 a>bd œ
"
/=
.
=#
=#
13. Using the fact that _c/+> 0 a>bd œ _c0 a>bd=p=+ ,
_" –
$x
a = #b
15. First consider the function
K a =b œ
It follows that
Hence
œ >$ /#> .
#a= "b
Þ
#= #
=#
Completing the square in the denominator,
K a =b œ
%—
#a= "b
a = "b # "
Þ
_" cKa=bd œ # /> -9= > Þ
_" /#= Ka=b‘ œ # /a>#b -9= a> #b ?# a>b Þ
16. The inverse transform of the function #Îa=# %b is 0 a>b œ =382 #> Þ Using the
><+8=6+>398 :<9:/<>C of the transform,
_" ”
# /#=
• œ =382 #a> #b † ?# a>b Þ
=# %
17. First consider the function
K a =b œ
=#
a = #b
Þ
%= $
Completing the square in the denominator,
________________________________________________________________________
page 277
—————————————————————————— CHAPTER 6. ——
K a =b œ
It follows that
a = #b
a = #b # "
Þ
_" cKa=bd œ /#> -9=2 > Þ
Hence
_" ”
a= #b/=
#a>"b
-9=2 a> "b ?" a>b Þ
•œ/
#
= %= $
18. Write the function as
J a =b œ
/=
/#=
/$=
/%=
Þ
=
=
=
=
It follows from the ><+8=6+>398 :<9:/<>C of the transform, that
_" ”
/= /#= /$= /%=
• œ ?" a>b ?# a>b ?$ a>b ?% a>b Þ
=
19a+b. By definition of the Laplace transform,
_c 0 a->bd œ (
_
!
/=> 0 a->b.> Þ
Making a change of variable, 7 œ -> , we have
_c 0 a->bd œ
Hence _c 0 a->bd œ
" _ =a7 Î-b
0 a 7 b. 7
( /
- !
" _ a=Î-b7
œ ( /
0 a 7 b. 7 Þ
- !
"
-
J ˆ -= ‰ , where =Î- + .
a,b. Using the result in Part a+b,
>
_” 0 Œ • œ 5 J a5=bÞ
5
Hence
_" cJ a5=bd œ
"
>
0 Œ Þ
5
5
________________________________________________________________________
page 278
—————————————————————————— CHAPTER 6. ——
a- b. From Part a,b,
_" cJ a+=bd œ
"
>
0 Œ Þ
+
+
Note that += , œ +a= ,Î+b. Using the fact that _c/-> 0 a>bd œ _c0 a>bd=p=- ,
_" cJ a+= ,bd œ /,>Î+
"
>
0Π.
+
+
20. First write
J a =b œ
8x
.
ˆ #= ‰8"
Let Ka=b œ 8xÎ=8" Þ Based on the results in Prob. "*,
in which 1a>b œ >8 . Hence
" "
=
_ ’KŠ ‹“ œ 1a#>b,
#
#
_" cJ a=bd œ # a#>b8 œ #8" >8 Þ
23. First write
/%a="Î#b
J a =b œ
Þ
#a= "Î#b
Now consider
Using the result in Prob. "*a,b,
K a =b œ
/#=
.
=
_" cKa#=bd œ
in which 1a>b œ ?# a>b. Hence _" cKa#=bd œ
_" cJ a=bd œ
"
#
"
>
1 Π,
# #
?# a>Î#b œ
" >Î#
/ ?% a >b .
#
"
#
?% a>b. It follows that
24. By definition of the Laplace transform,
________________________________________________________________________
page 279
—————————————————————————— CHAPTER 6. ——
_c 0 a>bd œ (
!
_
/=> ?" a>b.> Þ
That is,
_c 0 a>bd œ ( /=> .>
"
!
" /=
œ
Þ
=
25. First write the function as 0 a>b œ ?! a>b ?" a>b ?# a>b ?$ a>b . It follows that
_c 0 a>bd œ ( /=> .> ( /=> .> Þ
"
!
$
#
That is,
" /=
/#= /$=
_c 0 a>bd œ
=
=
" /= /#= /$=
œ
Þ
=
26. The transform may be computed directly. On the other hand, using the ><+8=6+>398
:<9:/<>C of the transform,
5=
" #8"
5 /
"
_c0 a>bd œ
a "b
= 5œ"
=
œ
" #8"
" a /= b5 —
= –5 œ !
" " a /= b#8#
.
œ
=
" /=
That is,
" a/#= b
_c0 a>bd œ
.
=a" /= b
8"
29. The given function is periodic, with X œ # . Using the result of Prob. #),
_c0 a>bd œ
#
"
"
"
=>
/
0
a
>
b
.>
œ
/=> .> .
(
(
" /#= !
" /#= !
That is,
________________________________________________________________________
page 280
—————————————————————————— CHAPTER 6. ——
_c0 a>bd œ
" /=
=a" /#= b
"
.
œ
=a" /= b
31. The function is periodic, with X œ " . Using the result of Prob. #),
_c0 a>bd œ
"
"
> /=> .> Þ
(
" /= !
It follows that
_c0 a>bd œ
" /= a" =b
Þ
=# a" /= b
32. The function is periodic, with X œ 1 . Using the result of Prob. #),
_c0 a>bd œ
1
"
=38 > † /=> .> Þ
(
" / 1 = !
We first calculate
=>
( =38 > † / .> œ
1
!
" / 1 =
Þ
" =#
Hence
_c0 a>bd œ
" / 1 =
Þ
a" /1= ba" =# b
33a+b.
_c0 a>bd œ _c"d _c?" a>bd
" /=
.
œ
=
=
________________________________________________________________________
page 281
—————————————————————————— CHAPTER 6. ——
a, b .
Let J a=b œ _c" ?" a>bd. Then
_”( c" ?" a7 bd. 7 • œ
>
!
"
" /=
J a= b œ
Þ
=
=#
a- b .
Let Ka=b œ _c1a>bd. Then
_c2a>bd œ Ka=b /= Ka=b
=
" /=
= " /
œ
/
=#
=#
= #
a" / b
.
œ
=#
________________________________________________________________________
page 282
—————————————————————————— CHAPTER 6. ——
34a+b.
a,b. The given function is periodic, with X œ # . Using the result of Prob. #),
_c0 a>bd œ
#
"
/=> :a>b.> Þ
(
" /#= !
Based on the piecewise definition of :a>b,
( /
#
!
=>
:a>b.> œ ( > /
"
!
=>
.> ( a# >b/=> .>
"
œ # a" /= b# Þ
=
Hence
_c:a>bd œ
#
"
a" /= b
Þ
=# a" /= b
a- b. Since :a>b satisfies the hypotheses of Theorem 'Þ#Þ" ,
_c: w a>bd œ = _c:a>bd :a!b .
Using the result of Prob. $!,
We note the :a!b œ ! , hence
_c: w a>bd œ
_c:a>bd œ
a" /= b
.
=a" /= b
" a" /= b
”
•.
= =a" /= b
________________________________________________________________________
page 283
—————————————————————————— CHAPTER 6. ——
Section 6.4
2. Let 2a>b be the forcing function on the right-hand-side. Taking the Laplace transform
of both sides of the ODE, we obtain
=# ] a=b = Ca!b C w a!b #c= ] a=b C a!bd # ] a=b œ _c2 a>bd.
Applying the initial conditions,
=# ] a=b #= ] a=b # ] a=b " œ _c2 a>bd.
The forcing function can be written as 2a>b œ ?1 a>b ?#1 a>b Þ Its transform is
_c2a>bd œ
/1= /#1=
.
=
Solving for ] a=b, the transform of the solution is
] a =b œ
"
/1= /#1=
.
=# #= # =a=# #= #b
First note that
=#
Using partial fractions,
=a =#
"
"
œ
.
#= #
a = "b # "
"
" " " a = "b "
œ
Þ
#= #b
# = # a = "b # "
Taking the inverse transform, term-by-term,
_”
"
"
œ
_
œ /> =38 > .
•
–
—
#
#
= #= #
a = "b "
Now let
K a =b œ
=a =#
"
.
#= #b
Then
_" cKa=bd œ
" " >
"
/ -9= > /> =38 > .
# #
#
Using Theorem 'Þ$Þ" ,
_" c/-= Ka=bd œ
"
"
?- a>b /a>-b c-9=a> - b =38a> - bd?- a>b .
#
#
Hence the solution of the IVP is
________________________________________________________________________
page 284
—————————————————————————— CHAPTER 6. ——
"
"
Ca>b œ /> =38 > ?1 a>b /a>1b c-9=a> 1b =38a> 1bd?1 a>b
#
#
"
" a>#1b
?#1 a>b /
c-9=a> #1b =38a> #1bd?#1 a>b .
#
#
That is,
"
"
Ca>b œ /> =38 > c?1 a>b ?#1 a>bd /a>1b c-9= > =38 >d ?1 a>b
#
#
" a>#1b
/
c-9= > =38 >d?#1 a>b .
#
The solution starts out as free oscillation, due to the initial conditions. The amplitude
increases, as long as the forcing is present. Thereafter, the solution rapidly decays.
4. Let 2a>b be the forcing function on the right-hand-side. Taking the Laplace transform
of both sides of the ODE, we obtain
=# ] a=b = Ca!b C w a!b % ] a=b œ _c2 a>bd.
Applying the initial conditions,
________________________________________________________________________
page 285
—————————————————————————— CHAPTER 6. ——
=# ] a=b % ] a=b œ _c2a>bd.
The transform of the forcing function is
_c2a>bd œ
"
/ 1 =
.
=# " = # "
Solving for ] a=b, the transform of the solution is
] a =b œ
"
/ 1 =
.
a=# %ba=# "b a=# %ba=# "b
Using partial fractions,
a =#
"
"
"
"
œ ” #
#
•.
#
%ba= "b
$ = " = %
It follows that
_" ”
a =#
"
"
"
• œ ”=38 > =38 #>•.
#
%ba= "b
$
#
Based on Theorem 'Þ$Þ" ,
_" ”
/ 1 =
"
"
• œ ”=38a> 1b =38a#> #1b•?1 a>b .
#
#
a= %ba= "b
$
#
Hence the solution of the IVP is
Ca>b œ
"
"
"
"
”=38 > =38 #>• ”=38 > =38 #>•?1 a>b .
$
#
$
#
________________________________________________________________________
page 286
—————————————————————————— CHAPTER 6. ——
Since there is no damping term, the solution follows the forcing function, after which
the response is a steady oscillation about C œ ! .
5. Let 0 a>b be the forcing function on the right-hand-side. Taking the Laplace transform
of both sides of the ODE, we obtain
=# ] a=b = Ca!b C w a!b $c= ] a=b C a!bd # ] a=b œ _c0 a>bd.
Applying the initial conditions,
=# ] a=b $= ] a=b # ] a=b œ _c0 a>bd.
The transform of the forcing function is
_c0 a>bd œ
" /"!=
.
=
=
Solving for the transform,
] a =b œ
"
/"!=
.
=a=# $= #b =a=# $= #b
Using partial fractions,
=a =#
"
" "
"
#
œ ”
•Þ
$= #b
# = =# ="
Hence
_" ”
"
" /#>
œ
/> .
•
=a=# $= #b
#
#
Based on Theorem 'Þ$Þ" ,
_" ”
"
/"!=
#a>"!b
#/a>"!b ‘?"! a>b .
• œ " /
#
=a= $= #b
#
Hence the solution of the IVP is
________________________________________________________________________
page 287
—————————————————————————— CHAPTER 6. ——
C a >b œ
"
/#>
"
c" ?"! a>bd
/> /a#>#!b #/a>"!b ‘?"! a>b .
#
#
#
The solution increases to a temporary steady value of C œ "Î#. After the forcing ceases,
the response decays exponentially to C œ ! .
6. Taking the Laplace transform of both sides of the ODE, we obtain
=# ] a=b = Ca!b C w a!b $c= ] a=b C a!bd # ] a=b œ
/#=
.
=
Applying the initial conditions,
=# ] a=b $= ] a=b # ] a=b " œ
/#=
.
=
Solving for the transform,
] a =b œ
"
/#=
.
=# $= # =a=# $= #b
Using partial fractions,
________________________________________________________________________
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—————————————————————————— CHAPTER 6. ——
=#
"
"
"
œ
$= #
=" =#
and
=a =#
"
" "
"
#
œ ”
•Þ
$= #b
# = =# ="
Taking the inverse transform. term-by-term, the solution of the IVP is
"
"
Ca>b œ /> /#> ” /a>#b /#a>#b •?# a>b Þ
#
#
Due to the initial conditions, the response has a transient overshoot, followed by an
exponential convergence to a steady value of C= œ "Î# .
7. Taking the Laplace transform of both sides of the ODE, we obtain
= # ] a = b = C a!b C w a!b ] a= b œ
/$1=
.
=
Applying the initial conditions,
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page 289
—————————————————————————— CHAPTER 6. ——
=# ] a=b ] a=b = œ
/$1=
.
=
Solving for the transform,
] a =b œ
=
/$1=
.
=# " = a= # " b
Using partial fractions,
"
"
=
œ
.
= a = # "b
= =# "
Hence
] a =b œ
=#
=
"
=
/$1= ” #
•.
"
= = "
Taking the inverse transform, the solution of the IVP is
Ca>b œ -9= > c" -9=a> $1bd?$1 a>b
œ -9= > c" -9= > d?$1 a>b Þ
Due to initial conditions, the solution temporarily oscillates about C œ ! . After the
forcing is applied, the response is a steady oscillation about C7 œ " .
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page 290
—————————————————————————— CHAPTER 6. ——
9. Let 1a>b be the forcing function on the right-hand-side. Taking the Laplace transform
of both sides of the ODE, we obtain
=# ] a=b = Ca!b C w a!b ] a=b œ _c1a>bd.
Applying the initial conditions,
=# ] a=b ] a=b " œ _c1a>bd.
The forcing function can be written as
1a>b œ
>
c" ?' a>bd $ ?' a>b
#
>
"
œ a> 'b?' a>b
# #
with Laplace transform
"
/'=
_c1a>bd œ
.
#=#
#=#
Solving for the transform,
] a =b œ
"
"
/'=
.
=# "
#=# a=# "b #=# a=# "b
Using partial fractions,
"
#=# a=#
"b
œ
" "
"
” # #
•Þ
# =
= "
Taking the inverse transform, and using Theorem 'Þ$Þ" , the solution of the IVP is
"
"
Ca>b œ =38 > c> =38 >d ca> 'b =38a> 'bd?' a>b
#
#
"
"
œ c> =38 >d ca> 'b =38a> 'bd?' a>bÞ
#
#
________________________________________________________________________
page 291
—————————————————————————— CHAPTER 6. ——
The solution increases, in response to the ramp input, and thereafter oscillates about a
mean value of C7 œ $ .
11. Taking the Laplace transform of both sides of the ODE, we obtain
= # ] a = b = C a!b C w a!b % ] a= b œ
/ 1 =
/$1=
.
=
=
Applying the initial conditions,
=# ] a=b % ] a=b œ
/ 1 =
/$1=
.
=
=
Solving for the transform,
] a =b œ
/ 1 =
/$1=
.
= a = # % b = a = # %b
Using partial fractions,
"
" "
=
œ ” #
•Þ
%b
% = = %
=a =#
Taking the inverse transform, and applying Theorem 'Þ$Þ" ,
C a >b œ
"
"
c" -9=a#> #1bd?1 a>b c" -9=a#> '1bd?$1 a>b
%
%
"
"
œ c?1 a>b ?$1 a>bd -9= #> † c?1 a>b ?$1 a>bd .
%
%
________________________________________________________________________
page 292
—————————————————————————— CHAPTER 6. ——
Since there is no damping term, the solution responds immediately to the forcing input.
There is a temporary oscillation about C œ "Î% Þ
12. Taking the Laplace transform of the ODE, we obtain
=% ] a=b =$ C a!b =# C w a!b = C ww a!b C www a!b ] a=b œ
/=
/#=
Þ
=
=
Applying the initial conditions,
=% ] a=b ] a= b œ
/=
/#=
.
=
=
Solving for the transform of the solution,
] a =b œ
/=
/#=
.
= a = % " b = a = % "b
Using partial fractions,
"
"
%
"
"
#=
œ
”
•Þ
= a = % "b
%
= = " = " =# "
It follows that
________________________________________________________________________
page 293
—————————————————————————— CHAPTER 6. ——
_" ”
"
"
>
>
• œ % / / # -9= >‘Þ
"b
%
=a =%
Based on Theorem 'Þ$Þ" , the solution of the IVP is
"
Ca>b œ c?" a>b ?# a>bd /a>"b /a>"b # -9=a> "b‘?" a>b
%
" a>#b
a
>#
b
/
/
# -9=a> #b‘?# a>b .
%
The solution increases without bound, exponentially.
13. Taking the Laplace transform of the ODE, we obtain
=% ] a=b =$ C a!b =# C w a!b = C ww a!b C www a!b
" / 1 =
&=# ] a=b = C a!b C w a!b‘ % ] a=b œ
Þ
=
=
Applying the initial conditions,
" / 1 =
= ] a=b &= ] a=b %] a=b œ
.
=
=
%
#
Solving for the transform of the solution,
________________________________________________________________________
page 294
—————————————————————————— CHAPTER 6. ——
] a =b œ
"
/ 1 =
.
=a=% &=# %b =a=% &=# %b
Using partial fractions,
"
" $
=
%=
œ
”
•Þ
=a=% &=# %b
"# = =# % =# "
It follows that
_" ”
=a =%
"
"
c$ -9= #> % -9= >dÞ
•œ
#
&= %b
"#
Based on Theorem 'Þ$Þ" , the solution of the IVP is
C a >b œ
"
"
c" ?1 a>bd c-9= #> % -9= >d
%
"#
"
c-9= #a> 1b % -9=a> 1bd?1 a>b .
"#
That is,
C a >b œ
"
"
c" ?1 a>bd c-9= #> % -9= >d
%
"#
"
c-9= #> % -9= >d?1 a>b .
"#
________________________________________________________________________
page 295
—————————————————————————— CHAPTER 6. ——
After an initial transient, the solution oscillates about C7 œ ! .
14. The specified function is defined by
Ú !ß
0 a >b œ Û
> ! bß
Ü 2ß
2
5 a>
! Ÿ > >!
>! Ÿ > > ! 5
> >! 5
which can conveniently be expressed as
0 a >b œ
2
2
a> >! b ?>! a>b a> >! 5 b ?>! 5 a>b Þ
5
5
15. The function is defined by
Ú
!ß
Ý
Ý2
a > > ! bß
1a>b œ Û 5 2
Ý
Ý 5 a> >! #5 bß
Ü !ß
! Ÿ > >!
>! Ÿ > > ! 5
>! 5 Ÿ > >! #5
> >! #5
which can also be written as
1a>b œ
2
#2
2
a> >! b ?>! a>b
a> >! 5 b ?>! 5 a>b a> >! #5 b ?>! #5 a>bÞ
5
5
5
16a. b. From Part a- b, the solution is
$
&
?a>b œ %5 ?$Î# a>b 2Œ> %5 ?&Î# a>b 2 Œ> ,
#
#
where
2a>b œ
" È( >Î)
$È( >
"
$È ( >
/
=38
/>Î) -9=
Þ
%
)%
)
%
)
Due to the damping term, the solution will decay to zero. The maximum will occur
________________________________________________________________________
page 296
—————————————————————————— CHAPTER 6. ——
shortly after the forcing ceases. By plotting the various solutions, it appears that the
solution will reach a value of C œ # , as long as 5 #Þ&" .
a/ bÞ
Based on the graph, and numerical calculation, k?a>bk !Þ" for > #&Þ'(($ .
17. We consider the initial value problem
C ww %C œ
with Ca!b œ C w a!b œ ! .
"
ca> &b ?& a>b a> & 5 b ?&5 a>bd,
5
a+b. The specified function is defined by
Ú !ß
"
5 a>
0 a >b œ Û
& bß
Ü "ß
!Ÿ>&
&Ÿ>&5
> &5
a,bÞ Taking the Laplace transform of both sides of the ODE, we obtain
________________________________________________________________________
page 297
—————————————————————————— CHAPTER 6. ——
/&=
/a&5 b=
= ] a = b = C a!b C a!b % ] a= b œ
.
5=#
5=#
#
w
Applying the initial conditions,
=# ] a=b % ] a= b œ
Solving for the transform,
] a =b œ
/&=
/a&5 b=
.
5=#
5=#
/&=
/a&5 b=
.
5=# a=# %b 5=# a=# %b
Using partial fractions,
"
" "
"
œ ” # #
•Þ
%b
% =
= %
=# a =#
It follows that
_" ”
"
"
"
• œ > =38 #> .
%b
%
)
=# a =#
Using Theorem 'Þ$Þ" , the solution of the IVP is
C a >b œ
"
c2a> &b ?& a>b 2a> & 5 b ?&5 a>bd ,
5
in which 2a>b œ "% > ") =38 #> .
a- b. Note that for > & 5 , the solution is given by
Ca>b œ
"
"
"
=38a#> "!b
=38a#> "! #5 b
% )5
)5
" =38 5
œ
-9=a#> "! 5 b Þ
%
%5
So for > & 5 , the solution oscillates about C7 œ "Î% , with an amplitude of
Eœ
k=38a5 bk
.
%5
________________________________________________________________________
page 298
—————————————————————————— CHAPTER 6. ——
________________________________________________________________________
page 299
—————————————————————————— CHAPTER 6. ——
18a+b.
a,b. The forcing function can be expressed as
0 5 a >b œ
"
c?%5 a>b ?%5 a>bdÞ
#5
Taking the Laplace transform of both sides of the ODE, we obtain
"
/a%5 b=
/a%5 b=
= ] a=b = Ca!b C a!b c= ] a=b C a!bd % ] a=b œ
.
$
#5=
#5=
#
w
Applying the initial conditions,
"
/a%5 b=
/a%5 b=
= # ] a = b = ] a = b % ] a= b œ
.
$
#5=
#5=
Solving for the transform,
] a =b œ
$ /a%5 b=
$ /a%5 b=
.
#5=a$=# = "#b #5=a$=# = "#b
Using partial fractions,
=a$=#
"
" "
" $=
œ
” #
•
= "#b
"# = $= = "#
" " " " 'ˆ= "' ‰
œ
Þ
"# – = ' ˆ= " ‰# "%$ —
'
$'
Let
"
" "
'
L a =b œ
–
"
)5 = ˆ= ‰#
'
"%$
$'
=
"
'
ˆ= "' ‰#
"%$
$'
—Þ
It follows that
________________________________________________________________________
page 300
—————————————————————————— CHAPTER 6. ——
2a>b œ _" cL a=bd œ
È"%$ >
È"%$ >
"
/>Î'
"
=38
-9=
'
' —Þ
)5
)5 – È"%$
Based on Theorem 'Þ$Þ" , the solution of the IVP is
Ca>b œ 2a> % 5 b ?%5 a>b 2a> % 5 b ?%5 a>b Þ
a- b .
As the parameter 5 decreases, the solution remains null for a longer period of time.
________________________________________________________________________
page 301
—————————————————————————— CHAPTER 6. ——
Since the magnitude of the impulsive force increases, the initial overshoot of the
response also increases. The duration of the impulse decreases. All solutions eventually
decay to C œ ! Þ
19a+b.
a- bÞ From Part a,b,
?a>b œ " -9= > #" a "b5 c" -9=a> 5 1bd?51 a>b Þ
8
5œ"
________________________________________________________________________
page 302
—————————————————————————— CHAPTER 6. ——
21a+b.
a,b. Taking the Laplace transform of both sides of the ODE, we obtain
=# Y a=b = ?a!b ? w a!b Y a=b œ
Applying the initial conditions,
= # Y a = b Y a =b œ
Solving for the transform,
Y a =b œ
8
"
a "b5 /5 1=
"
.
= 5œ"
=
8
"
a "b5 /5 1=
"
.
= 5œ"
=
8
"
a "b5 /5 1=
"
.
= a = # " b 5 œ " = a = # "b
Using partial fractions,
"
"
=
.
œ #
"b
= = "
=a =#
Let
________________________________________________________________________
page 303
—————————————————————————— CHAPTER 6. ——
2a>b œ _" ”
"
• œ " -9= > .
"b
=a =#
Applying Theorem 'Þ$Þ" , term-by-term, the solution of the IVP is
?a>b œ 2a>b " a "b5 2a> 5 1b ?51 a>b .
8
5œ"
Note that
2a> 5 1b œ ?! a> 5 1b -9=a> 5 1b
œ ?51 a>b a "b5 -9= > Þ
Hence
?a>b œ " -9= > " a "b5 ?51 a>b a-9= >b"?51 a>b Þ
8
8
5œ"
5œ"
a- b .
The ODE has no damping term. Each interval of forcing adds to the energy of the
system.
Hence the amplitude will increase. For 8 œ "& , 1a>b œ ! when > "&1 . Therefore the
oscillation will eventually become steady, with an amplitude depending on the values of
?a"&1b and ? w a"&1b .
a. b. As 8 increases, the interval of forcing also increases. Hence the amplitude of the
transient will increase with 8 . Eventually, the forcing function will be constant. In fact,
for large values of > ,
1a>b œ œ
"ß
!ß
8 even
8 odd
Further, for > 81 ,
________________________________________________________________________
page 304
—————————————————————————— CHAPTER 6. ——
?a>b œ " -9= > 8 -9= >
" a "b 8
Þ
#
Hence the steady state solution will oscillate about ! or " , depending on 8 , with an
amplitude of E œ 8 " .
In the limit, as 8 p _ , the forcing function will be a periodic function, with period #1 .
From Prob. #(, in Section 'Þ$ ,
_c1a>bd œ
"
.
=a" /= b
As 8 increases, the duration and magnitude of the transient will increase without bound.
22a+b. Taking the initial conditions into consideration, the transform of the ODE is
=# Y a=b !Þ" = Y a=b Y a=b œ
8
"
a "b5 /5 1=
"
.
= 5œ"
=
Solving for the transform,
8
"
a "b5 /5 1=
Y a =b œ
"
.
=a=# !Þ"= "b 5 œ " =a=# !Þ"= "b
Using partial fractions,
=a =#
"
"
= !Þ"
œ #
Þ
!Þ"= "b
= = !Þ"= "
Since the denominator in the second term is irreducible, write
=#
Let
= !Þ"
a= !Þ!&b !Þ!&
œ
Þ
!Þ"= "
a= !Þ!&b# Ð$**Î%!!Ñ
"
a= !Þ!&b
!Þ!&
2a>b œ _" –
#
= a= !Þ!&b Ð$**Î%!!Ñ a= !Þ!&b# Ð$**Î%!!Ñ —
œ " />Î#! –-9=
È$**
>
#!
È$**
"
=38
È$**
#!
>—Þ
Applying Theorem 'Þ$Þ" , term-by-term, the solution of the IVP is
?a>b œ 2a>b " a "b5 2a> 5 1b ?51 a>b .
8
5œ"
________________________________________________________________________
page 305
—————————————————————————— CHAPTER 6. ——
For odd values of 8, the solution approaches C œ ! .
For even values of 8, the solution approaches C œ " .
a,b. The solution is a sum of damped sinusoids, each of frequency = œ È$** Î#! ¸ " .
Each term has an 'initial' amplitude of approximately " Þ For any given 8 , the solution
contains 8 " such terms. Although the amplitude will increase with 8, the amplitude
will also be bounded by 8 " .
a- b. Suppose that the forcing function is replaced by 1a>b œ =38 > . Based on the
methods
in Chapter $, the general solution of the differential equation is
?a>b œ />Î#! –-" -9=
È$**
#!
> -# =38
È$**
#!
>— ?: a>b Þ
Note that ?: a>b œ E -9= > F =38 > . Using the method of undetermined coefficients,
E œ "! and F œ ! Þ Based on the initial conditions, the solution of the IVP is
________________________________________________________________________
page 306
—————————————————————————— CHAPTER 6. ——
?a>b œ "! />Î#! –-9=
È$**
#!
>
È$**
"
=38
È$**
#!
>— "! -9= > Þ
Observe that both solutions have the same frequency, = œ È$**Î#! ¸ " .
23a+b. Taking the initial conditions into consideration, the transform of the ODE is
= # Y a = b Y a =b œ
Solving for the transform,
Y a =b œ
8
"
a "b5 /a""5Î%b=
#"
.
=
=
5œ"
8
"
a "b5 /a""5Î%b=
"
#
.
= a = # "b
= a = # "b
5œ"
Using partial fractions,
"
"
=
œ #
.
"b
= = "
=a =#
Let
2a>b œ _" ”
"
• œ " -9= > .
= a = # "b
Applying Theorem 'Þ$Þ" , term-by-term, the solution of the IVP is
?a>b œ 2a>b # " a "b5 2 Œ>
8
5œ"
""5
?""5Î% a>b .
%
That is,
________________________________________________________________________
page 307
—————————————————————————— CHAPTER 6. ——
?a>b œ " -9= > # " a "b5 ”" -9=Œ>
8
5œ"
""5
• ?""5Î% a>b .
%
a, b .
a- b. Based on the plot, the 'slow period' appears to be )) . The 'fast period' appears to
be about ' . These values correspond to a 'slow frequency' of == œ !Þ!("% and a 'fast
frequency' =0 œ "Þ!%(# .
a. b. The natural frequency of the system is =! œ " Þ The forcing function is initially
periodic, with period X œ ""Î# œ &Þ& . Hence the corresponding forcing frequency is
A œ "Þ"%#% . Using the results in Section $Þ*, the 'slow frequency' is given by
== œ
and the 'fast frequency' is given by
=0 œ
k= =! k
œ !Þ!("#
#
k= =! k
œ "Þ!("# .
#
Based on theses values, the 'slow period' is predicted as ))Þ#%( and the 'fast period' is
given as &Þ)'&' .
________________________________________________________________________
page 308
—————————————————————————— CHAPTER 6. ——
Section 6.5
2. Taking the Laplace transform of both sides of the ODE, we obtain
=# ] a=b = Ca!b C w a!b %] a=b œ /1= /#1= .
Applying the initial conditions,
=# ] a=b %] a=b œ /1= /#1= .
Solving for the transform,
] a =b œ
/1= /#1=
/ 1 =
/#1=
œ
.
=# %
=# % = # %
Applying Theorem 'Þ$Þ" , the solution of the IVP is
C a >b œ
"
"
=38a#> #1b?1 a>b =38a#> %1b?#1 a>b
#
#
"
œ =38a#>bc ?1 a>b ?#1 a>b d Þ
#
4. Taking the Laplace transform of both sides of the ODE, we obtain
=# ] a=b = Ca!b C w a!b ] a=b œ #! /$= .
Applying the initial conditions,
=# ] a=b ] a=b = œ #! /$= .
Solving for the transform,
=
#! /$=
] a =b œ #
#
.
= "
= "
Using a table of transforms, and Theorem 'Þ$Þ" , the solution of the IVP is
Ca>b œ -9=2 > #! =382 a> $b?$ a>b .
________________________________________________________________________
page 309
—————————————————————————— CHAPTER 6. ——
6. Taking the initial conditions into consideration, the transform of the ODE is
=# ] a=b %] a=b =Î# œ /%1= .
Solving for the transform,
] a =b œ
=Î#
/%1=
.
=# % = # %
Using a table of transforms, and Theorem 'Þ$Þ" , the solution of the IVP is
C a >b œ
"
-9= #>
#
"
œ -9= #>
#
"
=38a#> )1b?%1 a>b
#
"
=38a#>b ?%1 a>b Þ
#
8. Taking the Laplace transform of both sides of the ODE, we obtain
=# ] a=b = Ca!b C w a!b %] a=b œ # /a1Î%b= .
Applying the initial conditions,
=# ] a=b %] a=b œ # /a1Î%b= .
Solving for the transform,
________________________________________________________________________
page 310
—————————————————————————— CHAPTER 6. ——
# /a1Î%b=
] a =b œ #
.
= %
Applying Theorem 'Þ$Þ" , the solution of the IVP is
Ca>b œ =38Š#>
1
‹?1Î% a>b œ -9=a#>b ?1Î% a>b Þ
#
9. Taking the initial conditions into consideration, the transform of the ODE is
=# ] a=b ] a= b œ
Solving for the transform,
] a =b œ
/a1Î#b=
/#1=
$ /a$1Î#b=
.
=
=
/a1Î#b=
$ /a$1Î#b=
/#1=
.
= a = # "b
=# "
= a= # " b
Using partial fractions,
"
"
=
œ
Þ
= a = # "b
= =# "
Hence
] a =b œ
/a1Î#b=
= /a1Î#b=
$ /a$1Î#b=
/#1=
= /#1=
#
Þ
=
= "
=# "
=
=# "
Based on Theorem 'Þ$Þ" , the solution of the IVP is
Ca>b œ ?1Î# a>b -9=Š>
1
$1
‹?1Î# a>b $ =38Œ>
?$1Î# a>b
#
#
?#1 a>b -9=a> #1b?#1 a>b .
That is,
Ca>b œ c" =38a>bd ?1Î# a>b $ -9=a>b ?$1Î# a>b c" -9=a>bd ?#1 a>b .
________________________________________________________________________
page 311
—————————————————————————— CHAPTER 6. ——
10. Taking the transform of both sides of the ODE,
#=# ] a=b =] a=b %] a=b œ (
!
_
/=> $ Š>
1
‹=38 > .>
'
"
œ /a1Î'b= Þ
#
Solving for the transform,
] a =b œ
/a1Î'b=
.
#a#=# = %b
First write
"
"
%
œ
#
"
#a#= = %b
ˆ= % ‰#
It follows that
Ca>b œ _" c] a=bd œ
$"
"'
.
È$"
" a>1Î'bÎ%
1
/
† =38
Š> ‹?1Î' a>b Þ
È$"
%
'
________________________________________________________________________
page 312
—————————————————————————— CHAPTER 6. ——
11. Taking the initial conditions into consideration, the transform of the ODE is
=
=# ] a=b #= ] a=b # ] a=b œ #
/a1Î#b= .
= "
Solving for the transform,
/a1Î#b=
] a =b œ #
.
a= "ba=# #= #b =# #= #
=
Using partial fractions,
a =#
=
"ba=#
#= #b
We can also write
œ
"
=
#
=%
#
#
” #
•Þ
& = " = " = #= #
=%
a = "b $
œ
Þ
=# #= #
a = "b # "
Let
]" a=b œ
a =#
=
"ba=#
#= #b
Þ
Then
_" c]" a=bd œ
"
#
"
-9= > =38 > /> c-9= > $ =38 >dÞ
&
&
&
Applying Theorem 'Þ$Þ" ,
_" –
/a1Î#b=
1
ˆ 1‰
œ / > # =38Š> ‹?1Î# a>b Þ
—
#
= #= #
#
Hence the solution of the IVP is
C a >b œ
"
#
"
-9= > =38 > /> c-9= > $ =38 >d
&
&
&
ˆ> 1# ‰
/
-9=a>b ?1Î# a>b Þ
________________________________________________________________________
page 313
—————————————————————————— CHAPTER 6. ——
12. Taking the initial conditions into consideration, the transform of the ODE is
=% ] a=b ] a=b œ /= .
Solving for the transform,
] a =b œ
/=
.
=% "
Using partial fractions,
=%
"
"
"
"
œ ” #
#
•Þ
"
# = " = "
It follows that
_" ”
=%
"
"
"
• œ =382 > =38 > Þ
"
#
#
Applying Theorem 'Þ$Þ" , the solution of the IVP is
C a >b œ
"
c=382a> "b =38a> "bd?" a>b Þ
#
________________________________________________________________________
page 314
—————————————————————————— CHAPTER 6. ——
14a+b. The Laplace transform of the ODE is
"
=# ] a=b = ] a=b ] a=b œ /= .
#
Solving for the transform of the solution,
] a =b œ
/=
.
=# =Î# "
First write
=#
"
"
œ
=Î# "
ˆ= "% ‰#
"&
"'
Þ
Taking the inverse transform and applying both shifting theorems,
Ca>b œ
È"&
% a>"bÎ%
/
=38
a> "b ?" a>b .
È"&
%
a,b. As shown on the graph, the maximum is attained at some >" #. Note that for
> #,
Ca>b œ
È"&
% a>"bÎ%
/
=38
a> "b Þ
È"&
%
Setting C w a>b œ ! , we find that >" ¸ #Þ$'"$ . The maximum value is calculated as
Ca#Þ$'"$b ¸ !Þ(""&$ Þ
a- b. Setting # œ "Î% , the transform of the solution is
] a =b œ
/=
.
=# =Î% "
Following the same steps, it follows that
________________________________________________________________________
page 315
—————————————————————————— CHAPTER 6. ——
Ca>b œ
) a>"bÎ)
$È(
/
=38
a> "b ?" a>b .
)
$È(
Once again, the maximum is attained at some >" # . Setting C w a>b œ ! , we find that
>" ¸ #Þ%&'* , with C a>" b ¸ !Þ)$$& Þ
a. b. Now suppose that ! # " . Then the transform of the solution is
] a =b œ
/=
.
=# # = "
First write
=#
It follows that
2a>b œ _" ”
Hence the solution is
=#
"
"
œ
Þ
#
#= "
a= # Î#b a" # # Î%b
"
#
/# >Î# =38ŠÈ" # # Î% † >‹Þ
•œ È
#= "
% ##
Ca>b œ 2a> "b ?" a>b .
The solution is nonzero only if > " , in which case C a>b œ 2a> "b Þ Setting C w a>b œ !
,
we obtain
"
>+8’È" # # Î% † a> "b“ œ È% # # ,
#
that is,
>+8È" # # Î% † a> "b‘
#
œ .
#
È" # Î%
#
________________________________________________________________________
page 316
—————————————————————————— CHAPTER 6. ——
As # p ! , we obtain the formal equation >+8a> "b œ _ . Hence >" p " 1# . Setting
> œ 1Î# in 2a>b, and letting # p ! , we find that C" p " Þ These conclusions agree with
the case # œ ! , for which it is easy to show that the solution is
Ca>b œ =38a> "b ?" a>b .
15a+b. See Prob. "% . It follows that the solution of the IVP is
C a >b œ
È"&
%5 a>"bÎ%
/
=38
a> "b ?" a>b .
È"&
%
This function is a multiple of the answer in Prob. "%a+b. Hence the peak value occurs at
>" ¸ #Þ$'"$ . The maximum value is calculated as C a#Þ$'"$b ¸ !Þ(""&$ 5 Þ We find
that the appropriate value of 5 is 5" œ #Î!Þ(""&$ ¸ #Þ)"!) .
a,b. Based on Prob. "%a- b, the solution is
Ca>b œ
) 5 a>"bÎ)
$È (
/
=38
a> "b ?" a>b .
)
$È(
Since this function is a multiple of the solution in Prob. "%a- b, we have >" ¸ #Þ%&'* ,
with Ca>" b ¸ !Þ)$$& 5 Þ The solution attains a value of C œ # , for 5" œ #Î!Þ)$$& ,
that is, 5" ¸ #Þ$**& .
a- b. Similar to Prob. "%a. b, for ! # " , the solution is
Ca>b œ 2a> "b ?" a>b ,
in which
2a>b œ
#5
/# >Î# =38ŠÈ" # # Î% † >‹Þ
È% # #
It follows that >" " p 1Î# . Setting > œ 1Î# in 2a>b, and letting # p ! , we find that
C" p 5 Þ Requiring that the peak value remains at C œ # , the limiting value of 5 is
5" œ # . These conclusions agree with the case # œ ! , for which it is easy to show
that the solution is
Ca>b œ 5 =38a> "b ?" a>b .
16a+b. Taking the initial conditions into consideration, the transformation of the ODE is
=# ] a=b ] a= b œ
" /a%5 b=
/a%5 b=
—.
#5 –
=
=
Solving for the transform of the solution,
________________________________________________________________________
page 317
—————————————————————————— CHAPTER 6. ——
] a =b œ
" /a%5 b=
/a%5 b=
Þ
#5 – =a=# "b =a=# "b —
Using partial fractions,
"
"
=
œ #
Þ
"b
= = "
=a =#
Now let
2a>b œ _" ”
"
• œ " -9= > Þ
"b
=a =#
Applying Theorem 'Þ$Þ" , the solution is
9a> ß 5 b œ
"
c2a> % 5 b ?%5 a>b 2 a> % 5 b ?%5 a>bdÞ
#5
That is,
9a> ß 5 b œ
"
c ?%5 a>b ?%5 a>bd
#5
"
c-9=a> % 5 b ?%5 a>b -9=a> % 5 b ?%5 a>bdÞ
#5
a,b. Consider various values of > . For any fixed > % , 9a> ß 5 b œ ! , as long as
% 5 >Þ If > % , then for % 5 > ,
9a> ß 5 b œ
It follows that
"
c-9=a> % 5 b -9=a> % 5 b dÞ
#5
-9=a> % 5 b -9=a> % 5 b
5Ä!
#5
œ =38a> %b Þ
lim 9a> ß 5 b œ lim
5Ä!
Hence
lim 9a> ß 5 b œ =38a> %b ?% a>b .
5Ä!
a- b. The Laplace transform of the differential equation
with Ca!b œ C w a!b œ ! , is
C ww C œ $a> %b,
=# ] a=b ] a=b œ /%= .
Solving for the transform of the solution,
________________________________________________________________________
page 318
—————————————————————————— CHAPTER 6. ——
] a =b œ
It follows that the solution is
/%=
Þ
=# "
9! a>b œ =38a> %b ?% a>b .
a. b .
18a,b. The transform of the ODE agiven the specified initial conditionsb is
= ] a=b ] a=b œ " a "b5" /5 1= .
#!
#
5œ"
Solving for the transform of the solution,
#!
"
"
] a =b œ #
a "b5" /5 1= .
= " 5œ"
Applying Theorem 'Þ$Þ" , term-by-term,
Ca>b œ " a "b5" =38a> 5 1b ?51 a>b
#!
5œ"
œ =38a>b † " ?51 a>b Þ
#!
5œ"
________________________________________________________________________
page 319
—————————————————————————— CHAPTER 6. ——
a- b .
19a,b. Taking the initial conditions into consideration, the transform of the ODE is
= ] a=b ] a=b œ " /a5 1Î#b= .
#!
#
5œ"
Solving for the transform of the solution,
#!
"
"
] a =b œ #
/a5 1Î#b= .
= " 5œ"
Applying Theorem 'Þ$Þ" , term-by-term,
Ca>b œ " =38Œ>
#!
5œ"
51
?51Î# a>b Þ
#
a- b .
________________________________________________________________________
page 320
—————————————————————————— CHAPTER 6. ——
20a,b. The transform of the ODE agiven the specified initial conditionsb is
= ] a=b ] a=b œ " a "b5 +1 /a5 1Î#b= .
#!
#
5œ"
Solving for the transform of the solution,
] a=b œ " a "b5 +1
#!
5œ"
/a5 1Î#b=
.
=# "
Applying Theorem 'Þ$Þ" , term-by-term,
Ca>b œ " a "b5" =38Œ>
#!
5œ"
51
?51Î# a>b Þ
#
a- b .
22a,b. Taking the initial conditions into consideration, the transform of the ODE is
=# ] a=b ] a=b œ " a "b5" /a""5Î%b= .
%!
5œ"
Solving for the transform of the solution,
] a=b œ " a "b5"
%!
5œ"
/a""5Î%b=
.
=# "
Applying Theorem 'Þ$Þ" , term-by-term,
________________________________________________________________________
page 321
—————————————————————————— CHAPTER 6. ——
Ca>b œ " a "b5" =38Œ>
%!
5œ"
""5
?""5Î% a>b Þ
%
a- b .
23a,bÞ The transform of the ODE agiven the specified initial conditionsb is
= ] a=b !Þ"= ] a=b ] a=b œ " a "b5" /5 1= .
#!
#
5œ"
Solving for the transform of the solution,
] a =b œ "
#!
/5 1=
.
=# !Þ"= "
5œ"
First write
=#
It follows that
_" ”
"
œ
!Þ"= "
ˆ=
"
" ‰#
#!
$**
%!!
.
È$**
"
#! >Î#!
œ
/
=38
• È
#! >Þ
=# !Þ"= "
$**
Applying Theorem 'Þ$Þ" , term-by-term,
________________________________________________________________________
page 322
—————————————————————————— CHAPTER 6. ——
Ca>b œ " a "b5" 2a> 5 1b ?51 a>b ,
#!
5œ"
in which
2a>b œ
È$**
#! >Î#!
/
=38
> Þ
È$**
#!
a- b .
24a,bÞ Taking the initial conditions into consideration, the transform of the ODE is
=# ] a=b !Þ"= ] a=b ] a=b œ "/a#5"b1= .
"&
5œ"
Solving for the transform of the solution,
/a#5"b1=
.
=# !Þ"= "
5œ"
] a =b œ "
"&
As shown in Prob. #$ ,
_" ”
=#
È$**
"
#! >Î#!
/
=38
> Þ
•œ È
!Þ"= "
#!
$**
Applying Theorem 'Þ$Þ" , term-by-term,
Ca>b œ " 2c> a#5 "b1d ?a#5"b1 a>b ,
"&
5œ"
in which
________________________________________________________________________
page 323
—————————————————————————— CHAPTER 6. ——
È$**
#! >Î#!
/
=38
> Þ
È$**
#!
2a>b œ
a- b .
25a+b. A fundamental set of solutions is C" a>b œ /> -9= > and C# a>b œ /> =38 > .
Based on Prob. ## , in Section $Þ( , a particular solution is given by
C: a>b œ (
>
!
In the given problem,
C: a>b œ (
> =>
/
!
C" a=bC# a>b C" a>bC# a=b
0 a=b.= .
[ aC" ß C# ba=b
c-9=a=b=38a>b =38a=b-9=a>bd
0 a=b.= .
/B:a #=b
œ ( /a>=b =38a> =b0 a=b.= .
>
!
Given the specified initial conditions,
Ca>b œ ( /a>=b =38a> =b0 a=b.= .
>
!
a,bÞ Let 0 a>b œ $ a> 1b . It is easy to see that if > 1 , C a>b œ ! . If > 1 ,
a>=b
=38a> =b$ a= 1b.= œ /a>1b =38a> 1b .
( /
>
!
Setting > œ 1 & , and letting & p ! , we find that C a1b œ ! . Hence
________________________________________________________________________
page 324
—————————————————————————— CHAPTER 6. ——
Ca>b œ /a>1b =38a> 1b ?1 a>b Þ
a- b. The Laplace transform of the solution is
] a =b œ
œ
/ 1 =
=# #= #
/ 1 =
a = "b # "
Þ
Hence the solutions agree.
________________________________________________________________________
page 325
—————————————————————————— CHAPTER 6. ——
Section 6.6
1a+b. The convolution integral is defined as
0 ‡ 1 a>b œ ( 0 a> 7 b1a7 b. 7 Þ
>
!
Consider the change of variable ? œ > 7 . It follows that
( 0 a> 7 b1a7 b. 7 œ ( 0 a?b1a> ?ba .?b
>
!
!
>
œ ( 1a> ?b0 a?b.?
>
œ 1 ‡ 0 a>b Þ
!
a,b. Based on the distributive property of the real numbers, the convolution is also
distributive.
a- b. By definition,
0 ‡ a1 ‡ 2ba>b œ ( 0 a> 7 bc1 ‡ 2 a7 bd. 7
>
!
œ ( 0 a> 7 b”( 1a7 (b2a(b. (•. 7
>
7
!
!
œ ( ( 0 a> 7 b1a7 (b2a(b . (. 7 Þ
>
!
7
!
The region of integration, in the double integral is the area between the straight lines
( œ ! , ( œ 7 and 7 œ > Þ Interchanging the order of integration,
( ( 0 a> 7 b1a7 (b2a(b . (. 7 œ ( ( 0 a> 7 b1a7 (b2a(b . 7 . (
>
!
7
>
!
>
!
(
œ ( ”( 0 a> 7 b1a7 (b. 7 •2a(b . ( Þ
>
>
!
(
Now let 7 ( œ ? . Then
( 0 a > 7 b1 a 7 ( b . 7 œ (
>
>(
0 a> ( ?b1a?b.?
œ 0 ‡ 1 a> (b .
(
!
Hence
( 0 a> 7 bc1 ‡ 2 a7 bd. 7 œ ( c0 ‡ 1 a> 7 bd2a7 b . 7 Þ
>
!
>
!
________________________________________________________________________
page 326
—————————————————————————— CHAPTER 6. ——
2. Let 0 a>b œ /> . Then
0 ‡ " a>b œ ( />7 † " . 7
>
!
œ / > ( / 7 . 7
>
!
œ /> " .
3. It follows directly that
0 ‡ 0 a>b œ ( =38a> 7 b =38a7 b . 7
>
!
" >
( c-9=a> #7 b -9=a>bd. 7
# !
"
œ c=38a>b > -9=a>bd.
#
œ
The range of the resulting function is ‘ Þ
5. We have _c/> d œ "Îa= "b and _c=38 >d œ "Îa=# "bÞ Based on Theorem 'Þ'Þ",
_”( /a>7 b =38a7 b . 7 • œ
>
"
"
† #
=" = "
"
œ
Þ
a= "ba=# "b
!
6. Let 1a>b œ > and 2a>b œ /> . Then 0 a>b œ 1 ‡ 2 a>b . Applying Theorem 'Þ'Þ" ,
_ ” ( 1 a > 7 b2 a 7 b . 7 • œ
>
"
"
†
#
= ="
"
œ #
Þ
= a = "b
!
7. We have 0 a>b œ 1 ‡ 2 a>b , in which 1a>b œ =38 > and 2a>b œ -9= > . The transform
of the convolution integral is
_ ” ( 1 a > 7 b2 a 7 b . 7 • œ
>
!
"
=
† #
" = "
=
œ
Þ
a = # "b #
=#
9. It is easy to see that
________________________________________________________________________
page 327
—————————————————————————— CHAPTER 6. ——
_" ”
"
>
•œ/
="
and _" ”
=#
=
• œ -9= #> .
%
Applying Theorem 'Þ'Þ" ,
_" ”
>
=
œ
/a>7 b -9= #7 . 7 Þ
(
•
a= "ba=# %b
!
10. We first note that
_" –
"
a = "b
#—
œ > />
and _" ”
=#
"
"
• œ =38 #> .
%
#
Based on the convolution theorem ,
_" –
" >
a>7 b
œ
=38 #7 . 7
( a > 7 b/
—
# #
#
a= "b a= %b
!
" >
œ ( 7 /7 =38a#> #7 b . 7 Þ
# !
"
11. Let 1a>b œ _" cKa=bd. Since _" c"Îa=# "bd œ =38 > , the inverse transform of
the product is
_" ”
>
K a =b
œ
1a> 7 b =38 7 . 7
(
•
=# "
!
œ ( =38a> 7 b 1a7 b . 7 Þ
>
!
12. Taking the initial conditions into consideration, the transform of the ODE is
=# ] a=b " =# ] a=b œ Ka=b.
Solving for the transform of the solution,
] a =b œ
=#
"
K a =b
#
.
#
=
= =#
As shown in a related situation, Prob. "" ,
_" ”
K a =b
" >
œ
( =38 =a> 7 b 1a7 b . 7 Þ
•
= !
=# =#
Hence the solution of the IVP is
________________________________________________________________________
page 328
—————————————————————————— CHAPTER 6. ——
C a >b œ
"
" >
=38 => ( =38 =a> 7 b 1a7 b . 7 Þ
=
= !
14. The transform of the ODE agiven the specified initial conditionsb is
%=# ] a=b %= ] a=b "( ] a=b œ Ka=b.
Solving for the transform of the solution,
] a =b œ
%=#
K a =b
.
%= "(
First write
"
"
%
œ
Þ
"
%=# %= "(
ˆ= # ‰# %
Based on the elementary properties of the Laplace transform,
_" ”
%=#
"
" >Î#
=38 #> Þ
•œ /
%= "(
)
Applying the convolution theorem, the solution of the IVP is
" > a>7 bÎ#
C a >b œ ( /
=38 #a> 7 b 1a7 b . 7 Þ
) !
16. Taking the initial conditions into consideration, the transform of the ODE is
=# ] a=b #= $ %c= ] a=b #d % ] a=b œ Ka=b.
Solving for the transform of the solution,
] a =b œ
#= &
a= #b
#
K a= b
a= #b#
.
We can write
#= &
a= #b
#
œ
#
"
Þ
= # a= #b#
It follows that
_" ”
#
#>
• œ #/
=#
and _" –
"
a = #b
#—
œ > /#> .
Based on the convolution theorem, the solution of the IVP is
________________________________________________________________________
page 329
—————————————————————————— CHAPTER 6. ——
Ca>b œ #/#> > /#> ( a> 7 b/#a>7 b 1a7 b . 7 Þ
>
!
18. The transform of the ODE agiven the specified initial conditionsb is
=% ] a=b ] a=b œ Ka=b.
Solving for the transform of the solution,
] a =b œ
K a =b
.
=% "
First write
=%
"
"
"
"
œ ” #
#
•Þ
"
# = "
= "
It follows that
_" ”
=4
"
"
• œ c=382 > =38 >d Þ
"
#
Based on the convolution theorem, the solution of the IVP is
C a >b œ
" >
( c=382a> 7 b =38a> 7 bd1a7 b . 7 Þ
# !
19. Taking the initial conditions into consideration, the transform of the ODE is
=% ] a=b =$ &=# ] a=b &= % ] a=b œ Ka=b.
Solving for the transform of the solution,
] a =b œ
=$ &=
K a= b
#
.
#
#
a= "ba= %b a= "ba=# %b
Using partial fractions, we find that
=$ &=
"
%=
=
œ ” #
#
•,
#
#
a= "ba= %b
$ = " = %
and
a =#
"
"
"
"
œ ” #
#
•Þ
#
"ba= %b
$ = " = %
It follows that
________________________________________________________________________
page 330
—————————————————————————— CHAPTER 6. ——
_" ”
= a = # &b
%
"
œ
-9=
>
-9= #>,
•
a=# "ba=# %b
$
$
and
_" ”
"
"
"
• œ =38 > =38 #> Þ
#
"ba= %b
$
'
a =#
Based on the convolution theorem, the solution of the IVP is
%
"
" >
Ca>b œ -9= > -9= #> ( c# =38a> 7 b =38 #a> 7 bd1a7 b . 7 Þ
$
$
' !
21a+b. Let 9a>b œ ? ww a>b . Substitution into the integral equation results in
? a>b ( a> 0b ? ww a0b . 0 œ =38 #> Þ
>
ww
!
Integrating by parts,
( a> 0 b ? a 0 b . 0 œ a> 0 b ? a 0 b º
>
ww
0œ>
w
( ? w a0 b . 0
>
œ > ? a!b ?a>b ?a!b Þ
!
0œ!
!
w
Hence
? ww a>b ?a>b > ? w a!b ?a!b œ =38 #> Þ
a,b. Substituting the given initial conditions for the function ?a>b,
? ww a>b ?a>b œ =38 #> Þ
Hence the solution of the IVP is equivalent to solving the integral equation in Part a+b.
a- b. Taking the Laplace transform of the integral equation, with Fa=b œ _c9a>bd ,
Fa=b
"
#
† Fa=b œ #
Þ
#
=
= %
Note that the convolution theorem was applied. Solving for the transform Fa=b ,
Fa=b œ
#=#
Þ
a=# "ba=# %b
Using partial fractions, we can write
________________________________________________________________________
page 331
—————————————————————————— CHAPTER 6. ——
#=#
#
%
"
œ
”
•Þ
a=# "ba=# %b
$ =# % = # "
Therefore the solution of the integral equation is
9a>b œ
%
#
=38 #> =38 > .
$
$
a. b. Taking the Laplace transform of the ODE, with Y a=b œ _c?a>bd ,
= # Y a = b Y a =b œ
=#
#
Þ
%
Solving for the transform of the solution,
Y a =b œ
a =#
#
Þ
"ba=# %b
Using partial fractions, we can write
a =#
#
"
#
#
œ ” #
#
•Þ
#
"ba= %b
$ = " = %
It follows that the solution of the IVP is
?a>b œ
#
"
=38 > =38 #> Þ
$
$
We find that ? ww a>b œ #$ =38 > %$ =38 #> Þ
22a+b. First note that
,
0 aC b
"
.C œ
‡ 0 a,bÞ
( È
ÈC
,C
!
Take the Laplace transformation of both sides of the equation. Using the convolution
theorem, with J a=b œ _c0 aC bd,
X!
"
"
œ
J a= b † _ –
Þ
È#1
ÈC —
=
It was shown in Prob. #(a- b, Section 'Þ" , that
_–
"
1
œÊ Þ
—
ÈC
=
Hence
________________________________________________________________________
page 332
—————————————————————————— CHAPTER 6. ——
X!
"
1
œ
J a= b † Ê ,
È#1
=
=
with
J a =b œ Ê
#1 X!
†
Þ
1 È=
Taking the inverse transform, we obtain
0 aC b œ
a,b. Combining equations a3b and a3@b,
X! #1
Þ
1ËC
#1 X!#
.B
œ"Œ Þ
#
1 C
.C
#
Solving for the derivative .BÎ.C ,
.B
#! C
œË
,
.C
C
in which ! œ 1X!# Î1# Þ
a- b. Consider the change of variable C œ #! =38# a)Î#bÞ Using the chain rule,
.C
.)
œ #! =38a)Î#b-9=a)Î#b †
.B
.B
and
.B
"
.B
œ
†
.
.C
#! =38a)Î#b-9=a)Î#b . )
It follows that
.B
-9=# a)Î#b
œ #! =38a)Î#b-9=a)Î#bË
.)
=38# a)Î#b
œ #! -9=# a)Î#b
œ ! ! -9= ) .
Direct integration results in
Ba) b œ ! ) ! =38 ) G .
Since the curve passes through the origin, we require Ca!b œ Ba!b œ ! Þ Hence G œ !,
and Ba) b œ ! ) ! =38 ) . We also have
________________________________________________________________________
page 333
—————————————————————————— CHAPTER 6. ——
Ca) b œ #! =38# a)Î#b
œ ! ! -9= ) Þ
________________________________________________________________________
page 334
—————————————————————————— CHAPTER 7. ——
Chapter Seven
Section 7.1
1. Introduce the variables B" œ ? and B# œ ? w . It follows that B"w œ B# and
B#w œ ? ww
œ #? !Þ& ? w .
In terms of the new variables, we obtain the system of two first order ODEs
B"w œ B#
B#w œ #B" !Þ& B# Þ
3. First divide both sides of the equation by ># , and write
? ww œ
" w
"
? Œ" # ? .
>
%>
Set B" œ ? and B# œ ? w . It follows that B"w œ B# and
B#w œ ? ww
œ
" w
"
? Œ" # ? .
>
%>
We obtain the system of equations
B"w œ B#
B#w œ Œ"
"
"
B " B # Þ
#
%>
>
6. One of the ways to transform the system is to assign the variables
C" œ B" , C# œ B"w , C$ œ B# , C% œ B#w Þ
Before proceeding, note that
"
c a5" 5# bB" 5# B# J" a>bd
7"
"
B#ww œ
c5# B" a5# 5$ bB# J# a>bd Þ
7#
B"ww œ
Differentiating the new variables, we obtain the system of four first order equations
________________________________________________________________________
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—————————————————————————— CHAPTER 7. ——
C"w œ C#
"
C#w œ
c a5" 5# bC" 5# C$ J" a>bd
7"
C$w œ C%
"
C%w œ
c5# C" a5# 5$ bC$ J# a>bd Þ
7#
7a+b. Solving the first equation for B# , we have B# œ B"w #B" . Substitution into the
second equation results in
aB"w #B" b œ B" #aB"w #B" b Þ
w
That is, B"ww % B"w $ B" œ ! . The resulting equation is a second order differential
equation with constant coefficients. The general solution is
B" a>b œ -" /> -# /$> .
With B# given in terms of B" , it follows that
B# a>b œ -" /> -# /$> .
a,b. Imposing the specified initial conditions, we obtain
-" - # œ #
-" - # œ $ ,
with solution -" œ &Î# and -# œ "Î# . Hence
B" a>b œ
& > " $>
&
"
/ /
and B# a>b œ /> /$> .
#
#
#
#
a- b .
10. Solving the first equation for B# , we obtain B# œ aB" B"w bÎ# . Substitution into
________________________________________________________________________
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—————————————————————————— CHAPTER 7. ——
the second equation results in
aB" B"w b Î# œ $ B" #aB" B"w b Þ
w
Rearranging the terms, the single differential equation for B" is
B"ww $ B"w # B" œ ! .
The general solution is
B" a>b œ -" /> -# /#> .
With B# given in terms of B" , it follows that
$
B# a>b œ -" /> -# /$> .
#
Invoking the specified initial conditions, -" œ ( and -# œ ' . Hence
B" a>b œ (/> ' /#> and B# a>b œ (/> * /$> .
11. Solving the first equation for B# , we have B# œ B"w Î# . Substitution into the
second equation results in
B"ww Î# œ # B" Þ
The resulting equation is B"ww % B" œ ! , with general solution
B" a>b œ -" -9= #> -# =38 #> .
With B# given in terms of B" , it follows that
B# a>b œ -" =38 #> -# -9= #> .
Imposing the specified initial conditions, we obtain -" œ $ and -# œ % . Hence
B" a>b œ $ -9= #> %=38 #> and B# a>b œ $=38 #> %-9= #> .
________________________________________________________________________
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—————————————————————————— CHAPTER 7. ——
12. Solving the first equation for B# , we obtain B# œ B"w Î# B" Î% . Substitution into
the second equation results in
B"ww Î# B"w Î% œ # B" aB"w Î# B" Î%bÎ# Þ
Rearranging the terms, the single differential equation for B" is
B"ww B"w
The general solution is
"(
B" œ ! .
%
B" a>b œ />Î# c-" -9= #> -# =38 #>d.
With B# given in terms of B" , it follows that
B# a>b œ />Î# c -" -9= #> -# =38 #>d .
Imposing the specified initial conditions, we obtain -" œ # and -# œ # . Hence
B" a>b œ />Î# c #-9= #> #=38 #>d and B# a>b œ />Î# c#-9= #> #=38 #>d .
13. Solving the first equation for Z , we obtain Z œ P † M w . Substitution into
the second equation results in
________________________________________________________________________
page 338
—————————————————————————— CHAPTER 7. ——
P † M ww œ
M
P w
M Þ
G
VG
Rearranging the terms, the single differential equation for M is
PVG † M ww P † M w V † M œ ! .
15. Direct substitution results in
a-" B" a>b -# B# a>bbw œ :"" a>bc-" B" a>b -# B# a>bd :"# a>bc-" C" a>b -# C# a>bd
a-" C" a>b -# C# a>bbw œ :#" a>bc-" B" a>b -# B# a>bd :## a>bc-" C" a>b -# C# a>bd Þ
Expanding the left-hand-side of the first equation,
-" B"w a>b -# B#w a>b œ -" c:"" a>bB" a>b :"# a>bC" a>bd
-# c:"" a>bB# a>b :"# a>bC# a>bd Þ
Repeat with the second equation to show that the system of ODEs is identically satisfied.
16. Based on the hypothesis,
B"w a>b œ :"" a>bB" a>b :"# a>bC" a>b 1" a>b
B#w a>b œ :"" a>bB# a>b :"# a>bC# a>b 1" a>b Þ
Subtracting the two equations,
B"w a>b B#w a>b œ :"" a>bcB"w a>b B#w a>bd :"# a>bcC"w a>b C#w a>bd Þ
Similarly,
C"w a>b C#w a>b œ :#" a>bcB"w a>b B#w a>bd :## a>bcC"w a>b C#w a>bd Þ
Hence the difference of the two solutions satisfies the homogeneous ODE.
17. For rectilinear motion in one dimension, Newton's second law can be stated as
"J œ 7 B ww Þ
The </=3=>381 force exerted by a linear spring is given by J= œ 5 $ , in which $ is the
displacement of the end of a spring from its equilibrium configuration. Hence, with
! B" B# , the first two springs are in tension, and the last spring is in compression.
The sum of the spring forces on 7" is
J=" œ 5" B" 5# aB# B" b Þ
The total force on 7" is
________________________________________________________________________
page 339
—————————————————————————— CHAPTER 7. ——
"J " œ 5" B" 5# aB# B" b J" a>b .
Similarly, the total force on 7# is
"J # œ 5# aB# B" b 5$ B# J# a>b .
18a+b. Taking a clockwise loop around each of the paths, it is easy to see that voltage
drops are given by Z" Z# œ ! , and Z# Z$ œ ! .
a,b. Consider the right node. The current in is given by M" M# . The current leaving
the node is M$ . Hence the current passing through the node is aM" M# b a M$ bÞ
Based on Kirchhoff's first law, M" M# M$ œ ! .
a- b. In the capacitor,
G Z"w œ M" Þ
In the resistor,
Z# œ V M # .
In the inductor,
P M$w œ Z$ Þ
a. b. Based on part a+b, Z$ œ Z# œ Z" . Based on part a,b,
G Z"w
"
Z# M $ œ ! Þ
V
It follows that
G Z"w œ
"
Z" M$ and P M$w œ Z" Þ
V
20. Let M" ß M# ß M$ ß and M% be the current through the resistors, inductor, and capacitor,
respectively. Assign Z" ß Z# ß Z$ ß and Z% as the respective voltage drops. Based on
Kirchhoff's second law, the net voltage drops, around each loop, satisfy
Z" Z$ Z% œ ! , Z" Z$ Z# œ ! and Z% Z# œ ! .
Applying Kirchhoff's first law to the upper-right node,
M $ aM # M % b œ ! Þ
Likewise, in the remaining nodes,
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—————————————————————————— CHAPTER 7. ——
M" M$ œ ! and M# M% M" œ ! .
That is,
Z% Z# œ ! , Z" Z$ Z% œ ! and M# M% M$ œ ! .
Using the current-voltage relations,
Z" œ V" M" , Z# œ V# M# , P M$w œ Z$ , G Z%w œ M% .
Combining these equations,
V" M$ P M$w Z% œ ! and G Z%w œ M$
Z%
.
V#
Now set M$ œ M and Z% œ Z , to obtain the system of equations
P M w œ V" M Z and G Z w œ M
Z
.
V#
22a+b.
Let U" a>b and U# a>b be the amount of salt in the respective tanks at time > . Note that
the volume of each tank remains constant. Based on conservation of mass, the rate of
increase of salt, in any given tank, is given by
<+>/ 90 38-</+=/ œ <+>/ 38 <+>/ 9?> .
For Tank ", the rate of salt flowing into Tank " is
________________________________________________________________________
page 341
—————————————————————————— CHAPTER 7. ——
<38 œ ’;"
U# 9D
9D
1+6
1+6
“’$
“”
•’" 738 “
1+6
738
"!! 1+6
œ $ ;"
U# 9D
Þ
"!! 738
The rate at which salt flow out of Tank " is
<9?> œ ”
U" 9D
U" 9D
1+6
Þ
’%
“œ
•
738
'! 1+6
"& 738
Hence
.U"
U#
U"
œ $ ;"
.
.>
"!!
"&
Similarly, for Tank #,
.U#
U"
$U#
.
œ ;#
.>
$!
"!!
The process is modeled by the system of equations
U"
U#
$ ;"
"&
"!!
U"
$U#
U#w œ
;# Þ
$!
"!!
U"w œ
The initial conditions are U" a!b œ U!" and U# a!b œ U!# Þ
a,b. The equilibrium values are obtain by solving the system
U"
U#
$ ;" œ !
"&
"!!
U"
$U#
;# œ ! Þ
$!
"!!
I
Its solution leads to UI
" œ &% ;" ' ;# and U# œ '! ;" %! ;# Þ
a- b. The question refers to possible solution of the system
&% ;" ' ;# œ '!
'! ;" %! ;# œ &! Þ
It is possible for formally solve the system of equations, but the unique solution gives
;" œ
( 9D
" 9D
and ;# œ
,
' 1+6
# 1+6
which is not physically possible.
________________________________________________________________________
page 342
—————————————————————————— CHAPTER 7. ——
a. b. We can write
UI
"
'
$
UI
;# œ ;" # ,
#
%!
;# œ * ;"
which are the equations of two lines in the ;" ;# -plane:
I
The intercepts of the first line are UI
" Î&% and U" Î' . The intercepts of the second
I
line are UI
# Î'! and U# Î%! . Therefore the system will have a unique solution, in the
I
I
I
first quadrant, as long as UI
" Î&% Ÿ U# Î'! or U# Î%! Ÿ U" Î' . That is,
"!
UI
#!
.
Ÿ #I Ÿ
*
$
U"
________________________________________________________________________
page 343
—————————————————————————— CHAPTER 7. ——
Section 7.2
2a+b.
a, b .
a- b .
a. b .
3.
A #B œ Œ
" 3 #3
$ #3 %
$A B œ Œ
$ $3 3
* '3 #
" #3 '
"3
œŒ
# 3 %3
" #3
$ '3 $
$ %3
œ
' $3 #3 Π"" '3
a" 3b3 #a " #3b
a$ #3b3 #a# 3b
$ &3 ( &3
œŒ
Þ
#3
( #3
AB œ Œ
a" 3b3 $a$ #3b
BA œ Œ
#a" 3b a #3ba$ #3b
) (3 % %3
œŒ
Þ
' %3
%
Î #
"
Ï #
Î "
$
œ
Ï &
AX BX œ
( #3
Þ
# $3
'3
Þ
' &3
$a" 3b a " #3ba #3b
$a$ #3b a# 3ba #3b
a " #3b3 $a# 3b
#a " #3b a #3ba# 3b
"
# Ñ Î"
!
" #
$
" Ò Ï$
%
!Ñ
" !
% "Ò
$
"
"
#Ñ
"
! Ò
œ aA BbX Þ
4a,b.
$ #3
AœŒ
#3
"3
Þ
# $3
X
a- b. By definition, A‡ œ ˆAX ‰ œ a Ab .
5.
________________________________________________________________________
page 344
—————————————————————————— CHAPTER 7. ——
Î&
#aA Bb œ # !
Ï"
# Ñ Î "!
&
!
œ
Ò
Ï
$
#
$
#
#
'
%
%
%Ñ
"! Þ
' Ò
7. Let A œ a+34 b and B œ a,34 b . The given operations in a+b a. b are performed
elementwise. That is,
a+ b .
a, b .
a- b .
a. b .
+34 ,34 œ ,34 +34 .
+34 a,34 -34 b œ a+34 ,34 b -34 .
!a+34 ,34 b œ ! +34 ! ,34 Þ
a! " b +34 œ ! +34 ! +34 Þ
In the following, let A œ a+34 b , B œ a,34 band C œ a-34 b .
a/b. Calculating the generic element,
aBCb34 œ " ,35 -54 Þ
8
5œ"
Therefore
œ " +3< " ,<5 -54
cAaBCbd34
8
<œ"
8
8
5œ"
œ " "+3< ,<5 -54
8
<œ" 5œ"
8
8
œ " –" +3< ,<5 -54 —Þ
5œ"
<œ"
The last summation is recognized as
" +3< ,<5 œ aABb35 ,
8
<œ"
which is the 35-th element of the matrix AB Þ
a0 b. Likewise,
________________________________________________________________________
page 345
—————————————————————————— CHAPTER 7. ——
cAaB Cbd34
œ " +35 a ,54 -54 b
8
5œ"
8
œ " +35 ,54 " +35 -54
8
œ aABb34 aACb34 Þ
5œ"
8a+b.
a, b .
a- b .
a. b .
5œ"
xX y œ #a " 3b #a$3b a" 3ba$ 3b œ %3 .
yX y œ a " 3b# ## a$ 3b# œ "# )3 Þ
ax , yb œ #a " 3b #a$3b a" 3ba$ 3b œ # #3 .
ay , yb œ a " 3ba " 3b ## a$ 3ba$ 3b œ "' .
9. Indeed,
xX y œ " B4 C4 œ yX x ,
8
4œ"
and
ax , yb œ " B4 C4 œ " C4 B4 œ " C4 B4 œ ay , xb Þ
8
8
8
4œ"
4œ"
4œ"
11. First augment the given matrix by the identity matrix:
cA l I d œ Œ
$
'
"
#
" !
Þ
! "
Divide the first row by $ , to obtain
Œ
"
'
"$
#
!
Þ
"
"
$
!
Adding ' times the first row to the second row results in
"
Œ
!
"$
%
"
$
#
!
Þ
"
Divide the s/-98. row by % , to obtain
"
!
"$
"
"
$
!
"
#
"
%
Þ
Finally, adding "Î$ times the second row to the first row results in
________________________________________________________________________
page 346
—————————————————————————— CHAPTER 7. ——
"
!
"
'
!
"
"
"#
"
%
"
#
Þ
Hence
$
Œ'
"
#
"
œ
"
#
Π'
"#
13. The augmented matrix is
Î"
#
Ï"
"
"
"
"
Þ
$
" ! !Ñ
! " ! Þ
! ! "Ò
"
"
#
Combining the elements of the first row with the elements of the second and third rows
results in
Î"
!
Ï!
"
$
!
"
! !Ñ
# " ! Þ
" ! "Ò
"
$
$
Divide the elements of the second row by $ , and the elements of the third row by $ Þ
Now subtracting the new second row from the first row yields
Î"
Ð!
Ï!
!
"
!
"
$
#
$
!
"
"
"
$
"
$
!
!Ñ
! ÓÞ
" Ò
"
$
$
Finally, combine the third row with the second row to obtain
Î"
Ð!
Ï!
!
"
!
"
$
"
$
!
!
"
"
$
"
$
!
"
$
!Ñ
" Ó
Þ
$
" Ò
$
Hence
Î"
#
Ï"
"
"
"
"Ñ
"
# Ò
"
"
"Î
"
œ
$Ï
"
"
"
!
!Ñ
" Þ
"Ò
15. Elementary row operations yield
________________________________________________________________________
page 347
—————————————————————————— CHAPTER 7. ——
Î#
!
Ï!
Î"
Ð!
Ï!
"
#
!
!
"
!
!
"
#
!
"
"
!
!
!
"
!
"
%
"
#
!Ñ Î"
! pÐ!
"Ò Ï!
"
#
!
!
!
"
#
"
!
"
"
#
!
!
!
!
! Ñ Î" !
"% Óp Ð ! "
" Ò Ï
! !
#
"
%
"
#
!
!Ñ
! Óp
" Ò
"
#
#
"%
!
"
"
#
!
!
!
"
#
"
%
! Ñ
"% ÓÞ
" Ò
#
Finally, combining the 0 3<=> and >23<. rows results in
Î"
Ð!
!
"
!
Ï!
!
!
"
"
#
!
!
!
Ñ
ÓÞ
" Ò
"
$
"
"
#
%
"
#
"
%
"
)
"
%
#
16. Elementary row operations yield
Î"
#
Ï$
"
"
#
Ï!
!
"
!
Î"
Ð!
" ! !Ñ Î"
! " ! p !
! ! "Ò Ï!
"
!
"
#
$
"!
$
"
$
"
$
#
$
(
$
"
$
"
$
!Ñ Î" !
! Óp Ð ! "
"Ò Ï! !
"
$
Finally, normalizing the last row results in
Î"
Ð!
Ï!
!
"
!
"
"!
!
!
"
$
"&
(
"!
$
"!
#
&
"
"!
"
! !Ñ
# " ! p
$ ! "Ò
"
"!
!
!
$
"&
($
"!
$
$
"!
#
&
Ñ
ÓÞ
" Ò
"
"!
"
$
"
&
Ñ
ÓÞ
$ Ò
"
"!
"
&
"!
17. Elementary row operations on the augmented matrix yield the row-reduced form of
the augmented matrix
Î"
Ð!
Ï!
!
"
!
$
(
!
"
(
!
!
"
"
(
%
(
#
Ñ
ÓÞ
"Ò
#
(
"
(
The left submatrix cannot be converted to the identity matrix. Hence the given matrix is
singular.
18. Elementary row operations on the augmented matrix yield
________________________________________________________________________
page 348
—————————————————————————— CHAPTER 7. ——
Î "
Ð !
Ð
"
Ï !
!
"
!
"
!
"
"
"
" " ! ! !Ñ Î"
!
! " ! !Ó Ð!
Óp Ð
!
! ! " !
!
"
! ! ! "Ò Ï!
!
"
!
!
!
"
"
!
"
!
"
"
"
!
"
!
Î"
Ð!
Ð
!
Ï!
! !Ñ Î"
! !Ó Ð!
Óp Ð
!
" !
! "Ò Ï!
!
"
!
"
!
"
!
!
!
"
!
"
!
!
"
!
" " ! ! !Ñ
!
! " ! !Ó
Óp
" " ! " !
"
! ! ! "Ò
!
"
"
"
!
!
!
"
"
"
"
!
"
!
"
"
!
"
"
!
"Ñ
"Ó
ÓÞ
"
"Ò
19. Elementary row operations on the augmented matrix yield
Î "
Ð "
Ð
"
Ï #
"
#
!
#
#
%
"
!
!
#
$
"
!
"
!
!
!
#
"
%
#
#
"
"
#
"
#
#
Î"
Ð!
Ð
!
Ï!
" ! ! !Ñ Î"
! " ! !Ó Ð!
Óp Ð
! ! " !
!
! ! ! "Ò Ï!
"
"
"
!
"
"
"
!
! !Ñ Î" ! !
! !Ó Ð! " !
Óp Ð
! ! "
" !
! "Ò Ï! ! !
#
#
"
%
!
#
$
"
"
! ! !Ñ
"
" ! !Ó
Óp
" ! " !
#
! ! "Ò
#
%
"
&
#
$
#
"!
"
"
"
%
!Ñ
!Ó
ÓÞ
!
"Ò
!
#
"
%
Normalizing the last row and combining it with the others results in
Î"
Ð!
Ð!
Ï!
!
"
!
!
!
!
"
!
#
%
"
"
#
$
#
#
"
"
"
%&
!
#
"
%
&
! Ñ Î"
! Ó Ð
Ð!
! Óp Ð !
"& Ò Ï !
!
"
!
!
!
!
"
!
!
!
!
"
'
&
!
#
"$
&
""
&
"
&
%
&
"
&
%
&
)
&
'
&
#
&
%
&
"
&
Ñ
Ó
ÓÞ
Ó
"& Ò
20. Suppose that A is nonsingular, and that there exist matrices B and C, such that
AB œ I and AC œ I Þ Based on the properties of matrices, it follows that
AaB Cb œ AY œ 08‚8 Þ
Write the difference of the two matrices, Y , in terms of its columns as
Y œ y" l y# lâ l y8 ‘.
The 4-th column of the product matrix, AY , can be expressed as A yj . Now since all
columns of the product matrix consist only of zeros, we end up with 8 homogeneous
systems of linear equations
A yj œ 08‚" , 4 œ "ß #ß âß 8 Þ
Since A is nonsingular, each system must have a trivial solution. That is, yj œ 08‚" ,
for 4 œ "ß #ß âß 8 . Hence Y œ 08‚8 and B œ C Þ
21a+b.
________________________________________________________________________
page 349
—————————————————————————— CHAPTER 7. ——
A $B œ
œ
>
Î /
#/>
Ï />
>
Î (/
/>
Ï )/>
#/>
/>
$/>
&/>
(/>
!
/#> Ñ Î '/>
/#> $/>
#/#> Ò Ï */>
"!/#> Ñ
#/#> Þ
/#> Ò
$/>
'/>
$/>
*/#> Ñ
$/#>
$/#> Ò
a,b. Based on the standard definition of matrix multiplication,
#>
$>
Î #/ # $/
AB œ
%/#> " $/$>
Ï #/#> $ '/$>
a- b .
a. b. Note that
" %/#> />
# #/#> />
" '/#> #/>
$/$> #/> /%> Ñ
Þ
'/$> /> /%>
$>
>
%> Ò
$/ $/ #/
/>
.A Î >
œ
#/
.>
Ï />
#/>
/>
$/>
#/#> Ñ
#/#> Þ
%/#> Ò
>
Î /
#/>
( Aa>b.> œ
Ï />
#/>
/>
$/>
/#> Î# Ñ
/#> Î# C .
/#> Ò
Therefore
#/"
/# Î# Ñ Î "
Î /
#
#/
/"
/# Î#
( Aa>b.> œ
!
Ï / $/"
Ò
Ï
#
"
/
"
#
/ Î# "Î# Ñ
Î / " # #/
"
œ #/ # " /
"Î# /# Î# .
Ï " / $ $/"
/# " Ò
"
The result can also be written as
Î "
a / "b Ð #
Ï "
#
/
"
/
$
/
#
"
$
"Î# Ñ
"Î#
" Ò
"
# a / "b Ñ
#" a/ "b ÓÞ
/"
Ò
23. First note that
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—————————————————————————— CHAPTER 7. ——
"
"
$/> #> />
x w œ Œ /> #Œ ˆ/> > /> ‰ œ Œ >
.
!
"
#/ #> />
We also have
#
Œ$
"
# "
" >
#
x œŒ
/ Œ
Œ
#
$ #
!
$
#
#
œ Œ /> Œ ˆ> /> ‰
$
#
>
>
#/ #> /
œŒ >
.
$/ #> />
"
# ˆ >‰
>/
Œ
#
#
It follows that
#
Œ$
"
"
$/> #> />
>
x
/
œ
Π"
Π#/> #> /> .
#
24. It is easy to see that
'/> Ñ
'Ñ
! Ñ
Î
Î
Î
)
%
xw œ
/>
/#> œ
)/> %/#> .
Ï % Ò
Ï %Ò
Ï %/> %/#> Ò
On the other hand,
Î"
#
Ï!
"
"
"
" Ñ
" x
" Ò
œ
œ
Î"
#
Ï!
"
"
"
" ÑÎ ' Ñ
Î"
"
) /> #
Ï!
" ÒÏ % Ò
Î ' Ñ > Î ! Ñ #>
)
%
/
/ .
Ï % Ò
Ï %Ò
26. Differentiation, elementwise, results in
Î
/>
Gw œ
%/>
Ï />
#/#>
#/#>
#/#>
"
"
"
" ÑÎ ! Ñ
"
#
/#>
ÒÏ
Ò
"
#
$/$> Ñ
'/$> Þ
$/$> Ò
On the other hand,
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—————————————————————————— CHAPTER 7. ——
Î"
$
Ï#
"
#
"
% Ñ
Î"
" G œ $
Ï#
"Ò
Î
"
#
"
/>
œ
%/>
Ï />
/#>
% ÑÎ />
"
%/> /#>
" ÒÏ />
/#>
#/#> $/$> Ñ
#/#>
'/$> Þ
#/#>
$/$> Ò
/$> Ñ
#/$>
/$> Ò
________________________________________________________________________
page 352
—————————————————————————— CHAPTER 7. ——
Section 7.3
4. The augmented matrix is
Î"
#
Ï"
#
"
"
"
"
#
!Ñ
! Þ
!Ò
l
l
l
Adding # times the first row to the second row and subtracting the first row from the
third row results in
Î"
!
Ï!
#
$
$
"
$
$
!Ñ
! Þ
!Ò
l
l
l
Adding the negative of the second row to the third row results in
Î"
!
Ï!
#
$
!
"
$
!
!Ñ
! Þ
!Ò
l
l
l
We evidently end up with an equivalent system of equations
B" #B# B$ œ !
B# B$ œ ! Þ
Since there is no unique solution, let B$ œ ! , where ! is arbitrary. It follows that
B# œ ! , and B" œ ! . Hence all solutions have the form
xœ!
5. The augmented matrix is
Î "
$
Ï "
!
"
"
Î "Ñ
"
Þ
Ï " Ò
"
"
#
l
l
l
!Ñ
! Þ
!Ò
Adding $ times the first row to the second row and adding the first row to the last row
yields
Î"
!
Ï!
!
"
"
" l
$
l
"
l
!Ñ
! Þ
!Ò
Now add the negative of the second row to the third row to obtain
________________________________________________________________________
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—————————————————————————— CHAPTER 7. ——
Î"
!
Ï!
!
"
!
!Ñ
! Þ
!Ò
" l
$
l
# l
We end up with an equivalent linear system
B" B$ œ !
B# $ B$ œ !
B$ œ ! Þ
Hence the unique solution of the given system of equations is B" œ B# œ B$ œ ! Þ
7. Write the given vectors as columns of the matrix
Xœ
Î#
"
Ï!
!
"
!
"Ñ
#
Þ
Ò
!
It is evident that ./>aXb œ !. Hence the vectors are linearly dependent. In order to find
a linear relationship between them, write -" xa"b -# xa#b -$ xa$b œ 0 Þ The latter
equation is equivalent to
Î#
"
Ï!
!
"
!
" ÑÎ -" Ñ Î ! Ñ
#
-# œ ! .
! ÒÏ -$ Ò Ï ! Ò
Performing elementary row operations,
Î#
"
Ï!
!
"
!
" l
#
l
!
l
!Ñ Î"
! p !
!Ò Ï!
!
"
!
"Î#
&Î#
!
l
l
l
!Ñ
! Þ
!Ò
We obtain the system of equations
-" -$ Î# œ !
-# &-$ Î# œ ! Þ
Setting -$ œ # , it follows that -" œ " and -$ œ & Þ Hence
xa"b &xa#b #xa$b œ 0 Þ
9. The matrix containing the given vectors as columns is
Î "
Ð #
XœÐ
"
Ï !
#
$
"
"
"
!
#
#
$ Ñ
"Ó
ÓÞ
"
$ Ò
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—————————————————————————— CHAPTER 7. ——
We find that ./>aXb œ (!. Hence the given vectors are linearly independent.
10. Write the given vectors as columns of the matrix
Î "
#
Xœ
Ï #
$
"
!
#
"
"
% Ñ
$
Þ
#Ò
The four vectors are necessarily linearly dependent. Hence there are nonzero scalars
such that -" xa"b -# xa#b -$ xa$b -% xa%b œ 0 Þ The latter equation is equivalent to
Î "
#
Ï #
$
"
!
#
"
"
Performing elementary row operations,
Î "
#
Ï #
$
"
!
#
"
"
%
$
#
!
% ÑÎ " Ñ Î Ñ
Ð # Ó Ð!Ó
$
Ð Ó œ Ð Ó.
!
Ò
# Ï $Ò Ï Ò
-%
!
l !Ñ Î" ! ! " l !Ñ
l ! p ! " ! " l ! Þ
l !Ò Ï! ! " ! l !Ò
We end up with an equivalent linear system
-" - % œ !
-# - % œ !
-$ œ ! Þ
Let -% œ " . Then -" œ " and -# œ " Þ Therefore we find that
xa"b xa#b xa%b œ 0 Þ
11. The matrix containing the given vectors as columns, X , is of size 8 ‚ 7 . Since
~
8 7, we can augment the matrix with 7 8 rows of zeros. The resulting matrix, X ,
~
is of size 7 ‚ 7 . Since X is square matrix, with at least one row of zeros, it follows
~
~
that ./>ˆX‰ œ ! Þ Hence the column vectors of X are linearly dependent. That is, there
~
is a nonzero vector, c , such that X c œ 07‚" . If we write only the first 8 rows of the
latter equation, we have X c œ 08‚" . Therefore the column vectors of X are linearly
dependent.
12. By inspection, we find that
xa"b a>b # xa#b a>b œ Œ
/>
Þ
!
Hence $ xa"b a>b ' xa#b a>b xa$b a>b œ 0 , and the vectors are linearly dependent.
13. Two vectors are linearly dependent if and only if one is a nonzero scalar multiple
________________________________________________________________________
page 355
—————————————————————————— CHAPTER 7. ——
of the other. However, there is no nonzero scalar, - , such that # =38 > œ - =38 > and
=38 > œ #- =38 > for all > − a _ ß _b. Therefore the vectors are linearly independent.
16. The eigenvalues - and eigenvectors x satisfy the equation
Œ
$%
B"
!
#
œ Œ Þ
Œ
"B#
!
For a nonzero solution, we must have a$ -ba " -b ) œ ! , that is,
- # #- & œ ! Þ
The eigenvalues are -" œ " # 3 and -# œ " # 3 Þ The components of the eigenvector
xa"b are solutions of the system
Œ
# #3
%
B"
!
#
œ Œ Þ
Œ
# #3
B#
!
Œ
# #3
%
B"
!
#
œ Œ ,
Œ
# #3
B#
!
The two equations reduce to a" 3bB" œ B# . Hence xa"b œ a" ß " 3bX Þ Now setting
- œ -# œ " # 3 , we have
with solution given by xa#b œ a" ß " 3bX Þ
17. The eigenvalues - and eigenvectors x satisfy the equation
Œ
#"
B"
!
"
œ Œ Þ
Œ
#B#
!
For a nonzero solution, we must have a # -ba # -b " œ ! , that is,
- # %- $ œ ! Þ
The eigenvalues are -" œ $ and -# œ " Þ For -" œ $ , the system of equations
becomes
"
Œ"
B"
!
"
œ Œ ,
Œ
"
B#
!
which reduces to B" B# œ ! . A solution vector is given by xa"b œ a" ß "bX Þ
Substituting - œ -# œ " , we have
"
Π"
B"
!
"
œ
.
" ΠB# Π!
The equations reduce to B" œ B# . Hence a solution vector is given by xa#b œ a" ß "bX Þ
19. The eigensystem is obtained from analysis of the equation
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—————————————————————————— CHAPTER 7. ——
È$
" È
$
"-
B"
!
Œ B œ Œ ! Þ
#
For a nonzero solution, the determinant of the coefficient matrix must be zero. That is,
-# % œ ! Þ
Hence the eigenvalues are -" œ # and -# œ # Þ Substituting the first eigenvalue,
- œ # , yields
$
È
$
È$
"
B"
!
Œ B œ Œ ! Þ
#
The system is equivalent to the equation È$ B" B# œ ! . A solution vector is given
X
by xa"b œ Š" ß È$ ‹ Þ Substitution of - œ # results in
"
È
$
È$
$
B"
!
Œ B œ Œ ! ,
#
which reduces to B" œ È$ B# . A corresponding solution vector is xa#b œ ŠÈ$ ß "‹ Þ
X
20. The eigenvalues - and eigenvectors x satisfy the equation
Œ
$&
B"
!
$Î%
œ Œ Þ
Œ
"B#
!
For a nonzero solution, we must have a $ -ba" -b "&Î% œ ! , that is,
-# #- $Î% œ ! Þ
Hence the eigenvalues are -" œ $Î# and -# œ "Î# Þ In order to determine the
eigenvector corresponding to -" , set - œ $Î# . The system of equations becomes
$Î#
Π&
B"
!
$Î%
œ
,
&Î# Œ B# Œ !
&Î#
Π&
B"
!
$Î%
œ Œ ,
Œ
$Î#
B#
!
which reduces to # B" B# œ ! . A solution vector is given by xa"b œ a" ß #bX Þ
Substitution of - œ -# œ "Î# results in
which reduces to "! B" œ $ B# . A corresponding solution vector is xa#b œ a$ ß "!bX Þ
22. The eigensystem is obtained from analysis of the equation
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—————————————————————————— CHAPTER 7. ——
Î$ "
Ï #
#
ÑÎ B" Ñ Î ! Ñ
"
B# œ ! .
ÒÏ
"B$ Ò Ï ! Ò
#
%%
The characteristic equation of the coefficient matrix is -$ '-# ""- ' œ ! , with
roots -" œ " , -# œ # and -$ œ $ . Setting - œ -" œ " , we have
Î #
"
Ï #
# ÑÎ B" Ñ Î ! Ñ
"
B# œ ! .
# ÒÏ B$ Ò Ï ! Ò
#
$
%
This system is reduces to the equations
B" B$ œ !
B# œ ! Þ
A corresponding solution vector is given by xa"b œ a" ß ! ß "bX Þ Setting - œ -# œ # ,
the reduced system of equations is
B" # B# œ !
B$ œ ! Þ
A corresponding solution vector is given by xa#b œ a # ß " ß !bX Þ Finally, setting
- œ -$ œ $ , the reduced system of equations is
B" œ !
B# B$ œ ! Þ
A corresponding solution vector is given by xa$b œ a! ß " ß "bX Þ
23. For computational purposes, note that if - is an eigenvalue of B , then - - is an
eigenvalue of the matrix A œ - B Þ Eigenvectors are unaffected, since they are only
determined up to a scalar multiple. So with
Bœ
Î ""
#
Ï )
#
#
"!
) Ñ
"! ,
& Ò
the associated characteristic equation is .$ ").# )". "%&) œ ! , with
roots ." œ * , .# œ * and .$ œ ") . Hence the eigenvalues of the given matrix, A ,
are -" œ " , -# œ " and -$ œ # . Setting - œ -" œ " , Ðwhich corresponds to
using ." œ * in the modified problemÑ the reduced system of equations is
# B" B$ œ !
B# B$ œ ! Þ
A corresponding solution vector is given by xa"b œ a" ß # ß #bX Þ Setting - œ -# œ " ,
the reduced system of equations is
________________________________________________________________________
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—————————————————————————— CHAPTER 7. ——
B" # B$ œ !
B# # B$ œ ! Þ
A corresponding solution vector is given by xa#b œ a# ß # ß "bX Þ Finally, setting
- œ -# œ " , the reduced system of equations is
B" B$ œ !
# B# B$ œ ! Þ
A corresponding solution vector is given by xa$b œ a# ß " ß #bX Þ
25. Suppose that Ax œ 0 , but that x Á 0 . Let A œ a+34 bÞ Using elementary row
operations, it is possible to transform the matrix into one that is not upper triangular.
If it were upper triangular, backsubstitution would imply that x œ 0 . Hence a linear
combination of all the rows results in a row containing only zeros. That is, there are
8 scalars, "3 , one for each row and not all zero, such that for each for column 4 ,
" "3 +34 œ ! .
8
3œ"
Now consider A‡ œ a,34 b. By definition, ,34 œ +43 , or +34 œ ,43 Þ It follows that
for each 4 ,
" "3 ,43 œ " ,45 "5 œ " ,45 "5 œ ! .
8
8
8
3œ"
5œ"
5œ"
Let y œ ˆ"" ß "# ß âß "8 ‰ . We therefore have nonzero vector, y , such that A‡ y œ 0 .
X
26. By definition,
aAx ß yb œ "aAxb3 C3
8
3œ!
8
œ " "+34 B4 C3 Þ
8
3œ! 4œ!
Let ,34 œ +43 , so that +34 œ ,43 Þ Now interchanging the order or summation,
aAx ß yb œ " B4 " +34 C3
8
8
4œ!
8
3œ!
8
4œ!
3œ!
œ " B4 " ,43 C3 Þ
Now note that
________________________________________________________________________
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—————————————————————————— CHAPTER 7. ——
" ,43 C3 œ " ,43 C3 œ aA‡ yb4 .
8
8
3œ!
3œ!
Therefore
aAx ß yb œ " B4 aA‡ yb4 œ ax ß A‡ yb .
8
4œ!
28. By linearity,
Aˆxa!b ! 0 ‰ œ Axa!b ! A0
œb0
œ bÞ
29. Let -34 œ +43 . By the hypothesis, there is a nonzero vector, y , such that
" -34 C4 œ " +43 C4 œ ! , 3 œ "ß #ß âß 8 .
8
8
4œ"
4œ"
Taking the conjugate of both sides, and interchanging the indices, we have
" +34 C3 œ ! Þ
8
3œ"
This implies that a linear combination of each row of A is equal to zero. Now consider
the augmented matrix cA lbd. Replace the last row by
" C3 c+3" ß +3# ß âß +38 ß ,3 d œ –! ß ! ß âß ! ß " C3 ,3 —Þ
8
8
3œ"
3œ"
We find that if ab ß yb œ !, then the last row of the augmented matrix contains only zeros.
Hence there are 8 " remaining equations. We can now set B8 œ ! , some parameter,
and solve for the other variables in terms of ! Þ Therefore the system of equations
Ax œ b has a solution.
30. If - œ ! is an eigenvalue of A , then there is a nonzero vector, x , such that
Ax œ - x œ 0 .
That is, Ax œ 0 has a nonzero solution. This implies that the mapping defined by A is
not 1-to-1, and hence not invertible. On the other hand, if A is singular, then
./>aAb œ !Þ
Thus, Ax œ 0 has a nonzero solution. The latter equation can be written as Ax œ ! x Þ
31. As shown in Prob. #' , aAx ß yb œ ax ß A‡ yb . By definition of a Hermitian matrix,
________________________________________________________________________
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—————————————————————————— CHAPTER 7. ——
A œ A‡ Þ
32a+b. Based on Prob. $" , aAx ß xb œ ax ß Axb .
a,b. Let x be an eigenvector corresponding to an eigenvalue - Þ It then follows that
aAx ß xb œ a-x ß xb and ax ß Axb œ ax ß -xb Þ Based on the properties of the inner product,
a-x ß xb œ -ax ß xb and ax ß -xb œ -ax ß xb . Then from Part a+b,
-ax ß xb œ -ax ß xb Þ
a- b. From Part a,b,
ˆ- -‰ax ß xb œ ! Þ
Based on the definition of an eigenvector, ax ß xb œ lxl# ! Þ Hence we must have
- - œ ! , which implies that - is real.
33. From Prob. $" ,
Hence
ˆAxa"b ß xa#b ‰ œ ˆxa"b ß Axa#b ‰ .
-" ˆxa"b ß xa#b ‰ œ -# ˆxa"b ß xa#b ‰ œ -# ˆxa"b ß xa#b ‰ ,
since the eigenvalues are real. Therefore
a-" -# bˆxa"b ß xa#b ‰ œ ! .
Given that -" Á -# , we must have ˆxa"b ß xa#b ‰ œ ! .
________________________________________________________________________
page 361
—————————————————————————— CHAPTER 7. ——
Section 7.4
3. Eq. a"%b states that the Wronskian satisfies the first order linear ODE
.[
œ a:"" :## â :88 b[ Þ
.>
The general solution is
[ a>b œ G /B:”( a:"" :## â :88 b.> •,
in which G is an arbitrary constant. Let X" and X# be matrices representing two sets
of fundamental solutions. It follows that
./>aX" b œ [" a>b œ G" /B:”( a:"" :## â :88 b.> •
./>aX# b œ [# a>b œ G# /B:”( a:"" :## â :88 b.> • Þ
Hence ./>aX" bÎ./>aX# b œ G" ÎG# . Note that G# Á !Þ
4. First note that :"" :## œ :a>bÞ As shown in Prob. a$b,
[ xa"b ß xa#b ‘ œ - /
' :a>b.>
Þ
For second order linear ODE, the Wronskian aas defined in Chap. $b satisfies the first
order differential equation [ w :a>b[ œ ! . It follows that
[ Ca"b ß Ca#b ‘ œ -" /
' :a>b.>
Þ
Alternatively, based on the hypothesis,
Ca"b œ !"" B"" !"# B"#
Ca#b œ !#" B"" !## B"# Þ
Direct calculation shows that
! B !"# B"# !#" B"" !## B"#
[ Ca"b ß Ca#b ‘ œ º "" ""w
!"" B"" !"# B"#w !#" B""w !## B"#w º
œ a!"" !## !"# !#" bB"" B"#w a!"" !## !"# !#" bB"# B""w
œ a!"" !## !"# !#" bB"" B## a!"" !## !"# !#" bB"# B#" Þ
Here we used the fact that B"w œ B# . Hence
[ Ca"b ß Ca#b ‘ œ a!"" !## !"# !#" b[ xa"b ß xa#b ‘ Þ
w
5. The particular solution satisfies the ODE xa:b ‘ œ Pa>bxa:b ga>b Þ Now let
________________________________________________________________________
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—————————————————————————— CHAPTER 7. ——
x œ 9a>b be any solution of the homogeneous equation. That is, x w œ Pa>bx . We know
that x œ x- , in which x- is a linear combination of some fundamental solution. By
linearity of the differential equation, it follows that x œ xa:b x- is a solution of the
ODE. Based on the uniqueness theorem, all solutions must have this form.
7a+b. By definition,
[ xa"b ß xa#b ‘ œ º
># />
ˆ#
‰ >
> º œ > #> / Þ
#> /
a,b. The Wronskian vanishes at >! œ ! and >! œ # . Hence the vectors are linearly
independent on W œ a _ ß !b a! ß #b a# ß _bÞ
a- b. It follows from Theorem (Þ%Þ$ that one or more of the coefficients of the ODE
must be discontinuous at >! œ ! and >! œ # . If not, the Wronskian would not vanish.
a. b. Let
>#
/>
x œ -" Œ - # Œ > Þ
#>
/
Then
x w œ -" Œ
#>
/>
# Œ > Þ
#
/
On the other hand,
:""
Œ:
#"
>#
/>
:"#
:
:"#
:
:"#
-# Π""
x œ -" Œ ""
Œ
Œ
:##
:#" :##
:#" :##
#>
/>
-" c:"" ># #:"# >d -# c:"" :"# d/>
œŒ
Þ
-" c:#" ># #:## >d -# c:#" :## d/>
Comparing coefficients, we find that
:"" ># #:"# > œ #>
:"" :"# œ "
#
:#" > #:## > œ #
:#" :## œ " Þ
Solution of this system of equations results in
# #>
># #
:"" a>b œ ! ß :"# a>b œ " ß :#" a>b œ #
ß :## a>b œ #
Þ
> #>
> #>
Hence the vectors are solutions of the ODE
________________________________________________________________________
page 363
—————————————————————————— CHAPTER 7. ——
xw œ
"
!
Œ
#
> #> # #>
># #>
x Þ
># #
8. Suppose that the solutions xa"b , xa#b ,â, xa7b are linearly dependent at > œ >! . Then
there are constants -" ß -# ß âß -7 anot all zerob such that
-" xa"b a>! b -# xa#b a>! b â -7 xa7b a>! b œ 0 .
Now let za>b œ -" xa"b a>b -# xa#b a> b â -7 xa7b a> b . Then clearly, za>b is a
solution of x w œ Pa>bx , with za>! b œ ! . Furthermore, ya>b ´ 0 is also a solution,
with ya>! b œ ! . By the uniqueness theorem, za>b œ ya>b œ 0 Þ Hence
-" xa"b a> b -# xa#b a> b â -7 xa7b a> b œ 0
on the entire interval ! > " . Going in the other direction is trivial.
9a+b. Let ya>b be any solution of x w œ Pa>bx . It follows that
za>b ya>b œ -" xa"b a> b -# xa#b a> b â -8 xa8b a> b ya>b
is also a solution. Now let >! − a! ß " b . Then the collection of vectors
xa"b a>! bß xa#b a>! bß âß xa8b a>! bß ya>! b
constitutes 8 " vectors, each with 8 components. Based on the assertion in Prob. "",
Section (Þ$ , these vectors are necessarily linearly dependent. That is, there are 8 "
constants ," ß ,# ß âß ,8 ß ,8" anot all zerob such that
," xa"b a>! b ,# xa#b a>! b â ,8 xa8b a>! b ,8" ya>! b œ 0 .
From Prob. ), we have
," xa"b a> b ,# xa#b a> b â ,8 xa8b a> b ,8" ya> b œ 0
for all > − a! ß " bÞ Now ,8" Á ! , otherwise that would contradict the fact that the
first 8 vectors are linearly independent. Hence
ya> b œ
"
,8"
ˆ," xa"b a> b ,# xa#b a> b â ,8 xa8b a> b‰,
and the assertion is true.
a,b. Consider za>b œ -" xa"b a> b -# xa#b a> b â -8 xa8b a> b, and suppose that we also
have
za>b œ 5" xa"b a> b 5# xa#b a> b â 58 xa8b a> b Þ
Based on the assumption,
________________________________________________________________________
page 364
—————————————————————————— CHAPTER 7. ——
a5" -" bxa"b a> b a5# -# bxa#b a> b â a58 -8 bxa8b a> b œ 0 .
The collection of vectors
xa"b a> bß xa#b a> bß âß xa8b a> b
is linearly independent on ! > " . It follows that 53 -3 œ ! , for 3 œ "ß #ß âß 8 .
________________________________________________________________________
page 365
—————————————————————————— CHAPTER 7. ——
Section 7.5
2. Setting x œ 0 /<> , and substituting into the ODE, we obtain the algebraic equations
Œ
"<
$
0"
!
#
œ Œ .
Œ
%<
0#
!
For a nonzero solution, we must have ./>aA < Ib œ <# $< # œ ! . The roots of the
characteristic equation are <" œ " and <# œ # . For < œ " , the two equations
reduce to 0" œ 0# . The corresponding eigenvector is 0 a"b œ a" ß "bX Þ Substitution of
< œ # results in the single equation $ 0" œ # 0# . A corresponding eigenvector is
0 a#b œ a# ß $bX Þ Since the eigenvalues are distinct, the general solution is
"
#
x œ -" Œ /> -# Œ /#> Þ
"
$
3. Setting x œ 0 /<> results in the algebraic equations
Œ
#<
$
0"
!
"
œ Œ .
Œ
#<
0#
!
For a nonzero solution, we must have ./>aA < Ib œ <# " œ ! . The roots of the
characteristic equation are <" œ " and <# œ " . For < œ " , the system of equations
reduces to 0" œ 0# . The corresponding eigenvector is 0 a"b œ a" ß "bX Þ Substitution of
< œ " results in the single equation $ 0" œ 0# . A corresponding eigenvector is
0 a#b œ a" ß $bX Þ Since the eigenvalues are distinct, the general solution is
"
"
x œ -" Œ /> -# Œ /> Þ
"
$
________________________________________________________________________
page 366
—————————————————————————— CHAPTER 7. ——
The system has an unstable eigendirection along 0 a"b œ a" ß "bX Þ Unless -" œ ! , all
solutions will diverge.
4. Solution of the ODE requires analysis of the algebraic equations
Œ
"<
%
0"
!
"
œ
.
# < Π0# Π!
For a nonzero solution, we must have ./>aA < Ib œ <# < ' œ ! . The roots of the
characteristic equation are <" œ # and <# œ $ . For < œ # , the system of equations
reduces to 0" œ 0# . The corresponding eigenvector is 0 a"b œ a" ß "bX Þ Substitution of
< œ $ results in the single equation % 0" 0# œ ! . A corresponding eigenvector is
0 a#b œ a" ß %bX Þ Since the eigenvalues are distinct, the general solution is
"
"
$>
x œ -" Œ /#> -# Œ
/ Þ
"
%
The system has an unstable eigendirection along 0 a"b œ a" ß "bX Þ Unless -" œ ! , all
solutions will diverge.
8. Setting x œ 0 /<> results in the algebraic equations
________________________________________________________________________
page 367
—————————————————————————— CHAPTER 7. ——
$<
Π"
0"
!
'
œ Œ .
Œ
#<
0#
!
For a nonzero solution, we must have ./>aA < Ib œ <# < œ ! . The roots of the
characteristic equation are <" œ " and <# œ ! . With < œ " , the system of equations
reduces to 0" $0# œ !. The corresponding eigenvector is 0 a"b œ a$ ß "bX Þ For the
case < œ !, the system is equivalent to the equation 0" # 0# œ ! . An eigenvector is
0 a#b œ a# ß "bX Þ Since the eigenvalues are distinct, the general solution is
x œ -" Œ
$
#
>
/ - # Œ
Þ
"
"
The entire line along the eigendirection 0 a#b œ a# ß "bX consists of equilibrium points.
All other solutions diverge. The direction field changes across the line B" #B# œ ! Þ
Eliminating the exponential terms in the solution, the trajectories are given by
B" $ B# œ -# .
10. The characteristic equation is given by
#<
º "
#3
œ <# a" 3b< 3 œ ! Þ
"3<º
The equation has complex roots <" œ " and <# œ 3. For < œ ", the components of the
solution vector must satisfy 0" a# 3b0# œ ! . Thus the corresponding eigenvector is
0 a"b œ a# 3 ß "bX Þ Substitution of < œ 3 results in the single equation 0" 0# œ !.
A corresponding eigenvector is 0 a#b œ a" ß "bX Þ Since the eigenvalues are distinct, the
general solution is
x œ -" Œ
#3 >
"
3>
/ - # Œ
/ Þ
"
"
11. Setting x œ 0 /<> results in the algebraic equations
________________________________________________________________________
page 368
—————————————————————————— CHAPTER 7. ——
Î" <
"
Ï #
"
#<
"
Î $
"
Ï #
"
#
"
# ÑÎ 0" Ñ Î ! Ñ
"
0# œ ! .
ÒÏ
"<
0$ Ò Ï ! Ò
For a nonzero solution, we must have ./>aA < Ib œ <$ %<# < % œ ! . The roots
of the characteristic equation are <" œ % , <# œ " and <$ œ " . Setting < œ % , we
have
# ÑÎ 0" Ñ Î ! Ñ
"
0# œ ! .
Ò
Ï
$
0$ Ò Ï ! Ò
This system is reduces to the equations
0" 0 $ œ !
0# 0 $ œ ! Þ
A corresponding solution vector is given by 0 a"b œ a" ß " ß "bX Þ Setting - œ " ,
the reduced system of equations is
0" 0 $ œ !
0# # 0 $ œ ! Þ
A corresponding solution vector is given by 0 a#b œ a" ß # ß "bX Þ Finally, setting
- œ " , the reduced system of equations is
0" 0 $ œ !
0# œ ! Þ
A corresponding solution vector is given by 0 a$b œ a" ß ! ß "bX Þ Since the eigenvalues
are distinct, the general solution is
x œ -"
Î " Ñ %>
Î " Ñ
Î " Ñ >
" / -# # / > - $
!
/ Þ
Ï"Ò
Ï " Ò
Ï "Ò
12. The eigensystem is obtained from analysis of the equation
Î$ <
#
Ï %
#
<
#
% ÑÎ 0" Ñ Î ! Ñ
#
0# œ ! .
$ < ÒÏ 0$ Ò Ï ! Ò
The characteristic equation of the coefficient matrix is <$ '<# "&< ) œ ! , with
roots <" œ ) , <# œ " and <$ œ " . Setting < œ <" œ ) , we have
________________________________________________________________________
page 369
—————————————————————————— CHAPTER 7. ——
Î &
#
Ï %
% ÑÎ 0" Ñ Î ! Ñ
#
0# œ ! .
Ò
Ï
&
0$ Ò Ï ! Ò
#
)
#
This system is reduced to the equations
0" 0 $ œ !
# 0# 0 $ œ ! Þ
A corresponding solution vector is given by 0 a"b œ a# ß " ß #bX Þ Setting < œ " , the
system of equations is reduced to the single equation
# 0" 0 # # 0 $ œ ! Þ
Two independent solutions are obtained as
0 a2b œ a" ß # ß !bX and 0 a$b œ a! ß # ß "bX Þ
Hence the general solution is
Î # Ñ )>
Î " Ñ >
Î ! Ñ >
x œ -" " / - # # / - $ # / Þ
Ï#Ò
Ï ! Ò
Ï " Ò
13. Setting x œ 0 /<> results in the algebraic equations
Î" <
#
Ï )
"
ÑÎ 0" Ñ Î ! Ñ
"
0# œ ! .
ÒÏ
$<
0$ Ò Ï ! Ò
"
"<
&
For a nonzero solution, we must have ./>aA < Ib œ <$ <# %< % œ ! . The roots
of the characteristic equation are <" œ # , <# œ # and <$ œ " . Setting < œ # , we
have
Î "
#
Ï )
" ÑÎ 0" Ñ Î ! Ñ
"
0# œ ! .
& ÒÏ 0$ Ò Ï ! Ò
"
"
&
This system is reduces to the equations
0" œ !
0# 0 $ œ ! Þ
A corresponding solution vector is given by 0 a"b œ a! ß " ß "bX Þ Setting - œ " ,
the reduced system of equations is
# 0" $ 0 $ œ !
0# # 0 $ œ ! Þ
________________________________________________________________________
page 370
—————————————————————————— CHAPTER 7. ——
A corresponding solution vector is given by 0 a#b œ a$ ß % ß #bX Þ Finally, setting
- œ # , the reduced system of equations is
( 0" % 0 $ œ !
(0# & 0$ œ ! Þ
A corresponding solution vector is given by 0 a$b œ a% ß & ß (bX Þ Since the
eigenvalues are distinct, the general solution is
x œ -"
Î ! Ñ #>
Î $ Ñ
Î % Ñ
"
/ -# % /> -$ & /#> Þ
Ï "Ò
Ï #Ò
Ï (Ò
15. Setting x œ 0 /<> results in the algebraic equations
Œ
&<
$
0"
!
"
œ Œ .
Œ
"<
0#
!
For a nonzero solution, we must have ./>aA < Ib œ <# '< ) œ ! . The roots of
the characteristic equation are <" œ % and <# œ # . With < œ % , the system of equations
reduces to 0" 0# œ !. The corresponding eigenvector is 0 a"b œ a" ß "bX Þ For the
case < œ # , the system is equivalent to the equation $ 0" 0# œ ! . An eigenvector is
0 a#b œ a" ß $bX Þ Since the eigenvalues are distinct, the general solution is
"
"
x œ -" Œ /%> -# Œ /#> Þ
"
$
Invoking the initial conditions, we obtain the system of equations
-" - # œ #
-" $ - # œ " Þ
Hence -" œ (Î# and -# œ $Î# , and the solution of the IVP is
xœ
( " %> $ " #>
Œ / Œ / Þ
# "
# $
17. Setting x œ 0 /<> results in the algebraic equations
Î" <
!
Ï "
"
#<
"
# ÑÎ 0" Ñ Î ! Ñ
#
0# œ ! .
$ < ÒÏ 0$ Ò Ï ! Ò
For a nonzero solution, we must have ./>aA < Ib œ <$ '<# ""< ' œ ! . The
roots of the characteristic equation are <" œ " , <# œ # and <$ œ $ . Setting < œ " ,
we have
________________________________________________________________________
page 371
—————————————————————————— CHAPTER 7. ——
Î !
!
Ï "
" # ÑÎ 0" Ñ Î ! Ñ
" #
0# œ ! .
ÒÏ
" #
0$ Ò Ï ! Ò
This system is reduces to the equations
0" œ !
0# # 0 $ œ ! Þ
A corresponding solution vector is given by 0 a"b œ a! ß # ß "bX Þ Setting - œ # , the
reduced system of equations is
0" 0 # œ !
0$ œ ! Þ
A corresponding solution vector is given by 0 a#b œ a" ß " ß !bX Þ Finally, upon setting
- œ $ , the reduced system of equations is
0" # 0 $ œ !
0# # 0 $ œ ! Þ
A corresponding solution vector is given by 0 a$b œ a# ß # ß "bX Þ Since the eigenvalues
are distinct, the general solution is
Î ! Ñ >
Î " Ñ #>
Î # Ñ $>
x œ -" # / - # " / - $ # / Þ
Ï " Ò
Ï!Ò
Ï"Ò
Invoking the initial conditions, the coefficients must satisfy the equations
-# # - $ œ #
# -" - # # - $ œ !
-" - $ œ " Þ
It follows that -" œ " , -# œ # and -$ œ ! . Hence the solution of the IVP is
xœ
Î ! Ñ >
Î"Ñ
# / # " /#> Þ
Ï " Ò
Ï!Ò
18. The eigensystem is obtained from analysis of the equation
Î <
#
Ï "
!
<
#
" ÑÎ 0" Ñ Î ! Ñ
!
0# œ ! .
% < ÒÏ 0$ Ò Ï ! Ò
The characteristic equation of the coefficient matrix is <$ %<# < % œ ! , with
roots <" œ " , <# œ " and <$ œ % . Setting < œ <" œ " , we have
________________________________________________________________________
page 372
—————————————————————————— CHAPTER 7. ——
Î "
#
Ï "
!
"
#
" ÑÎ 0" Ñ Î ! Ñ
!
0# œ ! .
Ò
Ï
$
0$ Ò Ï ! Ò
This system is reduced to the equations
0" 0 $ œ !
0# # 0 $ œ ! Þ
A corresponding solution vector is given by 0 a"b œ a" ß # ß "bX Þ Setting < œ " , the
system reduces to the equations
0" 0 $ œ !
0# # 0 $ œ ! Þ
The corresponding eigenvector is 0 a2b œ a" ß # ß "bX Þ Finally, upon setting < œ % ,
the system is equivalent to the equations
% 0" 0 $ œ !
) 0# 0 $ œ ! Þ
The corresponding eigenvector is 0 a$b œ a# ß " ß )bX Þ Hence the general solution is
x œ -"
Î " Ñ >
Î " Ñ >
Î # Ñ %>
# / -#
#
"
/ -$
/ Þ
Ï " Ò
Ï "Ò
Ï )Ò
Invoking the initial conditions,
-" - # # - $ œ (
# -" # - # - $ œ &
-" - # ) - $ œ & Þ
It follows that -" œ $ , -# œ ' and -$ œ " . Hence the solution of the IVP is
Î " Ñ >
Î " Ñ > Î # Ñ %>
#
"
xœ$ # / '
/
/ Þ
Ï " Ò
Ï "Ò
Ï )Ò
19. Set x œ 0 >< . Substitution into the system of differential equations results in
> † <><" 0 œ A 0 >< ,
which upon simplification yields is, A 0 <0 œ 0 Þ Hence the vector 0 and constant <
must satisfy aA <Ib0 œ 0 Þ
21. Setting x œ 0 >< results in the algebraic equations
________________________________________________________________________
page 373
—————————————————————————— CHAPTER 7. ——
Œ
&<
$
0"
!
"
œ Œ .
Œ
"<
0#
!
For a nonzero solution, we must have ./>aA < Ib œ <# '< ) œ ! . The roots of
the characteristic equation are <" œ % and <# œ # . With < œ % , the system of equations
reduces to 0" 0# œ !. The corresponding eigenvector is 0 a"b œ a" ß "bX Þ For the
case < œ # , the system is equivalent to the equation $ 0" 0# œ ! . An eigenvector is
0 a#b œ a" ß $bX Þ It follows that
"
"
xa"b œ Œ >% and xa#b œ Œ ># Þ
"
$
The Wronskian of this solution set is [ cxa"b ß xa#b d œ #>' . Thus the solutions are linearly
independent for > ! . Hence the general solution is
"
"
x œ - " Œ >% - # Œ > # Þ
"
$
22. As shown in Prob. "* , solution of the ODE requires analysis of the equations
Œ
%<
)
0"
!
$
œ Œ .
Œ
'<
0#
!
For a nonzero solution, we must have ./>aA < Ib œ <# #< œ ! . The roots of the
characteristic equation are <" œ ! and <# œ # . For < œ ! , the system of equations
reduces to % 0" œ $ 0# . The corresponding eigenvector is 0 a"b œ a$ ß %bX Þ Setting
< œ # results in the single equation # 0" 0# œ ! . A corresponding eigenvector is
0 a#b œ a" ß #bX Þ It follows that
$
"
xa"b œ Œ and xa#b œ Œ ># Þ
%
#
The Wronskian of this solution set is [ cxa"b ß xa#b d œ # ># . These solutions are
linearly independent for > ! . Hence the general solution is
$
"
x œ -" Œ -# Œ ># Þ
%
#
23. Setting x œ 0 >< results in the algebraic equations
Œ
$<
#
0"
!
#
œ
.
# < Π0# Π!
For a nonzero solution, we must have ./>aA < Ib œ <# < # œ ! . The roots of
the characteristic equation are <" œ # and <# œ " . Setting < œ # , the system of
equations reduces to 0" # 0# œ !. The corresponding eigenvector is 0 a"b œ a# ß "bX Þ
________________________________________________________________________
page 374
—————————————————————————— CHAPTER 7. ——
With < œ " , the system is equivalent to the equation # 0" 0# œ ! . An eigenvector
is 0 a#b œ a" ß #bX Þ It follows that
#
"
xa"b œ Œ ># and xa#b œ Œ >" Þ
"
#
The Wronskian of this solution set is [ cxa"b ß xa#b d œ $ >. Thus the solutions are linearly
independent for > ! . Hence the general solution is
#
"
x œ -" Œ ># -# Œ >" Þ
"
#
24a+b. The general solution is
x œ -" Œ
" >
" #>
/ - # Œ / Þ
#
#
a, b .
a- b .
________________________________________________________________________
page 375
—————————————————————————— CHAPTER 7. ——
26a+b. The general solution is
a, b .
x œ -" Œ
" >
" #>
/ - # Œ / Þ
#
#
________________________________________________________________________
page 376
—————————————————————————— CHAPTER 7. ——
a, b .
a- b .
28a+b. We note that aA <3 Ib0 a3b œ 0 , for 3 œ "ß # .
a,b. It follows that aA <# Ib0 a"b œ A 0 a"b <# 0 a"b œ <" 0 a"b <# 0 a"b .
a- b. Suppose that 0 a"b and 0 a#b are linearly dependent. Then there exist constants -"
and -# , not both zero, such that -" 0 a"b -# 0 a#b œ 0 Þ Assume that -" Á ! Þ It is clear
that aA <# Ibˆ-" 0 a"b -# 0 a#b ‰ œ 0 Þ On the other hand,
aA <# Ibˆ-" 0 a"b -# 0 a#b ‰ œ -" a<" <# b0 a"b 0
œ -" a<" <# b0 a"b .
Since <" Á <# , we must have -" œ ! , which leads to a contradiction.
a. b. Note that aA <" Ib0 a#b œ a<# <" b0 a#b .
________________________________________________________________________
page 377
—————————————————————————— CHAPTER 7. ——
a/b. Let 8 œ $, with <" Á <# Á <$ . Suppose that 0 a"b , 0 a#b and 0 a$b are indeed linearly
dependent. Then there exist constants -" , -# and -$ , not all zero, such that
-" 0 a"b -# 0 a#b -$ 0 a$b œ 0 Þ
Assume that -" Á ! Þ It is clear that aA <# Ibˆ-" 0 a"b -# 0 a#b -$ 0 a$b ‰ œ 0 Þ On the
other hand,
aA <# Ibˆ-" 0 a"b -# 0 a#b -$ 0 a$b ‰ œ -" a<" <# b0 a"b -$ a<$ <# b0 a$b Þ
It follows that -" a<" <# b0 a"b -$ a<$ <# b0 a$b œ 0 Þ Based on the result of Part a+b,
which is actually not dependent on the value of 8 , the vectors 0 a"b and 0 a$b are linearly
independent. Hence we must have -" a<" <# b œ -$ a<$ <# b œ ! , which leads to
a contradiction.
29a+b. Let B" œ C and B# œ C w . It follows that B"w œ B# and
B#w œ C ww
"
œ a- C , C w b.
+
In terms of the new variables, we obtain the system of two first order ODEs
B"w œ B#
"
B#w œ a- B" , B# b Þ
+
a,b. The coefficient matrix is given by
AœŒ
!
+-
"
Þ
+,
Setting x œ 0 /<> results in the algebraic equations
<
Œ+
"
,
+
0"
!
.
œ
< Π0# Π!
For a nonzero solution, we must have
,
./>aA < Ib œ <# < œ ! .
+
+
Multiplying both sides of the equation by + , we obtain + <# , < - œ ! Þ
30. Solution of the ODE requires analysis of the algebraic equations
Œ
"<
%
0"
!
"
.
œ
# < Π0# Π!
For a nonzero solution, we must have ./>aA < Ib œ ! . The characteristic equation is
________________________________________________________________________
page 378
—————————————————————————— CHAPTER 7. ——
)! <# #% < " œ ! , with roots <" œ "Î% and <# œ "Î#! . With < œ "Î% ,
the system of equations reduces to # 0" 0# œ ! . The corresponding eigenvector is
0 a"b œ a" ß #bX Þ Substitution of < œ "Î#! results in the equation # 0" $ 0# œ ! .
A corresponding eigenvector is 0 a#b œ a$ ß #bX Þ Since the eigenvalues are distinct, the
general solution is
x œ -" Œ
"
$
>Î%
-# Œ />Î#! Þ
/
#
#
Invoking the initial conditions, we obtain the system of equations
-" $ -# œ "(
# -" # -# œ #" Þ
Hence -" œ #*Î) and -# œ &&Î) , and the solution of the IVP is
xœ
#*
"
&& $
>Î%
Œ />Î#! Þ
Œ
/
) #
) #
a, b .
a- b. Both functions are monotone increasing. It is easy to show that !Þ& Ÿ B" a>b !
and !Þ& Ÿ B# a>b ! provided that > X ¸ (%Þ$* Þ
31a+b. For ! œ "Î# , solution of the ODE requires that
"<
Œ "Î#
0"
!
"
œ Œ .
Œ
"<
0#
!
The characteristic equation is # <# % < " œ ! , with roots <" œ " "ÎÈ# and
<# œ " "ÎÈ# . With < œ " "ÎÈ# , the system of equations reduces to
X
È# 0" # 0# œ ! . The corresponding eigenvector is 0 a"b œ Š È# ß "‹ Þ Substitution
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—————————————————————————— CHAPTER 7. ——
of < œ " "ÎÈ# results in the equation È# 0" # 0# œ ! . An eigenvector is
X
0 a#b œ ŠÈ# ß "‹ Þ The general solution is
x œ -"
È#
È#
Š#È#‹>Î#
Š#È#‹>Î#
/
/
Þ
#
"
"
The eigenvalues are distinct and both 8/1+>3@/Þ The equilibrium point is a stable node.
a,b. For ! œ # , the characteristic equation is given by <# # < " œ ! , with roots
<" œ " È# and <# œ " È# . With < œ " È# , the system of equations
X
reduces to È# 0" 0# œ ! . The corresponding eigenvector is 0 a"b œ Š" ß È#‹ Þ
Substitution of < œ " È# results in the equation È# 0" 0# œ ! . An eigenvector
X
is 0 a#b œ Š" ß È#‹ Þ The general solution is
x œ -" Œ
"
"
Š"È#‹>
Š"È#‹>
- # Œ È /
Þ
/
È
#
#
The eigenvalues are of opposite sign, hence the equilibrium point is a saddle point.
32. The system of differential equations is
"#
. M
œ
Π$
.> Z
#
"#
M
Þ
& Œ
Z
#
Solution of the system requires analysis of the eigenvalue problem
"
#
$
#
<
&
#
!
0"
Œ œ Œ .
!
0
<
"
#
#
The characteristic equation is <# $ < #, with roots <" œ " and <# œ #. With
< œ " , the equations reduce to 0" 0# œ ! . A corresponding eigenvector is given
by 0 a"b œ a" ß "bX Þ Setting < œ # , the system reduces to the equation $ 0" 0# œ ! .
An eigenvector is 0 a#b œ a" ß $bX Þ Hence the general solution is
Œ
M
" >
" #>
œ -" Œ / - # Œ / Þ
Z
"
$
a,bÞ The eigenvalues are distinct and both 8/1+>3@/Þ We find that the equilibrium point
a! ß !b is a stable node. Hence all solutions converge to a! ß !bÞ
33a+b. Solution of the ODE requires analysis of the algebraic equations
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—————————————————————————— CHAPTER 7. ——
V"
P
"
G
<
!
0"
Œ œ Œ .
<
!
0
"
P
"
GV#
#
The characteristic equation is
<# Œ
P GV" V#
V" V #
œ !Þ
<
PGV#
PGV#
The eigenvectors are real and distinct, provided that the discriminant is positive.
That is,
P GV" V#
V" V #
Œ
%Œ
!,
PGV#
PGV#
#
which simplifies to the condition
"
V"
%
!.
Œ
GV#
P
PG
#
a,b. The parameters in the ODE are all positive. Observe that the sum of the roots is
P GV" V#
!Þ
PGV#
Also, the product of the roots is
V" V #
!Þ
PGV#
It follows that both roots are negative. Hence the equilibrium solutio8 M œ ! ß Z œ !
represents a stable node, which attracts all solutions.
a- b. If the condition in Part a+b is not satisfied, that is,
"
V"
%
Ÿ !,
Œ
GV#
P
PG
#
then the real part of the eigenvalues is
V/a<"ß# b œ
P GV" V#
Þ
# PGV#
As long as the parameters are all positive, then the solutions will still converge to the
equilibrium point a! ß !b.
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page 381
—————————————————————————— CHAPTER 7. ——
Section 7.6
2. Setting x œ 0 /<> results in the algebraic equations
Œ
"<
"
0"
!
%
œ Œ .
Œ
"<
0#
!
For a nonzero solution, we require that ./>aA < Ib œ <# #< & œ !. The roots of
the characteristic equation are < œ " „ #3 . Substituting < œ " #3 , the two
equations reduce to 0" #3 0# œ ! . The two eigenvectors are 0 a"b œ a #3 ß "bX and
0 a#b œ a#3 ß "bX Þ Hence one of the complex-valued solutions is given by
xa"b œ Œ
#3 a"#3b>
/
"
#3 >
œŒ
/ a-9= #> 3 =38 #>b
"
# =38 #>
# -9= #>
>
œ /> Œ
3/ Œ
Þ
-9= #>
=38 #>
Based on the real and imaginary parts of this solution, the general solution is
x œ -" /> Œ
# =38 #>
> # -9= #>
-# / Œ
Þ
-9= #>
=38 #>
3. Solution of the ODEs is based on the analysis of the algebraic equations
Œ
#<
"
0"
!
&
œ Œ .
Œ
#<
0#
!
For a nonzero solution, we require that ./>aA < Ib œ <# " œ !. The roots of the
characteristic equation are < œ „3 . Setting < œ 3 , the equations are equivalent to
0" a# 3b0# œ ! . The eigenvectors are 0 a"b œ a# 3 ß "bX and 0 a#b œ a# 3 ß "bX Þ
Hence one of the complex-valued solutions is given by
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—————————————————————————— CHAPTER 7. ——
xa"b œ Œ
# 3 3>
/
"
#3
œŒ
a-9= > 3 =38 >b
"
# -9= > =38 >
-9= > # =38 >
œŒ
3Œ
Þ
-9= >
=38 >
Therefore the general solution is
x œ -" Œ
# -9= > =38 >
-9= > # =38 >
-# Œ
Þ
-9= >
=38 >
The solution may also be written as
x œ -" Œ
& -9= >
& =38 >
-# Œ
Þ
# -9= > =38 >
-9= > # =38 >
4. Setting x œ 0 /<> results in the algebraic equations
#<
Œ *Î&
0"
!
&Î#
œ Œ .
Œ
"<
0#
!
For a nonzero solution, we require that ./>aA < Ib œ <# < &# œ !. The roots of
the characteristic equation are < œ a" „ $3bÎ# . With < œ a" $ 3bÎ# , the equations
reduce to the single equation a$ $3b0" & 0# œ ! . The corresponding eigenvector is
given by 0 a"b œ a& ß $ $ 3bX Þ Hence one of the complex-valued solutions is
________________________________________________________________________
page 383
—————————————————————————— CHAPTER 7. ——
xa"b œ Œ
&
a"$3b>Î#
/
$ $3
# 3 >Î#
$
$
œŒ
/ Œ-9= > 3 =38 >
"
#
#
œ />Î#
$
$
# -9= $# > =38 $# >
>Î# -9= # > # =38 # >
3/
Þ
-9= $# >
=38 $# >
The general solution is
x œ -" /
>Î#
$
$
# -9= $# > =38 $# >
>Î# -9= # > # =38 # >
-# /
Þ
-9= $# >
=38 $# >
The solution may also be written as
x œ -" /
>Î#
& -9= $# >
& =38 $# >
>Î#
$ -9= $ > $=38 $ > -# / $ -9= $ > $=38 $ > Þ
#
#
#
#
5. Setting x œ 0 >< results in the algebraic equations
Œ
"<
&
0"
!
"
œ Œ .
Œ
$<
0#
!
The characteristic equation is <# # < # œ ! , with roots < œ " „ 3 . Substituting
< œ " 3 reduces the system of equations to a# 3b0" 0# œ ! . The eigenvectors
are 0 a"b œ a" ß # 3bX and 0 a#b œ a" ß # 3bX Þ Hence one of the complex-valued
solutions is given by
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page 384
—————————————————————————— CHAPTER 7. ——
xa"b œ Œ
"
a"3b>
/
#3
"
>
œŒ
/ a-9= > 3 =38 >b
#3
-9= >
=38 >
>
œ /> Œ
3/ Œ
Þ
# -9= > =38 >
-9= > #=38 >
The general solution is
x œ -" /> Œ
-9= >
=38 >
>
-# / Œ
Þ
# -9= > =38 >
-9= > #=38 >
6. Solution of the ODEs is based on the analysis of the algebraic equations
"<
Π&
0"
!
#
œ Œ .
Œ
"<
0#
!
For a nonzero solution, we require that ./>aA < Ib œ <# * œ !. The roots of the
characteristic equation are < œ „ $ 3 . Setting < œ $ 3 , the two equations reduce to
a" $ 3b0" #0# œ ! . The corresponding eigenvector is 0 a"b œ a # ß " $3bX Þ Hence
one of the complex-valued solutions is given by
xa"b œ Œ
#
$3>
/
" $3
#
œŒ
a-9= $> 3 =38 $>b
" $3
# -9= $>
# =38 $>
œŒ
3Œ
Þ
-9= $> $ =38 $>
$-9= $> =38 $>
The general solution is
________________________________________________________________________
page 385
—————————————————————————— CHAPTER 7. ——
x œ -" Œ
# -9= $>
# =38 $>
-# Œ
Þ
-9= $> $ =38 $>
$-9= $> =38 $>
8. The eigensystem is obtained from analysis of the equation
Î $<
"
Ï #
!
"<
"
# ÑÎ 0" Ñ Î ! Ñ
!
0# œ ! .
ÒÏ
<
0$ Ò Ï ! Ò
The characteristic equation of the coefficient matrix is <$ %<# (< ' œ ! , with
roots <" œ # , <# œ " È# 3 and <$ œ " È# 3 . Setting < œ # , the
equations reduce to
0" # 0 $ œ !
0" 0 # œ ! Þ
The corresponding eigenvector is 0 a"b œ a# ß # ß "bX Þ With < œ " È# 3 , the
system of equations is equivalent to
Š# 3È# ‹0" # 0$ œ !
0" 3È # 0 # œ ! Þ
An eigenvector is given by 0 a#b œ Š 3È# ß " ß " 3È# ‹ Þ Hence one of the
X
complex-valued solutions is given by
________________________________________________________________________
page 386
—————————————————————————— CHAPTER 7. ——
a#b
x
œ
3È # Ñ
È
Ó/Š"3 #‹3>
"
Ï " 3È # Ò
Î
Ð
3È # Ñ
Ó/> Š-9= È# > 3 =38 È# >‹
"
Ï " 3È # Ò
Î
œÐ
œ/
È# =38 È# >
È# -9= È# >
Ñ
Î
Ñ
> Ð
Ó
ÓÞ
È
3/
=38 È# >
-9= # >
Ï -9= È# > È# =38 È# > Ò
Ï È# -9= È# > =38 È# > Ò
Î
> Ð
The other complex-valued solution is xa$b œ 0 a#b /<$ > . The general solution is
Î # Ñ #>
x œ -" # /
Ï " Ò
-# /
È# =38 È# >
È# -9= È# >
Ñ
Î
Ñ
>
Ó
Ð
ÓÞ
È
È
/
-9= # >
=38 # >
$
Ï -9= È# > È# =38 È# > Ò
Ï È# -9= È# > =38 È# > Ò
Î
> Ð
It is easy to see that all solutions converge to the equilibrium point a! ß ! ß !b Þ
10. Solution of the system of ODEs requires that
Œ
$<
"
0"
!
#
œ Œ .
Œ
"<
0#
!
The characteristic equation is <# % < & œ ! , with roots < œ # „ 3 . Substituting
< œ # 3 , the equations are equivalent to 0" a" 3b0# œ ! . The corresponding
eigenvector is 0 a"b œ a" 3 ß "bX Þ One of the complex-valued solutions is given by
xa"b œ Œ
" 3 a#3b>
/
"
" 3 #>
œŒ
/ a-9= > 3 =38 >b
"
-9= > =38 >
-9= > =38 >
#>
œ /#> Œ
3/ Œ
Þ
-9= >
=38 >
Hence the general solution is
x œ -" /#> Œ
-9= > =38 >
-9= > =38 >
#>
-# / Œ
Þ
-9= >
=38 >
Invoking the initial conditions, we obtain the system of equations
-" - # œ "
-" œ # Þ
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page 387
—————————————————————————— CHAPTER 7. ——
Solving for the coefficients, the solution of the initial value problem is
x œ # /#> Œ
-9= > =38 >
-9= > =38 >
#>
$/ Œ
-9= >
=38 >
-9= > & =38 >
œ /#> Œ
Þ
# -9= > $ =38 >
11a+b. With xa!b œ a# ß #bX , the solution is
11a,b.
x œ />Î% Œ
# -9= > # =38 >
Þ
# -9= >
________________________________________________________________________
page 388
—————————————————————————— CHAPTER 7. ——
11a- b.
12. Solution of the ODEs is based on the analysis of the algebraic equations
%
&
<
"
'
&
#
0"
!
Œ œ Œ .
0#
!
<
The characteristic equation is #& <# "! < #' œ ! , with roots < œ "& „ 3 . Setting
< œ "Î& 3 , the two equations reduce to 0" a" 3b0# œ ! . The corresponding
eigenvector is 0 a"b œ a" 3 ß "bX Þ One of the complex-valued solutions is given by
" 3 ˆ " 3‰>
/ &
"
" 3 >Î&
œŒ
/ a-9= > 3 =38 >b
"
-9= > =38 >
-9= > =38 >
>Î&
œ />Î& Œ
3/ Œ
Þ
-9= >
=38 >
xa"b œ Œ
Hence the general solution is
x œ -" />Î& Œ
-9= > =38 >
-9= > =38 >
>Î&
-# / Œ
Þ
-9= >
=38 >
a,b. Let xa!b œ aB!" ß B!# bX Þ The solution of the initial value problem is
-9= > =38 >
-9= > =38 >
>Î&
!
!
aB# B" b/ Œ
Þ
-9= >
=38 >
B! -9= > a# B!# B!" b=38 >
œ />Î& Œ " !
Þ
B# -9= > aB!# B!" b=38 >
x œ B!# />Î& Œ
With xa!b œ a" ß #bX , the solution is
________________________________________________________________________
page 389
—————————————————————————— CHAPTER 7. ——
x œ />Î& Œ
-9= > $ =38 >
Þ
# -9= > =38 >
a- b .
________________________________________________________________________
page 390
—————————————————————————— CHAPTER 7. ——
a. b .
13a+b. The characteristic equation of the coefficient matrix is <# #!< " !# , with
roots < œ ! „ 3 .
a,b. When ! ! and ! ! , the equilibrium point a! ß !b is a stable spiral and an
unstable spiral, respectively. The equilibrium point is a center when ! œ ! Þ
a- b .
________________________________________________________________________
page 391
—————————————————————————— CHAPTER 7. ——
14a+b. The roots of the characteristic equation, <# ! < & œ ! , are
<"ß# œ
! "È #
! #! Þ
„
# #
a,b. Note that the roots are complex when È#! ! È#! . For the case when
! − Š È#! ß !‹, the equilibrium point a! ß !b is a stable spiral. On the other hand,
when ! − Š! ß È#! ‹, the equilibrium point is an unstable spiral. For the case ! œ !,
the roots are purely imaginary, so the equilibrium point is a center. When !# #! ,
the roots are real and distinct. The equilibrium point becomes a node, with its stability
dependent on the sign of ! Þ Finally, the case !# œ #! marks the transition from spirals
to nodes.
a- b .
________________________________________________________________________
page 392
—————————————————————————— CHAPTER 7. ——
17. The characteristic equation of the coefficient matrix is <# #< " ! œ ! , with
roots given formally as <"ß# œ " „ È ! . The roots are real provided that ! Ÿ ! Þ
First note that the sum of the roots is # and the product of the roots is " ! . For
negative values of ! , the roots are distinct, with one always negative. When ! ",
the roots have opposite signs. Hence the equilibrium point is a saddle. For the case
" ! ! , the roots are both negative, and the equilibrium point is a stable node.
! œ " represents a transition from saddle to node. When ! œ ! , both roots are
equal. For the case ! ! , the roots are complex conjugates, with negative real part.
Hence the equilibrium point is a stable spiral.
________________________________________________________________________
page 393
—————————————————————————— CHAPTER 7. ——
19. The characteristic equation for the system is given by
<# a% !b< "! %! œ !Þ
The roots are
<"ß# œ #
! È #
„ ! )! #% Þ
#
First note that the roots are complex when % #È"! ! % #È"! . We also
find that when % #È"! ! # , the equilibrium point is a stable spiral. For the
case ! œ # , the equilibrium point is a center. When # ! % #È"! , the
equilibrium point is an unstable spiral. For all other cases, the roots are real. When
! #Þ& , the roots have opposite signs, with the equilibrium point being a saddle. For
the case % #È"! ! #Þ& , the roots are both positive, and the equilibrium point
is an unstable node. Finally, when ! % #È"! , both roots are negative, with the
equilibrium point being a stable node.
________________________________________________________________________
page 394
—————————————————————————— CHAPTER 7. ——
20. The characteristic equation is <# # < a#% )!b œ ! , with roots
<"ß# œ " „ È#& )! Þ
The roots are complex when ! #&Î) . Since the real part is negative, the origin
is a stable spiral. Otherwise the roots are real. When #& ! $ , both roots
are negative, and hence the equilibrium point is a stable node. For ! $ , the roots
are of opposite sign and the origin is a saddle.
22. Based on the method in Prob. "* of Section (Þ& , setting x œ 0 >< results in the
________________________________________________________________________
page 395
—————————————————————————— CHAPTER 7. ——
algebraic equations
Œ
#<
"
0"
!
&
œ Œ .
Œ
#<
0#
!
The characteristic equation for the system is <# " œ ! , with roots <"ß# œ „ 3 . With
< œ 3 , the equations reduce to the single equation 0" a# 3b0# œ !. A corresponding
eigenvector is 0 a"b œ a# 3 ß "bX Þ One complex-valued solution is
xa"b œ Œ
#3 3
> Þ
"
We can write >3 œ /3 68 > Þ Hence
xa"b œ Œ
# 3 3 68 >
/
"
#3
œŒ
c-9=a68 >b 3 =38a68 >bd
"
# -9=a68 >b =38a68 >b
-9=a68 >b # =38a68 >b
œŒ
3Œ
Þ
-9=a68 >b
=38a68 >b
Therefore the general solution is
x œ -" Œ
# -9=a68 >b =38a68 >b
-9=a68 >b # =38a68 >b
-# Œ
Þ
-9=a68 >b
=38a68 >b
Other combinations are also possible.
24a+b. The characteristic equation of the system is
#
)"
"(
<$ <# <
œ !,
&
)!
"'!
with eigenvalues <" œ "Î"! , and <#ß$ œ "Î% „ 3 . For < œ "Î"!, simple calculations
reveal that a corresponding eigenvector is 0 a"b œ a! ß ! ß "bX Þ Setting < œ "Î% 3 ,
we obtain the system of equations
0" 3 0 # œ !
0$ œ ! Þ
A corresponding eigenvector is 0 a#b œ a3 ß " ß !bX Þ Hence one solution is
xa"b œ
Î ! Ñ >Î"!
! / Þ
Ï"Ò
Another solution, which is complex-valued, is given by
________________________________________________________________________
page 396
—————————————————————————— CHAPTER 7. ——
a#b
x
Î 3 Ñ ˆ " 3‰>
œ " / %
Ï!Ò
Î 3 Ñ >Î%
œ " /
a-9= > 3 =38>b
Ï!Ò
œ />Î%
Î =38 > Ñ
Î -9= > Ñ
-9= > 3/>Î% =38 > Þ
Ï ! Ò
Ï ! Ò
Using the real and imaginary parts of xa#b , the general solution is constructed as
Î ! Ñ >Î"!
Î =38 > Ñ
Î -9= > Ñ
>Î%
>Î%
-9= > -$ /
=38 > Þ
x œ -" ! /
-# /
Ï"Ò
Ï ! Ò
Ï ! Ò
a,b. Let xa!b œ aB!" ß B!# ß B!$ b . The solution can be written as
Î
!
!
! Ñ
Î B# =38 > B" -9= > Ñ
>Î%
!
!
xœ
/
B# -9= > B!" =38 > Þ
Ï B! />Î"! Ò
Ï
Ò
!
$
With xa!b œ a" ß " ß "b, the solution of the initial value problem is
Î ! Ñ
Î =38 > -9= > Ñ
>Î%
!
-9= > =38 > Þ
xœ
/
Ï />Î"! Ò
Ï
Ò
!
________________________________________________________________________
page 397
—————————————————————————— CHAPTER 7. ——
25a+b. Based on Probs. ") #! of Section (Þ" , the system of differential equations is
VP"
. M
Œ œ "
.> Z
G
With V" œ V# œ % ohms , G œ
"
#
#
"
#
P"
M
Þ
" Œ
GV#
Z
farads and P œ ) henrys , the eigenvalue problem is
<
"
#
0"
!
Œ œ Œ .
0
!
<
"
)
#
a,b. The characteristic equation of the system is <# <
<"ß# œ
"
#
œ ! , with eigenvalues
" "
„ 3Þ
# #
Setting < œ "Î# 3Î# , the algebraic equations reduce to %30" 0# œ ! . It follows
that 0 a"b œ a" ß %3bX Þ Hence one complex-valued solution is
________________________________________________________________________
page 398
—————————————————————————— CHAPTER 7. ——
M a"b
"
a"3b>Î#
Œ œŒ
/
Z
%3
"
>Î#
œŒ
c-9=a>Î#b 3 =38a>Î#bd
/
%3
-9=a>Î#b
=38a>Î#b
>Î#
œ />Î# Œ
3/
Œ
Þ
% =38a>Î#b
% -9=a>Î#b
Therefore the general solution is
Œ
M
-9=a>Î#b
=38a>Î#b
>Î#
>Î#
œ -" /
Œ
-# /
Œ
Þ
Z
% =38a>Î#b
% -9=a>Î#b
a- b. Imposing the initial conditions, we arrive at the equations -" œ # and -# œ $% ,
and
$
M
>Î# # -9=a>Î#b % =38a>Î#b
Œ œ/
Œ
Þ
Z
) =38a>Î#b $ -9=a>Î#b
a. b. Since the eigenvalues have negative real parts, all solutions converge to the origin.
26a+b. The characteristic equation of the system is
<#
"
"
<
œ !,
VG
GP
with eigenvalues
<"ß# œ
"
"
%V # G
Ê"
„
Þ
#VG #VG
P
The eigenvalues are real and different provided that
"
%V # G
!.
P
The eigenvalues are complex conjugates as long as
%V # G
"
!Þ
P
a,b. With the specified values, the eigenvalues are <"ß# œ " „ 3 Þ The eigenvector
corresponding to < œ " 3 is 0 a"b œ a" ß %3bX Þ Hence one complex-valued solution
is
________________________________________________________________________
page 399
—————————————————————————— CHAPTER 7. ——
M a"b
"
a"3b>
Œ œŒ
/
Z
"3
"
>
œŒ
/ a-9= > 3 =38 >b
"3
-9= >
=38 >
>
œ /> Œ
3/ Œ
Þ
-9= > =38 >
-9= > =38 >
Therefore the general solution is
Œ
M
-9= >
=38 >
>
>
œ -" / Œ
-# / Œ
Þ
Z
-9= > =38 >
-9= > =38 >
a- b. Imposing the initial conditions, we arrive at the equations
-" œ #
-" - # œ " ,
with -" œ # and -# œ $ . Therefore the solution of the IVP is
Œ
M
> # -9= > $ =38 >
œ/ Œ
Þ
Z
-9= > &=38 >
a. b. Since V/a<"ß# b œ " , all solutions converge to the origin.
27a+b. Suppose that -" a -# b œ 0 . Since a and b are the real and imaginary parts of
the vector 0 a"b , respectively, a œ ˆ0 a"b 0 a"b ‰Î# and b œ ˆ0 a"b 0 a"b ‰Î#3 . Hence
-" ˆ0 a"b 0 a"b ‰ 3-# ˆ0 a"b 0 a"b ‰ œ 0 ,
which leads to
a-" 3-# b0 a"b a-" 3-# b0 a"b œ 0 Þ
Now since 0 a"b and 0 a"b are linearly independent, we must have
-" 3-# œ !
-" 3-# œ ! Þ
It follows that -" œ -# œ ! .
a- b. Recall that
ua>b œ /-> aa -9= .> b =38 .>b
va>b œ /-> aa -9= .> b =38 .>b .
Consider the equation -" ua>! b -# va>! b œ 0 , for some >! . We can then write
________________________________________________________________________
page 400
—————————————————————————— CHAPTER 7. ——
-" /->! aa -9= .>! b =38 .>! b -# /->! aa -9= .>! b =38 .>! b œ 0 . a‡b
Rearranging the terms, and dividing by the exponential,
a-" -# b-9= .>! a a-# -" b=38 .>! b œ 0 .
From Part a,b, since a and b are linearly independent, it follows that
a-" -# b-9= .>! œ a-# -" b=38 .>! œ ! Þ
Without loss of generality, assume that the trigonometric factors are nonzero. Otherwise
proceed again from Equation a‡b, above. We then conclude that
-" -# œ ! and -# -" œ ! ,
which leads to -" œ -# œ !. Thus ua>! b and va>! b are linearly independent for some >! ,
and hence the functions are linearly independent at every point.
28a+b. Let B" œ ? and B# œ ? w . It follows that B"w œ B# and
B#w œ ? ww
œ
5
?.
7
In terms of the new variables, we obtain the system of two first order ODEs
B"w œ B#
B#w œ
5
B" Þ
7
a,b. The associated eigenvalue problem is
<
Œ 5Î7
0"
!
"
œ Œ .
Œ
<
0#
!
The characteristic equation is <# 5Î7 œ ! , with roots <"ß# œ „ 3È5Î7 Þ
a- b. Since the eigenvalues are purely imaginary, the origin is a center. Hence the phase
curves are ellipses, with a clockwise flow. For computational purposes, let 5 œ " and
7 œ #.
________________________________________________________________________
page 401
—————————————————————————— CHAPTER 7. ——
a. b. The general solution of the second order equation is
?a>b œ -" -9= Ê
5
5
> -# =38 Ê > .
7
7
The general solution of the system of ODEs is given by
________________________________________________________________________
page 402
—————————————————————————— CHAPTER 7. ——
5 Ñ
5 Ñ
ÎÈ7
ÎÈ7
É7
É7
>
>
5 =38
5 -9=
Ð
Ó
Ð
ÓÞ
x œ -"
-#
5
5
Ï -9=É 7 > Ò
Ï =38É 7 > Ò
It is evident that the natural frequency of the system is equal to M7a<"ß# b .
________________________________________________________________________
page 403
—————————————————————————— CHAPTER 7. ——
Section 7.7
1. The eigenvalues and eigenvectors were found in Prob. " , Section (Þ& .
"
#
<" œ " , 0 a"b œ Œ à <# œ # , 0 a#b œ Œ Þ
#
"
The general solution is
x œ -" Œ
/>
# /#>
#Œ
Þ
# />
/#>
Hence a fundamental matrix is given by
Ga>b œ Œ
/>
# />
# /#>
Þ
/#>
We now have
Ga!b œ Œ
"
#
" "
#
and G" a!b œ Œ
"
#
$
#
,
"
So that
Fa>b œ Ga>bG" a!b œ
" /> %/#>
Œ
$ #/> #/#>
#/> #/#>
Þ
%/> /#>
3. The eigenvalues and eigenvectors were found in Prob. $ , Section (Þ& . The general
solution of the system is
/>
/>
x œ -" Œ > -# Œ > Þ
/
$/
Given the initial conditions xa!b œ ea"b , we solve the equations
-" - # œ "
-" $-# œ ! ,
to obtain -" œ $Î# , -# œ "Î# . The corresponding solution is
x œ #$
/ "# />
Þ
> $ />
/
#
#
$ >
Given the initial conditions xa!b œ ea#b , we solve the equations
-" - # œ !
-" $-# œ " ,
to obtain -" œ "Î# , -# œ "Î# . The corresponding solution is
________________________________________________________________________
page 404
—————————————————————————— CHAPTER 7. ——
xœ
"# /> "# />
Þ
"# /> $# />
Therefore the fundamental matrix is
Fa>b œ
" $/> />
Œ
# $/> $/>
/> />
Þ
/> $/>
5. The general solution, found in Prob. $ , Section (Þ' , is given by
x œ -" Œ
& -9= >
& =38 >
-# Œ
Þ
# -9= > =38 >
-9= > # =38 >
Given the initial conditions xa!b œ ea"b , we solve the equations
&-" œ "
#-" -# œ ! ,
resulting in -" œ "Î& , -# œ #Î& . The corresponding solution is
xœŒ
-9= > # =38 >
Þ
=38 >
Given the initial conditions xa!b œ ea#b , we solve the equations
&-" œ !
#-" -# œ " ,
resulting in -" œ ! , -# œ " . The corresponding solution is
xœŒ
& =38 >
Þ
-9= > # =38 >
Therefore the fundamental matrix is
Fa>b œ Œ
-9= > # =38 >
=38 >
& =38 >
Þ
-9= > # =38 >
7. The general solution, found in Prob. "& , Section (Þ& , is given by
x œ -" Œ
/#>
/%>
#
Œ %> Þ
$/#>
/
Given the initial conditions xa!b œ ea"b , we solve the equations
-" - # œ "
$-" -# œ ! ,
________________________________________________________________________
page 405
—————————————————————————— CHAPTER 7. ——
resulting in -" œ "Î# , -# œ $Î# . The corresponding solution is
xœ
" /#> $/%>
Œ
Þ
# $/#> $/%>
The initial conditions xa!b œ ea#b require that
-" - # œ !
$-" -# œ " ,
resulting in -" œ "Î# , -# œ "Î# . The corresponding solution is
xœ
" /#> /%>
Œ
Þ
# $/#> /%>
Therefore the fundamental matrix is
Fa>b œ
" /#> $/%>
Œ
# $/#> $/%>
/#> /%>
Þ
$/#> /%>
8. The general solution, found in Prob. & , Section (Þ' , is given by
x œ -" /> Œ
-9= >
=38 >
>
-# / Œ
Þ
# -9= > =38 >
-9= > # =38 >
The specific solution corresponding to the initial conditions xa!b œ ea"b is
x œ /> Œ
-9= > # =38 >
Þ
&=38 >
x œ /> Œ
=38 >
Þ
-9= > #=38 >
For the initial conditions xa!b œ ea#b , the solution is
Therefore the fundamental matrix is
Fa>b œ /> Œ
-9= > # =38 >
&=38 >
=38 >
Þ
-9= > #=38 >
9. The general solution, found in Prob. "$ , Section (Þ& , is given by
#>
>
Î %/
Ñ
Î $/
Ñ
Î ! Ñ
>
#>
/#>
x œ -" &/
-# %/
-$
Þ
Ï (/#> Ò
Ï #/> Ò
Ï /#> Ò
Given the initial conditions xa!b œ ea"b , we solve the equations
________________________________________________________________________
page 406
—————————————————————————— CHAPTER 7. ——
%-" $-# œ "
&-" %-# -$ œ !
(-" #-# -$ œ ! ,
resulting in -" œ "Î# , -# œ " , -$ œ $Î# . The corresponding solution is
Î #/ $/
Ñ
x œ Ð & /#> %/> $ /#> ÓÞ
#>
>
Ï (# /#> #/> $# /#> Ò
#
#
The initial conditions xa!b œ ea#b , we solve the equations
%-" $-# œ !
&-" %-# -$ œ "
(-" #-# -$ œ ! ,
resulting in -" œ "Î% , -# œ "Î$ , -$ œ "$Î"# . The corresponding solution is
/#> />
Î
Ñ
x œ Ð & /#> % /> "$ /#> ÓÞ
Ï (% /#> #$ />
%
$
"#
"$ #>
"# /
The initial conditions xa!b œ ea$b , we solve the equations
Ò
%-" $-# œ !
&-" %-# -$ œ !
(-" #-# -$ œ " ,
resulting in -" œ "Î% , -# œ "Î$ , -$ œ "Î"# . The corresponding solution is
/#> />
Î
Ñ
& #>
% >
" #> Ó
Ð
/ / /
xœ
Þ
Ï (% /#> #$ />
%
$
"#
" #>
"# /
Ò
Therefore the fundamental matrix is
Î #/ $/
Fa>b œ Ð & /#> %/> $ /#>
#>
>
Ï (# /#> #/> $# /#>
#
#
/#> />
#>
%$ /> "$
"# /
"$ #>
#$ /> "#
/
& #>
%/
( #>
%/
/#> />
Ñ
" #> Ó
%$ /> "#
/
Þ
" #> Ò
#$ /> "#
/
& #>
%/
( #>
%/
12. The solution of the initial value problem is given by
________________________________________________________________________
page 407
—————————————————————————— CHAPTER 7. ——
x œ Fa>bxa!b
/> -9= #> #/> =38 #>
$
œ Œ " >
Œ
>
/ -9= #>
"
# / =38 #>
$-9= #> #=38 #>
œ /> Œ $
Þ
=38
#>
-9=
#>
#
13. Let
a"b
Î B" a>b
Ga>b œ
ã
Ï Ba"b a>b
8
It follows that
a"b
Î B" a>! b
Ga>! b œ
ã
Ï Ba"b a> b
!
8
Ba"8b a>b Ñ
Þ
ã
Ò
a8b
B8 a>b
â
â
Ba"8b a>! b Ñ
ã
a8b
B8 a>! b Ò
â
â
is a scalar matrix, which is invertible, since the solutions are linearly independent.
Let G" a>! b œ a-34 b. Then
a"b
Î B" a>b
Ga>bG a>! b œ
ã
Ï Ba"b a>b
8
Ba"8b a>b ÑÎ -""
ã
ã
ÒÏ
a8b
B8 a>b
8"
â
"
â
â
â
-"8 Ñ
ã Þ
-88 Ò
The 4-th column of the product matrix is
cGa>bG" a>! bd
a4b
œ " -54 xa5 b ,
8
5œ"
which is a solution vector, since it is a linear combination of solutions. Furthermore, the
columns are all linearly independent, since the vectors xa5 b areÞ Hence the product is
a fundamental matrix. Finally, setting > œ >! , Ga>! bG" a>! b œ I . This is precisely the
definition of Fa>b.
14. The fundamental matrix Fa>b for the system
xw œ Œ
"
%
"
x
"
is given by
Fa>b œ
" #/$> #/>
Œ
% %/$> %/>
/$> />
Þ
#/$> #/>
Direct multiplication results in
________________________________________________________________________
page 408
—————————————————————————— CHAPTER 7. ——
Fa>bFa=b œ
" #/$> #/>
/$> />
#/$= #/=
Π$>
Œ
%/$= %/=
"' %/ %/> #/$> #/>
"
)a/$>$= />= b %a/$>$= />= b
œ
Œ
Þ
"' "'a/$>$= />= b )a/$>$= />= b
Hence
Fa>bFa=b œ
" #/$a>=b #/a>=b
Œ
% %/$a>=b %/a>=b
/$= /=
#/$= #/=
/$a>=b /a>=b
Þ
#/$a>=b #/a>=b
15a+b. Let = be arbitrary, but fixed, and > variable. Similar to the argument in Prob. "$ ,
the columns of the matrix Fa>bFa=b are linear combinations of fundamental solutions.
Hence the columns of Fa>bFa=b are also solution of the system of equations. Further,
setting > œ ! , Fa!bFa=b œ I Fa=b œ Fa=b Þ That is, Fa>bFa=b is a solution of the
initial value problem Z w œ AZ , with Za!b œ Fa=b Þ Now consider the change of
variable 7 œ > = . Let Wa7 b œ Za7 =b. The given initial value problem can be
reformulated as
.
W œ AW , with Wa=b œ Fa=b Þ
.7
Since Fa>b is a fundamental matrix satisfying F w œ AF , with Fa!b œ I , it follows
that
Wa7 b œ Fa7 bF" a=b‘Fa=b
œ Fa 7 b Þ
That is, Fa> =b œ Fa7 b œ Wa7 b œ Za>b œ Fa>bFa=b .
a,b. Based on Part a+b, Fa>bFa >b œ Fa> a >bb œ Fa!b œ I . Hence
Fa >b œ F" a>bÞ
a- b. It also follows that Fa> =b œ Fa> a =bb œ Fa>bFa =b œ Fa>bF" a=bÞ
16. Let A be a diagonal matrix, with A œ c+" ea"b ß +# ea#b ß âß +8 ea8b d. Note that for any
positive integer, 5 ,
A5 œ +"5 ea"b ß +#5 ea#b ß âß +85 ea8b ‘.
It follows, from basic matrix algebra, that
________________________________________________________________________
page 409
—————————————————————————— CHAPTER 7. ——
>5
Î ! +"5 5x
Ð 5œ !
Ð
Ð
7
5
!
Ð
5>
"
I
A
œÐ
5x Ð
5œ"
Ð
ã
Ð
!
Ï
Ñ
Ó
Ó
Ó
Ó
ÓÞ
Ó
Ó
Ó
7
!
! +#5 >5
â
!
â
!
7
5x
5œ !
ã
ã
! +85 >5 Ò
5x
7
!
â
5œ !
It can be shown that the partial sums on the left hand side converge for all > . Taking the
limit aas 7 p _b on both sides of the equation, we obtain
Î/
Ð !
/B:aA>b œ Ð
ã
Ï !
+" >
!
/ +# >
ã
!
â
â
â
! Ñ
! Ó
ÓÞ
ã
/ +8 > Ò
Alternatively, consider the system x w œ Ax . Since ODEs are uncoupled, the vectors
xa4b œ /B:a+4 >b ea4b ß 4 œ "ß #ß â8 , are a set of linearly independent solutions. Hence
the matrix
X œ /B:a+" >b ea"b ß /B:a+# >b ea#b ß âß /B:a+8 >b ea8b ‘
is a fundamental matrix. Finally, since Xa!b œ I , it follows that
/B:a+" >b ea"b ß /B:a+# >b ea#b ß âß /B:a+8 >b ea8b ‘ œ Fa>b œ /B:aA>b .
17a+b. Assuming that x œ 9a>b is a solution, then 9 w œ A9, with 9a!b œ x! Þ Integrate
both sides of the equation to obtain
9a>b 9a!b œ ( A9a=b.= .
>
!
Hence
9a>b œ x! ( A9a=b.= .
>
!
a,b. Proceed with the iteration
9a3"b a>b œ x! ( A9a3b a=b.= .
>
!
With 9a!b a>b œ x! , and noting that A is a constant matrix,
________________________________________________________________________
page 410
—————————————————————————— CHAPTER 7. ——
9a"b a>b œ x! ( Ax! .=
>
!
œ x Ax! > Þ
!
That is, 9a"b a>b œ aI A>bx! .
a- b. We then have
9a#b a>b œ x! ( AaI A>bx! .=
>
!
# !>
œ x Ax > A x
!
!
#
#
#
>
œ ŒI A> A# x! .
#
Now suppose that
9a8b a>b œ ŒI A> A#
>#
>8
â A8 x! .
#
8x
It follows that
#
( AŒI A> A
>
!
>#
>8
â A8 x! .= œ
#
8x
$
8"
>#
#>
8 >
!
œ AŒI> A A
âA
x
#
$x
a8 "bx
#
$
>
>
>8
œ ŒA> A# A$ â A8" x! Þ
#
$x
8x
Therefore
9a8"b a>b œ ŒI A> A#
>#
>8"
!
â A8"
x .
#
a 8 "b x
By induction, the asserted form of 9a8b a>b is valid for all 8
!Þ
a. b. Define 9a_b a>b œ lim 9a8b a>b. It can be shown that the limit does exist. In fact,
8 Ä_
9a_b a>b œ /B:aA>bx! Þ
Term-by-term differentiation results in
________________________________________________________________________
page 411
—————————————————————————— CHAPTER 7. ——
. a_b
.
>#
>8
9 a>b œ ŒI A> A# â A8 x!
.>
.>
#
8x
>8"
œ ŒA A# > â A8
x!
a 8 "b x
#
>
>8"
œ AŒI A> A# â A8"
x! Þ
a 8 "b x
#
That is,
. a_b
9 a>b œ A9a_b a>b.
.>
Furthermore, 9a_b a!b œ x! . Based on uniqueness of solutions, 9a>b œ 9a_b a>b .
________________________________________________________________________
page 412
—————————————————————————— CHAPTER 7. ——
Section 7.8
2. Setting x œ 0 >< results in the algebraic equations
Œ
%<
)
0"
!
#
œ Œ .
Œ
%<
0#
!
The characteristic equation is <# œ ! , with the single root < œ ! . Substituting < œ !
reduces the system of equations to #0" 0# œ ! . Therefore the only eigenvector is
0 œ a" ß #bX Þ One solution is
"
xa"b œ Œ ,
#
which is a constant vector. In order to generate a second linearly independent solution,
we must search for a generalized eigenvector. This leads to the system of equations
%
Œ)
("
"
#
œ Œ .
Œ
%
(#
#
This system also reduces to a single equation, #(" (# œ "Î# . Setting (" œ 5 , some
arbitrary constant, we obtain (# œ #5 "Î# . A second solution is
"
5
xa#b œ Œ > Œ
#
#5 "Î#
"
!
"
œ Œ > Œ
5 Œ Þ
#
"Î#
#
Note that the last term is a multiple of xa"b and may be dropped. Hence
"
!
xa#b œ Œ > Œ
.
#
"Î#
The general solution is
"
"
!
x œ - " Œ - # ” Œ > Œ
•Þ
#
#
"Î#
________________________________________________________________________
page 413
—————————————————————————— CHAPTER 7. ——
All of the points on the line B# œ #B" are equilibrium points. Solutions starting at all
other points become unbounded.
3. Solution of the ODEs is based on the analysis of the algebraic equations
$
#
<
"
"
%
"
#
!
0"
Œ œ Œ .
!
0
<
#
The characteristic equation is <# # < " œ ! , with a single root < œ " . Setting
< œ " , the two equations reduce to 0" #0# œ ! . The corresponding eigenvector is
0 œ a# ß "bX Þ One solution is
#
xa"b œ Œ /> .
"
A second linearly independent solution is obtained by finding a generalized eigenvector.
We therefore analyze the system
"
#
"
%
"
"
#
"
("
Œ ( œ Œ # .
#
The equations reduce to the single equation (" #(# œ #Þ Let (" œ #5 . We obtain
(# œ " 5 , and a second linearly independent solution is
#
#5
>
xa#b œ Œ >/> Œ
/
"
"5
#
!
#
œ Œ >/> Œ /> 5 Œ /> Þ
"
"
"
Dropping the last term, the general solution is
#
#
!
x œ -" Œ /> -# ”Œ >/> Œ /> •Þ
"
"
"
________________________________________________________________________
page 414
—————————————————————————— CHAPTER 7. ——
4. Solution of the ODE requires analysis of the algebraic equations
$<
&
#
0"
!
Œ œ Œ .
0
!
#<
&
#
#
For a nonzero solution, we must have ./>aA < Ib œ <# < "% œ ! . The only root
is < œ "Î# , which is an eigenvalue of multiplicity two. Setting < œ "Î# is the
coefficient matrix reduces the system to the single equation 0" 0# œ ! . Hence the
corresponding eigenvector is 0 œ a" ß "bX Þ One solution is
"
xa"b œ Œ />Î# .
"
In order to obtain a second linearly independent solution, we find a solution of the system
&
#
&
#
&
#
&
#
("
"
Œ ( œ Œ " .
#
There equations reduce to &(" &(# œ # . Set (" œ 5 , some arbitrary constant. Then
(# œ 5 #Î& . A second solution is
"
5
>Î#
xa#b œ Œ >/>Î# Œ
/
"
5 #Î&
"
!
"
>Î#
œ Œ >/>Î# Œ
5 Œ />Î# Þ
/
"
#Î&
"
Dropping the last term, the general solution is
"
"
!
>Î#
x œ -" Œ />Î# -# ”Œ >/>Î# Œ
/
•Þ
"
"
#Î&
________________________________________________________________________
page 415
—————————————————————————— CHAPTER 7. ——
6. The eigensystem is obtained from analysis of the equation
Î <
"
Ï "
" ÑÎ 0" Ñ Î ! Ñ
"
0# œ ! .
< ÒÏ 0$ Ò Ï ! Ò
"
<
"
The characteristic equation of the coefficient matrix is <$ $< # œ ! , with
roots <" œ # and <#ß$ œ " . Setting < œ # , we have
Î #
"
Ï "
" ÑÎ 0" Ñ Î ! Ñ
"
0# œ ! .
Ò
Ï
#
0$ Ò Ï ! Ò
"
#
"
This system is reduced to the equations
0" 0 $ œ !
0# 0 $ œ ! Þ
A corresponding eigenvector vector is given by 0 a"b œ a" ß " ß "bX Þ Setting < œ " ,
the system of equations is reduced to the single equation
0" 0 # 0 $ œ ! Þ
An eigenvector vector is given by 0 a#b œ a" ß ! ß "bX Þ Since the last equation has two
free variables, a third linearly independent eigenvector aassociated with < œ "b is
0 a$b œ a! ß " ß "bX Þ Therefore the general solution may be written as
Î " Ñ #>
Î " Ñ >
Î ! Ñ >
!
"
x œ -" " / - #
/ -$
/ Þ
Ï"Ò
Ï "Ò
Ï "Ò
7. Solution of the ODE requires analysis of the algebraic equations
Œ
"<
%
0"
!
%
.
œ
( < Π0# Π!
For a nonzero solution, we must have ./>aA < Ib œ <# '< * œ ! . The only root
is < œ $ , which is an eigenvalue of multiplicity two. Substituting < œ $ into the
coefficient matrix, the system reduces to the single equation 0" 0# œ ! . Hence the
corresponding eigenvector is 0 œ a" ß "bX Þ One solution is
"
xa"b œ Œ /$> .
"
For a second linearly independent solution, we search for a generalized eigenvector.
Its components satisfy
________________________________________________________________________
page 416
—————————————————————————— CHAPTER 7. ——
%
Œ%
("
"
%
œ Œ ,
Œ
%
(#
"
that is, %(" %(# œ " . Let (# œ 5 , some arbitrary constant. Then (" œ 5 "Î% Þ
It follows that a second solution is given by
"
5 "Î% $>
xa#b œ Œ >/$> Œ
/
"
5
"
"Î% $>
" $>
œ Œ >/$> Œ
/ 5 Œ / Þ
"
!
"
Dropping the last term, the general solution is
"
"
"Î% $>
x œ -" Œ /$> -# ”Œ >/$> Œ
/ •Þ
"
"
!
Imposing the initial conditions, we require that
"
-" - # œ $
%
-" œ # ,
which results in -" œ # and -# œ % Þ Therefore the solution of the IVP is
$
%
x œ Œ /$> Œ >/$> Þ
#
%
________________________________________________________________________
page 417
—————————————————————————— CHAPTER 7. ——
8. Solution of the ODEs is based on the analysis of the algebraic equations
&
#
<
$
#
"
#
0"
!
Œ œ Œ .
0
!
<
$
#
#
The characteristic equation is <# # < " œ ! , with a single root < œ " . Setting
< œ ", the two equations reduce to 0" 0# œ !. The corresponding eigenvector is
0 œ a" ß "bX Þ One solution is
"
xa"b œ Œ /> .
"
A second linearly independent solution is obtained by solving the system
$
#
$
#
$
#
$
#
("
"
Œ ( œ Œ " .
#
The equations reduce to the single equation $(" $(# œ #Þ Let (" œ 5 . We obtain
(# œ #Î$ 5 , and a second linearly independent solution is
"
5
>
xa#b œ Œ >/> Œ
/
"
#Î$ 5
"
!
" >
>
œ Œ >/> Œ
/ 5 Œ / Þ
"
#Î$
"
Dropping the last term, the general solution is
"
"
!
>
x œ -" Œ /> -# ”Œ >/> Œ
/ •Þ
"
"
#Î$
Imposing the initial conditions, find that
________________________________________________________________________
page 418
—————————————————————————— CHAPTER 7. ——
-" œ $
#
-" - # œ " ,
$
so that -" œ $ and -# œ ' Þ Therefore the solution of the IVP is
xœŒ
$
'
>
>
/ Œ >/ Þ
"
'
10. The eigensystem is obtained from analysis of the equation
$<
Π"
0"
!
*
œ Œ .
Œ
$<
0#
!
The characteristic equation is <# œ ! , with a single root < œ ! . Setting < œ ! , the two
equations reduce to 0" $0# œ !. The corresponding eigenvector is 0 œ a $ ß "bX Þ
Hence one solution is
________________________________________________________________________
page 419
—————————————————————————— CHAPTER 7. ——
xa"b œ Œ
$
,
"
which is a constant vector. A second linearly independent solution is obtained from the
system
$
Π"
("
$
*
œŒ
Œ
.
$
(#
"
The equations reduce to the single equation (" $(# œ " Þ Let (# œ 5 . We obtain
(" œ " $5 , and a second linearly independent solution is
xa#b œ Œ
$
" $5
> Œ
"
5
$
"
$
œŒ
> Œ
5Œ
Þ
"
!
"
Dropping the last term, the general solution is
x œ -" Œ
$
$
"
-# ”Œ
> Œ
•Þ
"
"
!
Imposing the initial conditions, we require that
$-" -# œ #
-" œ % ,
which results in -" œ % and -# œ "% Þ Therefore the solution of the IVP is
#
$
x œ Œ "%Œ
>Þ
%
"
________________________________________________________________________
page 420
—————————————————————————— CHAPTER 7. ——
12. The characteristic equation of the system is ) <$ '! <# "#' < %* œ ! . The
eigenvalues are <" œ "Î# and <#ß$ œ (Î# . The eigenvector associated with <"
is 0 a"b œ a" ß " ß "bX Þ Setting < œ (Î# , the components of the eigenvectors must
satisfy the relation
0" 0 # 0 $ œ ! Þ
An eigenvector vector is given by 0 a#b œ a" ß ! ß "bX Þ Since the last equation has two
free variables, a third linearly independent eigenvector aassociated with < œ (Î#b
is 0 a$b œ a! ß " ß "bX Þ Therefore the general solution may be written as
x œ -"
Î " Ñ >Î#
Î " Ñ (>Î#
Î ! Ñ (>Î#
" /
!
"
-#
/
-$
/
Þ
Ï"Ò
Ï "Ò
Ï "Ò
Invoking the initial conditions, we require that
-" - # œ #
-" - $ œ $
-" - # - $ œ " Þ
Hence the solution of the IVP is
"
" Ñ
! Ñ
% Î Ñ >Î# # Î
&Î
(>Î#
" /
!
"
xœ
/
/(>Î# Þ
$Ï Ò
$Ï
$Ï
"
"Ò
"Ò
13. Setting x œ 0 >< results in the algebraic equations
Œ
$<
"
0"
!
%
œ Œ .
Œ
"<
0#
!
The characteristic equation is <# #< " œ ! , with a single root of <"ß# œ " . With
________________________________________________________________________
page 421
—————————————————————————— CHAPTER 7. ——
< œ " , the system reduces to a single equation 0" # 0# œ !. An eigenvector is given by
0 œ a# ß "bX Þ Hence one solution is
#
xa"b œ Œ > .
"
In order to find a second linearly independent solution, we search for a generalized
eigenvector whose components satisfy
#
Œ"
("
#
%
œ Œ .
Œ
#
(#
"
These equations reduce to (" # (# œ " . Let (# œ 5 , some arbitrary constant. Then
(" œ " #5 Þ Ò Before proceeding, note that if we set ? œ 68 > , the original equation is
transformed into a constant coefficient equation with independent variable ? . Recall that
a second solution is obtained by multiplication of the first solution by the factor ? . This
implies that we must multiply first solution by a factor of 68 > . Ó Hence a second linearly
independent solution is
#
" #5
xa#b œ Œ > 68 > Œ
>
"
5
#
"
#
œ Œ > 68 > Œ > 5 Œ >Þ
"
!
"
Dropping the last term, the general solution is
#
#
"
x œ -" Œ > -# ”Œ > 68 > Œ >•Þ
"
"
!
15. The characteristic equation is
<# a+ . b< +. ,- œ ! Þ
Hence the eigenvalues are
<"ß# œ
+. "
„ Éa+ . b# %a+. ,- b Þ
#
#
16a+b. Using the result in Prob. "& , the eigenvalues are
<"ß# œ
ÈP# %V # GP
"
„
Þ
#VG
#VGP
The discriminant vanishes when P œ %V # GP .
________________________________________________________________________
page 422
—————————————————————————— CHAPTER 7. ——
a,b. The system of differential equations is
. M
!
Œ œŒ
.> Z
"
"
%
"
Œ
M
Þ
Z
The associated eigenvalue problem is
<
Œ
"
"
%
"<
Œ
0"
!
œ Œ .
0#
!
The characteristic equation is <# < "Î% œ ! , with a single root of <"ß# œ "Î# .
Setting < œ "Î# , the algebraic equations reduce to #0" 0# œ !. An eigenvector is
given by 0 œ a" ß #bX Þ Hence one solution is
M a"b
"
>Î#
.
Œ œŒ
/
Z
#
A second solution is obtained from a generalized eigenvector whose components satisfy
"
"
#
("
"
Œ œŒ
.
(#
#
"
%
"
#
It follows that (" œ 5 and (# œ % #5 Þ A second linearly independent solution is
M
Œ
Z
a#b
œŒ
"
5
>Î#
>Î#
Œ
> /
/
#
% #5
"
!
"
>Î#
>Î#
œŒ
Œ />Î# 5 Œ
Þ
> /
/
#
%
#
Dropping the last term, the general solution is
Œ
M
"
"
!
>Î#
>Î#
-# ” Œ
Œ />Î# •Þ
œ -" Œ
/
> /
Z
#
#
%
Imposing the initial conditions, we require that
-" œ "
#-" %-# œ # ,
which results in -" œ " and -# œ " Þ Therefore the solution of the IVP is
Œ
M
" >Î#
"
>Î#
Œ
Þ
œ Œ /
> /
Z
#
#
18a+b. The eigensystem is obtained from analysis of the equation
________________________________________________________________________
page 423
—————————————————————————— CHAPTER 7. ——
Î& <
)
Ï %
$
&<
$
# ÑÎ 0" Ñ Î ! Ñ
%
0# œ ! .
ÒÏ
$<
0$ Ò Ï ! Ò
The characteristic equation of the coefficient matrix is <$ $<# $< " œ ! , with a
single root of multiplicity three, < œ " . Setting < œ " , we have
Î %
)
Ï %
# ÑÎ 0" Ñ Î ! Ñ
%
0# œ ! .
# ÒÏ 0$ Ò Ï ! Ò
$
'
$
The system of algebraic equations reduce to a single equation
%0" $0# #0$ œ ! Þ
An eigenvector vector is given by 0 a"b œ a" ß ! ß #bX Þ Since the last equation has two
free variables, a second linearly independent eigenvector aassociated with < œ "b is
0 a#b œ a! ß # ß $bX Þ Therefore two solutions are obtained as
xa"b œ
Î"Ñ >
Î ! Ñ >
! / and xa#b œ
#
/Þ
Ï#Ò
Ï $Ò
a,b. It follows directly that x w œ 0>/> 0/> ( /> Þ Hence the coefficient vectors must
satisfy 0 >/> 0/> ( /> œ A0>/> A(/> . Rearranging the terms , we have
0 /> œ aA Ib0>/> aA Ib( /> .
Given an eigenvector 0 , it follows that aA Ib( œ 0 .
a- b. Note that a linear combination of two eigenvectors, associated with the same
eigenvalue, is also an eigenvector. Consider the equation aA Ib( œ -" 0 a"b -# 0 a#b Þ
The augmented matrix is
Î %
)
Ï %
$
'
$
#
%
#
Using elementary row operations, we obtain
Î%
!
Ï!
$
!
!
#
!
!
l
l
l
l
l
l
-"
Ñ
#-#
Þ
Ò
#-" $-#
-"
Ñ
#-" #-# Þ
$-" $-# Ò
It is evident that a solution exists provided -" œ -# .
a. b. Let -" œ -# œ # . The components of the generalized eigenvector must satisfy
________________________________________________________________________
page 424
—————————————————————————— CHAPTER 7. ——
Î %
)
Ï %
# ÑÎ (" Ñ Î # Ñ
%
(# œ
%
.
ÒÏ
Ò
Ï
Ò
#
($
#
$
'
$
Based on Part a- b, the equations reduce to the single equation %(" $(# #($ œ # Þ
Let (" œ ! and (# œ #" , where ! and " are arbitrary constants. We then have
($ œ " #! $" ,
so that
Î
!
Ñ Î ! Ñ
Î"Ñ
Î ! Ñ
#"
!
#
(œ
œ
! ! "
Þ
Ï " #! $ " Ò Ï " Ò
Ï#Ò
Ï $Ò
Observe that ( œ ! 0 a"b " 0 a#b Þ Hence a third linearly independent solution is
Î # Ñ > Î ! Ñ >
%
!
œ
>/
/Þ
Ï #Ò
Ï "Ò
a$b
x
a/b. Given the three linearly independent solutions, a fundamental matrix is given by
>
Î /
Ga>b œ
!
Ï #/>
!
#/>
$/>
a0 b. We construct the transformation matrix
Tœ
with inverse
T
"
Î"
!
Ï#
Î"
œ !
Ï#
#
%
#
#> />
Ñ
Þ
%> />
>
>Ò
#> / /
! Ñ
!
,
Ò
"
! Ñ
!
Þ
"Ò
"Î#
"Î%
$Î#
The Jordan form of the matrix A is
Î"
J œ T AT œ !
Ï!
"
!
"
!
!Ñ
" Þ
"Ò
20a+b. Direct multiplication results in
________________________________________________________________________
page 425
—————————————————————————— CHAPTER 7. ——
Î#
J œ
!
Ï !
#
!
-#
!
! Ñ
-$
Î
#- ß J$ œ
!
Ï
#Ò
!
a,b. Suppose that
Î8
J œ
!
Ï !
8
Then
J8+1
!
-$
!
-%
! Ñ
Î
$-# ß J% œ
!
Ï
$ Ò
!
!
-8
!
! Ñ
8-8" Þ
-8 Ò
!
-%
!
! Ñ
%-$ Þ
-% Ò
8
!
! ÑÎ - ! ! Ñ
Î8
! - "
œ
! - 8-8"
8 ÒÏ !
Ï !
! -Ò
!
8
!
!
Î- † Ñ
8
8
œ
!
- † - - 8- † -8" Þ
Ï !
Ò
!
- † -8
Hence the result follows by mathematical induction.
a- b. Note that J is block diagonal. Hence each block may be exponentiated. Using the
result in Prob. a"*b,
Î/
/B:aJ>b œ
!
Ï !
->
!
/- >
!
! Ñ
>/-> Þ
/- > Ò
a. b. Setting - œ " , and using the transformation matrix T in Prob. a")b,
#
! ÑÎ /> ! ! Ñ
Î"
%
!
T/B:aJ>b œ !
! /> >/>
Ï # # " ÒÏ ! ! /> Ò
>
#/>
#> />
Î /
Ñ
>
œ
Þ
!
%/
%> />
Ï #/> #/> #> /> /> Ò
Based on the form of J , /B:aJ>b is the fundamental matrix associated with the solutions
ya"b œ 0 a"b /> , ya#b œ ˆ#0a"b #0a#b ‰/> and ya$b œ ˆ#0a"b #0a#b ‰>/> ( /> Þ
Hence the resulting matrix is the fundamental matrix associated with the solution set
________________________________________________________________________
page 426
—————————————————————————— CHAPTER 7. ——
˜0 a"b /> ß ˆ#0 a"b #0 a#b ‰/> ß ˆ#0 a"b #0 a#b ‰>/> ( /> ™,
as opposed to the solution set in Prob. a")b, given by
˜0 a"b /> ß 0 a#b /> ß ˆ#0 a"b #0 a#b ‰>/> ( /> ™.
21a+b. Direct multiplication results in
ÎJ œ
!
Ï !
#
#
#-#
!
$
" Ñ
Î$
#- ß J œ
!
Ï
#Ò
!
a,b. Suppose that
Then
8+1
J
Î -8
8
J œÐ !
Ï !
Î -8
œÐ !
Ï !
8-8"
-8
!
Î - † -8
œÐ !
Ï
!
$-#
-$
!
8-8"
-8
!
8a8"b 8#
#
8"
8-8
-8 8- † -8"
- † -8
!
%
$- Ñ
Î%
$-# ß J œ
!
Ï
$ Ò
!
8a8"b 8#
#
8"
8-8
ÑÎ Ó !
ÒÏ !
"
!
%-$
-%
!
'-# Ñ
%-$ Þ
-% Ò
Ñ
ÓÞ
Ò
!Ñ
"
-Ò
b
8-8" 8a8"
- † -8# Ñ
#
ÓÞ
-8 8- † -8"
Ò
- † -8
The result follows by noting that
8-8"
8a8 "b
8a8 "b 8"
- † -8# œ ”8
•#
#
8# 8 8"
œ
- Þ
#
a- b. We first observe that
________________________________________________________________________
page 427
—————————————————————————— CHAPTER 7. ——
" -8
_
8œ!
" 8-8"
_
>8
œ /->
8x
_
>8
>8"
"
œ>
-8"
œ > /->
8x
a8 "bx
8œ"
8a8 "b 8# >8
># _ 8# >8#
>#
"
œ "œ /-> Þ
#
8x
# 8œ#
#
a8 #bx
8œ!
8œ!
_
Therefore
Î /- >
/B:aJ>b œ Ð !
Ï !
>/->
/- >
!
># - >
#/
->
>/
/- >
Ñ
ÓÞ
Ò
a. b. Setting - œ # , and using the transformation matrix T in Prob. a"(b,
Î !
"
T/B:aJ>b œ
Ï "
Î !
œ Ð /#>
Ï /#>
" # ÑÎ /#>
" ! Ð !
! $ ÒÏ !
/#>
>/#> /#>
>/#>
>/#>
/#>
!
># #>
#/
#>
Ñ
Ó
>/
/#> Ò
>/#> #/#> Ñ
># #>
#> Ó
Þ
# / >/
#
> #>
#> Ò
/ $/
#
________________________________________________________________________
page 428
—————————————————————————— CHAPTER 7. ——
Section 7.9
5. As shown in Prob. # , Section (Þ) , the general solution of the homogeneous equation
is
"
>
x- œ -" Œ -# Œ
Þ
#
#> "#
An associated fundamental matrix is
Ga>b œ Œ
"
#
>
Þ
#> "#
The inverse of the fundamental matrix is easily determined as
G" a>b œ Œ
%> $
)> )
#> #
Þ
%> &
We can now compute
G" a>bga>b œ
" #># %> "
Œ
,
>$
#> %
and
"
( G a>bga>b .> œ Œ
"# ># %>" #68 >
Þ
#># #>"
Finally,
va>b œ Ga>b( G" a>bga>b .>,
where
"
@" a>b œ ># #>" # 68 > #
#
"
@# a>b œ &> % 68 > % Þ
Note that the vector a# ß %bX is a multiple of one of the fundamental solutions. Hence we
can write the general solution as
"
>
" "Î#
" #
"
x œ -" Œ - # Œ
Œ #68 >Œ Þ
" #Œ
#
#> #
>
!
> &
#
6. The eigenvalues of the coefficient matrix are <" œ ! and <# œ & . It follows that
the solution of the homogeneous equation is
________________________________________________________________________
page 429
—————————————————————————— CHAPTER 7. ——
"
#/&>
x- œ -" Œ -# Œ &> Þ
#
/
The coefficient matrix is symmetric. Hence the system is diagonalizable. Using the
normalized eigenvectors as columns, the transformation matrix, and its inverse, are
Tœ
"
"
È& Œ #
"
#
"
, T" œ
"
È& Œ #
#
Þ
"
Setting x œ Ty , and ha>b œ T" ga>b , the transformed system is given, in scalar form,
as
C"w œ
& )>
È& >
C#w œ &C#
%
Þ
È&
The solutions are readily obtained as
C" a>b œ È& 68 >
%
%
> -" and C# a>b œ -# /&>
Þ
È&
&È&
Transforming back to the original variables, we have x œ Ty , with
x œ
œ
"
"
È& Œ #
C" a>b
#
Œ
"
C# a>b
"
"
"
#
C" a>b
Œ
Œ
C# a>bÞ
È& #
È&
"
Hence the general solution is,
"
#/&>
"
% "
% #
x œ 5" Œ 5 # Œ &> Œ 68 > Œ > Œ
Þ
#
/
#
& #
#&
"
7. The solution of the homogeneous equation is
x- œ -" Œ
/>
/$>
#Œ
Þ
#/>
#/$>
Based on the simple form of the right hand side, we use the method of undetermined
coefficients. Set v œ a /> . Substitution into the ODE yields
Œ
+" >
"
/ œ Œ %
+#
+" >
#
"
>
/ Œ
Œ
/ Þ
"
+#
"
In scalar form, after canceling the exponential, we have
________________________________________________________________________
page 430
—————————————————————————— CHAPTER 7. ——
+" œ + " + # #
+# œ %+" +# " ,
with +" œ "Î% and +# œ # . Hence the particular solution is
vœŒ
"Î% >
/ ,
#
so that the general solution is
x œ -" Œ
/>
/$>
"
/>
#Œ
Œ
Þ
#/>
#/$>
% )/>
8. The eigenvalues of the coefficient matrix are <" œ " and <# œ " . It follows that
the solution of the homogeneous equation is
"
"
x- œ -" Œ /> -# Œ /> Þ
"
$
Use the method of undetermined coefficients. Since the right hand side is related to one
of the fundamental solutions, set v œ a >/> b /> . Substitution into the ODE yields
Œ
+"
," >
+"
# "
>
>
>
ˆ/ >/ ‰ Œ / œ Œ $ # Œ >/
+#
,#
+#
," >
"
# "
>
Œ
/ Œ
Œ
/ Þ
$ #
,#
"
In scalar form, we have
a+" ," b/> +" >/> œ a#+" +# b>/> a#," ,# b/> />
a+# ,# b/> +# >/> œ a$+" #+# b>/> a$," #,# b/> /> Þ
Equating the coefficients in these two equations, we find that
+"
+ " ,"
+#
+# ,#
œ #+" +#
œ #," ,# "
œ $+" #+#
œ $," #,# " Þ
It follows that +" œ +# . Setting +" œ +# œ + , the equations reduce to
," ,# œ + "
$," $,# œ " + Þ
Combining these equations, it is necessary that + œ # . As a result, ," œ ,# " .
Choosing +" œ +# œ # , and ,# œ 5 , some arbitrary constant, a particular solution is
________________________________________________________________________
page 431
—————————————————————————— CHAPTER 7. ——
#
5" >
#
" >
" >
>
v œ Œ >/> Œ
/ œ Œ >/ 5 Œ / Œ / Þ
#
5
#
"
!
Since the second vector is a fundamental solution, the general solution can be written as
"
"
#
"
x œ -" Œ /> -# Œ /> Œ >/> Œ /> Þ
"
$
#
!
9. Note that the coefficient matrix is symmetric. Hence the system is diagonalizable.
The eigenvalues and eigenvectors are given by
<" œ
"
"
"
, 0 a"b œ Œ and <# œ # , 0 a#b œ Œ
Þ
#
"
"
Using the normalized eigenvectors as columns, the transformation matrix, and its inverse,
are
Tœ
"
"
Œ
È# "
"
"
"
, T" œ
Œ
"
È# "
"
Þ
"
Setting x œ Ty , and ha>b œ T" ga>b , the transformed system is given, in scalar form,
as
"
" >
C"w œ C" È# >
/
È#
#
" >
C#w œ #C# È# >
/ Þ
È#
Using any elementary method for first order linear equations, the solutions are
È#
/ > %È # #È # >
$
" >
"
"
C# a>b œ 5 # /#>
/
>Þ
È#
$È#
#È#
C" a>b œ 5" />Î#
Transforming back to the original variables, x œ Ty , the general solution is
"
"
" "(
" &
" " >
#>
x œ -" Œ />Î# -# Œ
/ Œ Œ > Œ / Þ
"
"
% "&
# $
' $
10. Since the coefficient matrix is symmetric, the differential equations can be
decoupled.
The eigenvalues and eigenvectors are given by
________________________________________________________________________
page 432
—————————————————————————— CHAPTER 7. ——
È#
"
<" œ % , 0 a"b œ
and <# œ " , 0 a#b œ Œ È Þ
"
#
Using the normalized eigenvectors as columns, the transformation matrix, and its inverse,
are
Tœ
È#
"
È$ "
È#
"
"
"
,
T
œ
È#
È$ "
"
Þ
È#
Setting x œ Ty , and ha>b œ T" ga>b , the transformed system is given, in scalar form,
as
"
Š" È#‹/>
È$
"
C#w œ C#
Š" È#‹ /> Þ
È$
C"w œ %C"
The solutions are easily obtained as
"
Š" È#‹/>
È
$ $
"
C# a>b œ 5 # />
Š" È# ‹>/> Þ
È$
C" a>b œ 5" /%>
Transforming back to the original variables, the general solution is
x œ -"
È#
/
"
Note that
4>
"
" # È# $È$ > " " È#
-# Œ È /> È È
>/> Þ
/ È
* $ ' # "
$
# #
#
# È # $È $
# È#
"
È
œ
$È' È# " È# " $ $Œ È# Þ
The second vector is an eigenvector, hence the solution may be written as
x œ -"
È#
"
# È#
"
" " È#
4>
>
>
/
/
/
>/> Þ
#
Œ
È#
* È# "
$ È# #
"
11. Based on the solution of Prob. $ of Section (Þ' , a fundamental matrix is given by
Ga>b œ Œ
& -9= >
# -9= > =38 >
& =38 >
Þ
-9= > # =38 >
The inverse of the fundamental matrix is easily determined as
________________________________________________________________________
page 433
—————————————————————————— CHAPTER 7. ——
G" a>b œ
" -9= > #=38 >
Œ
& #-9= > =38 >
&=38 >
Þ
&-9= >
It follows that
G" a>bga>b œ Œ
-9= > =38 >
,
-9=# >
and
"
( G a>bga>b .> œ
Þ
"# -9= > =38 > "# >
"
#
# =38 >
A particular solution is constructed as
va>b œ Ga>b( G" a>bga>b .>,
where
@" a>b œ
&
&
-9= > =38 > -9=# > > "
#
#
"
"
@# a>b œ -9= > =38 > -9=# > > Þ
#
#
Hence the general solution is
x œ -" Œ
& -9= >
& =38 >
-# Œ
# -9= > =38 >
-9= > # =38 >
&Î#
!
> =38 >Œ
a> -9= > =38 >bŒ
Þ
"
"Î#
13a+b. As shown in Prob. #& of Section (Þ' , the solution of the homogeneous system is
Ba"-b
-9=a>Î#b
=38a>Î#b
>Î#
>Î#
Œ
-# /
Œ
Þ
Ba-b œ -" /
% =38a>Î#b
% -9=a>Î#b
#
Therefore the associated fundamental matrix is given by
Ga>b œ />Î# Œ
-9=a>Î#b
% =38a>Î#b
a,bÞ The inverse of the fundamental matrix is
/>Î# % -9=a>Î#b
G a>b œ
Œ % =38a>Î#b
%
"
=38a>Î#b
Þ
% -9=a>Î#b
=38a>Î#b
Þ
-9=a>Î#b
________________________________________________________________________
page 434
—————————————————————————— CHAPTER 7. ——
It follows that
G" a>bga>b œ
" -9=a>Î#b
Œ
,
# =38a>Î#b
and
"
( G a>bga>b .> œ Œ
=38a>Î#b
Þ
-9=a>Î#b
A particular solution is constructed as
va>b œ Ga>b( G" a>bga>b .>,
where
@" a>b œ !
@# a>b œ % />Î# Þ
Hence the general solution is
x œ -" />Î# Œ
-9=a>Î#b
=38a>Î#b
>Î#
>Î# !
-# /
Œ
%/
Œ Þ
% =38a>Î#b
% -9=a>Î#b
"
Imposing the initial conditions, we require that
-" œ !
%-# % œ ! ,
which results in -" œ ! and -# œ " Þ Therefore the solution of the IVP is
x œ />Î# Œ
=38a>Î#b
Þ
% % -9=a>Î#b
15. The general solution of the homogeneous problem is
Ba"-b
" "
# #
Ba-b œ -" Œ # > -# Œ " > ,
#
which can be verified by substitution into the system of ODEs. Since the vectors are
linearly independent, a fundamental matrix is given by
Ga>b œ Œ
>"
#>"
#>#
Þ
>#
The inverse of the fundamental matrix is
________________________________________________________________________
page 435
—————————————————————————— CHAPTER 7. ——
G" a>b œ
" >
Œ
$ #>#
#>
Þ
>#
Dividing both equations by > , we obtain
ga>b œ Œ
>$
#
.
>"
Proceeding with the method of variation of parameters,
G" a>bga>b œ
# %
#
#
$> $> $
,
"
% #
"$ >$
$> $>
and
"
( G a>bga>b .> œ
# &
" #
#
"& > $ > $ >
Þ
" #
% "
" #
>
>
>
'
$
'
Hence a particular solution is obtained as
"& >% $> "
vœ " %
Þ
$
"! > #> #
The general solution is
"
#
" # %
$
"
x œ œ -" Œ >" -# Œ ># Œ
> Œ > Œ
Þ
#
"
"!
"
#
$Î#
16. Based on the hypotheses,
9 w a>b œ Pa>b9a>b ga>b and v w a>b œ Pa>bva>b ga>b .
Subtracting the two equations results in
9 w a>b v w a>b œ Pa>b9a>b Pa>bva>b ,
that is,
c9a>b va>bd w œ Pa>bc9a>b va>bdÞ
It follows that 9a>b va>b is a solution of the homogeneous equation. According to
Theorem (Þ%Þ# ,
9a>b va>b œ -" xa"b a>b -# xa#b a>b â -8 xa8b a>bÞ
Hence
9a>b œ ua>b va>b,
________________________________________________________________________
page 436
—————————————————————————— CHAPTER 7. ——
in which ua>b is the general solution of the homogeneous problem.
17a+b. Setting >! œ ! in Eq. a$%b,
x œ Fa>bx! Fa>b( F" a=bga=b.=
>
!
œ Fa>bx! ( Fa>bF" a=bga=b.= Þ
>
!
It was shown in Prob. "&a- b in Section (Þ( that Fa>bF" a=b œ Fa> =bÞ Therefore
x œ Fa>bx! ( Fa> =bga=b.= Þ
>
!
a,b. The principal fundamental matrix is identified as Fa>b œ /B:aA>b. Hence
x œ /B:aA>bx! ( /B:cAa> =bdga=b.= Þ
>
!
In Prob. #' of Section $Þ(, the particular solution is given as
] a>b œ ( O a> =b1a=b.= ,
>
>!
in which the kernel O a>b depends on the nature of the fundamental solutions.
________________________________________________________________________
page 437
—————————————————————————— CHAPTER 8. ——
Chapter Eight
Section 8.1
2. The Euler formula for this problem is
C8" œ C8 2ˆ& >8 $ÈC8 ‰,
in which >8 œ >! 82 Þ Since >! œ ! , we can also write
C8" œ C8 &82# $2 ÈC8 ,
with C! œ # .
a+b. Euler method with 2 œ !Þ!& À
8œ#
!Þ"
"Þ&**)!
>8
C8
8œ%
!Þ#
"Þ#*#))
8œ'
!Þ$
"Þ!(#%#
8œ)
!Þ%
!Þ*$!"(&
8œ)
!Þ#
"Þ$"$'"
8 œ "#
!Þ$
"Þ"!!"#
8 œ "'
!Þ%
!Þ*'#&&#
a,b. Euler method with 2 œ !Þ!#& À
8œ%
!Þ"
"Þ'""#%
>8
C8
The backward Euler formula is
C8" œ C8 2ˆ& >8+1 $ÈC8+1 ‰,
in which >8 œ >! 82 Þ Since >! œ ! , we can also write
C8" œ C8 &a8 "b2# $2 ÈC8" ,
with C! œ # . Solving for C8" , and choosing the positive root, we find that
C8"
#
$
"È
#
œ” 2
a#!8 #*b2 %C8 • Þ
#
#
________________________________________________________________________
page 438
—————————————————————————— CHAPTER 8. ——
a- b. Backward Euler method with 2 œ !Þ!& À
>8
C8
8œ#
!Þ"
"Þ'%$$(
8œ%
!Þ#
"Þ$("'%
8œ'
!Þ$
"Þ"(('$
8œ)
!Þ%
"Þ!&$$%
8 œ "#
!Þ$
"Þ"&#'(
8 œ "'
!Þ%
"Þ!#%!(
a. b. Backward Euler method with 2 œ !Þ!#& À
>8
C8
8œ%
!Þ"
"Þ'$$!"
8œ)
!Þ#
"Þ$&#*&
3. The Euler formula for this problem is
C8" œ C8 2a# C8 $ >8 b,
in which >8 œ >! 82 Þ Since >! œ ! ,
C8" œ C8 #2C8 $82 # ,
with C! œ " .
a+b. Euler method with 2 œ !Þ!& À
>8
C8
8œ#
!Þ"
"Þ#!#&
8œ%
!Þ#
"Þ%"'!$
8œ'
!Þ$
"Þ'%#)*
8œ)
!Þ%
"Þ))&*!
8œ)
!Þ#
"Þ%"*$'
8 œ "#
!Þ$
"Þ'%)*'
8 œ "'
!Þ%
"Þ)*&(#
a,b. Euler method with 2 œ !Þ!#& À
>8
C8
8œ%
!Þ"
"Þ#!$))
The backward Euler formula is
C8" œ C8 2a# C8+1 $ >8" b,
in which >8 œ >! 82 Þ Since >! œ ! , we can also write
C8" œ C8 #2 C8" $a8 "b2 # ,
with C! œ " . Solving for C8" , we find that
C8" œ
C8 $a8 "b2#
Þ
" #2
________________________________________________________________________
page 439
—————————————————————————— CHAPTER 8. ——
a- b. Backward Euler method with 2 œ !Þ!& À
>8
C8
8œ#
!Þ"
"Þ#!)'%
8œ%
!Þ#
"Þ%$"!%
8œ'
!Þ$
"Þ'(!%#
8œ)
!Þ%
"Þ*$!('
8 œ "#
!Þ$
"Þ''#'&
8 œ "'
!Þ%
"Þ*")!#
a. b. Backward Euler method with 2 œ !Þ!#& À
>8
C8
4. The Euler formula is
8œ%
!Þ"
"Þ#!'*$
8œ)
!Þ#
"Þ%#')$
C8" œ C8 2c# >8 /B:a >8 C8 bdÞ
Since >8 œ >! 82 and >! œ ! , we can also write
C8" œ C8 #82# 2 /B:a 82 C8 b,
with C! œ " Þ
a+b. Euler method with 2 œ !Þ!& À
>8
C8
8œ#
!Þ"
"Þ"!#%%
8œ%
!Þ#
"Þ#"%#'
8œ'
!Þ$
"Þ$$%)%
8œ)
!Þ%
"Þ%'$**
8œ)
!Þ#
"Þ#"'&'
8 œ "#
!Þ$
"Þ$$)"(
8 œ "'
!Þ%
"Þ%')$#
a,b. Euler method with 2 œ !Þ!#& À
>8
C8
8œ%
!Þ"
"Þ"!$'&
The backward Euler formula is
C8" œ C8 2c# >8" /B:a >8" C8" bdÞ
Since >! œ ! and >8" œ a8 "b2 , we can also write
C8" œ C8 #2# a8 "b 2 /B:c a8 "b2 C8" d,
with C! œ " Þ This equation cannot be solved explicitly for C8" . At each step, given
the current value of C8 , the equation must be solved numerically for C8" Þ
________________________________________________________________________
page 440
—————————————————————————— CHAPTER 8. ——
a- b. Backward Euler method with 2 œ !Þ!& À
>8
C8
8œ#
!Þ"
"Þ"!(#!
8œ%
!Þ#
"Þ##$$$
8œ'
!Þ$
"Þ$%(*(
8œ)
!Þ%
"Þ%)""!
8 œ "#
!Þ$
"Þ$%%($
8 œ "'
!Þ%
"Þ%('))
a. b. Backward Euler method with 2 œ !Þ!#& À
>8
C8
8œ%
!Þ"
"Þ"!'!$
8œ)
!Þ#
"Þ##""!
6. The Euler formula for this problem is
C8" œ C8 2ˆ>#8 C8# ‰=38 C8 .
Here >! œ ! and >8 œ 82 . So that
C8" œ C8 2ˆ8# 2# C8# ‰=38 C8 ,
with C! œ " .
a+b. Euler method with 2 œ !Þ!& À
>8
C8
8œ#
!Þ"
!Þ*#!%*)
8œ%
!Þ#
!Þ)&(&$)
8œ'
!Þ$
!Þ)!)!$!
8œ)
!Þ%
!Þ((!!$)
8 œ "#
!Þ$
!Þ)#$!!
8 œ "'
!Þ%
!Þ((%*'&
a,b. Euler method with 2 œ !Þ!#& À
>8
C8
8œ%
!Þ"
!Þ*##&(&
8œ)
!Þ#
!Þ)'!*#$
________________________________________________________________________
page 441
—————————————————————————— CHAPTER 8. ——
The backward Euler formula is
#
‰=38 C8" Þ
C8" œ C8 2ˆ>#8" C8"
Since >! œ ! and >8" œ a8 "b2 , we can also write
# ‘
C8" œ C8 2a8 "b# 2 # C8"
=38 C8" ,
with C! œ " . Note that this equation cannot be solved explicitly for C8" . Given C8 ,
the transcendental equation
#
C8" 2 C8"
=38 C8" œ C8 2 a8 "b# 2 #
must be solved numerically for C8" Þ
a- b. Backward Euler method with 2 œ !Þ!& À
>8
C8
8œ#
!Þ"
!Þ*#)!&*
8œ%
!Þ#
!Þ)(!!&%
8œ'
!Þ$
!Þ)#%!#"
8œ)
!Þ%
!Þ())')'
8 œ "#
!Þ$
!Þ)#!#(*
8 œ "'
!Þ%
!Þ()%#(&
a. b. Backward Euler method with 2 œ !Þ!#& À
>8
C8
8œ%
!Þ"
!Þ*#'$%"
8. The Euler formula
8œ)
!Þ#
!Þ)'("'$
C8" œ C8 2ˆ& >8 $ÈC8 ‰,
in which >8 œ >! 82 Þ Since >! œ ! , we can also write
C8" œ C8 &82# $2 ÈC8 ,
with C! œ # .
a+b. Euler method with 2 œ !Þ!#& À
>8
C8
8 œ #!
!Þ&
!Þ)*")$!
8 œ %!
"Þ!
"Þ#&##&
8 œ '!
"Þ&
#Þ$()")
8 œ )!
#Þ!
%Þ!(#&(
________________________________________________________________________
page 442
—————————————————————————— CHAPTER 8. ——
a,b. Euler method with 2 œ !Þ!"#& À
8 œ %!
!Þ&
!Þ*!)*!#
>8
C8
8 œ )!
"Þ!
"Þ#')(#
8 œ "#!
"Þ&
#Þ$*$$'
8 œ "'!
#Þ!
%Þ!)(**
The backward Euler formula is
C8" œ C8 2ˆ& >8+1 $ÈC8+1 ‰,
in which >8 œ >! 82 Þ Since >! œ ! , we can also write
C8" œ C8 &a8 "b2# $2 ÈC8" ,
with C! œ # . Solving for C8" , and choosing the positive root, we find that
C8"
#
$
"È
#
œ” 2
a#!8 #*b2 %C8 • Þ
#
#
a- b. Backward Euler method with 2 œ !Þ!#& À
>8
C8
8 œ #!
!Þ&
!Þ*&)&'&
8 œ %!
"Þ!
"Þ$"()'
8 œ '!
"Þ&
#Þ%$*#%
8 œ )!
#Þ!
%Þ"$%(%
8 œ "#!
"Þ&
#Þ#%$)*
8 œ "'!
#Þ!
%Þ""*!)
a. b. Backward Euler method with 2 œ !Þ!"#& À
>8
C8
8 œ %!
!Þ&
!Þ*%##'"
8 œ )!
"Þ!
"Þ$!"&$
9. The Euler formula for this problem is
C8" œ C8 2È>8 C8 .
Here >! œ ! and >8 œ 82 . So that
C8" œ C8 2È82 C8 ,
with C! œ $ .
________________________________________________________________________
page 443
—————————————————————————— CHAPTER 8. ——
10. The Euler formula is
C8" œ C8 2c# >8 /B:a >8 C8 bdÞ
Since >8 œ >! 82 and >! œ ! , we can also write
C8" œ C8 #82# 2 /B:a 82 C8 b,
with C! œ " Þ
a+b. Euler method with 2 œ !Þ!#& À
>8
C8
8 œ #!
!Þ&
"Þ'!(#*
8 œ %!
"Þ!
#Þ%')$!
8 œ '!
"Þ&
$Þ(#"'(
8 œ )!
#Þ!
&Þ%&*'$
8 œ )!
"Þ!
#Þ%(%'!
8 œ "#!
"Þ&
$Þ($$&'
8 œ "'!
#Þ!
&Þ%(((%
a,b. Euler method with 2 œ !Þ!"#& À
>8
C8
8 œ %!
!Þ&
"Þ'!**'
The backward Euler formula is
C8" œ C8 2c# >8" /B:a >8" C8" bdÞ
Since >! œ ! and >8" œ a8 "b2 , we can also write
C8" œ C8 #2# a8 "b 2 /B:c a8 "b2 C8" d,
with C! œ " Þ This equation cannot be solved explicitly for C8" . At each step, given
the current value of C8 , the equation must be solved numerically for C8" Þ
a- b. Backward Euler method with 2 œ !Þ!#& À
>8
C8
8 œ #!
!Þ&
"Þ'"(*#
8 œ %!
"Þ!
#Þ%*$&'
8 œ '!
"Þ&
$Þ('*%!
8 œ )!
#Þ!
&Þ&$##$
________________________________________________________________________
page 444
—————————————————————————— CHAPTER 8. ——
a. b. Backward Euler method with 2 œ !Þ!"#& À
>8
C8
11. The Euler formula is
8 œ %!
!Þ&
"Þ'"&#)
8 œ )!
"Þ!
#Þ%)(#$
8 œ "#!
"Þ&
$Þ(&(%#
8 œ "'!
#Þ!
&Þ&"%!%
C8" œ C8 2a% >8 C8 bΈ" C8# ‰Þ
Since >8 œ >! 82 and >! œ ! , we can also write
C8" œ C8 ˆ%2 82 # C8 ‰Îˆ" C8# ‰,
with C! œ # Þ
a+b. Euler method with 2 œ !Þ!#& À
>8
C8
8 œ #!
!Þ&
"Þ%&)'&
8 œ %!
"Þ!
!Þ#"(&%&
8 œ '!
"Þ&
"Þ!&("&
8 œ )!
#Þ!
"Þ%"%)(
8 œ "#!
"Þ&
"Þ!&*!$
8 œ "'!
#Þ!
"Þ%"#%%
a,b. Euler method with 2 œ !Þ!"#& À
>8
C8
8 œ %!
!Þ&
"Þ%&$##
8 œ )!
"Þ!
!Þ")!)"$
The backward Euler formula is
# ‰
C8" œ C8 2a% >8" C8" bΈ" C8"
Þ
Since >! œ ! and >8" œ a8 "b2 , we can also write
# ‰
# ‰
C8" ˆ" C8"
œ C8 ˆ" C8"
%2 a8 "b2 # C8" ‘,
with C! œ # Þ This equation cannot be solved explicitly for C8" . At each step, given
the current value of C8 , the equation must be solved numerically for C8" Þ
________________________________________________________________________
page 445
—————————————————————————— CHAPTER 8. ——
a- b. Backward Euler method with 2 œ !Þ!#& À
8 œ #!
!Þ&
"Þ%$'!!
>8
C8
8 œ %!
"Þ!
!Þ!')"'&(
8 œ '!
"Þ&
"Þ!'%)*
8 œ )!
#Þ!
"Þ%!&(&
8 œ "#!
"Þ&
"Þ!'#*!
8 œ "'!
#Þ!
"Þ%!()*
a. b. Backward Euler method with 2 œ !Þ!"#& À
8 œ %!
!Þ&
"Þ%%"*!
>8
C8
8 œ )!
"Þ!
!Þ"!&($(
12. The Euler formula is
C8" œ C8 2ˆC8# # >8 C8 ‰Îˆ$ >8# ‰Þ
Since >8 œ >! 82 and >! œ ! , we can also write
C8" œ C8 ˆ2 C8# # 82 # C8 ‰Îˆ$ 8# 2 # ‰,
with C! œ !Þ& Þ
a+b. Euler method with 2 œ !Þ!#& À
>8
C8
8 œ #!
!Þ&
!Þ&)(*)(
8 œ %!
"Þ!
!Þ(*"&)*
8 œ '!
"Þ&
"Þ"%(%$
8 œ )!
#Þ!
"Þ(!*($
8 œ )!
"Þ!
!Þ(*&(&)
8 œ "#!
"Þ&
"Þ"&'*$
8 œ "'!
#Þ!
"Þ(#*&&
a,b. Euler method with 2 œ !Þ!"#& À
>8
C8
8 œ %!
!Þ&
!Þ&)*%%!
The backward Euler formula is
#
# ‰
C8" œ C8 2ˆC8"
# >8" C8" ‰Îˆ$ >8"
Þ
Since >! œ ! and >8" œ a8 "b2 , we can also write
#
C8" $ a8 "b# 2# ‘ 2 C8"
œ C8 $ a8 "b# 2 # ‘ #a8 "b2 # C8" ,
with C! œ !Þ& Þ Note that although this equation can be solved explicitly for C8" , it is
also possible to use a numerical equation solver. At each time step, given the current
________________________________________________________________________
page 446
—————————————————————————— CHAPTER 8. ——
value of C8 , the equation may be solved numerically for C8" Þ
a- b. Backward Euler method with 2 œ !Þ!#& À
>8
C8
8 œ #!
!Þ&
!Þ&*$*!"
8 œ %!
"Þ!
!Þ)!)("'
8 œ '!
"Þ&
"Þ")')(
8 œ )!
#Þ!
"Þ(*#*"
8 œ "#!
"Þ&
"Þ"(''%
8 œ "'!
#Þ!
"Þ(("""
a. b. Backward Euler method with 2 œ !Þ!"#& À
>8
C8
8 œ %!
!Þ&
!Þ&*#$*'
8 œ )!
"Þ!
!Þ)!%$"*
13. The Euler formula for this problem is
C8" œ C8 2a" >8 % C8 b,
in which >8 œ >! 82 Þ Since >! œ ! , we can also write
C8" œ C8 2 82 # %2 C8 ,
with C! œ " . With 2 œ !Þ!" , a total number of #!! iterations is needed to reach > œ # Þ
With 2 œ !Þ!!" , a total of #!!! iterations are necessary.
14. The backward Euler formula is
C8" œ C8 2a" >8" % C8" bÞ
Since the equation is linear, we can solve for C8" in terms of C8 À
C8" œ
C8 2 2 >8"
.
" %2
Here >! œ ! and C! œ " Þ With 2 œ !Þ!" , a total number of #!! iterations is needed to
reach > œ # Þ With 2 œ !Þ!!" , a total of #!!! iterations are necessary.
18. Let 9a>b be a solution of the initial value problem. The local truncation error for the
Euler method, on the interval >8 Ÿ > Ÿ >8" , is given by
/8" œ
" ww
9 a > 8 b2 # ,
#
where >8 >8 >8" Þ Since 9 w a>b œ ># c9a>bd# , it follows that
9 ww a>b œ #> # 9a>b9 w a>b
œ #> #># 9a>b #c9a>bd$ Þ
________________________________________________________________________
page 447
—————————————————————————— CHAPTER 8. ——
Hence
$ ‘ #
k/8" k Ÿ >8" >#8" Q8" Q8"
2 ,
in which Q8" œ 7+Be9a>b l >8 Ÿ > Ÿ >8" fÞ
20. Given that 9a>b is a solution of the initial value problem, the local truncation error
for the Euler method, on the interval >8 Ÿ > Ÿ >8" , is
/8" œ
" ww
9 a > 8 b2 # ,
#
where >8 >8 >8" Þ Based on the ODE, 9 w a>b œ È> 9a>b , and hence
" 9 w a >b
#È> 9a>b
"
"
œ
Þ
#
#È> 9a>b
9 ww a>b œ
Therefore
k/8" k Ÿ
"
"
"
2# Þ
–
È>8 9a >8 b —
%
21. Let 9a>b be a solution of the initial value problem. The local truncation error for the
Euler method, on the interval >8 Ÿ > Ÿ >8" , is given by
/8" œ
" ww
9 a > 8 b2 # ,
#
where >8 >8 >8" Þ Since 9 w a>b œ #> /B:c > 9a>bd, it follows that
9 ww a>b œ # #c9a>b > 9 w a>bd † /B:c > 9a>bd
œ # ˜9a>b #># > /B:c > 9a>bd™ † /B:c > 9a>bdÞ
Hence
/8" œ 2#
2#
#
š9a >8 b # >8 >8 /B:c >8 9a >8 bd› † /B:c >8 9a >8 bdÞ
#
22a+b. Direct integration yields 9a>b œ
"
&1 =38 &1>
"Þ
________________________________________________________________________
page 448
—————————————————————————— CHAPTER 8. ——
a, bÞ
>8
C8
8œ!
!Þ!
"Þ!
8œ"
!Þ#
"Þ#
8œ#
!Þ%
"Þ!
8œ$
!Þ'
"Þ#
>8
C8
8œ!
!Þ!
"Þ!
8œ#
!Þ#
"Þ"
8œ%
!Þ%
"Þ!
8œ'
!Þ'
"Þ"
a- bÞ
________________________________________________________________________
page 449
—————————————————————————— CHAPTER 8. ——
a. b. Since 9 ww a>b œ &1 =38 &1> , the local truncation error for the Euler method, on
the interval >8 Ÿ > Ÿ >8" , is given by
&12#
=38 &1 >8 Þ
#
/8" œ
In order to satisfy
k/8" k Ÿ
&12#
!Þ!& ,
#
it is necessary that
2
25a+b. The Euler formula is
>8
C8
È&!1
"
¸ !Þ!) Þ
C8" œ C8 2a" >8 % C8 b.
8œ#
!Þ"
"Þ&&
8œ%
!Þ#
#Þ$%
8œ'
!Þ$
$Þ%'
8œ)
!Þ%
&Þ!(
a,b. The Euler formula for this problem is
C8" œ C8 2a$ >8 C8 b.
>8
C8
8œ#
!Þ"
"Þ#!
8œ%
!Þ#
"Þ$*
8œ'
!Þ$
"Þ&(
8œ)
!Þ%
"Þ(%
a- b. The Euler formula is
________________________________________________________________________
page 450
—————————————————————————— CHAPTER 8. ——
C8" œ C8 2a# C8 $ >8 b.
>8
C8
26a+b.
"!!! † º
a, b .
a- b .
8œ#
!Þ"
"Þ#!
"!!! † º
'Þ!
#Þ!
8œ%
!Þ#
"Þ%#
8œ'
!Þ$
"Þ'&
8œ)
!Þ%
"Þ*!
")
œ "!!! † a!b œ ! Þ
'Þ! º
")Þ!
œ "!!!a!Þ!'b œ '! Þ
'Þ!! º
"!!! † º
'Þ!"
#Þ!!
'Þ!"!
#Þ!!%
")Þ!%
œ "!!!a !Þ!*#"'b œ *# Þ"' Þ
'Þ!!! º
27. Rounding to three digits, +a, - b ¸ !Þ##% . Likewise, to three digits, +, ¸ !Þ(!#
and +- ¸ !Þ%(( Þ It follows that +, +- ¸ !Þ##& Þ
________________________________________________________________________
page 451
—————————————————————————— CHAPTER 8. ——
Section 8.2
1. The improved Euler formula for this problem is
"
"
2#
C8" œ C8 2Œ$ >8 >8" C8 a$ >8 C8 bÞ
#
#
#
Since >8 œ >! 82 and >! œ ! , we can also write
C8" œ C8 2a$ C8 b
2#
82$
aC8 # #8b
,
#
#
with C! œ " .
a+b. 2 œ !Þ!& À
>8
C8
8œ#
!Þ"
"Þ"*&"#
8œ%
!Þ#
"Þ$)"#!
8œ'
!Þ$
"Þ&&*!*
8œ)
!Þ%
"Þ(#*&'
>8
C8
8œ%
!Þ"
"Þ"*&"&
8œ)
!Þ#
"Þ$)"#&
8 œ "#
!Þ$
"Þ&&*"'
8 œ "'
!Þ%
"Þ(#*'&
>8
C8
8œ)
!Þ"
"Þ"*&"'
8 œ "'
!Þ#
"Þ$)"#'
8 œ #%
!Þ$
"Þ&&*")
8 œ $#
!Þ%
"Þ(#*'(
a,b. 2 œ !Þ!#& À
a- b. 2 œ !Þ!"#& À
2. The improved Euler formula is
C8" œ C8
2
2
ˆ& >8 $ÈC8 ‰ Š& >8" $ÈO8 ‹,
#
#
in which O8 œ C8 2ˆ& >8 $ÈC8 ‰. Since >8 œ >! 82 and >! œ ! , we can also
write
C8" œ C8
2
2
ˆ& 82 $ÈC8 ‰ ’& a8 "b2 $ÈO8 “,
#
#
with C! œ # .
a+b. 2 œ !Þ!& À
________________________________________________________________________
page 452
—————————————————————————— CHAPTER 8. ——
>8
C8
8œ#
!Þ"
"Þ'##)$
8œ%
!Þ#
"Þ$$%'!
8œ'
!Þ$
"Þ"#)#!
8œ)
!Þ%
!Þ**&%%&
>8
C8
8œ%
!Þ"
"Þ'##%$
8œ)
!Þ#
"Þ$$$)'
8 œ "#
!Þ$
"Þ"#(")
8 œ "'
!Þ%
!Þ**%#"&
>8
C8
8œ)
!Þ"
"Þ'##$%
8 œ "'
!Þ#
"Þ$$$')
8 œ #%
!Þ$
"Þ"#'*$
8 œ $#
!Þ%
!Þ**$*#"
a,b. 2 œ !Þ!#& À
a- b. 2 œ !Þ!"#& À
3. The improved Euler formula for this problem is
C8" œ C8
2
a% C8 $ >8 $ >8" b 2 # a# C8 $ >8 bÞ
#
Since >8 œ >! 82 and >! œ ! , we can also write
C8" œ C8 #2 C8
2#
a% C8 $ '8b $82 $ ,
#
with C! œ " .
a+b. 2 œ !Þ!& À
>8
C8
8œ#
!Þ"
"Þ#!&#'
8œ%
!Þ#
"Þ%##($
8œ'
!Þ$
"Þ'&&""
8œ)
!Þ%
"Þ*!&(!
>8
C8
8œ%
!Þ"
"Þ#!&$$
8œ)
!Þ#
"Þ%##*!
8 œ "#
!Þ$
"Þ'&&%#
8 œ "'
!Þ%
"Þ*!'#"
a,b. 2 œ !Þ!#& À
a- b. 2 œ !Þ!"#& À
________________________________________________________________________
page 453
—————————————————————————— CHAPTER 8. ——
>8
C8
8œ)
!Þ"
"Þ#!&$%
8 œ "'
!Þ#
"Þ%##*%
8 œ #%
!Þ$
"Þ'&&&!
8 œ $#
!Þ%
"Þ*!'$%
5. The improved Euler formula is
C8"
C8# # >8 C8
O8# # >8" O8
,
œ C8 2
2
#a$ >#8 b
#ˆ$ >#8" ‰
in which
C8# # >8 C8
O8 œ C8 2
.
$ >#8
Since >8 œ >! 82 and >! œ ! , we can also write
C8" œ C8 2
C8# #82 C8
O8# #a8 "b2O8
2
,
#a$ 8# 2# b
#$ a8 "b# 2# ‘
with C! œ !Þ& .
a+b. 2 œ !Þ!& À
>8
C8
8œ#
!Þ"
!Þ&"!"'%
8œ%
!Þ#
!Þ&#%"#'
8œ'
!Þ$
!Þ&%!)$
8œ)
!Þ%
!Þ&'%#&"
>8
C8
8œ%
!Þ"
!Þ&"!"')
8œ)
!Þ#
!Þ&#%"$&
8 œ "#
!Þ$
!Þ&%#"!!
8 œ "'
!Þ%
!Þ&'%#((
>8
C8
8œ)
!Þ"
!Þ&"!'*
8 œ "'
!Þ#
!Þ&#%"$(
8 œ #%
!Þ$
!Þ&%#"!%
8 œ $#
!Þ%
!Þ&'%#)%
a,b. 2 œ !Þ!#& À
a- b. 2 œ !Þ!"#& À
6. The improved Euler formula for this problem is
C8" œ C8
2 #
2 #
ˆ>8 C8# ‰=38 C8 ˆ>8"
O8# ‰=38 O8 ,
#
#
________________________________________________________________________
page 454
—————————————————————————— CHAPTER 8. ——
in which
O8 œ C8 2 ˆ>#8 C8# ‰=38 C8 .
Since >8 œ >! 82 and >! œ ! , we can also write
C8" œ C8
2 # #
2
ˆ8 2 C8# ‰=38 C8 a8 "b# 2 # O8# ‘=38 O8 ,
#
#
with C! œ " .
a+b. 2 œ !Þ!& À
>8
C8
8œ#
!Þ"
!Þ*#%'&!
8œ%
!Þ#
!Þ)'%$$)
8œ'
!Þ$
!Þ)"''%#
8œ)
!Þ%
!Þ()!!!)
8œ%
!Þ"
!Þ*#%&&!
8œ)
!Þ#
!Þ)'%"((
8 œ "#
!Þ$
!Þ)"'%%#
8 œ "'
!Þ%
!Þ((*()"
8 œ "'
!Þ#
!Þ)'%"$)
8 œ #%
!Þ$
!Þ)"'$*$
8 œ $#
!Þ%
!Þ((*(#&
a,b. 2 œ !Þ!#& À
>8
C8
a- b. 2 œ !Þ!"#& À
>8
C8
8œ)
!Þ"
!Þ*#%&#&
7. The improved Euler formula for this problem is
C8" œ C8
2
a% C8 >8 >8" "b 2 # a# C8 >8 !Þ&bÞ
#
Since >8 œ >! 82 and >! œ ! , we can also write
C8" œ C8 2a# C8 !Þ&b 2 # a# C8 8b 82 $ ,
with C! œ " .
a+b. 2 œ !Þ!#& À
>8
C8
8 œ #!
!Þ&
#Þ*'("*
8 œ %!
"Þ!
(Þ))$"$
8 œ '!
"Þ&
#!Þ)""%
8 œ )!
#Þ!
&&Þ&"!'
________________________________________________________________________
page 455
—————————————————————————— CHAPTER 8. ——
a,b. 2 œ !Þ!"#& À
>8
C8
8 œ %!
!Þ&
#Þ*')!!
8 œ )!
"Þ!
(Þ))(&&
8 œ "#!
"Þ&
#!Þ)#*%
8 œ "'!
#Þ!
&&Þ&(&)
8. The improved Euler formula is
C8" œ C8
2
2
ˆ& >8 $ÈC8 ‰ Š& >8" $ÈO8 ‹,
#
#
in which O8 œ C8 2ˆ& >8 $ÈC8 ‰. Since >8 œ >! 82 and >! œ ! , we can also
write
C8" œ C8
2
2
ˆ& 82 $ÈC8 ‰ ’& a8 "b2 $ÈO8 “,
#
#
with C! œ # .
a+b. 2 œ !Þ!#& À
>8
C8
8 œ #!
!Þ&
!Þ*#'"$*
8 œ %!
"Þ!
"Þ#)&&)
8 œ '!
"Þ&
#Þ%!)*)
8 œ )!
#Þ!
%Þ"!$)'
>8
C8
8 œ %!
!Þ&
!Þ*#&)"&
8 œ )!
"Þ!
"Þ#)&#&
8 œ "#!
"Þ&
#Þ%!)'*
8 œ "'!
#Þ!
%Þ"!$&*
a,b. 2 œ !Þ!"#& À
9. The improved Euler formula for this problem is
C8" œ C8
2
2
È>8 C8 È>8" O8 ,
#
#
in which O8 œ C8 2È>8 C8 . Since >8 œ >! 82 and >! œ ! , we can also write
C8" œ C8
2È
2
82 C8 Èa8 "b2 O8 ,
#
#
with C! œ $ .
a+b. 2 œ !Þ!#& À
________________________________________________________________________
page 456
—————————————————————————— CHAPTER 8. ——
>8
C8
8 œ #!
!Þ&
$Þ*'#"(
8 œ %!
"Þ!
&Þ"!))(
8 œ '!
"Þ&
'Þ%$"$%
8 œ )!
#Þ!
(Þ*#$$#
>8
C8
8 œ %!
!Þ&
$Þ*'#")
8 œ )!
"Þ!
&Þ"!))*
8 œ "#!
"Þ&
'Þ%$"$)
8 œ "'!
#Þ!
(Þ*#$$(
a,b. 2 œ !Þ!"#& À
10. The improved Euler formula is
C8" œ C8
2
2
c# >8 /B:a >8 C8 bd c# >8" /B:a >8" O8 bd,
#
#
in which O8 œ C8 2c# >8 /B:a >8 C8 bd. Since >8 œ >! 82 and >! œ ! , we can
also write
C8" œ C8
2
2
c# 82 /B:a 82 C8 bd e#a8 "b2 /B:c a8 "b2O8 df,
#
#
with C! œ " .
a+b. 2 œ !Þ!#& À
>8
C8
8 œ #!
!Þ&
"Þ'"#'$
8 œ %!
"Þ!
#Þ%)!*(
8 œ '!
"Þ&
$Þ(%&&'
8 œ )!
#Þ!
&Þ%*&*&
>8
C8
8 œ %!
!Þ&
"Þ'"#'$
8 œ )!
"Þ!
#Þ%)!*#
8 œ "#!
"Þ&
$Þ(%&&!
8 œ "'!
#Þ!
&Þ%*&)*
a,b. 2 œ !Þ!"#& À
12. The improved Euler formula is
C8" œ C8 2
C8# # >8 C8
O8# # >8" O8
2
,
#a$ >#8 b
#ˆ$ >#8" ‰
in which
O8 œ C8 2
C8# # >8 C8
.
$ >#8
________________________________________________________________________
page 457
—————————————————————————— CHAPTER 8. ——
Since >8 œ >! 82 and >! œ ! , we can also write
C8" œ C8 2
C8# #82 C8
O8# #a8 "b2O8
,
2
#a$ 8# 2# b
#$ a8 "b# 2# ‘
with C! œ !Þ& .
a+b. 2 œ !Þ!#& À
>8
C8
8 œ #!
!Þ&
!Þ&*!)*(
8 œ %!
"Þ!
!Þ(***&!
8 œ '!
"Þ&
"Þ"''&$
8 œ )!
#Þ!
"Þ(%*'*
>8
C8
8 œ %!
!Þ&
!Þ&*!*!'
8 œ )!
"Þ!
!Þ(***))
8 œ "#!
"Þ&
"Þ"'''$
8 œ "'!
#Þ!
"Þ(%**#
a,b. 2 œ !Þ!"#& À
16. The exact solution of the initial value problem is 9a>b œ "# "# /#> Þ Based on the
result in Prob. "%a- b, the local truncation error for a linear differential equation is
/8" œ
" www
9 a > 8 b2 $ ,
'
where >8 >8 >8" Þ Since 9 www a>b œ % /#> , the local truncation error is
/8" œ
Furthermore, with ! Ÿ >8 Ÿ " ,
#
/B:a# >8 b2 $ .
$
k/8" k Ÿ
# # $
/ 2 .
$
It also follows that for 2 œ !Þ" ,
k/" k Ÿ
# !Þ#
"
/ a!Þ"b$ œ
/!Þ# .
$
"&!!
Using the improved Euler method, with 2 œ !Þ" , we have C" ¸ "Þ""!!! . The exact
value is given by 9a!Þ"b œ "Þ""!(!"% Þ
17. The exact solution of the initial value problem is given by 9a>b œ "# > /#> Þ Using
the modified Euler method, the local truncation error for a linear differential equation is
/8" œ
" www
9 a > 8 b2 $ ,
'
________________________________________________________________________
page 458
—————————————————————————— CHAPTER 8. ——
where >8 >8 >8" Þ Since 9 www a>b œ ) /#> , the local truncation error is
/8" œ
%
/B:a# >8 b2 $ .
$
Furthermore, with ! Ÿ >8 Ÿ " , the local error is bounded by
k/8" k Ÿ
% # $
/ 2 .
$
It also follows that for 2 œ !Þ" ,
k/" k Ÿ
% !Þ#
" !Þ#
/ a!Þ"b$ œ
/ .
$
(&!
Using the improved Euler method, with 2 œ !Þ" , we have C" ¸ "Þ#(!!! . The exact
value is given by 9a!Þ"b œ "Þ#("%!$ Þ
18. Using the Euler method,
C" œ " !Þ"a!Þ& ! # † "b
œ "Þ#& Þ
Using the improved Euler method,
C" œ " !Þ!&a!Þ& ! # † "b !Þ!&a!Þ& !Þ" # † "Þ#&b
œ "Þ#( Þ
The estimated error is /" ¸ "Þ#( "Þ#& œ !Þ!# Þ The step size should be adjusted by
a factor of È!Þ!!#&Î!Þ!# ¸ !Þ$&% . Hence the required step size is estimated as
2 ¸ a!Þ"ba!Þ$'b œ !Þ!$' Þ
________________________________________________________________________
page 459
—————————————————————————— CHAPTER 8. ——
20. Using the Euler method,
C" œ $ !Þ"È! $
œ $Þ"($#!& Þ
Using the improved Euler method,
C" œ $ !Þ!&È! $ !Þ!&È!Þ" $Þ"($#!&
œ $Þ"((!'$ Þ
The estimated error is /" ¸ $Þ"((!'$ $Þ"($#!& œ !Þ!!$)&) Þ The step size should
be adjusted by a factor of È!Þ!!#&Î!Þ!!$)&) ¸ !Þ)!& . Hence the required step size
is estimated as
2 ¸ a!Þ"ba!Þ)!&b œ !Þ!)!& Þ
21. Using the Euler method,
a!Þ&b# !
$!
œ !Þ&!)$$%
C" œ !Þ& !Þ"
Using the improved Euler method,
a!Þ&b# !
a!Þ&!)$$%b# #a!Þ"ba!Þ&!)$$%b
C" œ !Þ& !Þ!&
!Þ!&
$!
$ a!Þ"b#
œ !Þ&"!"%) Þ
The estimated error is /" ¸ !Þ&"!"%) !Þ&!)$$% œ !Þ!!") Þ The local truncation error
is less than the given tolerance. The step size can be adjusted by a factor of
È!Þ!!#&Î!Þ!!") ¸ "Þ"()& . Hence it is possible to use a step size of
2 ¸ a!Þ"ba"Þ"()&b ¸ !Þ""( Þ
22. Assuming that the solution has continuous derivatives at least to the third order,
9a>8" b œ 9a>8 b 9 w a>8 b2
9 ww a>8 b # 9 www a>8 b $
2
2 ,
#x
$x
where >8 >8 >8" Þ Suppose that C8 œ 9a>8 bÞ
a+b. The local truncation error is given by
/8" œ 9a>8" b C8" Þ
The modified Euler formula is defined as
________________________________________________________________________
page 460
—————————————————————————— CHAPTER 8. ——
"
"
C8" œ C8 2 0 ”>8 2 ß C8 2 0 a>8 ß C8 b•Þ
#
#
Observe that 9 w a>8 b œ 0 a>8 ß 9a>8 bb œ 0 a>8 ß C8 b . It follows that
/8" œ 9a>8" b C8"
9 ww a>8 b # 9 www a>8 b $
œ 2 0 a>8 ß C8 b
2
2
#x
$x
"
"
2 0 ”>8 2 ß C8 2 0 a>8 ß C8 b•Þ
#
#
a,b. As shown in Prob. "%a,b,
9 ww a>8 b œ 0> a>8 ß C8 b 0C a>8 ß C8 b0 a>8 ß C8 b .
Furthermore,
0 ”>8 2 ß C8 2 0 a>8 ß C8 b• œ 0 a>8 ß C8 b 0> a>8 ß C8 b
"
#
"
#
2
0C a>8 ß C8 b 5
#
" 2#
,
” 0>> 25 0>C 5 # 0CC •
#x %
>œ0 ß Cœ(
in which 5 œ "# 2 0 a>8 ß C8 b and >8 0 >8 2Î# , C8 ( C8 5 . Therefore
/8" œ
9 www a>8 b $ 2 2#
2 ” 0>> 25 0>C 5 # 0CC •
Þ
$x
#x %
>œ0 ß Cœ(
Note that each term in the brackets has a factor of 2# . Hence the local truncation error
is proportional to 2$ .
a- b. If 0 a>ß Cb is linear, then 0>> œ 0>C œ 0CC œ ! , and
/8" œ
9 www a>8 b $
2 Þ
$x
23. The modified Euler formula for this problem is
"
"
C8" œ C8 2œ$ >8 2 ”C8 2 a$ >8 C8 b•
#
#
#
2
œ C8 2a$ >8 C8 b aC8 >8 #bÞ
#
Since >8 œ >! 82 and >! œ ! , we can also write
C8"
2#
œ C8 2a$ 82 C8 b aC8 82 #b ,
#
with C! œ " . Setting 2 œ !Þ" , we obtain the following values À
________________________________________________________________________
page 461
—————————————————————————— CHAPTER 8. ——
>8
C8
8œ"
!Þ"
"Þ"*&!!
8œ#
!Þ#
"Þ$)!*)
8œ$
!Þ$
"Þ&&)()
8œ%
!Þ%
"Þ(#*#!
25. The modified Euler formula is
$
C8" œ C8 2”#C8 $>8 2 2 a#C8 $>8 b•
#
2#
œ C8 2a#C8 $>8 b a%C8 '>8 $bÞ
#
Since >8 œ >! 82 and >! œ ! , we can also write
C8"
2#
œ C8 2a#C8 $82 b a%C8 ' 82 $b ,
#
with C! œ " . Setting 2 œ !Þ" , we obtain À
>8
C8
8œ"
!Þ"
"Þ#!&!!
8œ#
!Þ#
"Þ%##"!
8œ$
!Þ$
"Þ'&$*'
8œ%
!Þ%
"Þ*!$)$
26. The modified Euler formula for this problem is
C8" œ C8 2œ#>8 2 /B:” Œ>8
2
O8 •,
#
in which O8 œ C8 2# c#>8 /B:a >8 C8 bdÞ Now >8 œ >! 82 , with >! œ ! and
C! œ " . Setting 2 œ !Þ" , we obtain the following values À
>8
C8
8œ"
!Þ"
"Þ"!%))&
8œ#
!Þ#
"Þ#")*#
8œ$
!Þ$
"Þ$%"&(
8œ%
!Þ%
"Þ%(#(#%
________________________________________________________________________
page 462
—————————————————————————— CHAPTER 8. ——
27. Let 0 a>ß C b be linear in both variables. The improved Euler formula is
"
C8" œ C8 2c0 a>8 ß C8 b 0 a>8 2ß C8 20 a>8 ß C8 bbd
#
"
"
"
œ C8 20 a>8 ß C8 b 20 a>8 ß C8 b 20 c2ß 20 a>8 ß C8 bd
#
#
#
"
œ 20 a2ß C8 b 20 c2ß 20 a>8 ß C8 bdÞ
#
The modified Euler formula is
"
"
C8" œ C8 20 ”>8 2ß C8 20 a>8 ß C8 b•
#
#
" "
œ C8 20 a>8 ß C8 b 20 ” 2ß 20 a>8 ß C8 b•Þ
# #
Since 0 a>ß C b is linear in both variables,
" "
"
0 ” 2ß 20 a>8 ß C8 b• œ 0 c2ß 20 a>8 ß C8 bdÞ
# #
#
________________________________________________________________________
page 463
—————————————————————————— CHAPTER 8. ——
Section 8.3
1. The ODE is linear, with 0 a>ß C b œ $ > C . The Runge-Kutta algorithm requires
the evaluations
5 8" œ 0 a>8 ß C8 b
"
"
5 8# œ 0 Œ>8 2 ß C8 25 8"
#
#
"
"
5 8$ œ 0 Œ>8 2 ß C8 25 8#
#
#
5 8% œ 0 a>8 2 ß C8 25 8$ b.
The next estimate is given as the weighted average
C8" œ C8
a+bÞ For 2 œ !Þ" À
a,bÞ For 2 œ !Þ!& À
2
a5 8" # 5 8# # 5 8$ 5 8% bÞ
'
>8
C8
8œ"
!Þ"
"Þ"*&"'
8œ#
!Þ#
"Þ$)"#(
8œ$
!Þ$
"Þ&&*")
8œ%
!Þ%
"Þ(#*')
>8
C8
8œ#
!Þ"
"Þ"*&"'
8œ%
!Þ#
"Þ$)"#(
8œ'
!Þ$
"Þ&&*")
8œ)
!Þ%
"Þ(#*')
The exact solution of the IVP is Ca>b œ # > /> Þ
2. In this problem, 0 a>ß C b œ &> $ÈC . At each time step, the Runge-Kutta algorithm
requires the evaluations
5 8" œ 0 a>8 ß C8 b
"
"
5 8# œ 0 Œ>8 2 ß C8 25 8"
#
#
"
"
5 8$ œ 0 Œ>8 2 ß C8 25 8#
#
#
5 8% œ 0 a>8 2 ß C8 25 8$ b.
The next estimate is given as the weighted average
C8" œ C8
2
a5 8" # 5 8# # 5 8$ 5 8% bÞ
'
________________________________________________________________________
page 464
—————————————————————————— CHAPTER 8. ——
a+bÞ For 2 œ !Þ" À
a,bÞ For 2 œ !Þ!& À
>8
C8
8œ"
!Þ"
"Þ'##$"
8œ#
!Þ#
"Þ$$$'#
8œ$
!Þ$
"Þ"#')'
8œ%
!Þ%
!Þ**$)$*
>8
C8
8œ#
!Þ"
"Þ'##$!
8œ%
!Þ#
"Þ$$$'#
8œ'
!Þ$
"Þ"#')&
8œ)
!Þ%
!Þ**$)#'
The exact solution of the IVP is given implicitly as
È#
"
œ
Þ
&"#
ˆ#ÈC &>‰& ˆ> ÈC ‰#
3. The ODE is linear, with 0 a>ß C b œ #C $> . The Runge-Kutta algorithm requires
the evaluations
5 8" œ 0 a>8 ß C8 b
"
"
5 8# œ 0 Œ>8 2 ß C8 25 8"
#
#
"
"
5 8$ œ 0 Œ>8 2 ß C8 25 8#
#
#
5 8% œ 0 a>8 2 ß C8 25 8$ b.
The next estimate is given as the weighted average
C8" œ C8
a+bÞ For 2 œ !Þ" À
a,bÞ For 2 œ !Þ!& À
2
a5 8" # 5 8# # 5 8$ 5 8% bÞ
'
>8
C8
8œ"
!Þ"
"Þ#!&$&
8œ#
!Þ#
"Þ%##*&
8œ$
!Þ$
"Þ'&&&$
8œ%
!Þ%
"Þ*!'$)
>8
C8
8œ#
!Þ"
"Þ#!&$&
8œ%
!Þ#
"Þ%##*'
8œ'
!Þ$
"Þ'&&&$
8œ)
!Þ%
"Þ*!'$)
The exact solution of the IVP is Ca>b œ /#> Î% $>Î# $Î% Þ
5. In this problem, 0 a>ß C b œ aC # #>C bÎa$ ># b . The Runge-Kutta algorithm
________________________________________________________________________
page 465
—————————————————————————— CHAPTER 8. ——
requires the evaluations
5 8" œ 0 a>8 ß C8 b
"
"
5 8# œ 0 Œ>8 2 ß C8 25 8"
#
#
"
"
5 8$ œ 0 Œ>8 2 ß C8 25 8#
#
#
5 8% œ 0 a>8 2 ß C8 25 8$ b.
The next estimate is given as the weighted average
C8" œ C8
a+bÞ For 2 œ !Þ" À
2
a5 8" # 5 8# # 5 8$ 5 8% bÞ
'
>8
C8
8œ"
!Þ"
!Þ&"!"(!
8œ#
!Þ#
!Þ&#%"$)
8œ$
!Þ$
!Þ&%#"!&
8œ%
!Þ%
!Þ&'%#)'
>8
C8
8œ#
!Þ"
!Þ&#!"'*
8œ%
!Þ#
!Þ&#%"$)
8œ'
!Þ$
!Þ&%#"!&
8œ)
!Þ%
!Þ&'%#)'
a,bÞ For 2 œ !Þ!& À
The exact solution of the IVP is Ca>b œ a$ ># bÎa' >b Þ
6. In this problem, 0 a>ß C b œ a># C# b=38 C . At each time step, the Runge-Kutta
algorithm requires the evaluations
5 8" œ 0 a>8 ß C8 b
"
"
5 8# œ 0 Œ>8 2 ß C8 25 8"
#
#
"
"
5 8$ œ 0 Œ>8 2 ß C8 25 8#
#
#
5 8% œ 0 a>8 2 ß C8 25 8$ b.
The next estimate is given as the weighted average
C8" œ C8
2
a5 8" # 5 8# # 5 8$ 5 8% bÞ
'
________________________________________________________________________
page 466
—————————————————————————— CHAPTER 8. ——
a+bÞ For 2 œ !Þ" À
>8
C8
8œ"
!Þ"
!Þ*#%&"(
8œ#
!Þ#
!Þ)'%"#&
8œ$
!Þ$
!Þ)"'$((
8œ%
!Þ%
!Þ((*(!'
>8
C8
8œ#
!Þ"
!Þ*#%&"(
8œ%
!Þ#
!Þ)'%"#&
8œ'
!Þ$
!Þ)"'$((
8œ)
!Þ%
!Þ((*(!'
a,bÞ For 2 œ !Þ!& À
7. a+bÞ For 2 œ !Þ" À
a,bÞ For 2 œ !Þ!& À
>8
C8
8œ&
!Þ&
#Þ*')#&
8 œ "!
"Þ!
(Þ))))*
8 œ "&
"Þ&
#!Þ)$%*
8 œ #!
#Þ!
&&Þ&*&(
>8
C8
8 œ "!
!Þ&
#Þ*')#)
8 œ #!
"Þ!
(Þ))*!%
8 œ $!
"Þ&
#!Þ)$&&
8 œ %!
#Þ!
&&Þ&*)!
The exact solution of the IVP is Ca>b œ /#> >Î# Þ
8. See Prob. # . for the exact solution.
a+bÞ For 2 œ !Þ" À
a,bÞ For 2 œ !Þ!& À
>8
C8
8œ&
!Þ&
!Þ*#&(#&
8 œ "!
"Þ!
"Þ#)&"'
8 œ "&
"Þ&
#Þ%!)'!
8 œ #!
#Þ!
%Þ"!$&!
>8
C8
8 œ "!
!Þ&
!Þ*#&(""
8 œ #!
"Þ!
"Þ#)&"&
8 œ $!
"Þ&
#Þ%!)'!
8 œ %!
#Þ!
%Þ"!$&!
________________________________________________________________________
page 467
—————————————————————————— CHAPTER 8. ——
9a+bÞ For 2 œ !Þ" À
a,bÞ For 2 œ !Þ!& À
>8
C8
8œ&
!Þ&
$Þ*'#"*
8 œ "!
"Þ!
&Þ"!)*!
8 œ "&
"Þ&
'Þ%$"$*
8 œ #!
#Þ!
(Þ*#$$)
>8
C8
8 œ "!
!Þ&
$Þ*'#"*
8 œ #!
"Þ!
&Þ"!)*!
8 œ $!
"Þ&
'Þ%$"$*
8 œ %!
#Þ!
(Þ*#$$)
The exact solution is given implicitly as
68”
#
• #È> C # +<->+82 È> C œ > #È$ # +<->+82 È$ Þ
C>"
10. See Prob. % .
a+bÞ For 2 œ !Þ" À
a,bÞ For 2 œ !Þ!& À
>8
C8
8œ&
!Þ&
"Þ'"#'#
8 œ "!
"Þ!
#Þ%)!*"
8 œ "&
"Þ&
$Þ(%&%)
8 œ #!
#Þ!
&Þ%*&)(
>8
C8
8 œ "!
!Þ&
"Þ'"#'#
8 œ #!
"Þ!
#Þ%)!*"
8 œ $!
"Þ&
$Þ(%&%)
8 œ %!
#Þ!
&Þ%*&)(
12. See Prob. & . for the exact solution.
a+bÞ For 2 œ !Þ" À
a,bÞ For 2 œ !Þ!& À
>8
C8
8œ&
!Þ&
!Þ&*!*!*
8 œ "!
"Þ!
!Þ)!!!!!
8 œ "&
"Þ&
"Þ"''''(
8 œ #!
#Þ!
"Þ(&!!!
>8
C8
8 œ "!
!Þ&
!Þ&*!*!*
8 œ #!
"Þ!
!Þ)!!!!!
8 œ $!
"Þ&
"Þ"''''(
8 œ %!
#Þ!
"Þ(&!!!
13. The ODE is linear, with 0 a>ß C b œ " > %C . The Runge-Kutta algorithm requires
________________________________________________________________________
page 468
—————————————————————————— CHAPTER 8. ——
the evaluations
5 8" œ 0 a>8 ß C8 b
"
"
5 8# œ 0 Œ>8 2 ß C8 25 8"
#
#
"
"
5 8$ œ 0 Œ>8 2 ß C8 25 8#
#
#
5 8% œ 0 a>8 2 ß C8 25 8$ b.
The next estimate is given as the weighted average
C8" œ C8
2
a5 8" # 5 8# # 5 8$ 5 8% bÞ
'
The exact solution of the IVP is Ca>b œ
a+bÞ For 2 œ !Þ" À
"* %>
"' /
"% >
$
"'
Þ
>8
C8
8œ&
!Þ&
)Þ(!*$"(&
8 œ "!
"Þ!
'%Þ)&)"!(
8 œ "&
"Þ&
%()Þ)"*#)
8 œ #!
#Þ!
$&$&Þ)''(
>8
C8
8 œ "!
!Þ&
)Þ("")!'!
8 œ #!
"Þ!
'%Þ)*%)(&
8 œ $!
"Þ&
%(*Þ##'(%
8 œ %!
#Þ!
$&$*Þ))!%
a,bÞ For 2 œ !Þ!& À
15a+b.
________________________________________________________________________
page 469
—————————————————————————— CHAPTER 8. ——
a,b. For the integral curve starting at a!ß !b, the slope becomes infinite near >Q ¸ "Þ& Þ
Note that the exact solution of the IVP is defined implicitly as
C$ %C œ >$ Þ
Using the classic Runge-Kutta algorithm, with 2 œ !Þ!" , we obtain the values
>8
C8
8 œ (!
!Þ(
!Þ!)&*"
8 œ )!
!Þ)
!Þ"#)&$
8 œ *!
!Þ*
!Þ")$)!
8 œ *&
!Þ*&
!Þ#"')*
a- b. Based on the direction field, the solution should decrease monotonically to the
limiting value C œ #ÎÈ$ Þ In the following table, the value of >Q corresponds to
the approximate time in the iteration process that the calculated values begin to increase.
2
!Þ"
!Þ!&
!Þ!#&
!Þ!"
>Q
"Þ*
"Þ'&
"Þ&&
"Þ%&&
a. b. Numerical values will continue to be generated, although they will not be associated
with the integral curve starting at a!ß !b. These values are approximations to nearby
integral curves.
a/b. We consider the solution associated with the initial condition Ca!b œ " . The exact
solution is given by
C$ %C œ >$ $ Þ
________________________________________________________________________
page 470
—————————————————————————— CHAPTER 8. ——
For the integral curve starting at a! ß "b, the slope becomes infinite near >Q ¸ #Þ! Þ In
the following table, the values of >Q corresponds to the approximate time in the iteration
process that the calculated values begin to increase.
2
!Þ"
!Þ!&
!Þ!#&
!Þ!"
>Q
"Þ)&
"Þ)&
"Þ)'
"Þ)$&
________________________________________________________________________
page 471
—————————————————————————— CHAPTER 8. ——
Section 8.4
1a+b. Using the notation 08 œ 0 a>8 ß C8 b , the predictor formula is
C8" œ C8
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
>8
C8
8œ!
!Þ!
"Þ!
8 œ %a:</b
!Þ%
"Þ(#*'('*!
8œ"
!Þ"
"Þ"*&"'
8 œ %a-9<b
!Þ%
"Þ(#*)')!"
8œ#
!Þ#
"Þ$)"#(
8œ$
!Þ$
"Þ&&*")
8 œ &a:</b
!Þ&
"Þ)*$%'%$'
8 œ &a-9<b
!Þ&
"Þ)*$%'*($
a,b. With 08" œ 0 a>8" ß C8" b, the fourth order Adams-Moulton formula is
C8" œ C8
2
a* 08" "* 08 & 08" 08# b Þ
#%
In this problem, 08" œ $ >8" C8" Þ Since the ODE is linear, we can solve for
C8" œ
"
c#% C8 #( 2 *2 >8" 2 a"* 08 & 08" 08# bd Þ
#% * 2
>8
C8
8œ%
!Þ%
"Þ(#*')!!
8œ&
!Þ&
"Þ)*$%'*&
a- b. The fourth order backward differentiation formula is
C8" œ
"
c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ
#&
In this problem, 08" œ $ >8" C8" Þ Since the ODE is linear, we can solve for
C8" œ
"
c$' 2 "#2 >8" %) C8 $' C8" "' C8# $ C8$ d Þ
#& "# 2
________________________________________________________________________
page 472
—————————————————————————— CHAPTER 8. ——
8œ%
!Þ%
"Þ(#*')!&
>8
C8
8œ&
!Þ&
"Þ)*$%(""
The exact solution of the IVP is Ca>b œ # > /> Þ
2a+b. Using the notation 08 œ 0 a>8 ß C8 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
>8
C8
8œ!
!Þ!
#Þ!
8 œ %a:</b
!Þ%
!Þ**$(&"
8œ"
!Þ"
"Þ'##$"
8 œ %a-9<b
!Þ%
!Þ**$)&#
8œ#
!Þ#
"Þ$$$'#
8œ$
!Þ$
"Þ"#')'
8 œ &a:</b
!Þ&
!Þ*#&%'*
8 œ &a-9<b
!Þ&
!Þ*#&('%
a,b. With 08" œ 0 a>8" ß C8" b, the fourth order Adams-Moulton formula is
C8" œ C8
2
a* 08" "* 08 & 08" 08# b Þ
#%
In this problem, 08" œ &>8" $ÈC8" Þ Since the ODE is 898linear, an equation
solver is needed to approximate the solution of
C8" œ C8
2
%&>8" #(ÈC8" "* 08 & 08" 08# ‘
#%
at each time step. We obtain the approximate values:
>8
C8
8œ%
!Þ%
!Þ**$)%(
8œ&
!Þ&
!Þ*#&(%'
a- b. The fourth order backward differentiation formula is
________________________________________________________________________
page 473
—————————————————————————— CHAPTER 8. ——
C8" œ
"
c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ
#&
Since the ODE is 898linear, an equation solver is used to approximate the solution of
C8" œ
"
%) C8 $' C8" "' C8# $ C8$ "#2 ˆ&>8" $ÈC8" ‰‘
#&
at each time step.
>8
C8
8œ%
!Þ%
!Þ**$)'*
8œ&
!Þ&
!Þ*#&)$(
The exact solution of the IVP is given implicitly by
È#
"
&
# œ &"# Þ
ˆ#ÈC &>‰ ˆ> ÈC ‰
3a+b. The predictor formula is
C8" œ C8
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
#%
Using the starting values generated by the Runge-Kutta method À
>8
C8
>8
C8
8œ!
!Þ!
"Þ!
8 œ %a:</b
!Þ%
"Þ*!'$%!
8œ"
!Þ"
"Þ#!&$&!
8 œ %a-9<b
!Þ%
"Þ*!'$)#
8œ#
!Þ#
"Þ%##*&%
8œ$
!Þ$
"Þ'&&&#(
8 œ &a:</b
!Þ&
#Þ"(*%&&
8 œ &a-9<b
!Þ&
#Þ"(*&'(
a,b. With 08" œ 0 a>8" ß C8" b, the fourth order Adams-Moulton formula is
C8" œ C8
2
a* 08" "* 08 & 08" 08# b Þ
#%
In this problem, 08" œ # C8" $ >8" Þ Since the ODE is linear, we can solve for
________________________________________________________________________
page 474
—————————————————————————— CHAPTER 8. ——
C8" œ
"
c#% C8 #(2 >8" 2 a"* 08 & 08" 08# bd Þ
#% ") 2
>8
C8
8œ%
!Þ%
"Þ*!'$)&
8œ&
!Þ&
#Þ"(*&('
a- b. The fourth order backward differentiation formula is
C8" œ
"
c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ
#&
In this problem, 08" œ # C8" $ >8" Þ Since the ODE is linear, we can solve for
C8" œ
"
c%) C8 $' C8" "' C8# $ C8$ $'2 >8" d Þ
#& #% 2
>8
C8
8œ%
!Þ%
"Þ*!'$*&
8œ&
!Þ&
#Þ"(*'""
The exact solution of the IVP is Ca>b œ /#> Î% $>Î# $Î% Þ
5a+b. The predictor formula is
C8" œ C8
2
a&& 08 &* 08" $( 08# * 08$ bÞ
#%
With 08" œ 0 a>8" ß C8" b, the corrector formula is
C8" œ C8
2
a* 08" "* 08 & 08" 08# b Þ
#%
________________________________________________________________________
page 475
—————————————————————————— CHAPTER 8. ——
Using the starting values generated by the Runge-Kutta method À
8œ!
!Þ!
!Þ&
>8
C8
>8
C8
8œ"
!Þ"
!Þ&"!"'*&!
8 œ %a:</b
!Þ%
!Þ&'%#)&$#
8œ#
!Þ#
!Þ&#%"$(*&
8 œ %a-9<b
!Þ%
!Þ&'%#)&((
8œ$
!Þ$
!Þ&%#"!&#*
8 œ &a:</b
!Þ&
!Þ&*!*!)"'
8 œ &a-9<b
!Þ&
!Þ&*!*!*")
a,b. With 08" œ 0 a>8" ß C8" b, the fourth order Adams-Moulton formula is
C8" œ C8
2
a* 08" "* 08 & 08" 08# b Þ
#%
In this problem,
08" œ
#
C8
" # >8" C8"
Þ
$ >#8"
Since the ODE is 898linear, an equation solver is needed to approximate the solution of
C8" œ C8
#
C8
2
" # >8" C8"
*
"* 08 & 08" 08# •
”
#%
$ >#8"
at each time step.
>8
C8
8œ%
!Þ%
!Þ&'%#)&()
8œ&
!Þ&
!Þ&*!*!*#!
a- b. The fourth order backward differentiation formula is
C8" œ
"
c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ
#&
Since the ODE is 898linear, an equation solver is needed to approximate the solution of
C8" œ
#
C8
"
" # >8" C8"
%)
C
$'
C
"'
C
$
C
"#2
8
8"
8#
8$
”
•
#&
$ >#8"
at each time step. We obtain the approximate values:
>8
C8
8œ%
!Þ%
!Þ&'%#)&))
8œ&
!Þ&
!Þ&*!*!*&#
________________________________________________________________________
page 476
—————————————————————————— CHAPTER 8. ——
The exact solution of the IVP is Ca>b œ a$ ># bÎa' >b Þ
6a+b. The predictor formula is
C8" œ C8
2
a&& 08 &* 08" $( 08# * 08$ bÞ
#%
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
>8
C8
8œ!
!Þ!
"Þ!
8œ"
!Þ"
!Þ*#%&"(
8 œ %a:</b
!Þ%
!Þ((*)$#
8œ#
!Þ#
!Þ)'%"#&
8 œ %a-9<b
!Þ%
!Þ((*'*$
8œ$
!Þ$
!Þ)"'$((
8 œ &a:</b
!Þ&
!Þ(&$$""
8 œ &a-9<b
!Þ&
!Þ(&$"$&
a,b. With 08" œ 0 a>8" ß C8" b, the fourth order Adams-Moulton formula is
C8" œ C8
2
a* 08" "* 08 & 08" 08# b Þ
#%
#
b=38 C8" Þ Since the ODE is 898linear, we obtain
In this problem, 08" œ a>#8" C8"
the implicit equation
C8" œ C8
2
# ‰
*ˆ>#8" C8"
=38 C8" "* 08 & 08" 08# ‘Þ
#%
>8
C8
8œ%
!Þ%
!Þ((*(!!
8œ&
!Þ&
!Þ(&$"%%
________________________________________________________________________
page 477
—————————————————————————— CHAPTER 8. ——
a- b. The fourth order backward differentiation formula is
C8" œ
"
c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ
#&
Since the ODE is 898linear, we obtain the implicit equation
C8" œ
"
# ‰
%) C8 $' C8" "' C8# $ C8$ "#2 ˆ>#8" C8"
=38 C8" ‘Þ
#&
8œ%
!Þ%
!Þ((*')!
>8
C8
8a+b. The predictor formula is
C8" œ C8
8œ&
!Þ&
!Þ(&$!)*
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!
8œ"
!Þ!&
"Þ(**'#*'
>8
C8
8 œ "!
!Þ&
!Þ*#&("$$
8œ#
!Þ"
"Þ'##$!%#
8 œ #!
"Þ!
"Þ#)&"%)
8 œ $!
"Þ&
#Þ%!)&*&
8œ$
!Þ"&
"Þ%'(#&!$
8 œ %!
#Þ!
%Þ"!$%*&
a,b. Since the ODE is 898linear, an equation solver is needed to approximate the
solution of
C8" œ C8
2
%&>8" #(ÈC8" "* 08 & 08" 08# ‘
#%
at each time step. We obtain the approximate values:
>8
C8
8 œ "!
!Þ&
!Þ*#&("#&
8 œ #!
"Þ!
"Þ#)&"%)
8 œ $!
"Þ&
#Þ%!)&*&
8 œ %!
#Þ!
%Þ"!$%*&
________________________________________________________________________
page 478
—————————————————————————— CHAPTER 8. ——
a- b. The fourth order backward differentiation formula is
C8" œ
"
c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ
#&
Since the ODE is 898linear, an equation solver is needed to approximate the solution of
C8" œ
"
%) C8 $' C8" "' C8# $ C8$ "#2 ˆ&>8" $ÈC8" ‰‘
#&
at each time step.
8 œ "!
!Þ&
!Þ*#&(#%)
>8
C8
8 œ #!
"Þ!
"Þ#)&"&)
8 œ $!
"Þ&
#Þ%!)&*%
8 œ %!
#Þ!
%Þ"!$%*$
The exact solution of the IVP is given implicitly by
È#
"
œ
Þ
&"#
ˆ#ÈC &>‰& ˆ> ÈC ‰#
9a+b. The predictor formula is
C8" œ C8
2
a&& 08 &* 08" $( 08# * 08$ bÞ
#%
With 08" œ 0 a>8" ß C8" b, the corrector formula is
C8" œ C8
2
a* 08" "* 08 & 08" 08# b Þ
#%
Using the starting values generated by the Runge-Kutta method À
>8
C8
>8
C8
8œ!
!Þ!
$Þ!
8œ"
!Þ!&
$Þ!)(&)'
8 œ "!
!Þ&
$Þ*'#")'
8 œ #!
"Þ!
&Þ"!)*!$
8œ#
!Þ"
$Þ"(("#(
8 œ $!
"Þ&
'Þ%$"$*!
8œ$
!Þ"&
$Þ#')'!*
8 œ %!
#Þ!
(Þ*#$$)&
a,b. With 08" œ 0 a>8" ß C8" b, the fourth order Adams-Moulton formula is
C8" œ C8
2
a* 08" "* 08 & 08" 08# b Þ
#%
________________________________________________________________________
page 479
—————————————————————————— CHAPTER 8. ——
In this problem, 08" œ È>8" C8" Þ Since the ODE is 898linear, an equation
solver must be implemented in order to approximate the solution of
C8" œ C8
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*#$$)&
a- b. The fourth order backward differentiation formula is
C8" œ
"
c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ
#&
Since the ODE is 898linear, an equation solver is needed to approximate the solution of
C8" œ
"
%) C8 $' C8" "' C8# $ C8$ "#2 È>8" C8" ‘
#&
at each time step.
>8
C8
8 œ "!
!Þ&
$Þ*'#")'
8 œ #!
"Þ!
&Þ"!)*!$
8 œ $!
"Þ&
'Þ%$"$*!
8 œ %!
#Þ!
(Þ*#$$)&
The exact solution is given implicitly by
68”
#
• #È> C # +<->+82 È> C œ > #È$ # +<->+82 È$ Þ
C>"
________________________________________________________________________
page 480
—————————————————————————— 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
2
e*c# >8" /B:a >8" C8" bd "* 08 & 08" 08# f
#%
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
"
c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ
#&
Since the ODE is 898linear, we obtain the implicit equation
C8" œ
"
e%) C8 $' C8" "' C8# $ C8$ "#2 c# >8" /B:a >8" C8" bdfÞ
#&
________________________________________________________________________
page 481
—————————————————————————— CHAPTER 8. ——
>8
C8
8 œ "!
!Þ&
"Þ'"#'#$
8 œ #!
"Þ!
#Þ%)!*!&
8 œ $!
"Þ&
$Þ(%&"%($
8 œ %!
#Þ!
&Þ%*&)'*
11a+b. The predictor formula is
C8" œ C8
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
#%
Using the starting values generated by the Runge-Kutta method À
>8
C8
8œ!
!Þ!
#Þ!
8œ"
!Þ!&
"Þ*&))$$
>8
C8
8 œ "!
!Þ&
"Þ%%('$*
8œ#
!Þ"
"Þ*"&##"
8 œ #!
"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" C8"
Þ
#
" C8
"
Since the differential equation is 898linear, an equation solver is used to approximate
the solution of
C8" œ C8
2
% >8" C8"
"* 08 & 08" 08# •
”*
#
#%
" C8
"
at each time step.
>8
C8
8 œ "!
!Þ&
"Þ%%('$)
8 œ #!
"Þ!
!Þ"%$'('(
8 œ $!
"Þ&
"Þ!'!*"$
8 œ %!
#Þ!
"Þ%"!"!$
________________________________________________________________________
page 482
—————————————————————————— CHAPTER 8. ——
a- b. The fourth order backward differentiation formula is
C8" œ
"
c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ
#&
Since the ODE is 898linear, an equation solver must be implemented in order to
approximate the solution of
C8" œ
"
% >8" C8"
”%) C8 $' C8" "' C8# $ C8$ "#2
•
#
#&
" C8
"
at each time step.
8 œ "!
!Þ&
"Þ%%('#"
>8
C8
8 œ #!
"Þ!
!Þ"%%('"*
8 œ $!
"Þ&
"Þ!'!("(
8 œ %!
#Þ!
"Þ%"!!#(
12a+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
>8
C8
8œ!
!Þ!
!Þ&
8œ"
!Þ!&
!Þ&!%'#")
8 œ "!
!Þ&
!Þ&*!*!*"
8œ#
!Þ"
!Þ&"!"'*&
8 œ #!
"Þ!
!Þ)!!!!!!
8 œ $!
"Þ&
"Þ"''''(
8œ$
!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,
________________________________________________________________________
page 483
—————————————————————————— CHAPTER 8. ——
08" œ
#
C8
" # >8" C8"
Þ
$ >#8"
Since the ODE is 898linear, an equation solver is needed to approximate the solution of
C8" œ C8
#
C8
2
" # >8" C8"
*
"* 08 & 08" 08# •
”
#%
$ >#8"
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
"
c%) C8 $' C8" "' C8# $ C8$ "#208" dÞ
#&
Since the ODE is 898linear, we obtain the implicit equation
C8"
#
C8
"
" # >8" C8"
œ
”%) C8 $' C8" "' C8# $ C8$ "#2
•Þ
#&
$ >#8"
>8
C8
8 œ "!
!Þ&
!Þ&*!*!*#
8 œ #!
"Þ!
!Þ)!!!!!#
8 œ $!
"Þ&
"Þ"''''(
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
________________________________________________________________________
page 485
—————————————————————————— CHAPTER 8. ——
08" #08 08"
# 2#
a#8 "b08" %808 a" #8b08"
"œ
#2
#
8 8
8# 8
#œ
08" ˆ" 8# ‰08
08" Þ
#
#
!œ
We then have
C8" C8 œ (
>8"
>8
!># " > # ‘.>
"
"
œ !2$ Œ8# 8 " 2 # Œ8 # 2 ,
$
#
which results in
C8" C8 œ
2
a&08" )08 08" b Þ
"#
________________________________________________________________________
page 486
—————————————————————————— CHAPTER 8. ——
Section 8.5
1a+b. The general solution of the ODE is Ca>b œ - /> # > Þ Imposing the initial
condition, Ca!b œ # , the solution of the IVP is 9" a>b œ # > .
a,b. If instead, the initial condition Ca!b œ #Þ!!" is given, the solution of the IVP is
9# a>b œ !Þ!!" /> # > . We then have 9# a>b 9" a>b œ !Þ!!" /> Þ
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
________________________________________________________________________
page 487
—————————————————————————— 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
________________________________________________________________________
page 487
—————————————————————————— CHAPTER 8. ——
for various values of - À
-œ"À
> œ !Þ#&
> œ !Þ&
> œ !Þ(&
> œ "Þ!
9a>b
!Þ!'#&
!Þ#&
!Þ&'#&
"Þ!
C8
!Þ!'#%***ÞÞ
!Þ#&
!Þ&'#&
"Þ!
kC8 9a>8 bk
# ‚ "!""
!
!
!
> œ !Þ#&
> œ !Þ&
> œ !Þ(&
> œ "Þ!
9a>b
!Þ!'#&
!Þ#&
!Þ&'#&
"Þ!
C8
!Þ!'#%**)ÞÞ
!Þ#%***(
!Þ&'#%'%
!Þ***&'%
kC8 9a>8 bk
#Þ#"& ‚ "!(
#Þ*#! ‚ "!'
$Þ&(* ‚ "!&
%Þ$'# ‚ "!%
> œ !Þ#&
> œ !Þ&
> œ !Þ(&
> œ "Þ!
9a>b
!Þ!'#&
!Þ#&
!Þ&'#&
"Þ!
C8
!Þ!'#))*ÞÞ
!Þ#%)$%#
!Þ$"'%&)
$&Þ&"$*
kC8 9a>8 bk
"Þ"! ‚ "!&
"Þ'&) ‚ "!$
!Þ#%'!%#
$'Þ&"$*
- œ "! À
- œ #! À
- œ &! À
> œ !Þ#&
> œ !Þ&
> œ !Þ(&
> œ "Þ!
9a>b
!Þ!'#&
!Þ#&
!Þ&'#&
"Þ!
C8
!Þ!%%)!$ÞÞ
#)''*Þ&&
(Þ''!"% ‚ "!*
#Þ!%'') ‚ "!"&
kC8 9a>8 bk
!Þ"!($!$
#)''*Þ)!%
(Þ''!"% ‚ "!*
#Þ!%'') ‚ "!"&
The following table shows the calculated value, C" , at the first time step À
9a>b
"!%
C" a- œ "b
*Þ***** ‚ "!&
C" a- œ "!b
*Þ***(* ‚ "!&
C" a- œ #!b
*Þ**)$$ ‚ "!&
C" a- œ &!b
*Þ*($*' ‚ "!&
a- b. Referring back to the exact solution given in Parta+b, if a nonzero initial condition,
say Ca!b œ & , is specified, the solution of the IVP becomes
9& a>b œ & /-> ># .
We then have k9a> b 9& a>bk œ k&k /-> Þ It is evident that for any > ! ,
________________________________________________________________________
page 488
—————————————————————————— CHAPTER 8. ——
lim k9a> b 9& a>bk œ _ Þ
-Ä_
This implies that virtually any error introduced early in the calculations will be magnified
as - p _ . The initial value problem is inherently unstable.
________________________________________________________________________
page 489
—————————————————————————— CHAPTER 8. ——
Section 8.6
1. In vector notation, the initial value problem can be written as
. B
BC>
"
Œ œŒ
, xa!b œ Œ Þ
.> C
%B #C
!
a+b. The Euler formula is
Œ
B8"
B8
B8 C8 >8
œ Œ 2Œ
Þ
C8"
C8
%B8 #C8
That is,
B8" œ B8 2aB8 C8 >8 b
C8" œ C8 2a%B8 #C8 bÞ
With 2 œ !Þ" , we obtain the values
>8
B8
C8
8œ#
!Þ#
"Þ#'
!Þ('
8œ%
!Þ%
"Þ(("%
"Þ%)#%
8œ'
!Þ'
#Þ&)**"
#Þ$(!$
8œ)
!Þ)
$Þ)#$(%
$Þ'!%"$
8 œ "!
"Þ!
&Þ'%#%'
&Þ$)))&
a,b. The Runge-Kutta method uses the following intermediate calculations:
k8" œ aB8 C8 >8 ß %B8 #C8 bX
X
2 "
2 #
2
2 "
2 #
k8# œ ”B8 58" C8 58" >8 ß %ŒB8 58" #ŒC8 58" •
#
#
#
#
#
X
2
2
2
2
2
k8$ œ ”B8 58" # C8 58# # >8 ß %ŒB8 58" # #ŒC8 58# # •
#
#
#
#
#
X
"
#
"
#
k8% œ cB8 258$ C8 258$ >8 2 ß %aB8 258$ b #aC8 258$ bd Þ
With 2 œ !Þ# , we obtain the values:
>8
B8
C8
8œ"
!Þ#
"Þ$#%*$
!Þ(&)*$$
8œ#
!Þ%
"Þ*$'(*
"Þ&(*"*
8œ$
!Þ'
#Þ*$%"%
#Þ''!**
8œ%
!Þ)
%Þ%)$")
%Þ##'$*
8œ&
"Þ!
'Þ)%#$'
'Þ&'%&#
a- bÞ With 2 œ !Þ" , we obtain
________________________________________________________________________
page 490
—————————————————————————— CHAPTER 8. ——
>8
B8
C8
8œ#
!Þ#
"Þ$#%)*
!Þ(&*&"'
8œ%
!Þ%
"Þ*$'*
"Þ&(***
8œ'
!Þ'
#Þ*$%&*
#Þ''#!"
8œ)
!Þ)
%Þ%)%##
%Þ##()%
8 œ "!
"Þ!
'Þ)%%%
'Þ&'')%
The exact solution of the IVP is
#
"
#
Ba>b œ /#> /$> >
*
$
*
)
#
"
Ca>b œ /#> /$> > Þ
*
$
*
3a+b. The Euler formula is
Œ
That is,
B8"
B8
>8 B8 C8 "
œ Œ 2Œ
Þ
C8"
C8
B8
B8" œ B8 2a >8 B8 C8 "b
C8" œ C8 2aB8 bÞ
With 2 œ !Þ" , we obtain the values
>8
B8
C8
8œ#
!Þ#
!Þ&)#
"Þ")
8œ%
!Þ%
!Þ""(*'*
"Þ#($%%
8œ'
!Þ'
!Þ$$'*"#
"Þ#($)#
8œ)
!Þ)
!Þ($!!!(
"Þ")&(#
8 œ "!
"Þ!
"Þ!#"$%
"Þ!#$("
a,b. The Runge-Kutta method uses the following intermediate calculations:
k8" œ a >8 B8 C8 "ß B8 bX
X
2
2 "
2 #
2 "
k8# œ ” Œ>8 ŒB8 58" ŒC8 58" " ß B8 58" •
#
#
#
#
X
2
2 "
2 #
2 "
k8$ œ ” Œ>8 ŒB8 58# ŒC8 58# " ß B8 58# •
#
#
#
#
"
#
" X
k8% œ c a>8 2baB8 258$ b aC8 258$ b " ß B8 258$ d Þ
________________________________________________________________________
page 491
—————————————————————————— CHAPTER 8. ——
With 2 œ !Þ# , we obtain the values:
>8
B8
C8
8œ"
!Þ#
!Þ&')%&"
"Þ"&((&
8œ#
!Þ%
!Þ"!*(('
"Þ##&&'
8œ$
!Þ'
!Þ$##!)
"Þ#!$%(
8œ%
!Þ)
!Þ')"#*'
"Þ"!"'#
8œ&
"Þ!
!Þ*$()&#
!Þ*$()&#
8œ'
!Þ'
!Þ$##!)"
"Þ#!$%(
8œ)
!Þ)
!Þ')"#*"
"Þ"!"'"
8 œ "!
"Þ!
!Þ*$()%"
!Þ*$()%
a- bÞ With 2 œ !Þ" , we obtain
>8
B8
C8
8œ#
!Þ#
!Þ&')%&
"Þ"&((&
8œ%
!Þ%
!Þ"!*(($
"Þ##&&(
4a+b. The Euler formula gives
B8" œ B8 2aB8 C8 B8 C8 b
C8" œ C8 2a$B8 #C8 B8 C8 bÞ
With 2 œ !Þ" , we obtain the values
>8
B8
C8
8œ#
!Þ#
!Þ"*)
!Þ'")
8œ%
!Þ%
!Þ$()(*'
!Þ#)$#*
a,b. Given
8œ'
!Þ'
!Þ&"*$#
!Þ!$#"!#&
8œ)
!Þ)
!Þ&*%$#%
!Þ$#')!"
8 œ "!
"Þ!
!Þ&))#()
!Þ&(&%&
0 a>ß Bß C b œ B C B C
1a>ß Bß C b œ $B #C B C ,
the Runge-Kutta method uses the following intermediate calculations:
k8" œ c0 a>8 ß B8 ß C8 bß 1a>8 ß B8 ß C8 bdX
"
#
"
#
k8# œ ”0 Œ>8 ß B8 58"
ß C8 58"
ß 1Œ>8 ß B8 58" ß C8 58" •
#
#
#
#
#
#
2
2
2
2
2
2
"
#
"
#
k8$ œ ”0 Œ>8 ß B8 58#
ß C8 58#
ß 1Œ>8 ß B8 58# ß C8 58# •
2
#
2
#
2
#
2
#
2
#
2
#
X
X
"
# ‰ ˆ
"
# ‰‘
k8% œ 0 ˆ>8 2ß B8 258$
Þ
ß C8 258$
ß 1 >8 2ß B8 258$
ß C8 258$
X
________________________________________________________________________
page 492
—————————————————————————— CHAPTER 8. ——
With 2 œ !Þ# , we obtain the values:
>8
B8
C8
8œ"
!Þ#
!Þ"*'*!%
!Þ'$!*$'
8œ#
!Þ%
!Þ$(#'%$
!Þ#*))))
8œ$
!Þ'
!Þ&!"$!#
!Þ!"""%#*
8œ%
!Þ)
!Þ&'"#(!
!Þ#))*%$
8œ&
"Þ!
!Þ&%(!&$
!Þ&!)$!$
8œ'
!Þ'
!Þ&!"$%&
!Þ!""#")%
8œ)
!Þ)
!Þ&'"#*#
!Þ#)*!(
8 œ "!
"Þ!
!Þ&%(!$"
!Þ&!)%#(
a- bÞ With 2 œ !Þ" , we obtain
>8
B8
C8
8œ#
!Þ#
!Þ"*'*$&
!Þ'$!*$*
8œ%
!Þ%
!Þ$(#')(
!Þ#*))''
5a+b. The Euler formula gives
B8" œ B8 2cB8 a" !Þ& B8 !Þ& C8 bd
C8" œ C8 2cC8 a !Þ#& !Þ& B8 bdÞ
With 2 œ !Þ" , we obtain the values
>8
B8
C8
8œ#
!Þ#
#Þ*'##&
"Þ$%&$)
a,b. Given
8œ%
!Þ%
#Þ$%""*
"Þ'("#"
8œ'
!Þ'
"Þ*!#$'
"Þ*("&)
8œ)
!Þ)
"Þ&''!#
#Þ#$)*&
8 œ "!
"Þ!
"Þ#*(')
#Þ%'($#
0 a>ß Bß C b œ Ba" !Þ& B !Þ& C b
1a>ß Bß C b œ C a !Þ#& !Þ& Bb ,
the Runge-Kutta method uses the following intermediate calculations:
k8" œ c0 a>8 ß B8 ß C8 bß 1a>8 ß B8 ß C8 bdX
"
#
"
#
k8# œ ”0 Œ>8 ß B8 58"
ß C8 58"
ß 1Œ>8 ß B8 58" ß C8 58" •
#
#
#
#
#
#
2
2
2
2
2
2
"
#
"
#
k8$ œ ”0 Œ>8 ß B8 58#
ß C8 58#
ß 1Œ>8 ß B8 58# ß C8 58# •
2
#
2
#
2
#
2
#
2
#
2
#
X
X
"
# ‰ ˆ
"
# ‰‘
k8% œ 0 ˆ>8 2ß B8 258$
Þ
ß C8 258$
ß 1 >8 2ß B8 258$
ß C8 258$
X
________________________________________________________________________
page 493
—————————————————————————— CHAPTER 8. ——
With 2 œ !Þ# , we obtain the values:
>8
B8
C8
8œ"
!Þ#
$Þ!'$$*
"Þ$%)&)
8œ#
!Þ%
#Þ%%%*(
"Þ')'$)
8œ$
!Þ'
"Þ**""
#Þ!!!$'
8œ%
!Þ)
"Þ'$)")
#Þ#(*)"
8œ&
"Þ!
"Þ$&&&
#Þ&"(&
8œ%
!Þ%
#Þ%%%'&
"Þ')'**
8œ'
!Þ'
"Þ**!(&
#Þ!!"!(
8œ)
!Þ)
"Þ'$()"
#Þ#)!&(
8 œ "!
"Þ!
"Þ$&&"%
#Þ&")#(
a- bÞ With 2 œ !Þ" , we obtain
>8
B8
C8
8œ#
!Þ#
$Þ!'$"%
"Þ$%)**
6a+b. The Euler formula gives
B8" œ B8 2c/B:a B8 C8 b -9= B8 d
C8" œ C8 2c=38aB8 $ C8 bdÞ
With 2 œ !Þ" , we obtain the values
>8
B8
C8
8œ#
!Þ#
"Þ%#$)'
#Þ")*&(
8œ%
!Þ%
"Þ)##$%
#Þ$'(*"
8œ'
!Þ'
#Þ#"(#)
#Þ&$$#*
8œ)
!Þ)
#Þ'""")
#Þ')('$
8 œ "!
"Þ!
#Þ**&&
#Þ)$$&%
a,b. The Runge-Kutta method uses the following intermediate calculations:
k8" œ c0 a>8 ß B8 ß C8 bß 1a>8 ß B8 ß C8 bdX
"
#
"
#
k8# œ ”0 Œ>8 ß B8 58"
ß C8 58"
ß 1Œ>8 ß B8 58" ß C8 58" •
2
#
2
#
2
#
2
#
2
#
2
#
"
#
"
#
k8$ œ ”0 Œ>8 ß B8 58#
ß C8 58#
ß 1Œ>8 ß B8 58# ß C8 58# •
#
#
#
#
#
#
2
2
2
2
2
2
X
X
"
# ‰ ˆ
"
# ‰‘
k8% œ 0 ˆ>8 2ß B8 258$
Þ
ß C8 258$
ß 1 >8 2ß B8 258$
ß C8 258$
X
________________________________________________________________________
page 494
—————————————————————————— CHAPTER 8. ——
With 2 œ !Þ# , we obtain the values:
>8
B8
C8
8œ"
!Þ#
"Þ%"&"$
#Þ")'**
8œ#
!Þ%
"Þ)"#!)
#Þ$'#$$
8œ$
!Þ'
#Þ#!'$&
#Þ&#&)
8œ%
!Þ)
#Þ&*)#'
#Þ'(*%
8œ&
"Þ!
#Þ*()!'
#Þ)#%)(
8œ%
!Þ%
"Þ)"#!*
#Þ$'#$$
8œ'
!Þ'
#Þ#!'$&
#Þ&#&)"
8œ)
!Þ)
#Þ&*)#'
#Þ'(*%"
8 œ "!
"Þ!
#Þ*()!'
#Þ)#%))
a- bÞ With 2 œ !Þ" , we obtain
>8
B8
C8
8œ#
!Þ#
"Þ%"&"$
#Þ")'**
7. The Runge-Kutta method uses the following intermediate calculations:
k8" œ cB8 % C8 ß B8 C8 dX
"
#
"
#
k8# œ ”B8 58"
%ŒC8 58"
ß ŒB8 58" C8 58" •
2
#
2
#
2
#
2
#
"
#
"
#
k8$ œ ”B8 58#
%ŒC8 58#
ß ŒB8 58# C8 58# •
2
#
2
#
2
#
2
#
X
X
"
# ‰
" ‰
# ‘
k8% œ B8 258$
Þ
%ˆC8 258$
ß ˆB8 258$
C8 258$
X
Using 2 œ !Þ!% , we obtain the following values:
>8
B8
C8
8œ&
!Þ#
"Þ$#!%
!Þ#&!)&
8 œ "!
!Þ%
"Þ**&#
!Þ''#%&
8 œ "&
!Þ'
$Þ#**#
"Þ$(&#
8 œ #!
!Þ)
&Þ($'#
#Þ'%$&
8 œ #&
"Þ!
"!Þ##(
%Þ*#*%
The exact solution is given by
/> /$>
/> /$>
9a>b œ
, <a>b œ
,
#
%
and the associated tabulated values:
>8
9a>8 b
<a>8 b
8œ&
!Þ#
"Þ$#!%
!Þ#&!)&
8 œ "!
!Þ%
"Þ**&#
!Þ''#%&
8 œ "&
!Þ'
$Þ#**#
"Þ$(&#
8 œ #!
!Þ)
&Þ($'#
#Þ'%$&
8 œ #&
"Þ!
"!Þ##(
%Þ*#*%
8. Let C œ B w . The second order ODE can be transformed into the first order system
________________________________________________________________________
page 495
—————————————————————————— CHAPTER 8. ——
Bw œ C
C w œ > $B ># C ,
with initial conditions Ba!b œ " , Ca!b œ # . Given
0 a>ß Bß C b œ C
1a>ß Bß C b œ > $B ># C ,
the Runge-Kutta method uses the following intermediate calculations:
k8" œ C8 ß >8 $ B8 >#8 C8 ‘
X
k8# œ ”C8
k 8$
k 8%
2 #
2
2 "
2 #
58" ß 1Œ>8 ß B8 58"
ß C8 58"
•
#
#
#
#
X
2 #
2
2 "
2 #
œ ”C8 58# ß 1Œ>8 ß B8 58# ß C8 58# •
#
#
#
#
X
#
"
#
œ C8 258$ ß 1ˆ>8 2ß B8 258$ ß C8 258$ ‰‘ Þ
X
With 2 œ !Þ" , we obtain the following values:
>8
B8
C8
8œ&
!Þ&
"Þ&%$
"Þ"%(%$
8 œ "!
"Þ!
!Þ!(!(&
"Þ$))&
9. The predictor formulas are
2
a&& 08 &* 08" $( 08# * 08$ b
#%
2
œ C8 a&& 18 &* 18" $( 18# * 18$ bÞ
#%
B8" œ B8
C8"
With 08" œ B8" % C8" and 18" œ B8" C8" , the corrector formulas are
2
a* 08" "* 08 & 08" 08# b
#%
2
œ C8 a* 18" "* 18 & 18" 18# bÞ
#%
B8" œ B8
C8"
________________________________________________________________________
page 496
—————————————————————————— CHAPTER 8. ——
We use the starting values from the exact solution À
>8
B8
C8
8œ!
!
"Þ!
!Þ!
8œ"
!Þ"
"Þ"#))$
!Þ""!&(
8œ#
!Þ#
"Þ$#!%#
!Þ#&!)%(
8œ$
!Þ$
"Þ'!!#"
!Þ%#*'*'
One time step using the predictor-corrector method results in the approximate values:
>8
B8
C8
8 œ %a:</b
!Þ%
"Þ**%%&
!Þ''#!'%
8 œ %a-9<b
!Þ%
"Þ**&#"
!Þ''#%%#
________________________________________________________________________
page 497
—————————————————————————— CHAPTER 9. ——
Chapter Nine
Section 9.1
2a+bÞ Setting x œ 0 /<> results in the algebraic equations
Œ
&<
$
0"
!
"
œ Œ .
Œ
"<
0#
!
For a nonzero solution, we must have ./>aA < Ib œ <# ' < ) œ ! . The roots of
the characteristic equation are <" œ # and <# œ % . For < œ #, the system of equations
reduces to $0" œ 0# . The corresponding eigenvector is 0 a"b œ a" ß $bX Þ Substitution of
< œ % results in the single equation 0" œ 0# . A corresponding eigenvector is
0 a#b œ a" ß "bX Þ
a,b. The eigenvalues are real and positive, hence the critical point is an unstable node.
a-ß . b.
________________________________________________________________________
page 498
—————————————————————————— CHAPTER 9. ——
3a+b. Solution of the ODE requires analysis of the algebraic equations
Œ
#<
$
0"
!
"
œ Œ .
Œ
#<
0#
!
For a nonzero solution, we must have ./>aA < Ib œ <# " œ ! . The roots of the
characteristic equation are <" œ " and <# œ " . For < œ " , the system of equations
reduces to 0" œ 0# . The corresponding eigenvector is 0 a"b œ a" ß "bX Þ Substitution of
< œ " results in the single equation $ 0" 0# œ ! . A corresponding eigenvector is
0 a#b œ a" ß $bX Þ
a,b. The eigenvalues are real, with <" <# ! . Hence the critical point is a saddle.
a-ß . b.
________________________________________________________________________
page 499
—————————————————————————— CHAPTER 9. ——
5a+b. The characteristic equation is given by
º
"<
"
&
œ <# # < # œ ! Þ
$<º
The equation has complex roots <" œ " 3 and <# œ " 3. For < œ " 3,
the components of the solution vector must satisfy 0" a# 3b0# œ ! . Thus the
corresponding eigenvector is 0 a"b œ a# 3 ß "bX Þ Substitution of < œ " 3 results
in the single equation 0" a# 3b0# œ ! . A corresponding eigenvector is
0 a#b œ a# 3 ß "bX Þ
a,b. The eigenvalues are complex conjugates, with negative real part. Hence the origin
is a stable spiral.
________________________________________________________________________
page 500
—————————————————————————— CHAPTER 9. ——
a-ß . b.
6a+b. Solution of the ODEs is based on the analysis of the algebraic equations
Œ
#<
"
0"
!
&
œ Œ .
Œ
#<
0#
!
For a nonzero solution, we require that ./>aA < Ib œ <# " œ !. The roots of the
characteristic equation are < œ „3 . Setting < œ 3 , the equations are equivalent to
0" a# 3b0# œ ! . The eigenvectors are 0 a"b œ a# 3 ß "bX and 0 a#b œ a# 3 ß "bX Þ
a,b. The eigenvalues are purely imaginary. Hence the critical point is a center.
________________________________________________________________________
page 501
—————————————————————————— CHAPTER 9. ——
a-ß . b.
7a+b. Setting x œ 0 /<> results in the algebraic equations
Œ
$<
%
0"
!
#
œ
.
" < Π0# Π!
For a nonzero solution, we require that ./>aA < Ib œ <# #< & œ !. The roots
of the characteristic equation are < œ " „ #3 . Substituting < œ " #3 , the two
equations reduce to a" 3b0" 0# œ ! . The two eigenvectors are 0 a"b œ a" ß " 3bX
and 0 a#b œ a" ß " 3bX Þ
a,b. The eigenvalues are complex conjugates, with positive real part. Hence the origin
is an unstable spiral.
________________________________________________________________________
page 502
—————————————————————————— CHAPTER 9. ——
a-ß . b.
8a+b. The characteristic equation is given by
º
"<
"
&
œ a< "ba< !Þ#&b œ ! ,
$<º
with roots <" œ " and <# œ !Þ#& . For < œ ", the components of the solution
vector must satisfy 0# œ ! . Thus the corresponding eigenvector is 0 a"b œ a" ß !bX Þ
Substitution of < œ !Þ#& results in the single equation !Þ(& 0" 0# œ ! . A
corresponding eigenvector is 0 a#b œ a% ß $bX Þ
a,b. The eigenvalues are real and both negative. Hence the critical point is a stable
________________________________________________________________________
page 503
—————————————————————————— CHAPTER 9. ——
nodeÞ
a-ß . b.
9a+b. Solution of the ODEs is based on the analysis of the algebraic equations
Œ
$<
"
0"
!
%
œ Œ .
Œ
"<
0#
!
For a nonzero solution, we require that ./>aA < Ib œ <# # < " œ !. The single
root of the characteristic equation is < œ " . Setting < œ " , the components of the
solution vector must satisfy 0" # 0# œ ! . A corresponding eigenvector is
0 œ a # ß "b X Þ
a,b. Since there is only one linearly independent eigenvector, the critical point is an
?8=>+,6/, improper node.
________________________________________________________________________
page 504
—————————————————————————— CHAPTER 9. ——
a-ß . bÞ
10a+b. The characteristic equation is given by
"<
º &
#
œ <# * œ ! Þ
"<º
The equation has complex roots <"ß# œ „ $3. For < œ $3, the components of the
solution vector must satisfy & 0" a" $3b0# œ ! . Thus the corresponding eigenvector
is 0 a"b œ a" $3 ß &bX Þ Substitution of < œ $3 results in & 0" a" $3b0# œ ! . A
corresponding eigenvector is 0 a#b œ a" $3 ß &bX Þ
a,b. The eigenvalues are purely imaginary, hence the critical point is a center.
________________________________________________________________________
page 505
—————————————————————————— CHAPTER 9. ——
a-ß . bÞ
11a+b. The characteristic equation is a< "b# œ ! , with double root < œ " . It is
easy to see that the two linearly independent eigenvectors are 0 a"b œ a" ß !bX and
0 a#b œ a! ß "bX Þ
a,bÞ Since there are two linearly independent eigenvectors, the critical point is a stable
proper node.
a-ß . b.
________________________________________________________________________
page 506
—————————————————————————— CHAPTER 9. ——
12a+b. Setting x œ 0 /<> results in the algebraic equations
#<
Œ *Î&
0"
!
&Î#
œ Œ .
Œ
"<
0#
!
For a nonzero solution, we require that ./>aA < Ib œ <# < &Î# œ !. The roots
of the characteristic equation are < œ "Î# „ $3Î# . Substituting < œ "Î# $3Î# , the
equations reduce to a$ $3b0" & 0# œ ! . Therefore the two eigenvectors are
0 a"b œ a& ß $ $3bX and 0 a#b œ a& ß $ $3bX Þ
a,b. Since the eigenvalues are complex, with positive real part, the critical point is an
unstable spiral.
a-ß . b.
________________________________________________________________________
page 507
—————————————————————————— CHAPTER 9. ——
14. Setting x w œ 0 , that is,
#
Π"
#
"
xœŒ
,
#
"
we find that the critical point is x! œ a "ß !bX Þ With the change of dependent variable,
x œ x! u , the differential equation can be written as
.u
#
œŒ
"
.>
"
u.
#
The critical point for the transformed equation is the origin. Setting u œ 0 /<> results in
the algebraic equations
Œ
#<
"
0"
!
"
œ Œ .
Œ
#<
0#
!
For a nonzero solution, we require that ./>aA < Ib œ <# %< $ œ !. The roots
of the characteristic equation are < œ $ , " . Hence the critical point is a stable
node.
15. Setting x w œ 0 , that is,
"
Π#
"
"
xœŒ
,
"
&
we find that the critical point is x! œ a #ß "bX Þ With the change of dependent variable,
x œ x! u , the differential equation can be written as
.u
"
œŒ
#
.>
"
u.
"
The characteristic equation is ./>aA < Ib œ <# #< $ œ !, with complex conjugate
roots < œ " „ 3È# Þ Since the real parts of the eigenvalues are negative, the critical
point is a stable spiral.
16. The critical point x! satisfies the system of equations
________________________________________________________________________
page 508
—————————————————————————— CHAPTER 9. ——
!
Œ$
!
"
xœŒ
Þ
!
#
It follows that B! œ # Î$ and C ! œ !Î" . Using the transformation, x œ x! u , the
differential equation can be written as
.u
!
œŒ
$
.>
"
u.
!
The characteristic equation is ./>aA < Ib œ <# " $ œ !Þ Since " $ ! , the roots
are purely imaginary, with < œ „ 3È"$ Þ Hence the critical point is a center.
20Þ The system of ODEs can be written as
.x
+
œ Œ ""
+#"
.>
+"#
x.
+##
The characteristic equation is <# : < ; œ !Þ The roots are given by
<"ß# œ
: „ È:# %;
: „ È?
œ
.
#
#
The results can be verified using Table *Þ"Þ" .
21a+b. If ; ! and : ! , then the roots are either complex conjugates with negative
real parts, or both real and negative.
a,b. If ; ! and : œ ! , then the roots are purely imaginary.
a- b. If ; ! , then the roots are real, with <" † <# ! . If : ! , then either the roots
are real, with <" † <# ! or the roots are complex conjugates with positive real parts.
________________________________________________________________________
page 509
—————————————————————————— CHAPTER 9. ——
Section 9.2
2. The differential equations can be combined to obtain a related ODE
.C
#C
œ
Þ
.B
B
The equation is separable, with
.C
# .B
œ
Þ
C
B
The solution is given by C œ G B# Þ Note that the system is uncoupled, and hence we
also have B œ B! /> and C œ C! /#> .
In order to determine the direction of motion along the trajectories, observe that for
positive initial conditions, B will decrease, whereas C will increase.
4. The trajectories of the system satisfy the ODE
.C
,B
œ
Þ
.B
+C
The equation is separable, with
+C .C œ ,B .B Þ
Hence the trajectories are given by , B# + C # œ G # , in which G is arbitrary. Evidently,
the trajectories are ellipses. Invoking the initial condition, we find that G # œ +, . The
system of ODEs can also be written as
.x
!
œŒ
,
.>
+
x.
!
Using the methods in Chapter (, it is easy to show that
B œ È+ -9= È+, >
C œ È, =38 È+, > Þ
________________________________________________________________________
page 510
—————————————————————————— CHAPTER 9. ——
Note that for positive initial conditions, B will 38crease, whereas C will ./crease.
5a+b. The critical points are given by the solution set of the equations
Ba" Cb œ !
Ca" #Bb œ ! Þ
Clearly, a! ß !b is a solution. If B Á ! , then C œ " and B œ "Î# . Hence the critical
points are a! ß !b and a "Î# ß "b.
a, b .
a- b. Based on the phase portrait, all trajectories starting near the origin diverge. Hence
the critical point a! ß !b is unstable. Examining the phase curves near the critical point
a "Î# ß "b,
________________________________________________________________________
page 511
—————————————————————————— CHAPTER 9. ——
the equilibrium point has the properties of a saddle, and hence it is unstable.
6a+b. The critical points are solutions of the equations
" #C œ !
" $B# œ ! Þ
There are two equilibrium points, Š "ÎÈ$ ß "Î#‹ and Š"ÎÈ$ ß "Î#‹Þ
a, bÞ
a- b. Locally, the trajectories near the point Š "ÎÈ$ ß "Î#‹ resemble the behavior near
a saddle. Hence the critical point is unstable. Near the point Š"ÎÈ$ ß "Î#‹, the
solutions are periodic. Therefore the second critical point is stable.
8a+b. The critical points are solutions of the equations
aB C ba" B C b œ !
Ba# Cb œ ! Þ
If B œ C , then B œ C œ ! or B œ C œ # Þ If B œ " C , then B œ ! and C œ " , or
B œ $ and C œ # Þ It follows that the critical points are a! ß !bß a # ß #bß a! ß "b
________________________________________________________________________
page 512
—————————————————————————— CHAPTER 9. ——
and a$ ß #bÞ
a, bÞ
a- b. Near the origin, the trajectories resemble those of a saddle, and hence it is unstable.
Near the critical point a! ß "b , the trajectories resemble those of a stable spiral. Hence the
equilibrium point is asymptotically stable.
Based on the global phase portrait, it is evident that the other critical points are nodes.
Closer examination reveals that the point a # ß #b is asymptotically stable, whereas
the point a$ ß #b is unstable.
________________________________________________________________________
page 513
—————————————————————————— CHAPTER 9. ——
9a+b. The critical points are given by the solution set of the equations
C a# B C b œ !
B C #BC œ ! Þ
Clearly, a! ß !b is a critical point. If B œ # C , then it follows that C aC #b œ " . The
additional critical points are Š" È# ß " È# ‹ and Š" È# ß " È# ‹Þ
a, b .
a- b. The behavior near the origin is that of a stable spiral. Hence the point a! ß !b is
asymptotically stable.
At the critical point Š" È# ß " È# ‹, the trajectories resemble those near a saddle.
Hence the critical point is unstable.
________________________________________________________________________
page 514
—————————————————————————— CHAPTER 9. ——
Near the point Š" È# ß " È# ‹, the trajectories resemble those near a saddle.
Hence the critical point is also unstable.
10a+bÞ The critical points are solutions of the equations
a# BbaC Bb œ !
Cˆ# B B# ‰ œ ! Þ
The origin is evidently a critical point. If B œ # , then C œ ! . If B œ C , then either
C œ ! or B œ C œ " or B œ C œ # . Hence the other critical points are a # ß !b,
a " ß "b and a# ß #b.
________________________________________________________________________
page 515
—————————————————————————— CHAPTER 9. ——
a, b .
a- b. Based on the global phase portrait, the critical points a! ß !b and a # ß !b have the
characteristics of a saddle. Hence these points are unstable. The behavior near the
remaining two critical points resembles those near a stable spiral. Hence the critical
points a " ß "b and a# ß #b are asymptotically stable.
11a+b. The critical points are given by the solution set of the equations
Ba" #Cb œ !
C B# C # œ ! Þ
If B œ ! , then either C œ ! or C œ " . If C œ "Î# , then B œ „"Î# . Hence the critical
points are at a! ß !b, a! ß "b, a "Î# ß "Î#b and a"Î# ß "Î#b.
a, b .
a- b. The trajectories near the critical points a "Î# ß "Î#b and a"Î# ß "Î#b are closed
curves. Hence the critical points have the characteristics of a center, which is stable.
The trajectories near the critical points a! ß !b and a!ß "b resemble those near a saddle.
Hence these critical points are unstable.
________________________________________________________________________
page 516
—————————————————————————— CHAPTER 9. ——
13a+b. The critical points are solutions of the equations
a# BbaC Bb œ !
a% BbaC Bb œ ! Þ
If C œ B , then either B œ C œ ! or B œ C œ % Þ If B œ # , then C œ # . If B œ C ,
then C œ # or C œ ! . Hence the critical points are at a! ß !b, a% ß %b and a # ß #b.
a, b .
a- b. The critical point at a% ß %b is evidently a stable spiral, which is asymptotically
stable. Closer examination of the critical point at a! ß !b reveals that it is a saddle,
which is unstable.
The trajectories near the critical point a # ß #b resemble those near an unstable node.
________________________________________________________________________
page 517
—————————————————————————— CHAPTER 9. ——
14a+b. The critical points consist of the solution set of the equations
#‰
Cœ!
ˆ" B C B œ ! Þ
It is easy to see that the only critical point is at a! ß !b.
a, b .
a- b. The origin is an unstable spiral.
16a+b. The trajectories are solutions of the differential equation
.C
%B
œ
,
.B
C
which can also be written as %B .B C .C œ ! . Integrating, we obtain
%B# C # œ G # .
Hence the trajectories are ellipses.
________________________________________________________________________
page 518
—————————————————————————— CHAPTER 9. ——
a, b .
Based on the differential equations, the direction of motion on each trajectory is
clockwise.
17a+b. The trajectories of the system satisfy the ODE
.C
#B C
œ
,
.B
C
which can also be written as a#B Cb.B C.C œ ! . This differential equation is
homogeneous. Setting C œ B @aBb, we obtain
@B
.@
#
œ ",
.B
@
that is,
B
.@
# @ @#
œ
Þ
.B
@
The resulting ODE is separable, with solution B$ a@ "ba@ #b# œ G . Reverting back
to the original variables, the trajectories are level curves of
L aB ß C b œ aB C baC #Bb# Þ
________________________________________________________________________
page 519
—————————————————————————— CHAPTER 9. ——
a, b .
The origin is a saddle. Along the line C œ #B , solutions increase without bound. Along
the line C œ B , solutions converge toward the origin.
18a+b. The trajectories are solutions of the differential equation
.C
BC
œ
,
.B
BC
which is homogeneous. Setting C œ B @aBb, we obtain
@B
.@
B B@
œ
,
.B
B B@
that is,
B
.@
" @#
œ
Þ
.B
"@
The resulting ODE is separable, with solution
+<->+8a@b œ 68kBkÈ" @# Þ
Reverting back to the original variables, the trajectories are level curves of
L aB ß C b œ +<->+8aCÎBb 68ÈB# C # Þ
________________________________________________________________________
page 520
—————————————————————————— CHAPTER 9. ——
a, b .
The origin is a stable spiral.
20a+b. The trajectories are solutions of the differential equation
.C
#BC # ' BC
œ #
,
.B
#B C $B# %C
which can also be written as a#BC# ' BCb.B a#B# C $B# %C b.C œ ! . The
resulting ODE is exact, with
`L
`L
œ #BC# ' BC and
œ #B# C $B# %C .
`B
`C
Integrating the first equation, we find that L aB ß C b œ B# C # $B# C 0 aC b. It follows
that
`L
œ #B# C $B# 0 w aC b.
`C
Comparing the two partial derivatives, we obtain 0 aCb œ #C # - . Hence
L aB ß C b œ B# C # $B# C #C # Þ
________________________________________________________________________
page 521
—————————————————————————— CHAPTER 9. ——
a, b .
The associated direction field shows the direction of motion along the trajectories.
22a+b. The trajectories are solutions of the differential equation
.C
' B B$
,
œ
.B
'C
which can also be written as a' B B$ b.B ' C.C œ ! . The resulting ODE is exact,
with
`L
`L
œ ' B B$ and
œ 'C.
`B
`C
Integrating the first equation, we have L aB ß Cb œ $B# B% Î% 0 aC b. It follows that
`L
œ 0 w aCb.
`C
Comparing the two partial derivatives, we conclude that 0 aCb œ $C # - . Hence
________________________________________________________________________
page 522
—————————————————————————— CHAPTER 9. ——
a, b .
L aB ß C b œ $B#
B%
$C # Þ
%
________________________________________________________________________
page 523
—————————————————————————— CHAPTER 9. ——
Section 9.3
1. Write the system in the form x w œ Ax gaxb. In this case, it is evident that
. B
"
Œ œ Œ"
.> C
B
C#
!
Þ
# ΠC ΠB#
That is, gaxb œ a C # ß B# bX Þ Using polar coordinates, lgaxbl œ <# È=38% ) -9=% )
and lxl œ < . Hence
lgaxbl
œ lim <È=38% ) -9=% ) œ ! ,
< Ä ! lxl
<Ä!
lim
and the system is almost linear. The origin is an isolated critical point of the linear
system
. B
"
Œ œ Œ"
.> C
B
!
Œ
Þ
#
C
The characteristic equation of the coefficient matrix is <# < # œ ! , with roots
<" œ " and <# œ # . Hence the critical point is a saddle, which is unstable.
2. The system can be written as
. B
"
Œ œŒ %
.> C
B
#BC
"
Þ
" ΠC ΠB# C #
Following the discussion in Example $ , we note that J aB ß Cb œ B C #BC and
KaB ß Cb œ %B C B# C # . Both of the functions J and K are twice differentiable,
hence the system is almost linear. Furthermore,
JB œ " #C , JC œ " #B , KB œ % #B , KC œ " #C .
The origin is an isolated critical point, with
J B a ! ß !b
Œ K a ! ß !b
B
J C a ! ß !b
"
œ
KC a! ß !b Œ %
"
.
"
The characteristic equation of the associated linear system is <# # < & œ ! , with
complex conjugate roots <"ß# œ " „ #3 Þ The origin is a stable spiral, which is
asymptotically stable.
5a+b. The critical points consist of the solution set of the equations
a# BbaC Bb œ !
a% BbaC Bb œ ! Þ
As shown in Prob. "$ of Section *Þ# , the only critical points are at a! ß !b, a% ß %b and
a # ß #b .
________________________________________________________________________
page 524
—————————————————————————— CHAPTER 9. ——
a,ß - b. First note that J aB ß C b œ a# BbaC Bb and KaB ß C b œ a% BbaC Bb . The
Jacobian matrix of the vector field is
JœŒ
JB a B ß C b
KB aB ß Cb
J C aB ß C b
# #B C
œŒ
KC a B ß C b
% C #B
#B
.
% B
At the origin, the coefficient matrix of the linearized system is
Ja! ß !b œ Œ
#
%
#
,
%
with eigenvalues <" œ " È"( and <# œ " È"( . The eigenvalues are real, with
opposite sign. Hence the critical point is a saddle, which is unstable. At the equilibrium
point a # ß #b, the coefficient matrix of the linearized system is
Ja # ß #b œ Œ
%
'
!
,
'
with eigenvalues <" œ % and <# œ ' . The eigenvalues are real, unequal and positive,
hence the critical point is an unstable node. At the point a% ß %b, the coefficient matrix
of the linearized system is
Ja% ß %b œ Œ
'
)
'
,
!
with complex conjugate eigenvalues <"ß# œ $ „ 3È$* . The critical point is a stable
spiral, which is asymptotically stable.
Based on Table *Þ$Þ" , the nonlinear terms do not affect the stability and type of each
critical point.
a. b .
7a+bÞ The critical points are solutions of the equations
________________________________________________________________________
page 525
—————————————————————————— CHAPTER 9. ——
"C œ!
aB CbaB Cb œ ! Þ
The first equation requires that C œ " . Based on the second equation, B œ „ " . Hence
the critical points are a " ß "b and a" ß "b.
a,ß - b. J aB ß C b œ " C and KaB ß C b œ B# C # . The Jacobian matrix of the vector
field is
JœŒ
JB a B ß C b
KB aB ß Cb
J C aB ß C b
!
œ
KC aB ß Cb Œ #B
"
.
#C
At the critical point a " ß "b, the coefficient matrix of the linearized system is
Ja " ß "b œ Œ
!
#
"
,
#
with eigenvalues <" œ " È$ and <# œ " È$ . The eigenvalues are real, with
opposite sign. Hence the critical point is a saddle, which is unstable. At the equilibrium
point a" ß "b, the coefficient matrix of the linearized system is
Ja" ß "b œ Œ
!
#
"
,
#
with complex conjugate eigenvalues <"ß# œ " „ 3 . The critical point is a stable
spiral, which is asymptotically stable.
a. b .
Based on Table *Þ$Þ" , the nonlinear terms do not affect the stability and type of each
critical point.
8a+b. The critical points are given by the solution set of the equations
________________________________________________________________________
page 526
—————————————————————————— CHAPTER 9. ——
Ba" B Cb œ !
Ca# C $Bb œ ! Þ
If B œ ! , then either C œ ! or C œ # . If C œ ! , then B œ ! or B œ " . If C œ " B ,
then either B œ "Î# or B œ ". If C œ # $B , then B œ ! or B œ "Î# . Hence the
critical points are at a! ß !b, a! ß #b, a" ß !b and a"Î# ß "Î#bÞ
a,ß - b. Note that J aB ß C b œ B B# BC and KaB ß C b œ a#C C # $BC bÎ% . The
Jacobian matrix of the vector field is
JœŒ
JB a B ß C b
KB aB ß Cb
J C aB ß C b
" #B C
œŒ
KC a B ß C b
$CÎ%
B
.
"Î# CÎ# $BÎ%
At the origin, the coefficient matrix of the linearized system is
Ja! ß !b œ Œ
"
!
!
"
#
,
with eigenvalues <" œ " and <# œ "Î# . The eigenvalues are real and both positive.
Hence the critical point is an unstable node. At the equilibrium point a! ß #b, the
coefficient matrix of the linearized system is
Ja! ß #b œ Œ
!
,
"#
"
$#
with eigenvalues <" œ " and <# œ "Î# . The eigenvalues are both negative, hence
the critical point is a stable node. At the point a" ß !b, the coefficient matrix
of the linearized system is
Ja" ß !b œ Œ
"
,
"%
"
!
with eigenvalues <" œ " and <# œ "Î% . Both of the eigenvalues are negative, and
hence the critical point is a stable node. At the critical point a"Î# ß "Î#b, the coefficient
matrix of the linearized system is
Ja"Î# ß "Î#b œ
"
#
$
)
"#
,
")
with eigenvalues <" œ &Î"' È&( Î"' and <# œ &Î"' È&( Î"' . The
eigenvalues are real, with opposite sign. Hence the critical point is a saddle, which is
unstable.
________________________________________________________________________
page 527
—————————————————————————— CHAPTER 9. ——
a. b .
Based on Table *Þ$Þ" , the nonlinear terms do not affect the stability and type of each
critical point.
9a+b. Based on Prob. ) , in Section *Þ# , the critical points are at a! ß !bß a # ß #bß
a! ß "b and a$ ß #bÞ
a,ß - bÞ First note that J aB ß C b œ aB C ba" B C b and KaB ß C b œ Ba# C b . The
Jacobian matrix of the vector field is
JœŒ
#B "
#C
" #C
.
B
At the origin, the coefficient matrix of the linearized system is
________________________________________________________________________
page 528
—————————————————————————— CHAPTER 9. ——
Ja! ß !b œ Œ
"
,
!
"
#
with eigenvalues <" œ " and <# œ # . The eigenvalues are real, with opposite sign.
Hence the critical point is a saddle, which is unstable. At the critical point a! ß "b,
the coefficient matrix of the linearized system is
Ja! ß "b œ Œ
"
,
!
"
$
with complex conjugate eigenvalues <"ß# œ "Î# „ 3È"" Î# . The critical point is a
stable spiral, which is asymptotically stable. At the point a # ß #b, the coefficient
matrix of the linearized system is
Ja # ß #b œ Œ
&
!
&
,
#
with eigenvalues <" œ # and <# œ & . The eigenvalues are unequal and negative,
hence the critical point is a stable node. At the point a$ ß #b, the coefficient matrix
of the linearized system is
Ja$ ß #b œ Œ
&
!
&
,
$
with eigenvalues <" œ $ and <# œ & . The eigenvalues are unequal and positive, hence
the critical point is an unstable node.
a. b .
Based on Table *Þ$Þ" , the nonlinear terms do not affect the stability and type of each
critical point.
________________________________________________________________________
page 529
—————————————————————————— CHAPTER 9. ——
11a+b. The critical points are solutions of the equations
#B C BC $ œ !
B #C BC œ ! Þ
Substitution of C œ BÎaB #b into the first equation results in
$B% "$B$ #)B# #!B œ ! .
One root of the resulting equation is B œ ! . The only other real root of the equation is
Bœ
"Î$
"Î$
"
"$•.
”Š#)( ")È#!"* ‹ )$Š#)( ")È#!"* ‹
*
Hence the critical points are a! ß !b and a "Þ"*$%&ÞÞÞ ß "Þ%(*(ÞÞÞbÞ
a,ß - b. J aB ß C b œ B B# BC and KaB ß C b œ a#C C # $BC bÎ% . The Jacobian
matrix of the vector field is
JœŒ
JB a B ß C b
KB aB ß Cb
J C aB ß C b
# C$
œŒ
KC a B ß C b
"C
" $BC #
.
#B
At the origin, the coefficient matrix of the linearized system is
Ja! ß !b œ Œ
#
"
"
,
#
with eigenvalues <" œ È& and <# œ È& . The eigenvalues are real and of opposite
sign. Hence the critical point is a saddle, which is unstable. At the equilibrium point
a "Þ"*$%&ÞÞÞ ß "Þ%(*(ÞÞÞb, the coefficient matrix of the linearized system is
Ja "Þ"*$%& ß "Þ%(*(b œ Œ
"Þ#$**
#Þ%(*(
'Þ)$*$
,
!Þ)!'&
with complex conjugate eigenvalues <"ß# œ "Þ!#$# „ %Þ""#& 3 . The critical point is
a stable spiral, which is asymptotically stable.
________________________________________________________________________
page 530
—————————————————————————— CHAPTER 9. ——
a. b .
In both cases, the nonlinear terms do not affect the stability and type of the critical point.
12a+b. The critical points are given by the solution set of the equations
a" Bb=38 C œ !
" B -9= C œ ! Þ
If B œ " , then we must have -9= C œ # , which is impossible. Therefore =38 C œ ! ,
which implies that C œ 81 , 8 œ ! ß „ " ß # ß ÞÞÞ Þ Based on the second equation,
B œ " -9= 81 Þ
It follows that the critical points are located at a! ß #5 1b and a# ß a#5 "b1b , where
5 œ ! ß „ " ß # ß ÞÞÞ Þ
a,ß - b. Given that J aB ß Cb œ a" Bb=38 C and KaB ß C b œ " B -9= C , the
Jacobian matrix of the vector field is
JœŒ
=38 C
"
a" Bb-9= C
.
=38 C
At the critical points a! ß #51b, the coefficient matrix of the linearized system is
Ja! ß #5 1b œ Œ
!
"
"
,
!
with purely complex eigenvalues <"ß# œ „ 3 . The critical points of the associated linear
systems are centers, which are stable. Note that Theorem *Þ$Þ# does not provide a
definite conclusion regarding the relation between the nature of the critical points of the
nonlinear systems and their corresponding linearizations. At the points a# ß a#5 "b1b,
the coefficient matrix of the linearized system is
Jc# ß a#5 "b1d œ Œ
!
"
$
,
!
________________________________________________________________________
page 531
—————————————————————————— CHAPTER 9. ——
with eigenvalues <" œ È$ and <# œ È$ . The eigenvalues are real, with opposite
sign. Hence the critical points of the associated linear systems are saddles, which are
unstable.
a. b .
As asserted in Theorem *Þ$Þ# , the trajectories near the critical points a# ß a#5 "b1b
resemble those near a saddle.
Upon closer examination, the critical points a! ß #51b are indeed centers.
13a+b. The critical points are solutions of the equations
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page 532
—————————————————————————— CHAPTER 9. ——
B C# œ !
C B# œ ! Þ
Substitution of C œ B# into the first equation results in
B B% œ ! ,
with real roots B œ ! , " . Hence the critical points are at a! ß !b and a" ß "b.
a,ß - b. In this problem, J aB ß C b œ B C # and KaB ß C b œ C B# . The Jacobian
matrix of the vector field is
JœŒ
#C
.
"
"
#B
At the origin, the coefficient matrix of the linearized system is
Ja! ß !b œ Œ
"
!
!
,
"
with repeated eigenvalues <" œ " and <# œ " . It is easy to see that the corresponding
eigenvectors are linearly independent. Hence the critical point is an unstable proper
node. Theorem *Þ$Þ# does not provide a definite conclusion regarding the relation
between the nature of the critical point of the nonlinear system and the corresponding
linearization. At the critical point a" ß "b, the coefficient matrix of the linearized system
is
Ja" ß "b œ Œ
"
#
#
,
"
with eigenvalues <" œ $ and <# œ " . The eigenvalues are real, with opposite sign.
Hence the critical point is a saddle, which is unstable.
a. b .
Closer examination reveals that the critical point at the origin is indeed a proper node.
________________________________________________________________________
page 533
—————————————————————————— CHAPTER 9. ——
14a+b. The critical points are given by the solution set of the equations
" BC œ !
B C$ œ ! Þ
After multiplying the second equation by C , it follows that C œ „ " . Hence the critical
points of the system are at a" ß "b and a " ß "b.
a,ß - b. Note that J aB ß C b œ " BC and KaB ß C b œ B C $ . The Jacobian matrix of
the vector field is
JœŒ
C
"
B
.
$C#
At the critical point a" ß "b, the coefficient matrix of the linearized system is
Ja" ß "b œ Œ
"
,
$
"
"
with eigenvalues <" œ # and <# œ # . The eigenvalues are real and equal. It is
easy to show that there is only one linearly independent eigenvector. Hence the critical
point is a stable improper node. Theorem *Þ$Þ# does not provide a definite conclusion
regarding the relation between the nature of the critical point of the nonlinear system and
the corresponding linearization. At the point a " ß "b, the coefficient matrix of the
linearized system is
Ja " ß "b œ Œ
"
"
"
,
$
with eigenvalues <" œ " È& and <# œ " È& . The eigenvalues are real,
with opposite sign. Hence the critical point of the associated linear system is a saddle,
which is unstable.
________________________________________________________________________
page 534
—————————————————————————— CHAPTER 9. ——
a. b .
Closer examination reveals that the critical point at a" ß "b is indeed a stable improper
node, which is asymptotically stable.
15a+b. The critical points are given by the solution set of the equations
#B C BˆB# C # ‰ œ !
B C CˆB# C # ‰ œ ! Þ
It is clear that the origin is a critical point. Solving the first equation for C , we find that
Cœ
" „È" )B# %B%
Þ
#B
Substitution of these relations into the second equation results in two equations of the
form 0" aBb œ ! and 0# aBb œ ! . Plotting these functions, we note that only 0" aBb œ !
has real roots given by B ¸ „ !Þ$$!(' Þ It follows that the additional critical points are
at a !Þ$$!(' ß "Þ!*#%b and a !Þ$$!(' ß "Þ!*#%bÞ
________________________________________________________________________
page 535
—————————————————————————— CHAPTER 9. ——
a,ß - b. Given that
J aB ß Cb œ #B C BˆB# C # ‰
KaB ß Cb œ B C C ˆB# C # ‰,
the Jacobian matrix of the vector field is
JœŒ
" #BC
.
" B# $C #
# $B# C #
" #BC
At the critical point a! ß !b, the coefficient matrix of the linearized system is
Ja! ß !b œ Œ
#
"
"
,
"
with complex conjugate eigenvalues <"ß# œ Š $ „ 3È$ ‹Î# . Hence the critical point
is a stable spiral, which is asymptotically stable. At the point a !Þ$$!(' ß "Þ!*#%b,
the coefficient matrix of the linearized system is
Ja !Þ$$!(' ß "Þ!*#%b œ Œ
$Þ&#"'
!Þ#(($&
!Þ#(($&
,
#Þ')*&
with eigenvalues <" œ $Þ&!*# and <# œ #Þ'((" . The eigenvalues are real, with
opposite sign. Hence the critical point of the associated linear system is a saddle,
which is unstable. Identical results hold for the point at a !Þ$$!(' ß "Þ!*#%b .
a. b .
A closer look at the origin reveals a spiral:
________________________________________________________________________
page 536
—————————————————————————— CHAPTER 9. ——
Near the point a !Þ$$!(' ß "Þ!*#%b the nature of the critical point is evident:
Based on Table *Þ$Þ" , the nonlinear terms do not affect the stability and type of each
critical point.
16a+b. The critical points are solutions of the equations
C Bˆ" B# C # ‰ œ !
B Cˆ" B# C # ‰ œ ! Þ
Multiply the first equation by C and the second equation by B . The difference of the
two equations gives B# C # œ ! . Hence the only critical point is at the origin.
a,ß - b. With J aB ß C b œ C Ba" B# C # b and KaB ß C b œ B C a" B# C # b,
the Jacobian matrix of the vector field is
JœŒ
" $B# C #
" #BC
" #BC
.
" B# $C #
At the origin, the coefficient matrix of the linearized system is
Ja! ß !b œ Œ
"
"
"
,
"
with complex conjugate eigenvalues <"ß# œ " „ 3 . Hence the origin is an unstable
________________________________________________________________________
page 537
—————————————————————————— CHAPTER 9. ——
spiral.
a. bÞ
17a+b. The Jacobian matrix of the vector field is
JœŒ
!
" 'B#
"
.
!
At the origin, the coefficient matrix of the linearized system is
Ja! ß !b œ Œ
"
!
!
,
"
with eigenvalues <" œ " and <# œ " . The eigenvalues are real, with opposite sign.
Hence the critical point is a saddle point.
a,b. The trajectories of the linearized system are solutions of the differential equation
.C
B
œ ,
.B
C
which is separable. Integrating both sides of the equation B .B C .C œ ! , the solution
is B# C # œ G . The trajectories consist of a family of hyperbolas.
________________________________________________________________________
page 538
—————————————————————————— CHAPTER 9. ——
It is easy to show that the general solution is given by Ba>b œ -" /> -# /> and
Ca>b œ -" /> -# /> Þ The only bounded solutions consist of those for which -" œ ! .
In that case, Ba>b œ -# /> œ C a>b.
a- b. The trajectories of the given system are solutions of the differential equation
.C
B #B$
œ
,
.B
C
which can also be written as aB #B$ b.B C .C œ ! . The resulting ODE is exact,
with
`L
`L
œ B #B$ and
œ C.
`B
`C
Integrating the first equation, we find that L aB ß C b œ B# Î# B% Î# 0 aC b. It follows
that
`L
œ 0 w aCb.
`C
Comparing the partial derivatives, we obtain 0 aCb œ C # Î# - . Hence the solutions
are level curves of the function
L aB ß Cb œ B# Î# B% Î# C # Î# Þ
The trajectories approaching to, or diverging from, the origin are no longer straight lines.
________________________________________________________________________
page 539
—————————————————————————— CHAPTER 9. ——
19a+b. The solutions of the system of equations
Cœ!
= =38 B œ !
#
consist of the points a„ 81 ß !b , 8 œ ! ß " ß # ß â Þ The functions J aB ß C b œ C and
KaB ß Cb œ =# =38 B are analytic on the entire plane. It follows that the system is
almost linear near each of the critical points.
a,b. The Jacobian matrix of the vector field is
JœŒ
!
"
.
= -9= B !
#
At the origin, the coefficient matrix of the linearized system is
Ja! ß !b œ Œ
!
=#
"
,
!
with purely complex eigenvalues <"ß# œ „ 3= . Hence the origin is a center. Since the
eigenvalues are purely complex, Theorem *Þ$Þ# gives no definite conclusion about the
critical point of the nonlinear system. Physically, the critical point corresponds to the
state ) œ ! , ) w œ ! Þ That is, the rest configuration of the pendulum.
a- b. At the critical point a1 ß !b, the coefficient matrix of the linearized system is
Ja1 ß !b œ Œ
!
=#
"
,
!
with eigenvalues <"ß# œ „ = . The eigenvalues are real and of opposite sign. Hence the
critical point is a saddle. Theorem *Þ$Þ# asserts that the critical point for the nonlinear
system is also a saddle, which is unstable. This critical point corresponds to the state
) œ 1 , ) w œ ! Þ That is, the upright rest configuration.
________________________________________________________________________
page 540
—————————————————————————— CHAPTER 9. ——
a. b. Let =# œ " . The following is a plot of the phase curves near a! ß !bÞ
The local phase portrait shows that the origin is indeed a center.
a/ b .
It should be noted that the phase portrait has a periodic pattern, since ) œ B 79. #1 .
20a+b. The trajectories of the system in Problem "* are solutions of the differential
equation
.C
=# =38 B
œ
,
.B
C
which can also be written as =# =38 B .B C .C œ ! . The resulting ODE is exact,
with
`L
`L
œ =# =38 B and
œ C.
`B
`C
Integrating the first equation, we find that L aB ß C b œ =# -9= B 0 aC b. It follows
that
________________________________________________________________________
page 541
—————————————————————————— CHAPTER 9. ——
`L
œ 0 w aCb.
`C
Comparing the partial derivatives, we obtain 0 aCb œ C # Î# G . Hence the solutions
are level curves of the function
L aB ß C b œ =# -9= B C # Î# Þ
Adding an arbitrary constant, say =# , to the function L aB ß C b does not change the nature
of the level curves. Hence the trajectories are can be written as
" #
C =# a" -9= Bb œ - ,
#
in which - is an arbitrary constant.
a,b. Multiplying by 7P# and reverting to the original physical variables, we obtain
"
.)
7P# Œ 7P# =# a" -9= ) b œ 7P# - Þ
#
.>
#
Since =# œ 1ÎP , the equation can be written as
"
.)
7P# Œ 71Pa" -9= ) b œ I ,
#
.>
#
in which I œ 7P# - .
a- b. The absolute velocity of the point mass is given by @ œ P . )Î.> . The kinetic
energy of the mass is X œ 7@# Î# . Choosing the rest position as the datum, that is, the
level of zero potential energy, the gravitational potential energy of the point mass is
Z œ 71Pa" -9= ) bÞ
It follows that the total energy, X Z , is constant along the trajectories.
21a+b. E œ !Þ#&
________________________________________________________________________
page 542
—————————————————————————— CHAPTER 9. ——
Since the system is undamped, and Ca!b œ ! , the amplitude is !Þ#& . The period is
estimated at 7 ¸ $Þ"' .
a, b .
E œ !Þ&
E œ "Þ!
E œ "Þ&
E œ #Þ!
V
!Þ&
"Þ!
"Þ&
#Þ!
7
$Þ#!
$Þ$&
$Þ'$
%Þ"(
a- bÞ Since the system is conservative, the amplitude is equal to the initial amplitude. On
________________________________________________________________________
page 543
—————————————————————————— CHAPTER 9. ——
the other hand, the period of the pendulum is a monotone increasing function of the
initial
position E .
It appears that as E p ! , the period approaches 1, the period of the corresponding linear
pendulum a#1Î=b.
a. b .
The pendulum is released from rest, at an inclination of % 1 radians from the vertical.
Based on conservation of energy, the pendulum will swing past the lower equilibrium
position a) œ #1b and come to rest, momentarily, at a maximum rotational displacement
of )7+B œ $1 a% 1b œ %1 % . The transition between the two dynamics occurs
at E œ 1 , that is, once the pendulum is released beyond the upright configuration.
24a+b. It is evident that the origin is a critical point of each system. Furthermore, it is
easy to see that the corresponding linear system, in each case, is given by
.B
œC
.>
.C
œ BÞ
.>
The eigenvalues of the coefficient matrix are <"ß# œ „ 3 . Hence the critical point of the
________________________________________________________________________
page 544
—————————————————————————— CHAPTER 9. ——
linearized system is a center.
a,b. Using polar coordinates, it is also easy to show that
lgaxbl
œ !.
< Ä ! lxl
lim
Alternatively, the nonlinear terms are analytic in the entire plane. Hence both systems
are
almost linear near the origin.
a- b. For system a33b, note that
B
.B
.C
C
œ BC B# ˆB# C # ‰ BC C # ˆB# C # ‰Þ
.>
.>
Converting to polar coordinates, and differentiating the equation <# œ B# C # with
respect to > , we find that
<
.<
.B
.C
#
œB
C
œ ˆB# C # ‰ œ <% Þ
.>
.>
.>
That is, < w œ <$ Þ It follows that <# œ "Îa#> - b, where - œ "Î<!# . Since < p ! as
> p ! , regardless of the value of <! , the origin is an +=C7:>9>3-+66C =>+,6/ equilibrium
point.
On the other hand, for system a3b,
<
.<
.B
.C
#
œB
C
œ ˆB# C # ‰ œ <% Þ
.>
.>
.>
That is, < w œ <$ Þ Solving the differential equation results in
<# œ
- #>
a#> - b#
.
Imposing the initial condition <a!b œ <! , we obtain a specific solution
<# œ
<!#
.
# <!# > "
Since the solution becomes unbounded as > p "Î#<!# , the critical point is unstable.
25. The characteristic equation of the coefficient matrix is <# " œ ! , with complex
roots <"ß# œ „ 3 . Hence the critical point at the origin is a center. The characteristic
equation of the perturbed matrix is <# # % < " %# œ ! , with complex conjugate
roots <"ß# œ % „ 3 . As long as % Á ! , the critical point of the perturbed system is a
spiral point. Its stability depends on the sign of % Þ
26. The characteristic equation of the coefficient matrix is a< "b# œ ! , with roots
________________________________________________________________________
page 545
—————————————————————————— CHAPTER 9. ——
<" œ <# œ " . Hence the critical point is an asymptotically stable node. On the
other hand, the characteristic equation of the perturbed system is <# #< " % œ ! ,
with roots <"ß# œ " „È % . If % ! , then <"ß# œ " „ 3È% are complex roots.
The critical point is a stable spiral. If % ! , then <"ß# œ " „ Èk%k are real and
both negative ak%k ¥ "b. The critical point remains a stable node.
27a. b. Set 5 œ =38a!Î#b œ =38aEÎ#b and 1ÎP œ % .
________________________________________________________________________
page 546
—————————————————————————— CHAPTER 9. ——
Section 9.4
1a+b.
a,bÞ The critical points are solutions of the system of equations
Ba"Þ& B !Þ& C b œ !
Ca# C !Þ(& Bb œ ! Þ
The four critical points are a! ß !b, a! ß #b, a"Þ& ß !b and a!Þ) ß "Þ%b.
a- b. The Jacobian matrix of the vector field is
JœŒ
$Î# #B CÎ#
$CÎ%
BÎ#
.
# $BÎ% #C
At the critical point a! ß !b, the coefficient matrix of the linearized system is
Ja! ß !b œ Œ
$Î#
!
!
.
#
The eigenvalues and eigenvectors are
"
!
<" œ $Î# , 0 a"b œ Œ ; <# œ # , 0 a#b œ Œ Þ
!
"
The eigenvalues are positive, hence the origin is an unstable node.
At the critical point a! ß #b, the coefficient matrix of the linearized system is
Ja! ß #b œ Œ
"Î#
$Î#
!
.
#
The eigenvalues and eigenvectors are
________________________________________________________________________
page 547
—————————————————————————— CHAPTER 9. ——
<" œ "Î# , 0 a"b œ Œ
"
!
a#b
; <# œ # , 0 œ Œ Þ
!Þ'
"
The eigenvalues are of opposite sign. Hence the critical point is a saddle, which is
unstable.
At the critical point a"Þ& ß !b, the coefficient matrix of the linearized system is
Ja"Þ& ß !b œ Œ
"Þ&
!
!Þ(&
.
!Þ)(&
The eigenvalues and eigenvectors are
"
!Þ$"&(*
<" œ "Þ& , 0 a"b œ Œ ; <# œ !Þ)(& , 0 a#b œ Œ
Þ
!
"
The eigenvalues are of opposite sign. Hence the critical point is also a saddle, which is
unstable.
At the critical point a!Þ) ß "Þ%b, the coefficient matrix of the linearized system is
Ja!Þ) ß "Þ%b œ Œ
!Þ)
"Þ!&
!Þ%
.
"Þ%
The eigenvalues and eigenvectors are
<" œ
"" È&"
"!
"!
"
"
"" È&"
, 0 a"b œ $È&" ; <# œ
, 0 a#b œ $È&" Þ
"!
"!
%
%
The eigenvalues are both negative. Hence the critical point is a stable node, which is
asymptotically stable.
a.ß /b.
a0 b. Except for initial conditions lying on the coordinate axes, almost all trajectories
________________________________________________________________________
page 548
—————————————————————————— CHAPTER 9. ——
converge to the stable node at a!Þ) ß "Þ%b.
2a+b.
a,bÞ The critical points are the solution set of the system of equations
Ba"Þ& B !Þ& C b œ !
Ca# !Þ& C "Þ& Bb œ ! Þ
The four critical points are a! ß !b, a! ß %b, a"Þ& ß !b and a" ß "b.
a- b. The Jacobian matrix of the vector field is
JœŒ
$Î# #B CÎ#
$CÎ#
BÎ#
.
# $BÎ# C
At the origin, the coefficient matrix of the linearized system is
Ja! ß !b œ Œ
$Î#
!
!
.
#
The eigenvalues and eigenvectors are
"
!
<" œ $Î# , 0 a"b œ Œ ; <# œ # , 0 a#b œ Œ Þ
!
"
The eigenvalues are positive, hence the origin is an unstable node.
At the critical point a! ß %b, the coefficient matrix of the linearized system is
Ja! ß %b œ Œ
"Î#
'
!
.
#
The eigenvalues and eigenvectors are
________________________________________________________________________
page 549
—————————————————————————— CHAPTER 9. ——
<" œ "Î# , 0 a"b œ Œ
"
!
a#b
; <# œ # , 0 œ Œ Þ
%
"
The eigenvalues are both negative, hence the critical point a! ß %b is a stable node, which
is asymptotically stable.
At the critical point a$Î# ß !b, the coefficient matrix of the linearized system is
Ja$Î# ß !b œ Œ
$Î#
!
$Î%
.
"Î%
The eigenvalues and eigenvectors are
"
$
<" œ $Î# , 0 a"b œ Œ ; <# œ "Î% , 0 a#b œ Œ
Þ
!
&
The eigenvalues are both negative, hence the critical point is a stable node, which is
asymptotically stable.
At the critical point a "ß "b, the coefficient matrix of the linearized system is
Ja" ß "b œ Œ
"
$Î#
"Î#
.
"Î#
The eigenvalues and eigenvectors are
<" œ
$ È"$
%
, 0 a"b œ
"
"È"$
#
; <# œ
$ È"$
%
!
, 0 a#b œ "È"$ Þ
#
The eigenvalues are of opposite sign, hence a "ß "b is a saddle, which is unstable.
a.ß /b.
a0 b. Trajectories approaching the critical point a" ß "b form a separatrix. Solutions on
either side of the separatrix approach either a! ß %b or a"Þ& ß !b.
________________________________________________________________________
page 550
—————————————————————————— CHAPTER 9. ——
4a+b.
a,bÞ The critical points are solutions of the system of equations
Ba"Þ& !Þ& B C b œ !
Ca!Þ(& C !Þ"#& Bb œ ! Þ
The four critical points are a! ß !b, a! ß $Î%b, a$ ß !b and a# ß "Î#b.
a- b. The Jacobian matrix of the vector field is
JœŒ
$Î# B C
CÎ)
B
.
$Î% BÎ) #C
At the origin, the coefficient matrix of the linearized system is
Ja! ß !b œ Œ
$Î#
!
!
.
$Î%
The eigenvalues and eigenvectors are
"
!
<" œ $Î# , 0 a"b œ Œ ; <# œ $Î% , 0 a#b œ Œ Þ
!
"
The eigenvalues are positive, hence the origin is an unstable node.
At the critical point a! ß $Î%b, the coefficient matrix of the linearized system is
Ja! ß $Î%b œ Œ
$Î%
$Î$#
!
.
$Î%
The eigenvalues and eigenvectors are
<" œ $Î% , 0 a"b œ Œ
"'
!
a#b
; <# œ $Î% , 0 œ Œ Þ
"
"
The eigenvalues are of opposite sign, hence the critical point a! ß $Î%b is a saddle, which
is unstable.
________________________________________________________________________
page 551
—————————————————————————— CHAPTER 9. ——
At the critical point a$ ß !b, the coefficient matrix of the linearized system is
Ja$ ß !b œ Œ
$Î#
!
$
.
$Î)
The eigenvalues and eigenvectors are
"
)
<" œ $Î# , 0 a"b œ Œ ; <# œ $Î) , 0 a#b œ Œ
Þ
!
&
The eigenvalues are of opposite sign, hence the critical point a! ß $Î%b is a saddle, which
is unstable.
At the critical point a# ß "Î#b, the coefficient matrix of the linearized system is
Ja# ß "Î#b œ Œ
"
"Î"'
#
.
"Î#
The eigenvalues and eigenvectors are
<" œ
$ È$
%
, 0 a"b œ
"
"È$
)
; <# œ
$ È$
%
!
, 0 a#b œ "È$ Þ
)
The eigenvalues are negative, hence the critical point a# ß "Î#b is a stable node, which
is asymptotically stable.
a.ß /bÞ
a0 b. Except for initial conditions along the coordinate axes, almost all solutions
converge
to the stable node a# ß "Î#b.
________________________________________________________________________
page 552
—————————————————————————— CHAPTER 9. ——
7. It follows immediately that
a5" \ 5# ] b# %5" 5# \] œ 5"# \ # #5" 5# \] 5## ] # %5" 5# \]
œ a5" \ 5# ] b# Þ
Since all parameters and variables are positive, it follows that
a5" \ 5# ] b# %a5" 5# !" !# b\]
Hence the radicand in Eq.a$*b is nonnegative.
!Þ
10a+b. The critical points consist of the solution set of the equations
Ba%" 5" B !" C b œ !
Ca%# 5# C !# Bb œ ! Þ
If B œ ! , then either C œ ! or C œ %# Î5# . If %" 5" B !" C œ ! , then solving for B
results in B œ a%" !" C bÎ5" . Substitution into the second equation yields
a5" 5# !" !# bC# a5" % # % " !# bC œ ! Þ
Based on the hypothesis, it follows that a5" % # % " !# bC œ ! Þ So if 5" % # % " !# Á ! ,
then C œ ! , and the critical points are located at a! ß !b, a! ß %# Î5# b and a%" Î5" ß !bÞ
For the case 5" % # % " !# œ !, C can be arbitrary. From the relation B œ a%" !" C bÎ5" ,
we conclude that all points on the line 5" B !" C œ %" are critical points, in addition to
the point a! ß !b.
a,b. The Jacobian matrix of the vector field is
JœŒ
%" #5" B !" C
!# C
!" B
.
%# #5# C !# B
At the origin, the coefficient matrix of the linearized system is
Ja! ß !b œ Œ
%"
!
!
,
%#
with eigenvalues <" œ %" and <# œ %# . Since both eigenvalues are positive, the origin
is an unstable node.
At the point a! ß %# Î5# b, the coefficient matrix of the linearized system is
Ja! ß %# Î5# b œ Œ
a%" !# 5" %# bÎ!#
%# !# Î5#
!
,
%#
with eigenvalues <" œ a%" !# 5" %# bÎ!# and <# œ %# . If 5" %# %" !# ! , then
both eigenvalues are negative. Hence the point a! ß %# Î5# b is a stable node, which is
asymptotically stable. If 5" %# %" !# ! , then the eigenvalues are of opposite sign.
Hence the point a! ß %# Î5# b is a saddle, which is unstable.
________________________________________________________________________
page 553
—————————————————————————— CHAPTER 9. ——
At the point a%" Î5" ß !b, the coefficient matrix of the linearized system is
Ja%" Î5" ß !b œ Œ
%" !" Î5"
,
a5" %# %" !# bÎ5"
%"
!
with eigenvalues <" œ a5" %# %" !# bÎ5" and <# œ %" . If 5" %# %" !# ! , then
the eigenvalues are of opposite sign. Hence the point a%" Î5" ß !b is a saddle, which is
unstable. If 5" %# %" !# ! , then both eigenvalues are negative. In that case the point
a%" Î5" ß !b is a stable node, which is asymptotically stable.
a- b. As shown in Part a+b, when 5" %# %" !# œ ! , the set of critical points consists of
a! ß !b and all of the points on the straight line 5" B !" C œ %" . Based on Part a,b, the
origin is still an unstable node. Setting C œ a%" 5" BbÎ!" , the Jacobian matrix of the
vector field, along the given straight line, is
JœŒ
5" B
!# a%" 5" BbÎ!"
!" B
.
!# B %" !# Î5"
The characteristic equation of the matrix is
<# ”
%" !# !# 5" B 5"# B
•< œ ! .
5"
Using the given hypothesis, a%" !# !# 5" B 5"# BbÎ5" œ % # !# B 5" B. Hence the
characteristic equation can be written as
<# c% # !# B 5" Bd< œ ! .
First note that ! Ÿ B Ÿ %" Î5" . Since the coefficient in the quadratic equation is linear,
and
% # !# B 5" B œ œ
% # at B œ !
% " at B œ %" Î5" ,
it follows that the coefficient is positive for ! Ÿ B Ÿ %" Î5" . Therefore, along the straight
line 5" B !" C œ %" , one eigenvalue is zero and the other one is negative. Hence the
continuum of critical points consists of stable nodes, which are asymptotically stable.
11a+b. The critical points are solutions of the system of equations
Ba" B Cb $ + œ !
Ca!Þ(& C !Þ& Bb $ , œ ! Þ
Assume solutions of the form
B œ B! B" $ B# $ # â
C œ C! C " $ C # $ # â .
Substitution of the series expansions results in
________________________________________________________________________
page 554
—————————————————————————— CHAPTER 9. ——
B! a" B! C! b aB" #B" B! B! C" B" C! +b$ ⠜ !
C! a!Þ(& C! !Þ& B! b a!Þ(& C" #C! C" B" C! Î# B! C" Î# , b$ â œ ! Þ
a,b. Taking a limit as $ p ! , the equations reduce to the original system of equations.
It follows that B! œ C! œ !Þ& .
a- b. Setting the coefficients of the linear terms equal to zero, we find that
C" Î# B" Î# + œ !
B" Î% C" Î# , œ ! ,
with solution B" œ %+ %, and C" œ #+ %, .
a. b. Consider the +, - parameter space . The collection of points for which , +
represents an increase in the level of species ". At points where , + , B" $ ! Þ
Likewise, the collection of points for which #, + represents an increase in the level
of species # . At points where #, + , C" $ ! Þ
It follows that if , + #, , the level of both species will increase. This condition is
represented by the wedge-shaped region on the graph. Otherwise, the level of one
species
will increase, whereas the level of the other species will simultaneously decrease. Only
for + œ , œ ! will both populations remain the same.
13a+b. The critical points consist of the solution set of the equations
C œ!
# C BaB !Þ"&baB #b œ ! Þ
Setting C œ ! , the second equation becomes BaB !Þ"&baB #b œ ! , with roots B œ ! ,
!Þ"& and # . Hence the critical points are located at a! ß !b, a!Þ"& ß !b and a# ß !b. The
Jacobian matrix of the vector field is
________________________________________________________________________
page 555
—————————————————————————— CHAPTER 9. ——
JœŒ
!
$B %Þ$ B !Þ$
#
"
.
#
At the origin, the coefficient matrix of the linearized system is
Ja! ß !b œ Œ
with eigenvalues
<"ß# œ
!
!Þ$
"
,
#
#
"
„ È#& # # $! .
# "!
Regardless of the value of # , the eigenvalues are real and of opposite sign. Hence a! ß !b
is a saddle, which is unstable.
At the critical point a!Þ"& ß !b, the coefficient matrix of the linearized system is
Ja!Þ"& ß !b œ Œ
with eigenvalues
<"ß# œ
!
!Þ#((&
"
,
#
#
"
„ È"!! # # """ .
# #!
If "!! # # """ ! , then the eigenvalues are real. Furthermore, since <" <# œ !Þ#((& ,
both eigenvalues will have the same sign. Therefore the critical point is a node, with its
stability dependent on the sign of # . If "!! # # """ ! , the eigenvalues are complex
conjugates. In that case the critical point a!Þ"& ß !b is a spiral, with its stability dependent
on the sign of # .
At the critical point a# ß !b, the coefficient matrix of the linearized system is
Ja# ß !b œ Œ
with eigenvalues
<"ß# œ
!
$Þ(
"
,
#
#
"
„ È#& # # $(! .
# "!
Regardless of the value of # , the eigenvalues are real and of opposite sign. Hence a# ß !b
is a saddle, which is unstable.
________________________________________________________________________
page 556
—————————————————————————— CHAPTER 9. ——
a, b .
It is evident that for # œ !Þ) , the critical point a!Þ"& ß !b is a stable spiral.
Closer examination shows that for # œ "Þ& , the critical point a!Þ"& ß !b is a stable node.
a- b. Based on the phase portraits in Part a,b, it is apparent that the required value of #
satisfies !Þ) # "Þ& . Using the initial condition Ba!b œ # and C a!b œ !Þ!" , it is
possible to solve the initial value problem for various values of # . A reasonable first
guess is # œ È"Þ"" . This value marks the change in qualitative behavior of the critical
________________________________________________________________________
page 557
—————————————————————————— CHAPTER 9. ——
point a!Þ"& ß !b. Numerical experiments show that the solution remains positive for
# ¸ "Þ#! .
________________________________________________________________________
page 558
—————————————————————————— CHAPTER 9. ——
Section 9.5
1a+b.
a,bÞ The critical points are solutions of the system of equations
Ba"Þ& !Þ& C b œ !
Ca !Þ& Bb œ ! Þ
The two critical points are a! ß !b and a!Þ& ß $b.
a- b. The Jacobian matrix of the vector field is
JœŒ
BÎ#
.
"Î# B
$Î# CÎ#
C
At the critical point a! ß !b, the coefficient matrix of the linearized system is
Ja! ß !b œ Œ
$Î#
!
!
.
"Î#
The eigenvalues and eigenvectors are
"
!
<" œ $Î# , 0 a"b œ Œ ; <# œ "Î# , 0 a#b œ Œ Þ
!
"
The eigenvalues are of opposite sign, hence the origin is a saddle, which is unstable.
At the critical point a!Þ& ß $b, the coefficient matrix of the linearized system is
Ja!Þ& ß $b œ Œ
!
$
"Î%
.
!
The eigenvalues and eigenvectors are
<" œ 3
È$
#
, 0 a"b œ Œ
È$
"
"
a#b
<
œ
3
œ
;
,
0
#
Œ
Þ
# 3È $
#
# 3È $
________________________________________________________________________
page 559
—————————————————————————— CHAPTER 9. ——
The eigenvalues are purely imaginary. Hence the critical point is a center, which is
stable.
a.ß /b.
a0 b. Except for solutions along the coordinate axes, almost all trajectories are closed
curves about the critical point a!Þ& ß $b.
2a+b.
a,b. The critical points are the solution set of the system of equations
Ba" !Þ& C b œ !
Ca !Þ#& !Þ& Bb œ ! Þ
The two critical points are a! ß !b and a!Þ& ß #b.
a- b. The Jacobian matrix of the vector field is
JœŒ
" CÎ#
CÎ#
BÎ#
.
"Î% BÎ#
At the critical point a! ß !b, the coefficient matrix of the linearized system is
________________________________________________________________________
page 560
—————————————————————————— CHAPTER 9. ——
Ja! ß !b œ Œ
"
!
!
.
"Î%
The eigenvalues and eigenvectors are
"
!
<" œ " , 0 a"b œ Œ ; <# œ "Î% , 0 a#b œ Œ Þ
!
"
The eigenvalues are of opposite sign, hence the origin is a saddle, which is unstable.
At the critical point a!Þ& ß #b, the coefficient matrix of the linearized system is
Ja!Þ& ß #b œ Œ
!
"
"Î%
.
!
The eigenvalues and eigenvectors are
<" œ 3Î# , 0 a"b œ Œ
"
"
a#b
; <# œ 3Î# , 0 œ Œ Þ
#3
#3
The eigenvalues are purely imaginary. Hence the critical point is a center, which is
stable.
a.ß /b.
a0 b. Except for solutions along the coordinate axes, almost all trajectories are closed
curves about the critical point a!Þ& ß #b.
________________________________________________________________________
page 561
—————————————————————————— CHAPTER 9. ——
4a+b.
a,b. The critical points are the solution set of the system of equations
Ba*Î) B CÎ#b œ !
Ca " Bb œ ! Þ
The three critical points are a! ß !b, a*Î) ß !b and a" ß "Î%b.
a- b. The Jacobian matrix of the vector field is
JœŒ
*Î) #B CÎ#
C
BÎ#
.
" B
At the critical point a! ß !b, the coefficient matrix of the linearized system is
Ja! ß !b œ Œ
*Î)
!
!
.
"
The eigenvalues and eigenvectors are
"
!
<" œ *Î) , 0 a"b œ Œ ; <# œ " , 0 a#b œ Œ Þ
!
"
The eigenvalues are of opposite sign, hence the origin is a saddle, which is unstable.
At the critical point a*Î) ß !b, the coefficient matrix of the linearized system is
Ja*Î) ß !b œ Œ
*Î)
!
*Î"'
.
"Î)
The eigenvalues and eigenvectors are
<" œ
*
"
"
*
, 0 a"b œ Œ ; <# œ , 0 a#b œ Œ
Þ
)
!
)
#!
________________________________________________________________________
page 562
—————————————————————————— CHAPTER 9. ——
The eigenvalues are of opposite sign, hence the critical point a*Î) ß !b is a saddle, which
is unstable.
At the critical point a" ß "Î%b, the coefficient matrix of the linearized system is
Ja" ß "Î%b œ Œ
"
"Î%
"Î#
.
!
The eigenvalues and eigenvectors are
<" œ
# È#
%
, 0 a"b œ
# È#
# È#
# È#
a#b
;
<
œ
,
0
œ
#
Þ
%
"
"
The eigenvalues are both negative. Hence the critical point is a stable node, which is
asymptotically stable.
a.ß /b.
a0 b. Except for solutions along the coordinate axes, all solutions converge to the critical
point a" ß "Î%b.
5a+b.
________________________________________________________________________
page 563
—————————————————————————— CHAPTER 9. ——
a,b. The critical points are solutions of the system of equations
Bˆ " #Þ& B !Þ$ C B# ‰ œ !
Ca "Þ& Bb œ ! Þ
The four critical points are a! ß !b, a"Î# ß !b, a# ß !b and a$Î# ß &Î$b.
a- b. The Jacobian matrix of the vector field is
JœŒ
" &B $B# $CÎ"!
C
$BÎ"!
.
$Î# B
At the critical point a! ß !b, the coefficient matrix of the linearized system is
Ja! ß !b œ Œ
"
!
!
.
$Î#
The eigenvalues and eigenvectors are
"
!
<" œ " , 0 a"b œ Œ ; <# œ $Î# , 0 a#b œ Œ Þ
!
"
The eigenvalues are both negative, hence the critical point a! ß !b is a stable node, which
is asymptotically stable.
At the critical point a"Î# ß !b, the coefficient matrix of the linearized system is
Ja"Î# ß !b œ Œ
$Î%
!
$Î#!
.
"
The eigenvalues and eigenvectors are
<" œ
$
"
$
, 0 a"b œ Œ ; <# œ " , 0 a#b œ Œ Þ
%
!
$&
The eigenvalues are of opposite sign, hence the critical point a"Î# ß !b is a saddle, which
is unstable.
At the critical point a# ß !b, the coefficient matrix of the linearized system is
Ja# ß !b œ Œ
$
!
$Î&
.
"Î#
The eigenvalues and eigenvectors are
"
'
<" œ $ , 0 a"b œ Œ ; <# œ "Î# , 0 a#b œ Œ
Þ
!
$&
The eigenvalues are of opposite sign, hence the critical point a# ß !b is a saddle, which
is unstable.
________________________________________________________________________
page 564
—————————————————————————— CHAPTER 9. ——
At the critical point a$Î# ß &Î$b, the coefficient matrix of the linearized system is
Ja$Î# ß &Î$b œ Œ
$Î%
&Î$
*Î#!
.
!
The eigenvalues and eigenvectors are
<" œ
$ 3È$*
)
, 0
a"b
œ
Î *3 $È$* Ñ
Ï
%!
"
Ò
; <# œ
$ 3È$*
)
, 0
a#b
œ
Î *3 $È$* Ñ
Ï
%!
"
Ò
Þ
The eigenvalues are complex conjugates. Hence the critical point a$Î# ß &Î$b is a stable
spiral, which is asymptotically stable.
a.ß /b.
a0 b. The single solution curve that converges to the node at a"Î# ß !b is a separatrix.
Except for initial conditions on the coordinate axes, trajectories on either side of the
separatrix converge to the node at a! ß !b or the stable spiral at a$Î# ß &Î$b.
6. Given that > is measured from the time that B is a maximum, we have
-O
-9=ˆÈ+- >‰
#
#
+
+ C œ O Ê =38ˆÈ+- >‰ Þ
!
! !
Bœ
The period of oscillation is evidently X œ #1ÎÈ+- Þ Both populations oscillate about
a mean value. The following is based on the properties of the -9= and =38 functions
The prey population aBb is maximum at > œ ! and > œ X . It is a minimum at > œ X Î# Þ
Its rate of increase is greatest at > œ $ X Î%. The rate of decrease of the prey population
is greatest at > œ X Î% .
The predator population aCb is maximum at > œ X Î% . It is a minimum at > œ $ X Î% Þ
________________________________________________________________________
page 565
—————————————————————————— CHAPTER 9. ——
The rate of increase of the predator population is greatest at > œ ! and > œ X . The rate
of decrease of the predator population is greatest at > œ X Î# .
In the following example, the system in Problem # is solved numerically with the initial
conditions Ba!b œ !Þ( and Ca!b œ # . The critical point of interest is at a!Þ& ß #b.
Since + œ " and - œ "Î% , it follows that the period of oscillation is X œ % 1 Þ
8a+b. The period of oscillation for the linear system is X œ #1ÎÈ+- Þ In system a#b,
+ œ " and - œ !Þ(& . Hence the period is estimated as X œ #1ÎÈ!Þ(& ¸ (Þ#&&# .
a,b. The estimated period appears to agree with the graphic in Figure *Þ&Þ$ .
a- b. The critical point of interest is at a$ ß #bÞ The system is solved numerically, with
Ca!b œ # and Ba!b œ $Þ& ß %Þ! ß %Þ& ß &Þ! Þ The resulting periods are shown in the table:
X
Ba!b œ $Þ&
(Þ#'
Ba!b œ %Þ!
(Þ#*
Ba!b œ %Þ&
(Þ$%
Ba!b œ &Þ!
(Þ%#
The actual amplitude steadily increases as the amplitude increases.
________________________________________________________________________
page 566
—————————————————————————— CHAPTER 9. ——
9. The system
.B
C
œ + BŠ" ‹
.>
#
.C
B
œ , CŠ " ‹
.>
$
is solved numerically for various values of the parameters. The initial conditions are
Ba!b œ & , Ca!b œ # .
a+b. + œ " and , œ " À
The period is estimated by observing when the trajectory becomes a closed curve. In this
case, X ¸ 'Þ%& .
a,b. + œ $ and + œ "Î$ , with , œ " À
For + œ $ , X ¸ $Þ'* . For + œ "Î$ , X ¸ ""Þ%% .
________________________________________________________________________
page 567
—————————————————————————— CHAPTER 9. ——
a- bÞ , œ $ and , œ "Î$ , with + œ " À
For , œ $ , X ¸ $Þ)# . For , œ "Î$ , X ¸ ""Þ!' .
a. b. It appears that if one of the parameters is fixed, the period varies inversely with
the other parameter. Hence one might postulate the relation
X œ
5
Þ
0 a+ ß , b
10a+b. Since X œ #1ÎÈ+- , we first note that
(
EX
E
-9=ˆÈ+- > 9‰.> œ (
EX
E
=38ˆÈ+- > 9‰.> œ ! .
Hence
" EX " EX +
+
Bœ (
.> œ and C œ (
.> œ .
X E
#
#
X E
!
!
a,b. One way to estimate the mean values is to find a horizontal line such that the area
above the line is approximately equal to the area under the line. From Figure *Þ&Þ$ , it
appears that B ¸ $Þ#& and C ¸ #Þ! . In Example " , + œ " , - œ !Þ(& , ! œ !Þ& and
# œ !Þ#& . Using the result in Part a+b, B œ $ and C œ # Þ
a- b. The system
.B
C
œ BŠ" ‹
.>
#
.C
$ B
œ CŒ
.>
% %
is solved numerically for various initial conditionsÞ
Ba!b œ $ and Ca!b œ #Þ& À
________________________________________________________________________
page 568
—————————————————————————— CHAPTER 9. ——
Ba!b œ $ and Ca!b œ $Þ! À
Ba!b œ $ and Ca!b œ $Þ& À
________________________________________________________________________
page 569
—————————————————————————— CHAPTER 9. ——
Ba!b œ $ and Ca!b œ %Þ! À
It is evident that the mean values increase as the amplitude increases. That is, the mean
values increase as the initial conditions move farther from the critical point.
12. The system of equations in model a"b is given by
.B
œ Ba+ ! C b
.>
.C
œ C a - # Bb .
.>
Based on the hypothesis, let the death rate of the insect population and the predators be
: B and ; C , respectively. The modified system of equations becomes
.B
œ Ba+ ! C b : B
.>
.C
œ C a - # Bb ; C ,
.>
in which : ! , ; ! . The critical points are solutions of the system of equations
Ba+ : ! C b œ !
C a - ; # Bb œ ! Þ
+:
It is easy to see that the critical points are now at a! ß !b and Š -;
# ß ! ‹Þ Furthermore,
since a- ; bÎ# -Î# , the equilibrium level of the insect population has increased.
On the other hand, since a+ :bÎ! +Î! , equilibrium level of the predators has
decreased. Indeed, the introduction of insecticide creates a potential to significantly
affect the predator population a+ ¸ :b.
________________________________________________________________________
page 570
—————————————————————————— CHAPTER 9. ——
Section 9.6
2. We consider the function Z aBß Cb œ + B# - C # . The rate of change of Z along any
trajectory is
Þ
.B
.C
Z œ ZB
ZC
.>
.>
" $
œ #+BŒ B #BC # #-C ˆ C $ ‰
#
œ +B% %+B# C # #-C % Þ
Let ? œ B# , @ œ C # , ! œ + , " œ %+ , and # œ #- . We then have
+B% %+B# C # #-C % œ ! ?# " ?@ # @# .
If + ! and - ! , then Z aBß Cb is positive definite. Furthermore, ! ! . Recall that
Theorem *Þ'Þ% asserts that if %!# " # œ )+- "' +# ! , then the function
! ?# " ?@ # @#
Þ
is negative definite. Hence if - #+ , then Z aBß C b is negative definite. One such
example is Z aBß Cb œ B# $ C # Þ It follows from Theorem *Þ'Þ" that the origin is an
asymptotically stable critical point.
4. Given Z aBß Cb œ + B# - C # , the rate of change of Z along any trajectory is
Þ
.B
.C
Z œ ZB
ZC
.>
.>
$
$‰
ˆ
œ #+B B C #-C ˆ#BC # %B# C #C $ ‰
œ #+ B% a%- #+bBC $ )- B# C # %- C % Þ
Setting + œ #- ,
Þ
Z œ %- B% )- B# C # %- C %
%- B% %- C % Þ
Þ
As long as + œ #- ! , the function Z aBß C b is positive definite and Z aBß C b is also
positive definite. It follows from Theorem *Þ'Þ# that a! ß !b is an unstable critical point.
5. Given Z aBß Cb œ - aB# C # b , the rate of change of Z along any trajectory is
Þ
.B
.C
Z œ ZB
ZC
.>
.>
œ #- BcC B0 aBß C bd #-C c B C0 aBß C bd
œ #- ˆB# C # ‰0 aBß C b Þ
If - ! , then Z aBß Cb is positive definite
. Furthermore, if 0 aBß C b is positive in some
Þ
neighborhood of the origin, then Z aBß Cb is negative definite. Theorem *Þ'Þ" asserts that
________________________________________________________________________
page 571
—————————————————————————— CHAPTER 9. ——
the origin is an asymptotically stable critical point.
On the other hand, if 0 aBß C b is negative in some neighborhood of the origin, then
Z aBßÞCb
and Z aBß Cb are both positive definite. It follows from Theorem *Þ'Þ# that the origin is an
unstable critical point.
9a+b. Letting B œ ? and C œ ? w , we obtain the system of equations
.B
œC
.>
.C
œ 1aBb C Þ
.>
Since 1a!b œ ! , it is evident that a! ß !b is a critical point of the system. Consider the
function
B
" #
Z aBß Cb œ C ( 1a=b.= Þ
#
!
It is clear that Z a!ß !b œ ! . Since 1a?b is an odd function in a neighborhood of ? œ ! ,
( 1a=b.= ! for B ! ,
B
!
and
( 1a=b.= œ ( 1a=b.= ! for B ! .
B
!
!
Therefore Z aBß Cb is positive definite.
B
The rate of change of Z along any trajectory is
Þ
.B
.C
Z œ ZB
ZC
.>
.>
œ 1aBb † aCb C c 1aBb Cd
œ C# Þ
Þ
It follows that Z aBß Cb is only negative semidefinite Þ Hence the origin is a stable critical
point.
a,b. Given
Z aBß Cb œ
B
" # "
C C =38aBb ( =38a=b.= ,
#
#
!
It is easy to see that Z a! ß !b œ ! . The rate of change of Z along any trajectory is
________________________________________________________________________
page 572
—————————————————————————— CHAPTER 9. ——
Þ
.B
.C
Z œ ZB
ZC
.>
.>
C
"
œ ’=38 B -9= B“aC b ”C =38 B•c =38 B Cd
#
#
"
"
C
œ C# -9= B =38# B =38 B C # Þ
#
#
#
For 1Î# B 1Î# , we can write =38 B œ B ! B$ Î' and -9= B œ " " B# Î# ,
in which ! œ !aBb , " œ " aBb Þ Note that ! ! ß " " . Then
#
Þ
C#
" B#
"
! B$
C
! B$
#
Z aB ß C b œ
"
B
B
Œ
Œ
Œ
C Þ
#
#
#
'
#
'
Using polar coordinates,
Þ
<#
Z a< ß ) b œ c" =38 ) -9= ) 2 a< ß ) bd
#
<#
"
œ ”" =38 #) 2a< ß ) b• .
#
#
It is easy to show that
k 2 a< ß ) b k Ÿ
" #
"
< <% .
#
(#
So if < is sufficiently small, then k2a< ß ) bk "Î# and ¸ "# =38 #) 2a< ß )b¸ " Þ Hence
Þ
Z aB ß Cb is negative definite.
Now we show that Z aBß Cb is positive definite. Since 1a?b œ =38 ? ,
Z aBß Cb œ
" # "
C C =38aBb " -9= B .
#
#
This time we set
-9= B œ "
B#
B%
#
.
#
#%
Note that ! # " for 1Î# B 1Î# . Converting to polar coordinates,
Z a<ß ) b œ
<#
<#
<#
$
%
”" =38 ) -9= ) =38 ) -9= ) # -9= ) •
#
"#
#%
<#
"
<#
<#
œ ”" =38 #) =38 ) -9=$ ) # -9=% ) •Þ
#
#
"#
#%
________________________________________________________________________
page 573
—————————————————————————— CHAPTER 9. ——
Now
<#
<#
"
=38 ) -9=$ ) # -9=% ) for < " .
"#
#%
)
It follows that when < ! ,
Z a<ß ) b
<# ( "
” =38 #) •
# ) #
$ <#
!.
"'
Therefore Z aBß Cb is indeed positive definite, and by Theorem *Þ'Þ" , the origin is an
asymptotically stable critical point.
12a+b. We consider the linear system
B
+""
Œ œ Œ+
C
#"
w
Let Z aBß Cb œ EB# FBC GC # , in which
B
+"#
Œ
.
+##
C
#
#
+#"
+##
a+"" +## +"# +#" b
#?
+"# +## +"" +#"
Fœ
?
#
#
+""
+"#
a+"" +## +"# +#" b
Gœ
,
#?
Eœ
and ? œ a+"" +## ba+"" +## +"# +#" b. Based on the hypothesis, the coefficients E and
F are negative. Therefore, except for the origin, Z aBß C b is negative on each
of the coordinate axes. Along each trajectory,
Þ
Z œ a#EB FC ba+"" B +"# Cb a#GC FBba+#" B +## C b
œ B# C # Þ
Þ
Hence Z aBß Cb is negative definite. Theorem *Þ'Þ# asserts that the origin is an unstable
critical point.
a,b. We now consider the system
B w
+""
Œ œ Œ+
C
#"
B
J " aB ß C b
+"#
Œ
Œ
,
+##
C
K" aB ß C b
in which J" aB ß CbÎ< p ! and K" aB ß C bÎ< p ! as < p ! . Let
Z aBß Cb œ EB# FBC GC # ,
in which
________________________________________________________________________
page 574
—————————————————————————— CHAPTER 9. ——
#
#
+#"
+##
a+"" +## +"# +#" b
#?
+"# +## +"" +#"
Fœ
?
#
#
+ +"# a+"" +## +"# +#" b
G œ ""
,
#?
Eœ
and ? œ a+"" +## ba+"" +## +"# +#" b. Based on the hypothesis, E , F ! . Except
for the origin, Z aBß Cb is positive on each of the coordinate axes. Along each trajectory,
Þ
Z œ B# C # a#EB FC bJ" aB ß Cb a#GC FBbK" aB ß C b .
Converting to polar coordinates, for < Á ! ,
Þ
Z œ <# <a#E-9= ) F=38 ) b J" <a#G=38 ) F-9= )b K"
J"
K"
œ <# <# ”a#E-9= ) F=38 ) b
a#G=38 ) F-9= )b
•Þ
<
<
Since the system is almost linear, there is an V such that
ºa#E-9= ) F=38 ) b
J"
K"
"
a#G=38 ) F-9= )b
º ,
<
<
#
and hence
a#E-9= ) F=38 ) b
J"
K"
"
a#G=38 ) F-9= )b
<
<
#
for < V . It follows that
Þ
"
Z <#
#
Þ
as long as ! < V . Hence Z is positive definite on the domain
H œ ˜ aB ß C b l B # C # V # ™ .
By Theorem *Þ'Þ# , the origin is an unstable critical point.
________________________________________________________________________
page 575
—————————————————————————— CHAPTER 9. ——
Section 9.7
3. The equilibrium solutions of the ODE
.<
œ <a< "ba< $b
.>
are given by <" œ ! , <# œ " and <$ œ $ . Note that
.<
! for ! < " and < $ ;
.>
.<
! for " < $ .
.>
< œ ! corresponds to an unstable critical point. The equilibrium solution <# œ " is
asymptotically stable, whereas the equilibrium solution <$ œ $ is unstable. Since the
critical values are isolated, a limit cycle is given by
< œ " , ) œ > >!
which is asymptotically stable. Another periodic solution is found to be
< œ $ , ) œ > >!
which is unstable.
5. The equilibrium solutions of the ODE
.<
œ =38 1<
.>
are given by < œ 8 , 8 œ ! ß " ß # ß â . Based on the sign of < w in the neighborhood of
each critical value, the equilibrium solutions < œ #5 , 5 œ " ß # ß â correspond to
unstable periodic solutions, with ) œ > >! . The equilibrium solutions < œ #5 " ,
5 œ ! ß " ß # ß â correspond to stable limit cycles, with ) œ > >! . The solution < œ !
represents an unstable critical point.
10. Given J aB ß Cb œ +"" B +"# C and KaB ß C b œ +#" B +## C , it follows that
JB KC œ +"" +## .
Based on the hypothesis, JB KC is either positive or negative on the entire plane.
By Theorem *Þ(Þ# , the system cannot have a nontrivial periodic solution.
12. Given that J aB ß Cb œ #B $C BC # and KaB ß C b œ C B$ B# C ,
J B KC œ " B # C # .
Since JB KC ! on the entire plane, Theorem *Þ(Þ# asserts that the system cannot
have a nontrivial periodic solution.
________________________________________________________________________
page 576
—————————————————————————— CHAPTER 9. ——
14a+b. Based on the given graphs, the following table shows the estimated values:
. œ !Þ#
. œ "Þ!
. œ &Þ!
X ¸ 'Þ#*
X ¸ 'Þ''
X ¸ ""Þ'!
a,bÞ The initial conditions were chosen as Ba!b œ # , Ca!b œ ! Þ
X ¸ 'Þ$) .
X ¸ (Þ'& .
________________________________________________________________________
page 577
—————————————————————————— CHAPTER 9. ——
X ¸ )Þ)' .
X ¸ "!Þ#& .
a- b. The period, X , appears to be a quadratic function of . .
15a+b. Setting B œ ? and C œ ? w , we obtain the system of equations
.B
œC
.>
.C
"
œ B .Œ" C # C .
.>
$
a,b. Evidently, C œ ! . It follows that B œ ! . Hence the only critical point of the system
is at a! ß !b. The components of the vector field are infinitely differentiable everywhere.
Therefore the system is almost linear.
________________________________________________________________________
page 578
—————————————————————————— CHAPTER 9. ——
The Jacobian matrix of the vector field is
JœŒ
!
"
"
.
. .C #
At the critical point a! ß !b, the coefficient matrix of the linearized system is
Ja! ß !b œ Œ
with eigenvalues
<"ß# œ
!
"
"
,
.
. "È #
. % Þ
„
# #
If . œ ! , the equation reduces to the ODE for a simple harmonic oscillator. For the case
! . # , the eigenvalues are complex, and the critical point is an unstable spiral. For
. # , the eigenvalues are real, and the origin is an unstable node.
a- b. The initial conditions were chosen as Ba!b œ # , Ca!b œ ! Þ
E ¸ #Þ"' and X ¸ 'Þ'& .
a. bÞ
________________________________________________________________________
page 579
—————————————————————————— CHAPTER 9. ——
E ¸ #Þ!! and X ¸ 'Þ$! .
E ¸ #Þ!% and X ¸ 'Þ$) .
E ¸ #Þ' and X ¸ (Þ'# .
________________________________________________________________________
page 580
—————————————————————————— CHAPTER 9. ——
E ¸ %Þ$( and X ¸ ""Þ'" .
a/ b .
. œ !Þ#
. œ !Þ&
. œ "Þ!
. œ #Þ!
. œ &Þ!
E
#Þ!!
#Þ!%
#Þ"'
#Þ'
%Þ$(
X
'Þ$!
'Þ$)
'Þ'&
(Þ'#
""Þ'"
________________________________________________________________________
page 581
—————————————————————————— CHAPTER 9. ——
Section 9.8
6. < œ #) , with initial point a& ß & ß &b:
< œ #) , with initial point a&Þ!" ß & ß &b:
________________________________________________________________________
page 582
—————————————————————————— CHAPTER 9. ——
7. < œ #) À
________________________________________________________________________
page 583
—————————————————————————— CHAPTER 9. ——
9a+b. < œ "!! , initial point a & ß "$ ß && b À
The period appears to be X ¸ "Þ"# .
a,b. < œ **Þ*% , initial point a & ß "$ ß && b À
The periodic trajectory appears to have split into two strands, indicative of a perioddoubling. Closer examination reveals that the peak values of D a>b are slightly different:
________________________________________________________________________
page 584
—————————————————————————— CHAPTER 9. ——
< œ **Þ( , initial point a & ß "$ ß && b À
________________________________________________________________________
page 585
—————————————————————————— CHAPTER 9. ——
a- b. < œ **Þ' , initial point a & ß "$ ß && b À
The strands again appear to have split.
Closer examination reveals that the peak values of D a>b are different:
________________________________________________________________________
page 586
—————————————————————————— CHAPTER 9. ——
10a+b. < œ "!!Þ& , initial point a & ß "$ ß && b À
________________________________________________________________________
page 587
—————————————————————————— CHAPTER 9. ——
< œ "!!Þ( , initial point a & ß "$ ß && b À
________________________________________________________________________
page 588
—————————————————————————— CHAPTER 9. ——
a,b. < œ "!!Þ) , initial point a & ß "$ ß && b À
< œ "!!Þ)" , initial point a & ß "$ ß && b À
The strands of the periodic trajectory are beginning to split apart.
________________________________________________________________________
page 589
—————————————————————————— CHAPTER 9. ——
< œ "!!Þ)# , initial point a & ß "$ ß && b À
________________________________________________________________________
page 590
—————————————————————————— CHAPTER 9. ——
< œ "!!Þ)$ , initial point a & ß "$ ß && b À
________________________________________________________________________
page 591
—————————————————————————— CHAPTER 9. ——
< œ "!!Þ)% , initial point a & ß "$ ß && b À
________________________________________________________________________
page 592
—————————————————————————— CHAPTER 10. ——
Chapter Ten
Section 10.1
1. The general solution of the ODE is CaBb œ -" -9= B -# =38 B Þ Imposing the first
boundary condition, it is necessary that -" œ ! . Therefore CaBb œ -# =38 B . Taking its
derivative, C w aBb œ -# -9= B . Imposing the second boundary condition, we require that
-# -9= 1 œ " . The latter equation is satisfied only if -# œ " . Hence the solution of
the boundary value problem is CaBb œ =38 B .
4. The general solution of the differential equation is CaBb œ -" -9= B -# =38 B Þ It
follows that C w aBb œ -" =38 B -# -9= B Þ Imposing the first boundary condition, we
find that -# œ " . Therefore CaBb œ -" -9= B =38 B Þ Imposing the second boundary
condition, we require that -" -9= P =38 P œ ! . If -9= P Á ! , that is, as long as
P Á a#5 "b1Î# , with 5 an integer, then -" œ >+8P . The solution of the boundary
value problem is
CaBb œ >+8P -9= B =38 B Þ
If -9= P œ ! , the boundary condition results in =38 P œ ! . The latter two equations
are inconsistent, which implies that the BVP has no solution.
5. The general solution of the homogeneous differential equation is
CaBb œ -" -9= B -# =38 B Þ
Using any of a number of methods, including the method of undetermined coefficients, it
is easy to show that a particular solution is ] aBb œ B . Hence the general solution of
the given differential equation is CaBb œ -" -9= B -# =38 B B Þ The first boundary
condition requires that -" œ ! . Imposing the second boundary condition, it is necessary
that -# =38 1 1 œ ! Þ The resulting equation has no solution. We conclude that the
boundary value problem has no solution.
6. Using the method of undetermined coefficients, it is easy to show that the general
solution of the ODE is CaBb œ -" -9=È# B -# =38È# B BÎ# Þ Imposing the first
boundary condition, we find that -" œ ! . The second boundary condition requires that
-# =38È# 1 1Î# œ ! Þ It follows that -# œ 1Î#=38È# 1 Þ Hence the solution of
the boundary value problem is
1
B
CaBb œ
=38È# B Þ
#
#=38È# 1
8. The general solution of the homogeneous differential equation is
CaBb œ -" -9= #B -# =38 #B Þ
Using the method of undetermined coefficients, a particular solution is ] aBb œ =38 BÎ$ .
________________________________________________________________________
page 593
—————————————————————————— CHAPTER 10. ——
Hence the general solution of the given differential equation is
"
CaBb œ -" -9= #B -# =38 #B =38 B Þ
$
The first boundary condition requires that -" œ ! . The second boundary requires that
-# =38 #1 "$ =38 1 œ ! Þ The latter equation is valid for all values of -# . Therefore the
solution of the boundary value problem is
"
CaBb œ -# =38 #B =38 B Þ
$
9. Using the method of undetermined coefficients, it is easy to show that the general
solution of the ODE is CaBb œ -" -9= #B -# =38 #B -9= BÎ$ Þ It follows that
C w aBb œ #-" =38 #B #-# -9= #B =38 BÎ$ . Imposing the first boundary condition,
we find that -# œ ! . The second boundary condition requires that
"
#-" =38 #1 =38 1 œ ! Þ
$
The resulting equation is satisfied for all values of -" Þ Hence the solution is the family of
functions
"
CaBb œ -" -9= #B -9= B Þ
$
10. The general solution of the differential equation is
"
CaBb œ -" -9=È$ B -# =38È$ B -9= B Þ
#
Its derivative is C w aBb œ È$ -" =38È$ B È$ -# -9=È$ B =38 BÎ# . The first
boundary condition requires that -# œ ! . Imposing the second boundary condition, we
obtain È$ -" =38È$ 1 œ ! . It follows that -" œ ! . Hence the solution of the BVP
is CaBb œ -9= BÎ# .
12. Assuming that - ! , we can set - œ .# . The general solution of the differential
equation is
CaBb œ -" -9= .B -# =38 .B ,
so that C w aBb œ .-" =38 .B .-# -9= .B . Imposing the first boundary condition, it
follows that -# œ ! . Therefore CaBb œ -" -9= .B . The second boundary condition
requires that -" -9= .1 œ ! . For a nontrivial solution, it is necessary that -9= .1 œ ! ,
that is, .1 œ a#8 "b1Î# , with 8 œ "ß #ß â . Therefore the eigenvalues are
-8 œ
a#8 "b#
, 8 œ "ß #ß â Þ
%
________________________________________________________________________
page 594
—————————————————————————— CHAPTER 10. ——
The corresponding eigenfunctions are given by
C8 œ -9=
a#8 "bB
, 8 œ "ß #ß â Þ
#
Assuming that - ! , we can set - œ .# . The general solution of the differential
equation is
CaBb œ -" -9=2 .B -# =382 .B ,
so that C w aBb œ .-" =382 .B .-# -9=2 .B . Imposing the first boundary condition, it
follows that -# œ ! . Therefore CaBb œ -" -9=2 .B . The second boundary condition
requires that -" -9=2 .1 œ ! , which results in -" œ ! . Hence the only solution is the
trivial solution. Finally, with - œ ! , the general solution of the ODE is
CaBb œ -" B -# .
It is easy to show that the boundary conditions require that -" œ -# œ ! . Therefore all of
the eigenvalues are positive.
13. Assuming that - ! , we can set - œ .# . The general solution of the differential
equation is
CaBb œ -" -9= .B -# =38 .B ,
so that C w aBb œ .-" =38 .B .-# -9= .B . Imposing the first boundary condition, it
follows that -# œ ! . The second boundary condition requires that -" =38 .1 œ ! . For a
nontrivial solution, we must have .1 œ 81 , 8 œ "ß #ß â . It follows that the eigenvalues
are
-8 œ 8# , 8 œ "ß #ß â ,
and the corresponding eigenfunctions are
C8 œ -9= 8B , 8 œ "ß #ß â Þ
Assuming that - ! , we can set - œ .# . The general solution of the differential
equation is
CaBb œ -" -9=2 .B -# =382 .B ,
so that C w aBb œ .-" =382 .B .-# -9=2 .B . Imposing the first boundary condition, it
follows that -# œ ! . The second boundary condition requires that -" =382 .1 œ ! . The
latter equation is satisfied only for -" œ ! .
Finally, for - œ ! , the solution is CaBb œ -" B -# . Imposing the boundary conditions,
we find that CaBb œ -# . Therefore - œ ! is also an eigenvalue, with corresponding
eigenfunction C! aBb œ " Þ
________________________________________________________________________
page 595
—————————————————————————— CHAPTER 10. ——
14. It can be shown, as in Prob. "# , that - ! . Setting - œ .# , the general solution
of the resulting ODE is
CaBb œ -" -9= .B -# =38 .B ,
with C w aBb œ .-" =38 .B .-# -9= .B . Imposing the first boundary condition, we
find that -# œ ! . Therefore CaBb œ -" -9= .B . The second boundary condition requires
that -" -9= .P œ ! . For a nontrivial solution, it is necessary that -9= .P œ ! , that is,
. œ a#8 "b1Îa#Pb , with 8 œ "ß #ß â . Therefore the eigenvalues are
a#8 "b# 1#
-8 œ
, 8 œ "ß #ß â Þ
%P#
The corresponding eigenfunctions are given by
C8 œ -9=
a#8 "b1B
, 8 œ "ß #ß â Þ
#P
16. Assuming that - ! , we can set - œ .# . The general solution of the differential
equation is
CaBb œ -" -9=2 .B -# =382 .B .
The first boundary condition requires that -" œ ! . Therefore CaBb œ -# =382 .B and
C w aBb œ -# -9=2 .B Þ Imposing the second boundary condition, it is necessary that
-# -9=2 .P œ ! Þ The latter equation is valid only for -# œ ! . The only solution is the
trivial solution.
Assuming that - ! , we set - œ .# . The general solution of the resulting ODE is
CaBb œ -" -9= .B -# =38 .B .
Imposing the first boundary condition, we find that -" œ ! . Hence CaBb œ -# =38 .B and
C w aBb œ -# -9= .B . In order to satisfy the second boundary condition, it is necessary that
-# -9= .P œ ! . For a nontrivial solution, . œ a#8 "b1Îa#Pb , with 8 œ "ß #ß â .
Therefore the eigenvalues are
a#8 "b# 1#
-8 œ
, 8 œ "ß #ß â Þ
%P#
The corresponding eigenfunctions are given by
C8 œ =38
a#8 "b1B
, 8 œ "ß #ß â Þ
#P
Finally, for - œ ! , the general solution is linear. Based on the boundary conditions, it
follows that CaBb œ ! . Therefore all of the eigenvalues are negative.
________________________________________________________________________
page 596
—————————————————————————— CHAPTER 10. ——
17a+b. Setting - œ .# , write the general solution of the ODE C ww .# C œ ! as
CaBb œ 5" /3.B 5 # / 3.B Þ
Imposing the boundary conditions Ca!b œ Ca1b œ ! , we obtain the system of equations
5" 5 # œ !
5" /3.1 5 # / 3.1 œ ! Þ
The system has a nontrivial solution if and only if the coefficient matrix is singular. Set
the determinant equal to zero to obtain
/ 3.1 / 3.1 œ ! .
a,b. Let . œ / 35 . Then 3.1 œ 3/1 51 , and the previous equation can be written
as
/51 / 3/1 /51 / 3/1 œ ! .
Using Euler's relation, /3/1 œ -9= /1 3 =38 /1 , we obtain
/51 a-9= / 3 =38 / b /51 a-9= / 3 =38 / b œ ! .
Equating the real and imaginary parts of the equation,
a/51 /51 b-9= /1 œ !
a/51 /51 b=38 /1 œ ! Þ
a- b. Based on the second equation, / œ 8 , 8 − ˆ . Since -9= 81 Á ! , it follows that
/51 œ /51 , or /#51 œ " . Hence 5 œ ! , and . œ 8 , 8 − ˆ .
________________________________________________________________________
page 597
—————————————————————————— CHAPTER 10. ——
Section 10.2
1. The period of the function =38 !B is X œ #1Î! . Therefore the function =38 &B has
period X œ #1Î& .
2. The period of the function -9= !B is also X œ #1Î! . Therefore the function -9=
#1B
has period X œ #1Î#1 œ " .
4. Based on Prob. " , the period of the function =38 1BÎP is X œ #1Îa1ÎPb œ #P .
6. Let X ! and consider the equation aB X b# œ B# Þ It follows that #X B X # œ !
and #B X œ ! . Since the latter equation is not an identity, the function B# cannot be
periodic with finite period.
8. The function is defined on intervals of length a#8 "b a#8 "b œ # . On any two
consecutive intervals, 0 aBb is identically equal to " on one of the intervals and alternates
between " and " on the other. It follows that the period is X œ % .
9. On the interval P B #P , a simple shift to the right results in
0 aBb œ aB #Pb œ #P B .
On the interval $P B #P , a simple shift to the left results in
0 aBb œ aB #Pb œ #P B .
11. The next fundamental period to the left is on the interval #P B ! . Hence the
interval P B ! is the second half of a fundamental period. A simple shift to the
left results in
0 aBb œ P aB #Pb œ P B .
12. First note that
-9=
and
-9=
71B
81 B
"
a7 8 b 1 B
a7 8 b 1 B
-9=
œ ”-9=
-9=
•
P
P
#
P
P
71B
81 B
"
a8 7b1B
a7 8 b 1 B
=38
œ ”=38
=38
•Þ
P
P
#
P
P
________________________________________________________________________
page 598
—————————————————————————— CHAPTER 10. ——
It follows that
(
P
-9=
P
71B
81 B
" P
a7 8 b 1 B
a7 8 b 1 B
-9=
.B œ ( ”-9=
-9=
•.B
P
P
# P
P
P
" P =38ca7 8b1BÎPd =38ca7 8b1BÎPd P
œ
œ
º
#1
78
78
P
œ !,
as long as 7 8 and 7 8 are not zero. For the case 7 œ 8 ,
81B #
" P
#81B
‹ .B œ ( ”" -9=
( Š-9=
•.B
P
# P
P
P
P
"
=38a#81BÎPb P
B
œ
º
#
#81ÎP
P
œ P.
œ
Likewise,
(
P
P
-9=
a8 7b1B
a7 8 b 1 B
71B
81 B
" P
=38
.B œ ( ”=38
=38
•.B
P
P
# P
P
P
" P -9=ca8 7b1BÎPd -9=ca7 8b1BÎPd P
œ
œ
º
#1
78
78
P
œ !,
as long as 7 8 and 7 8 are not zero. For the case 7 œ 8 ,
71B
81B
" P
#81B
=38
.B œ ( =38
.B
( -9=
P
P
# P
P
P
" -9=a#81BÎPb P
œ œ
º
#
#81ÎP
P
œ !.
P
________________________________________________________________________
page 599
—————————————————————————— CHAPTER 10. ——
14a+b. For P œ " ,
a,b. The Fourier coefficients are calculated using the Euler-Fourier formulas:
" P
( 0 aBb.B
P P
" !
œ ( .B
P P
œ "Þ
+! œ
For 8 ! ,
" P
81 B
.B
( 0 aBb-9=
P P
P
" !
81 B
œ ( -9=
.B
P P
P
œ !Þ
+8 œ
Likewise,
" P
81B
.B
( 0 aBb=38
P P
P
" !
81B
œ ( =38
.B
P P
P
" a "b 8
œ
Þ
81
,8 œ
It follows that ,#5 œ ! and ,#5" œ #Îca#5 "b1d, 5 œ "ß #ß $ß â . Therefore the
Fourier series for the given function is
0 aBb œ
"
# _
"
a#5 "b1B
"
=38
Þ
# 1 5 œ " #5 "
P
________________________________________________________________________
page 600
—————————————————————————— CHAPTER 10. ——
16a+b.
a,b. The Fourier coefficients are calculated using the Euler-Fourier formulas:
+! œ
" P
( 0 aBb.B
P P
œ ( aB "b.B ( a" Bb.B
!
"
"
!
œ "Þ
For 8 ! ,
+8 œ
" P
81B
.B
( 0 aBb-9=
P P
P
œ ( aB "b-9= 81B .B ( a" Bb-9= 81B .B
!
" a "b 8
œ #
Þ
8# 1 #
"
"
!
It follows that + #5 œ ! and + #5" œ %Îa#5 "b# 1# ‘, 5 œ "ß #ß $ß â . Likewise,
" P
81B
,8 œ ( 0 aBb=38
.B
P P
P
œ ( aB "b=38 81B .B ( a" Bb=38 81B .B
!
"
"
!
œ !Þ
Therefore the Fourier series for the given function is
"
% _
"
0 aBb œ # "
-9=a#5 "b1B Þ
# 1 5 œ " a#5 "b#
________________________________________________________________________
page 601
—————————————————————————— CHAPTER 10. ——
17a+b. For P œ " ,
a,b. The Fourier coefficients are calculated using the Euler-Fourier formulas:
" P
( 0 aBb.B
P P
" !
" P
œ ( aB Pb.B ( P .B
P P
P !
œ $PÎ# Þ
+! œ
For 8 ! ,
" P
81 B
.B
( 0 aBb-9=
P P
P
" !
81B
" P
81 B
œ ( aB Pb-9=
.B ( P -9=
.B
P P
P
P !
P
Pa" -9= 81b
œ
Þ
8# 1 #
+8 œ
Likewise,
" P
81B
.B
( 0 aBb=38
P P
P
" !
81B
" P
81 B
œ ( aB Pb=38
.B ( P =38
.B
P P
P
P !
P
P -9= 81
œ
Þ
81
,8 œ
Note that -9= 81 œ a "b8 . It follows that the Fourier series for the given function is
________________________________________________________________________
page 602
—————————————————————————— CHAPTER 10. ——
0 aBb œ
$P
P _
#
a#8 "b1B a "b8 1
81 B
# "–
-9=
=38
Þ
#
%
1 8 œ " a#8 "b
P
8
P —
18a+b.
a,b. The Fourier coefficients are calculated using the Euler-Fourier formulas:
" P
+! œ ( 0 aBb.B
P P
" "
œ ( B .B
# "
œ !Þ
For 8 ! ,
" P
81 B
.B
( 0 aBb-9=
P P
P
" "
81 B
œ ( B -9=
.B
# "
P
œ !Þ
+8 œ
Likewise,
" P
81 B
.B
( 0 aBb=38
P P
P
" "
81B
œ ( B =38
.B
# "
P
#
81
81
œ # # Š# =38
81 -9= ‹Þ
8 1
#
#
,8 œ
________________________________________________________________________
page 603
—————————————————————————— CHAPTER 10. ——
Therefore the Fourier series for the given function is
0 aBb œ " ”
_
8œ"
%
8# 1 #
=38
81
#
81
81B
-9= •=38
Þ
#
81
#
#
19a+b.
a,b. The Fourier cosine coefficients are given by
" P
81 B
+8 œ ( 0 aBb-9=
.B
P P
P
" !
81B
" #
81 B
œ ( -9=
.B ( -9=
.B
# #
#
# !
#
œ !Þ
The Fourier sine coefficients are given by
" P
81B
.B
( 0 aBb=38
P P
P
" !
81 B
" #
81 B
œ ( =38
.B ( =38
.B
# #
#
# !
#
" -9= 81
œ#
Þ
81
,8 œ
Therefore the Fourier series for the given function is
0 aBb œ
% _
"
a#8 "b1B
"
=38
Þ
1 8 œ " #8 "
#
________________________________________________________________________
page 604
—————————————————————————— CHAPTER 10. ——
a- b .
20a+b.
a,b. The Fourier cosine coefficients are given by
" P
81 B
+8 œ ( 0 aBb-9=
.B
P P
P
œ ( B -9= 81B.B
"
"
œ !Þ
The Fourier sine coefficients are given by
________________________________________________________________________
page 605
—————————————————————————— CHAPTER 10. ——
" P
81B
,8 œ ( 0 aBb=38
.B
P P
P
œ ( B =38 81B.B
"
"
œ #
-9= 81
Þ
81
Therefore the Fourier series for the given function is
# _ a "b 8
0 aBb œ "
=38 81B Þ
1 8œ" 8
a- b .
________________________________________________________________________
page 606
—————————————————————————— CHAPTER 10. ——
22a+b.
a,b. The Fourier cosine coefficients are given by
" P
+! œ ( 0 aBb.B
P P
" !
" #
œ ( aB #b.B ( a# #Bb.B
# #
# !
œ ",
and for 8 ! ,
" P
81B
+8 œ ( 0 aBb-9=
.B
P P
P
" !
81 B
" #
81 B
œ ( aB #b-9=
.B ( a# #Bb-9=
.B
# #
#
# !
#
a" -9= 81b
œ'
Þ
8# 1 #
The Fourier sine coefficients are given by
" P
81 B
.B
( 0 aBb=38
P P
P
" !
81B
" #
81 B
œ ( aB #b=38
.B ( a# #Bb=38
.B
# #
#
# !
#
-9= 81
œ#
Þ
81
,8 œ
Therefore the Fourier series for the given function is
________________________________________________________________________
page 607
—————————————————————————— CHAPTER 10. ——
" "# _
"
a#8 "b1B
# _ a "b8
81 B
"
"
0 aBb œ #
-9=
=38
Þ
#
# 1 8 œ " a#8 "b
#
1 8œ" 8
#
a- b .
23a+b.
a,b. The Fourier cosine coefficients are given by
________________________________________________________________________
page 608
—————————————————————————— CHAPTER 10. ——
" P
+! œ ( 0 aBb.B
P P
" !
B
" #
"
œ ( Š ‹.B ( Œ#B B# .B
# #
#
# !
#
œ ""Î' ,
and for 8 ! ,
" P
81B
+8 œ ( 0 aBb-9=
.B
P P
P
" !
B
81 B
" #
"
81 B
œ ( Š ‹-9=
.B ( Œ#B B# -9=
.B
# #
#
#
# !
#
#
a& -9= 81b
œ
Þ
8# 1 #
The Fourier sine coefficients are given by
" P
81B
.B
( 0 aBb=38
P P
P
" !
B
81 B
" #
"
81 B
œ ( Š ‹=38
.B ( Œ#B B# =38
.B
# #
#
#
# !
#
#
% a% 8# 1# b-9= 81
œ
Þ
8$ 1 $
,8 œ
Therefore the Fourier series for the given function is
0 aBb œ
""
" _ c a "b 8 & d
81 B
# "
-9=
#
"# 1 8 œ "
8
#
" _ c% a% 8# 1# ba "b8 d
81 B
"
=38
Þ
$
$
1 8œ"
8
#
________________________________________________________________________
page 609
—————————————————————————— CHAPTER 10. ——
a- b .
24a+b.
a,b. The Fourier cosine coefficients are given by
" P
( 0 aBb.B
P P
" $
œ ( B# a$ Bb.B
$ !
œ *Î% ,
+! œ
and for 8 ! ,
________________________________________________________________________
page 610
—————————————————————————— CHAPTER 10. ——
" P
81B
+8 œ ( 0 aBb-9=
.B
P P
P
" $
81B
œ ( B# a$ Bb-9=
.B
$ !
$
a' ' -9= 81 8# 1# -9= 81b
œ #(
Þ
8% 1 %
The Fourier sine coefficients are given by
" P
81B
,8 œ ( 0 aBb=38
.B
P P
P
" $
81B
œ ( B# a$ Bb=38
.B
$ !
$
" # -9= 81
œ &%
Þ
8$ 1 $
Therefore the Fourier series for the given function is
_
*
'c" a "b8 d a "b8
81 B
"
0 aBb œ #(
-9=
”
•
)
8% 1 %
8# 1 #
$
8œ"
&% _ c" #a "b8 d
81 B
$ "
=38
Þ
$
1 8œ"
8
$
a- bÞ
________________________________________________________________________
page 611
—————————————————————————— CHAPTER 10. ——
26.
It is evident that k/7 aBbk is greatest at B œ „ $ . Increasing the number of terms in the
partials sums, we find that if 7 #( , then k/7 aBbk Ÿ !Þ" , for all B − c $ ß $d .
________________________________________________________________________
page 612
—————————————————————————— CHAPTER 10. ——
Graphing the partial sum =#( aBb, the convergence is as predicted:
28. Let B œ X + , for some + − c! ß X d . First note that for any value of 2 ,
0 aB 2b 0 aBb œ 0 aX + 2 b 0 aX +b
œ 0 a+ 2b 0 a+b Þ
Since 0 is differentiable,
0 aB 2b 0 aBb
2Ä!
2
0 a+ 2b 0 a+b
œ lim
2Ä!
2
w
œ 0 a+ b .
0 w aBb œ lim
That is, 0 w a+ X b œ 0 w a+b . By induction, it follows that 0 w a+ X b œ 0 w a+b for every
value of + .
On the other hand, if 0 aBb œ " -9= B , then the function
J aBb œ (
B
!
c" -9= >d.>
œ B =38 B
is not periodic, unless its definition is restricted to a specific interval.
29a+b. Based on the hypothesis, the vectors v" , v# and v$ are a basis for ‘$ . Given any
vector u − ‘$ , it can be expressed as a linear combination u œ +" v" +# v# +$ v$ .
Taking the inner product of both sides of this equation with v3 , we have
u † v 3 œ a + " v " + # v # + $ v $ b † v3
œ +3 v3 † v3 ,
________________________________________________________________________
page 613
—————————————————————————— CHAPTER 10. ——
since the basis vectors are mutually orthogonal. Hence
u † v3
+3 œ
, 3 œ "ß #ß $ .
v3 † v3
Recall that u † v3 œ ? @3 -9= ) , in which ) is the angle between u and v3 . Therefore
+3 œ
? -9= )
Þ
@3
Here ? -9= ) is interpreted as the magnitude of the projection of u in the direction of v3 .
a,b. Assuming that a Fourier series converges to a periodic function, 0 aBb ,
0 aBb œ
_
_
+!
9! aBb " +7 97 aBb " ,7 <7 aBb Þ
#
7œ"
7œ"
Taking the inner product, defined by
a? ß @ b œ (
P
P
?aBb@aBb.B ,
of both sides of the series expansion with the specified trigonometric functions, we have
_
_
+!
"
"
a0 ß 98 b œ a9! ß 98 b
+7 a97 ß 98 b
,7 a<7 ß 98 b
#
7œ"
7œ"
for 8 œ !ß "ß #ß â .
a- b. It also follows that
a0 ß < 8 b œ
_
_
+!
a9! ß <8 b " +7 a97 ß <8 b " ,7 a<7 ß <8 b
#
7œ"
7œ"
for 8 œ "ß #ß â . Based on the orthogonality conditions,
a97 ß 98 b œ P $78 , a<7 ß <8 b œ P $78 ,
and a<7 ß 98 b œ P $78 . Note that a9! ß 9! b œ #P . Therefore
+! œ
and
+8 œ
#a0 ß 9! b
" P
œ ( 0 aBb9! aBb.B
a9! ß 9! b
P P
a0 ß 98 b
" P
œ ( 0 aBb98 aBb.B , 8 œ "ß #ß â .
a98 ß 98 b
P P
________________________________________________________________________
page 614
—————————————————————————— CHAPTER 10. ——
Likewise,
,8 œ
a0 ß < 8 b
" P
œ ( 0 aBb<8 aBb.B , 8 œ "ß #ß â .
a< 8 ß < 8 b
P P
________________________________________________________________________
page 615
—————————————————————————— CHAPTER 10. ——
Section 10.3
1a+b. The given function is assumed to be periodic with #P œ # . The Fourier cosine
coefficients are given by
" P
+! œ ( 0 aBb.B
P P
œ ( a "b.B ( a"b.B
!
"
œ !,
"
!
and for 8 ! ,
" P
81 B
+8 œ ( 0 aBb-9=
.B
P P
P
œ ( -9= 81B .B ( -9= 81B .B
!
"
"
!
œ !Þ
The Fourier sine coefficients are given by
,8 œ
" P
81 B
.B
( 0 aBb=38
P P
P
œ ( =38 81B .B ( =38 81B .B
!
"
"
!
" -9= 81
œ#
Þ
81
Therefore the Fourier series for the specified function is
0 aBb œ
% _
"
"
=38 a#8 "b1B .
1 8 œ " #8 "
________________________________________________________________________
page 616
—————————————————————————— CHAPTER 10. ——
a, b .
The function is piecewise continuous on each finite interval. The points of discontinuity
are at integer values of B . At these points, the series converges to
k0 aB b 0 aB bk œ ! .
3a+b. The given function is assumed to be periodic with X œ #P . The Fourier cosine
coefficients are given by
" P
( 0 aBb.B
P P
" !
" P
œ ( aP Bb.B ( aP Bb.B
P P
P !
œ P,
+! œ
and for 8 ! ,
" P
81 B
.B
( 0 aBb-9=
P P
P
" !
81 B
" P
81 B
œ ( aP Bb-9=
.B ( aP Bb-9=
.B
P P
P
P !
P
" -9= 81
œ #P
Þ
8# 1 #
+8 œ
The Fourier sine coefficients are given by
" P
81B
.B
( 0 aBb=38
P P
P
" !
81 B
" P
81 B
œ ( aP Bb=38
.B ( aP Bb=38
.B
P P
P
P !
P
œ !Þ
,8 œ
________________________________________________________________________
page 617
—————————————————————————— CHAPTER 10. ——
Therefore the Fourier series of the specified function is
0 aBb œ
P %P _
"
a#8 "b1B
# "
-9=
.
#
#
1 8 œ " a#8 "b
P
a,b. For P œ " ,
Note that 0 aBb is continuous. Based on Theorem "!Þ$Þ" , the series converges to the
continuous function 0 aBb.
5a+b. The given function is assumed to be periodic with #P œ #1 . The Fourier cosine
coefficients are given by
" P
+! œ ( 0 aBb.B
P P
" 1Î#
œ (
a"b.B
1 1Î#
œ ",
and for 8 ! ,
" P
81 B
.B
( 0 aBb-9=
P P
P
" 1Î#
œ (
a"b-9= 8B .B
1 1Î#
#
81
œ
=38Š ‹Þ
81
#
+8 œ
The Fourier sine coefficients are given by
________________________________________________________________________
page 618
—————————————————————————— CHAPTER 10. ——
" P
81B
,8 œ ( 0 aBb=38
.B
P P
P
" 1Î#
a"b=38 8B .B
œ (
1 1Î#
œ !Þ
Observe that
=38Š
! , 8 œ #5
81
‹œœ
5+1
#
a "b , 8 œ #5 "
, 5 œ "ß #ß â .
Therefore the Fourier series of the specified function is
0 aBb œ
"
# _ a "b 8
"
-9= a#8 "bB .
# 1 8 œ " #8 "
a, b .
The given function is piecewise continuous, with discontinuities at odd multiples of 1Î# .
At B. œ a#5 "b1Î# , 5 œ !ß "ß #ß â , the series converges to
k0 aB. b 0 aB. bk œ "Î# .
6a+b. The given function is assumed to be periodic with #P œ # . The Fourier cosine
coefficients are given by
+! œ
" P
( 0 aBb.B
P P
œ ( B# .B
"
!
œ "Î$ ,
and for 8 ! ,
________________________________________________________________________
page 619
—————————————————————————— CHAPTER 10. ——
" P
81 B
+8 œ ( 0 aBb-9=
.B
P P
P
œ ( B# -9= 81B .B
"
!
# -9= 81
œ
Þ
8# 1 #
The Fourier sine coefficients are given by
" P
81B
,8 œ ( 0 aBb=38
.B
P P
P
œ ( B# =38 81B .B
"
!
œ
# # -9= 81 8# 1# -9= 81
Þ
8$ 1 $
Therefore the Fourier series for the specified function is
"
# _ a "b 8
0 aBb œ # "
-9= 81B
' 1 8 œ " 8#
"”
_
8œ"
#c" a "b8 d a "b8
•=38 81B .
8$ 1 $
81
a, b .
The given function is piecewise continuous, with discontinuities at the odd integers .
________________________________________________________________________
page 620
—————————————————————————— CHAPTER 10. ——
At B. œ #5 " , 5 œ !ß "ß #ß â , the series converges to
k0 aB. b 0 aB. bk œ "Î# .
8a+b. As shown in Problem "' of Section "!Þ# ,
a, b .
0 aBb œ
"
% _
"
# "
-9=a#8 "b1B Þ
# 1 8 œ " a#8 "b#
________________________________________________________________________
page 621
—————————————————————————— CHAPTER 10. ——
a- b .
9a+b. As shown in Problem #! of Section "!Þ# ,
# _ a "b 8
0 aBb œ "
=38 81B Þ
1 8œ" 8
a, b .
________________________________________________________________________
page 622
—————————————————————————— CHAPTER 10. ——
a- bÞ The given function is discontinuous at B œ „ " . At these points, the series will
converge to a value of zero. The error can never be made arbitrarily small.
10a+b. As shown in Problem ## of Section "!Þ# ,
0 aBb œ
" "# _
"
a#8 "b1B
# _ a "b8
81 B
"
# "
-9=
=38
Þ
#
# 1 8 œ " a#8 "b
#
1 8œ" 8
#
a, b .
________________________________________________________________________
page 623
—————————————————————————— CHAPTER 10. ——
a- b. The given function is discontinuous at B œ „ # . At these points, the series will
converge to a value of " . The error can never be made arbitrarily small.
11a+b. As shown in Problem ' , above ,
"
# _ a "b 8
0 aBb œ # "
-9= 81B
' 1 8 œ " 8#
"”
_
8œ"
#c" a "b8 d a "b8
•=38 81B .
8$ 1 $
81
a, bÞ
________________________________________________________________________
page 624
—————————————————————————— CHAPTER 10. ——
a- b. The given function is piecewise continuous, with discontinuities at the odd integers .
At B. œ #5 " , 5 œ !ß "ß #ß â , the series converges to
k0 aB. b 0 aB. bk œ "Î# .
At these points the error can never be made arbitrarily small.
13. The solution of the homogenous differential equation is
C- a>b œ -" -9= => -# =38 => Þ
Given that =# Á 8# , we can use the method of undetermined coefficients to find a
particular solution
] a>b œ
=#
"
=38 8> .
8#
Hence the general solution of the ODE is
Ca>b œ -" -9= => -# =38 =>
=#
"
=38 8> .
8#
Imposing the initial conditions, we obtain the equations
-" œ !
8
= -# #
œ !.
= 8#
It follows that -# œ 8Îc=a=# 8# bdÞ The solution of the IVP is
Ca>b œ
=#
"
8
=38 8>
=38 => .
#
#
8
=a= 8# b
If =# œ 8# , then the forcing function is also one of the fundamental solutions of the
ODE.
The method of undetermined coefficients may still be used, with a more elaborate trial
solution. Using the method of variation of parameters, we obtain
________________________________________________________________________
page 625
—————————————————————————— CHAPTER 10. ——
] a>b œ -9= 8>(
=38# 8>
-9= 8> =38 8>
.> =38 8>(
.>
8
8
=38 8> 8> -9= 8>
œ
Þ
#8#
In this case, the general solution is
Ca>b œ -" -9= 8> -# =38 8>
>
-9= 8> .
#8
Invoking the initial conditions, we obtain -" œ ! and -# œ "Î#8# . Therefore the
solution
of the IVP is
C a >b œ
"
>
=38 8>
-9= 8> .
#
#8
#8
16. Note that the function 0 a>b and the function given in Problem ) have the same
Fourier
series. Therefore
0 a >b œ
"
% _
"
# "
-9=a#8 "b1> Þ
# 1 8 œ " a#8 "b#
The solution of the homogeneous problem is
C- a>b œ -" -9= => -# =38 => Þ
Using the method of undetermined coefficients, we assume a particular solution of the
form
] a>b œ E! " E8 -9= 81> .
_
8œ"
Substitution into the ODE and equating like terms results in E! œ "Î#=# and
+8
E8 œ #
.
= 8# 1 #
It follows that the general solution is
Ca>b œ -" -9= => -# =38 =>
Setting Ca!b œ " , we find that
"
% _
-9=a#8 "b1>
"
Þ
#
#
#=
1 8 œ " a#8 "b# =# a#8 "b# 1# ‘
"
% _
-9=a#8 "b1>
-" œ " # # "
Þ
#=
1 8 œ " a#8 "b# =# a#8 "b# 1# ‘
________________________________________________________________________
page 626
—————————————————————————— CHAPTER 10. ——
Invoking the initial condition C w a!b œ ! , we obtain -# œ ! . Hence the solution of the
initial value problem is
"
"
% _
-9=a#8 "b1> -9= =>
Ca>b œ -9= => # -9= => # # "
Þ
#=
#=
1 8 œ " a#8 "b# =# a#8 "b# 1# ‘
17. Let
_
+!
81 B
81 B
"
0 aBb œ
,8 =38
’+8 -9=
“Þ
# 8œ"
P
P
Squaring both sides of the equation, we formally have
k0 aBbk# œ
_
_
+!#
81 B
81 B
81 B
81 B
" ’+8# -9=#
,8# =38#
,8 =38
“ +! " ’+8 -9=
“
%
P
P
P
P
8œ"
8œ"
" ’-78 -9=
7Á8
71B
81B
=38
“Þ
P
P
Integrating both sides of the last equation, and using the orthogonality conditions,
(
P
P
_
P
P
+!#
81 B
81 B
.B " –( +8# -9=#
.B ( ,8# =38#
.B—
P
P
P
P %
8 œ " P
k0 aBbk# .B œ (
œ
P
_
+!#
P " +8# P ,8# P‘Þ
#
8œ"
Therefore,
_
" P
+!#
#
" ˆ+8# ,8# ‰Þ
( k0 aBbk .B œ
P P
#
8œ"
19a+b. As shown in the Example, the Fourier series of the function
0 aBb œ œ
is given by
0 aBb œ
!,
P,
PB!
! B P,
P #P _
"
a#8 "b1B
"
=38
Þ
#
1 8 œ " #8 "
P
Setting P œ " ,
"
# _
"
0 aBb œ "
=38a#8 "b1B Þ
# 1 8 œ " #8 "
________________________________________________________________________
page 627
—————————————————————————— CHAPTER 10. ——
It follows that
"
"
1
"
=38a#8 "b1B œ ”0 aBb •.
#8 "
#
#
8œ"
a33b
_
a,b. Given that
1aBb œ "
_
#8 "
8œ""
a#8 "b#
and subtracting Eq.a33b from Eq.a3b, we find that
=38a#8 "b1B ,
a 3b
_
1
"
#8 "
"
1aBb ”0 aBb • œ
=38a#8 "b1B
#
#
#
8 œ " " a#8 "b
"
"
=38a#8 "b1B Þ
#8 "
8œ"
_
Based on the fact that
#8 "
" a#8 "b
#
"
"
,
œ
#8 "
a#8 "b" a#8 "b# ‘
and the fact that we can combine the two series, it follows that
1aBb œ
_
1
"
=38a#8 "b1B
"
0
a
B
b
.
”
•
#‘
#
#
8 œ " a#8 "b " a#8 "b
________________________________________________________________________
page 628
—————————————————————————— CHAPTER 10. ——
Section 10.4
1. Since the function contains only odd powers of B , the function is odd.
2. Since the function contains both odd and even powers of B , the function is neither
even nor odd.
4. We have =/- B œ "Î-9= B . Since the quotient of two even functions is even, the
function is even.
5. We can write kBk$ œ kBk † kBk# œ kBk † B# . Since both factors are even, it follows that
the function is even.
8. P œ # .
________________________________________________________________________
page 629
—————————————————————————— CHAPTER 10. ——
9. P œ # .
11. P œ # .
________________________________________________________________________
page 630
—————————————————————————— CHAPTER 10. ——
12. P œ " .
16. P œ # . For an odd extension of the function, the cosine coefficients are zero. The
sine coefficients are given by
________________________________________________________________________
page 631
—————————————————————————— CHAPTER 10. ——
# P
81B
,8 œ ( 0 aBb=38
.B
P !
P
"
#
81B
81 B
œ ( B =38
.B ( =38
.B
#
#
!
"
# =38 8#1 81 -9= 81
œ#
Þ
8# 1 #
Observe that
=38Š
! , 8 œ #5
81
‹œœ
5+1
#
a "b , 8 œ #5 "
, 5 œ "ß #ß â .
Likewise,
" , 8 œ #5
-9= 81 œ œ
" , 8 œ #5 "
, 5 œ "ß #ß â .
Therefore the Fourier sine series of the specified function is
0 aBb œ
" _ "
# _ #a "b8" a#8 "b1
a#8 "b1B
" =38 81B # "
=38
.
#
1 8œ" 8
1 8œ"
#
a#8 "b
17. P œ 1 . For an even extension of the function, the sine coefficients are zero.
The cosine coefficients are given by
# P
( 0 aBb.B
P !
# 1
œ ( a"b.B
1 !
œ #,
+! œ
________________________________________________________________________
page 632
—————————————————————————— CHAPTER 10. ——
and for 8 ! ,
# P
81B
.B
( 0 aBb-9=
P !
P
# 1
œ ( a"b-9= 8B .B
1 !
œ !Þ
+8 œ
The even extension of the given function is a constant function. As expected, the Fourier
cosine series is
+!
0 aBb œ
œ ".
#
19. P œ $1 . For an odd extension of the function, the cosine coefficients are zero. The
sine coefficients are given by
# P
81B
.B
( 0 aBb=38
P !
P
#1
$1
#
8B
#
8B
œ
=38
.B
(
( # =38 .B
$1 1
$
$1 #1
$
81
#81
# -9= 81 -9= $ -9= $
œ #
Þ
81
,8 œ
Therefore the Fourier sine series of the specified function is
# _ "
81
#81
8B
0 aBb œ " ”-9=
-9=
# -9= 81• =38 .
1 8œ" 8
$
$
$
________________________________________________________________________
page 633
—————————————————————————— CHAPTER 10. ——
21. Extend the function over the interval c P ß Pd as
0 aBb œ œ
B P,
P B,
PŸB!
! Ÿ B Ÿ P.
Since the extended function is even, the sine coefficients are zero. The cosine
coefficients
are given by
# P
( 0 aBb.B
P !
# P
œ ( aP Bb.B
P !
œ P,
+! œ
and for 8 ! ,
# P
81B
.B
( 0 aBb-9=
P !
P
# P
81B
œ ( aP Bb-9=
.B
P !
P
" -9= 81
œ #P
Þ
8# 1 #
+8 œ
Therefore the Fourier cosine series of the extended function is
0 aBb œ
P %P _
"
a#8 "b1B
# "
-9=
.
#
#
1 8 œ " a#8 "b
P
In order to compare the result with Example " of Section "!Þ# , set P œ # . The cosine
series converges to the function graphed below:
This function is a shift of the function in Example " of Section "!Þ# Þ
________________________________________________________________________
page 634
—————————————————————————— CHAPTER 10. ——
22. Extend the function over the interval c P ß Pd as
0 aBb œ œ
B P,
P B,
PŸB!
! B Ÿ P,
with 0 a!b œ ! . Since the extended function is odd, the cosine coefficients are zero. The
sine coefficients are given by
# P
81B
,8 œ ( 0 aBb=38
.B
P !
P
# P
81B
œ ( aP Bb=38
.B
P !
P
#P
œ
Þ
81
Therefore the Fourier cosine series of the extended function is
#P _ "
81 B
" =38
0 aBb œ
.
1 8œ" 8
P
________________________________________________________________________
page 635
—————————————————————————— CHAPTER 10. ——
Setting P œ # , for example, the series converges to the function graphed below:
23a+b. P œ #1 . For an even extension of the function, the sine coefficients are zero.
The cosine coefficients are given by
# P
( 0 aBb.B
P !
" 1
œ ( B .B
1 !
œ 1Î# ,
+! œ
and for 8 ! ,
# P
81B
.B
( 0 aBb-9=
P !
P
" 1
8B
œ ( B -9=
.B
1 !
#
# -9=ˆ 8#1 ‰ 81=38ˆ 8#1 ‰ #
œ#
Þ
8# 1
+8 œ
Therefore the Fourier cosine series of the given function is
0 aBb œ
1
# _ 1
81
#
81
8B
" ” =38
# Š-9=
"‹•-9= .
%
1 8œ" 8
#
8
#
#
Observe that
=38Š
! , 8 œ #5
81
‹œœ
5+1
#
a "b , 8 œ #5 "
, 5 œ "ß #ß â .
Likewise,
________________________________________________________________________
page 636
—————————————————————————— CHAPTER 10. ——
-9=Š
81
a "b5 , 8 œ #5
‹œœ
#
! , 8 œ #5 "
, 5 œ "ß #ß â .
a, b .
a- b .
24a+b. P œ 1 . For an odd extension of the function, the cosine coefficients are zero.
Note that 0 aBb œ B on ! Ÿ B 1 . The sine coefficients are given by
# P
81B
.B
( 0 aBb=38
P !
P
# 1
œ ( B =38 8B .B
1 !
# -9= 81
œ
Þ
8
,8 œ
Therefore the Fourier sine series of the given function is
________________________________________________________________________
page 637
—————————————————————————— CHAPTER 10. ——
a "b 8
=38 8B.
8
8œ"
0 aBb œ # "
_
a, b .
a- b .
26a+b. P œ % . For an even extension of the function, the sine coefficients are zero. The
cosine coefficients are given by
# P
+! œ ( 0 aBb.B
P !
" %
œ ( ˆB# #B‰.B
# !
œ )Î$ ,
and for 8 ! ,
________________________________________________________________________
page 638
—————————————————————————— CHAPTER 10. ——
# P
81B
+8 œ ( 0 aBb-9=
.B
P !
P
" %
81 B
œ ( ˆB# #B‰-9=
.B
# !
%
" $ -9= 81
œ "'
Þ
8# 1 #
Therefore the Fourier cosine series of the given function is
% "' _ " $a "b8
81 B
0 aBb œ # "
-9=
.
#
$ 1 8œ"
8
%
a, b .
a- b .
________________________________________________________________________
page 639
—————————————————————————— CHAPTER 10. ——
27a+b.
a,b. P œ $ . For an even extension of the function, the cosine coefficients are given by
# P
( 0 aBb.B
P !
# $
œ ( a$ Bb.B
$ !
œ $,
+! œ
and for 8 ! ,
# P
81B
.B
( 0 aBb-9=
P !
P
# $
81B
œ ( a$ Bb-9=
.B
$ !
$
" -9= 81
œ'
Þ
8# 1 #
+8 œ
Therefore the Fourier cosine series of the given function is
1aBb œ
$
' _ " a "b 8
81 B
# "
-9=
.
#
# 1 8œ"
8
$
For an odd extension of the function, the sine coefficients are given by
# P
81 B
,8 œ ( 0 aBb=38
.B
P !
P
# $
81 B
œ ( a$ Bb=38
.B
$ !
$
'
œ
Þ
81
Therefore the Fourier sine series of the given function is
________________________________________________________________________
page 640
—————————————————————————— CHAPTER 10. ——
' _ "
81B
2aBb œ " =38
.
1 8œ" 8
$
a- b. For the even extension:
For the odd extension:
a. b. Since the even extension is continuous, the series converges uniformly. On the
other
hand, the odd extension is discontinuous. Gibbs' phenomenon results in a finite error for
all values of 8 .
________________________________________________________________________
page 641
—————————————————————————— CHAPTER 10. ——
29a+b.
a,b. P œ # . For an even extension of the function, the cosine coefficients are given by
# P
( 0 aBb.B
P !
#
%B# %B $
œ( ”
•.B
%
!
œ &Î' ,
+! œ
and for 8 ! ,
# P
81B
.B
( 0 aBb-9=
P !
P
#
%B# %B $
81 B
œ( ”
.B
•-9=
%
#
!
" $ -9= 81
œ%
Þ
8# 1 #
+8 œ
Therefore the Fourier cosine series of the given function is
1aBb œ
&
% _ " $a "b 8
81 B
# "
-9=
.
#
"# 1 8 œ "
8
#
For an odd extension of the function, the sine coefficients are given by
# P
81B
,8 œ ( 0 aBb=38
.B
P !
P
#
%B# %B $
81 B
œ( ”
.B
•=38
%
#
!
$# $8# 1# &8# 1# -9= 81 $# -9= 81
œ
Þ
# 8$ 1 $
Therefore the Fourier sine series of the given function is
________________________________________________________________________
page 642
—————————————————————————— CHAPTER 10. ——
" _ $#a" -9= 81b 8# 1# a$ & -9= 81b
81 B
2aBb œ $ "
=38
.
$
#1 8 œ "
8
#
a- b. For the even extension:
For the 9.. extension:
a. b. Since the even extension is continuous, the series converges uniformly. On the
other
hand, the odd extension is discontinuous. Gibbs' phenomenon results in a finite error for
all values of 8 .
________________________________________________________________________
page 643
—————————————————————————— CHAPTER 10. ——
30a+b.
a,b. P œ $ . For an even extension of the function, the cosine coefficients are given by
# P
( 0 aBb.B
P !
# $
œ ( ˆB$ &B# &B "‰.B
$ !
œ "Î# ,
+! œ
and for 8 ! ,
# P
81B
.B
( 0 aBb-9=
P !
P
# $
81B
œ ( ˆB$ &B# &B "‰-9=
.B
$ !
$
"'# "& 8# 1# ' 8# 1# -9= 81 "'# -9= 81
œ#
Þ
8% 1 %
+8 œ
Therefore the Fourier cosine series of the given function is
1aBb œ
"
# _ "'#a" -9= 81b $ 8# 1# a& # -9= 81b
81 B
% "
-9=
.
% 1 8œ"
8%
$
For an odd extension of the function, the sine coefficients are given by
# P
81 B
.B
( 0 aBb=38
P !
P
# $
81B
œ ( ˆB$ &B# &B "‰=38
.B
$ !
$
*! 8# 1# # 8# 1# -9= 81 (# -9= 81
œ#
Þ
8$ 1 $
,8 œ
Therefore the Fourier sine series of the given function is
________________________________________________________________________
page 644
—————————————————————————— CHAPTER 10. ——
# _ ")a& % -9= 81b 8# 1# a" # -9= 81b
81 B
2aBb œ $ "
=38
.
$
1 8œ"
8
$
a- b. For the even extension:
For the odd extension:
a. b. Since the even extension is continuous, the series converges uniformly. On the
other
hand, the odd extension is discontinuous. Gibbs' phenomenon results in a finite error for
all values of 8 ; particularly at B œ „ $ .
33. Let 0 aBb be a differentiable even function. For any B in its domain,
0 a B 2b 0 a Bb œ 0 aB 2 b 0 aBb .
It follows that
________________________________________________________________________
page 645
—————————————————————————— CHAPTER 10. ——
0 a B 2 b 0 a Bb
2Ä!
2
0 aB 2b 0 aBb
œ lim
2Ä!
2
0 aB 2b 0 aBb
œ lim
.
2Ä!
a 2b
0 w a Bb œ lim
Setting 2 œ $ , we have
0 aB $ b 0 aBb
2Ä!
$
0 aB $ b 0 aBb
œ lim
$ Ä !
$
w
œ 0 aBb .
0 w a Bb œ lim
Therefore 0 w a Bb œ 0 w aBb .
If 0 aBb is a differentiable odd function, for any B in its domain,
0 a B 2b 0 a Bb œ 0 aB 2 b 0 aBb .
It follows that
0 a B 2 b 0 a Bb
2Ä!
2
0 aB 2b 0 aBb
œ lim
2Ä!
2
0 aB 2b 0 aBb
œ lim
.
2Ä!
a 2b
0 w a Bb œ lim
Setting 2 œ $ , we have
0 aB $ b 0 aBb
2Ä!
$
0 aB $ b 0 aBb
œ lim
$ Ä !
$
w
œ 0 aBb .
0 w a Bb œ lim
Therefore 0 w a Bb œ 0 w aBb .
36. From Example " of Section "!Þ# , the function
0 aBb œ œ
B,
B,
aP œ #b has a convergent Fourier series
#ŸB!
! Ÿ B #,
________________________________________________________________________
page 646
—————————————————————————— CHAPTER 10. ——
0 aBb œ "
) _
"
a#8 "b1B
"
-9=
.
#
#
1 8 œ " a#8 "b
#
Since 0 aBb is continuous, the series converges everywhere. In particular, at B œ ! ,
we have
! œ 0 a!b œ "
) _
"
"
.
1# 8 œ " a#8 "b#
It follows immediately that
_
1#
"
"
"
"
œ "
œ " # # # â.
#
)
$
&
(
8 œ " a#8 "b
40. Since one objective is to obtain a Fourier series containing only cosine terms, any
extension of 0 aBb should be an even function. Another objective is to derive a series
containing only the terms
-9=
a#8 "b1B
, 8 œ "ß #ß â .
#P
First note that the functions
-9=
81B
, 8 œ "ß #ß â
P
are symmetric about B œ P . Indeed,
________________________________________________________________________
page 647
—————————————————————————— CHAPTER 10. ——
-9=
81a#P Bb
81 B
œ -9=Š#81
‹
P
P
81B
œ -9=Š
‹
P
81B
œ -9=
.
P
It follows that if 0 aBb is extended into aP ß #Pb as an antisymmetric function about
B œ P,
that is, 0 a#P Bb œ 0 aBb for ! Ÿ B Ÿ #P , then
(
#P
!
0 aBb-9=
81B
.B œ ! .
P
This follows from the fact that the integrand is antisymmetric function about B œ P.
Now
extend the function 0 aBb to obtain
0 aBb œ œ
µ
0 aBb, ! Ÿ B P
0 a#P Bb, P B #P .
Finally, extend the resulting function into a #P ß !b as an even function, and then as a
periodic function of period %P.
By construction, the Fourier series will contain only cosine terms. We first note that
#P µ
#
+! œ
( 0 aBb .B
#P !
" P
" #P
œ ( 0 aBb.B ( 0 a#P Bb.B
P !
P P
P
"
" P
œ ( 0 aBb.B ( 0 a?b.?
P !
P !
œ !.
For 8 ! ,
#P µ
#
81B
.B
( 0 aBb-9=
#P !
#P
" P
81B
" #P
81 B
œ ( 0 aBb-9=
.B ( 0 a#P Bb-9=
.B .
P !
#P
P P
#P
+8 œ
For the second integral, let ? œ #P B . Then
-9=
81B
81a#P ?b
81 ?
œ -9=
œ a "b8 -9=
#P
#P
#P
________________________________________________________________________
page 648
—————————————————————————— CHAPTER 10. ——
and therefore
(
#P
P
0 a#P Bb-9=
Hence
P
81B
81 ?
.B œ a "b8 ( 0 a?b-9=
.? .
#P
#P
!
" a "b 8 P
81 B
+8 œ
.B .
( 0 aBb-9=
P
#P
!
It immediately follows that +8 œ ! for 8 œ #5 , 5 œ !ß "ß #ß â , and
+#5"
# P
a#5 "b1B
œ ( 0 aBb-9=
.B , for 5 œ "ß #ß â .
P !
#P
The associated Fourier series representation
0 aBb œ " +#8" -9=
_
8œ!
a#8 "b1B
#P
converges almost everywhere on a #P ß #Pb and hence on a! ß Pb.
For example, if 0 aBb œ B for ! Ÿ B Ÿ P œ " , the graph of the extended function is:
________________________________________________________________________
page 649
—————————————————————————— CHAPTER 10. ——
Section 10.5
1. We consider solutions of the form ?aB ß >b œ \ aBbX a>b. Substitution into the partial
differential equation results in
B\ ww X \X w œ ! .
Divide both sides of the differential equation by the product \X to obtain
B
\ ww
Xw
œ!ß
\
X
so that
B
\ ww
Xw
œ
.
\
X
Since both sides of the resulting equation are functions of different variables, each must
be equal to a constant, say - . We obtain the ordinary differential equations
B\ ww -\ œ ! and X w -X œ ! .
2. In order to apply the method of separation of variables, we consider solutions of the
form ?aB ß >b œ \ aBbX a>b. Substituting the assumed form of the solution into the partial
differential equation, we obtain
> \ ww X B \X w œ ! .
Divide both sides of the differential equation by the product B>\X to obtain
\ ww
Xw
œ!ß
B\
>X
so that
\ ww
Xw
œ
.
B\
>X
Since both sides of the resulting equation are functions of different variables, it follows
that
\ ww
Xw
œ
œ -.
B\
>X
Therefore \ aBb and X a>b are solutions of the ordinary differential equations
\ ww -B \ œ ! and X w -> X œ ! .
________________________________________________________________________
page 650
—————————————————————————— CHAPTER 10. ——
4. Assume that the solution of the PDE has the form ?aB ß >b œ \ aBbX a>b. Substitution
into the partial differential equation results in
c:aBb\ w d X <aBb\ X ww œ ! .
w
Divide both sides of the differential equation by the product <aBb\X to obtain
c:aBb\ w d w
X ww
œ!ß
<aBb\
X
that is,
c:aBb\ w d w
X ww
œ
.
<aBb\
X
Since both sides of the resulting equation are functions of different variables, each must
be equal to a constant, say - . We obtain the ordinary differential equations
c:aBb\ w d -<aBb\ œ ! and X ww -X œ ! .
w
6. We consider solutions of the form ?aB ß Cb œ \ aBb] aCb. Substitution into the partial
differential equation results in
\ ww ] \] ww B\] œ ! .
Divide both sides of the differential equation by the product \] to obtain
\ ww
] ww
Bœ!ß
\
]
that is,
\ ww
] ww
Bœ
Þ
\
]
Since both sides of the resulting equation are functions of different variables, it follows
that
\ ww
] ww
Bœ
œ -.
\
]
We obtain the ordinary differential equations
\ ww aB -b\ œ ! and ] ww -] œ ! .
7. The heat conduction equation, "!! ?BB œ ?> , and the given boundary conditions are
homogeneous. We consider solutions of the form ?aB ß >b œ \ aBbX a>b. Substitution
into
the partial differential equation results in
________________________________________________________________________
page 651
—————————————————————————— CHAPTER 10. ——
"!! \ ww X œ \X w .
Divide both sides of the differential equation by the product \X to obtain
\ ww
Xw
œ
Þ
\
"!! X
Since both sides of the resulting equation are functions of different variables, it follows
that
\ ww
Xw
œ
œ -.
\
"!! X
Therefore \ aBb and X a>b are solutions of the ordinary differential equations
\ ww - \ œ ! and X w "!!- X œ ! .
The general solution of the spatial equation is \ œ -" -9= -"Î# B -# =38 -"Î# B . In order
to satisfy the homogeneous boundary conditions, we require that -" œ ! , and
-"Î# œ 81 .
Hence the eigenfunctions are \8 œ =38 81B , with associated eigenvalues -8 œ 8# 1# .
We thus obtain the family of equations X w "!!-8 X œ ! . Solution are given by
X8 œ /"!!-8 > Þ
Hence the fundamental solutions of the PDE are
?8 aB ß >b œ /"!!8 1 > =38 81B ,
# #
which yield the general solution
? aB ß >b œ " -8 /"!!8 1 > =38 81B Þ
_
# #
8œ"
Finally, the initial condition ?aB ß !b œ =38 #1B =38 &1B must be satisfied. Therefore
is it necessary that
" -8 =38 81B œ =38 #1B =38 &1B .
_
8œ"
It follows from the othogonality conditions that -# œ -& œ " , with all other -8 œ ! .
Therefore the solution of the given heat conduction problem is
? aB ß >b œ /%!!1 > =38 #1B /#&!!1 > =38 &1B .
#
#
________________________________________________________________________
page 652
—————————————————————————— CHAPTER 10. ——
9. The heat conduction problem is formulated as
?BB œ ?> ,
?a! ß >b œ ! ,
?aB ß !b œ &! ,
! B %! , > ! à
?a%! ß >b œ ! , > ! à
! B %! .
Assume a solution of the form ?aB ß >b œ \ aBbX a>b . Following the procedure in this
section, we obtain the eigenfunctions \8 œ =38 81BÎ%! , with associated eigenvalues
-8 œ 8# 1# /"'!! . The solutions of the temporal equations are
X8 œ /-8 > Þ
Hence the general solution of the given problem is
? aB ß >b œ " -8 /8 1 >Î"'!! =38
_
8œ"
# #
81B
Þ
%!
The coefficients -8 are the Fourier sine coefficients of ?aB ß !b œ &! . That is,
# P
81B
.B
( 0 aBb=38
P !
P
& %!
81B
œ ( =38
.B
# !
%!
" -9= 81
œ "!!
Þ
81
-8 œ
The sine series of the initial condition is
&! œ
"!! _ " -9= 81
81 B
"
=38
.
1 8œ"
8
%!
Therefore the solution of the given heat conduction problem is
"!! _ " -9= 81 8# 1# >Î"'!!
81 B
"
? a B ß >b œ
/
=38
Þ
1 8œ"
8
%!
11. Refer to Prob. * for the formulation of the problem. In this case, the initial condition
is given by
Ú !ß
? aB ß !b œ Û &! ß
Ü !ß
! Ÿ B "! ß
"! Ÿ B Ÿ $! ß
$! B Ÿ %! .
All other data being the same, the solution of the given problem is
________________________________________________________________________
page 653
—————————————————————————— CHAPTER 10. ——
?aB ß >b œ " -8 /8 1 >Î"'!! =38
_
# #
8œ"
81B
Þ
%!
The coefficients -8 are the Fourier sine coefficients of ?aB ß !b . That is,
# P
81B
.B
( 0 aBb=38
P !
P
& $!
81 B
œ ( =38
.B
# "!
%!
-9= 8%1 -9= $8%1
œ "!!
Þ
81
-8 œ
Therefore the solution of the given heat conduction problem is
?aB ß >b œ
"!! _ -9= 8%1 -9= $8%1 8# 1# >Î"'!!
81B
"
/
=38
Þ
1 8œ"
8
%!
12. Refer to Prob. * for the formulation of the problem. In this case, the initial condition
is given by
?aB ß !b œ B , ! B %! .
All other data being the same, the solution of the given problem is
? aB ß >b œ " -8 /8 1 >Î"'!! =38
_
8œ"
# #
81B
Þ
%!
The coefficients -8 are the Fourier sine coefficients of ?aB ß !b œ B . That is,
# P
81B
-8 œ ( 0 aBb=38
.B
P !
P
" %!
81B
œ
.B
( B =38
#! !
%!
-9= 81
œ )!
Þ
81
Therefore the solution of the given heat conduction problem is
? a B ß >b œ
)! _ a "b8" 8# 1# >Î"'!!
81 B
"
/
=38
Þ
1 8œ"
8
%!
________________________________________________________________________
page 654
—————————————————————————— CHAPTER 10. ——
13. Substituting B œ #! , into the solution, we have
? a#! ß >b œ
We can also write
Therefore,
"!! _ " -9= 81 8# 1# >Î"'!!
81
"
/
=38
Þ
1 8œ"
8
#
#!! _ a "b5" a#5"b# 1# >Î"'!!
"
? a#! ß >b œ
/
Þ
1 5 œ " #5 "
#!! _ a "b5" a#5"b# 1# Î$#!
"
? a#! ß &b œ
/
Þ
1 5 œ " #5 "
Let
E5 œ
a "b8" #!! a#5"b# 1# Î$#!
/
Þ
1a#5 "b
It follows that kE5 k !Þ!!& for 5
unaffected by additional terms.
For > œ #! ,
? a#! ß #!b œ
Let
* . So for 8 œ #5 "
"(, the summation is
#!! _ a "b5" a#5"b# 1# Î)!
"
/
Þ
1 5 œ " #5 "
a "b8" #!! a#5"b# 1# Î)!
E5 œ
/
Þ
1a#5 "b
It follows that kE5 k !Þ!!$ for 5
unaffected by additional terms.
For > œ )! ,
? a#! ß )!b œ
Let
E5 œ
& . So for 8 œ #5 "
*, the summation is
#!! _ a "b5" a#5"b# 1# Î#!
"
/
Þ
1 5 œ " #5 "
a "b8" #!! a#5"b# 1# Î#!
/
Þ
1a#5 "b
It follows that kE5 k !Þ!!!!& for 5
unaffected by additional terms.
$ . So for 8 œ #5 "
&, the summation is
________________________________________________________________________
page 655
—————————————————————————— CHAPTER 10. ——
The series solution converges faster as > increases.
14a+b. The solution of the given heat conduction problem is
"!! _ " -9= 81 8# 1# >Î"'!!
81 B
"
? a B ß >b œ
/
=38
Þ
1 8œ"
8
%!
Setting > œ &ß "!ß #!ß %!ß "!!ß #!! À
a,b. Setting B œ &ß "!ß "&ß #! À
________________________________________________________________________
page 656
—————————————————————————— CHAPTER 10. ——
a- b. Surface plot of ?aB ß >b À
a. b. ! Ÿ ?aB ß >b Ÿ " for >
'(& =/- .
16a+b. The solution of the given heat conduction problem is
"!! _ -9= 8%1 -9= $8%1 8# 1# >Î"'!!
81B
"
?aB ß >b œ
/
=38
Þ
1 8œ"
8
%!
________________________________________________________________________
page 657
—————————————————————————— CHAPTER 10. ——
Setting > œ &ß "!ß #!ß %!ß "!!ß #!! À
a,bÞ Setting B œ &ß "!ß "&ß #! À
a- b. Surface plot of ?aB ß >b À
________________________________________________________________________
page 658
—————————————————————————— CHAPTER 10. ——
a. b. ! Ÿ ?aB ß >b Ÿ " for >
'"& =/- .
17a+b. The solution of the given heat conduction problem is
? a B ß >b œ
)! _ a "b8" 8# 1# >Î"'!!
81 B
"
/
=38
Þ
1 8œ"
8
%!
Setting > œ &ß "!ß #!ß %!ß "!!ß #!! À
a,b. Analyzing the individual plots, we find that the 'hot spot' varies with time:
>
B2
&
$$
"!
$"
#!
#*
%!
#'
"!!
##
#!!
#"
________________________________________________________________________
page 659
—————————————————————————— CHAPTER 10. ——
Location of the 'hot spot', B2 , versus time À
Evidently, the location of the greatest temperature migrates to the center of the rod.
a- bÞ Setting B œ &ß "!ß "&ß #! À
a. b. Surface plot of ?aB ß >b À
________________________________________________________________________
page 660
—————————————————————————— CHAPTER 10. ——
a/b. ! Ÿ ?aB ß >b Ÿ " for >
&#& =/- .
19. The solution of the given heat conduction problem is
? a B ß >b œ
#!! _ " -9= 81 8# 1# !# >Î%!!
81 B
"
/
=38
Þ
1 8œ"
8
#!
Setting B œ "! -7 ,
#!! _ " -9= 81 8# 1# !# >Î%!!
81
"
? a"! ß >b œ
/
=38
Þ
1 8œ"
8
#
A two-term approximation is given by
? a"! ß >b ¸
%!!
# #
# #
’$ /1 ! >Î%!! /*1 ! >Î%!! “Þ
$1
From Table "!Þ&Þ" À
silver
aluminum
cast iron
!#
"Þ("
!Þ)'
!Þ"#
________________________________________________________________________
page 661
—————————————————————————— CHAPTER 10. ——
a+b. !# œ "Þ(" À
a,b. !# œ !Þ)' À
a- b. !# œ !Þ"# À
________________________________________________________________________
page 662
—————————————————————————— CHAPTER 10. ——
21a+b. Given the partial differential equation
+ ?BB , ?> - ? œ ! ,
in which + , , , and - are constants, set ?aB ß >b œ /$> AaB ß >b. Substitution into the PDE
results in
+ /$ > ABB ,ˆ$ /$ > A /$ > A> ‰ - /$ > A œ ! .
Dividing both sides of the equation by /$> , we obtain
+ ABB , A> a- , $ b A œ ! .
As long as , Á ! , choosing $ œ -Î, yields
+
ABB A> œ ! ,
,
which is the heat conduction equation with dependent variable A .
23. The heat conduction equation in polar coordinates is given by
!# ”?<<
"
"
?< # ?)) • œ ?> .
<
<
We consider solutions of the form ?a< ß ) ß >b œ V a<b@a)bX a>b. Substitution into the
PDE
results in
!# ”V ww @X
" w
"
V @X # V @ ww X • œ V @X w .
<
<
Dividing both sides of the equation by the factor V @X , we obtain
V ww
" Vw
" @ ww
Xw
#
œ # .
V
< V
< @
! X
Since both sides of the resulting differential equation depend on different variables, each
side must be equal to a constant, say - . That is,
V ww
" Vw
" @ ww
Xw
#
œ # œ -# .
V
< V
< @
! X
It follows that X w !# -# X œ ! , and
V ww
" Vw
" @ ww
#
œ -# .
V
< V
< @
Multiplying both sides of this differential equation by <# , we find that
<#
V ww
Vw
@ ww
<
œ -# < # ,
V
V
@
________________________________________________________________________
page 663
—————————————————————————— CHAPTER 10. ——
which can be written as
<#
V ww
Vw
@ ww
<
-# < # œ
.
V
V
@
Once again, since both sides of the resulting differential equation depend on different
variables, each side must be equal to a constant. Hence
<#
V ww
Vw
@ ww
<
-# <# œ .# and
œ .# .
V
V
@
The resulting ordinary equations are
<# V ww <V w ˆ-# <# .# ‰V œ !
@ ww .# @ œ !
X w !# -# X œ ! Þ
________________________________________________________________________
page 664
—————————————————————————— CHAPTER 10. ——
Section 10.6
1. The steady-state solution, @aBb, satisfies the boundary value problem
@ ww aBb œ ! , ! B &! , @a!b œ "! , @a&!b œ %! .
The general solution of the ODE is @aBb œ EB F . Imposing the boundary conditions,
we have
@aBb œ
%! "!
$B
B "! œ
"! Þ
&!
&
2. The steady-state solution, @aBb, satisfies the boundary value problem
@ ww aBb œ ! , ! B %! , @a!b œ $! , @a%!b œ #! .
The solution of the ODE is linear. Imposing the boundary conditions, we have
@aBb œ
#! $!
&B
B $! œ
$! Þ
%!
%
4. The steady-state solution is also a solution of the boundary value problem given by
@ ww aBb œ ! , ! B P , and the conditions @ w a!b œ ! , @aPb œ X . The solution of the
ODE is @aBb œ EB F . The boundary condition @ w a!b œ ! requires that E œ ! . The
other condition requires that F œ X Þ Hence @aBb œ X Þ
5. As in Prob. % , the steady-state solution has the form @aBb œ EB F . The boundary
condition @a!b œ ! requires that F œ ! . The boundary condition @ w aPb œ ! requires
that E œ ! . Hence @aBb œ ! Þ
6. The steady-state solution has the form @aBb œ EB F . The first boundary
condition, @a!b œ X , requires that F œ X . The other boundary condition, @ w aPb œ ! ,
requires that E œ ! . Hence @aBb œ X Þ
8. The steady-state solution, @aBb, satisfies the differential equation @ ww aBb œ ! , along
with the boundary conditions
@a!b œ X , @ w aPb @aPb œ ! .
The general solution of the ODE is @aBb œ EB F . The boundary condition @ w a!b œ !
requires that F œ X . It follows that @aBb œ EB X , and
@ w aPb @aPb œ E EP X .
The second boundary condition requires that E œ X Îa" Pb . Therefore
@aBb œ
XB
X .
"P
________________________________________________________________________
page 665
—————————————————————————— CHAPTER 10. ——
10a+b. Based on the symmetry of the problem, consider only left half of the bar. The
steady-state solution satisfies the ODE @ ww aBb œ ! , along with the boundary conditions
@a!b œ ! and @a&!b œ "!! . The solution of this boundary value problem is @aBb œ #B .
It follows that the steady-state temperature is the entire rod is given by
0 aBb œ œ
#B , ! Ÿ B Ÿ &!
#!! #B , &! Ÿ B Ÿ "!! Þ
a,b. The heat conduction problem is formulated as
!# ?BB œ ?> ,
?a! ß >b œ #! ,
?aB ß !b œ 0 aBb ,
! B "!! , > ! à
?a"!! ß >b œ ! , > ! à
! B "!! .
First express the solution as ?aB ß >b œ 1aBb AaB ß >b , where 1aBb œ BÎ& #! and
A satisfies the heat conduction problem
!# ABB œ A> ,
! B "!! , > ! à
A a! ß > b œ ! ,
Aa"!! ß >b œ ! , > ! à
AaB ß !b œ 0 aBb 1aBb ,
! B "!! .
Based on the results in Section "!Þ& ,
AaB ß >b œ " -8 /8 1 ! >Î"!!!! =38
_
8œ"
# # #
81B
,
"!!
in which the coefficients -8 are the Fourier sine coefficients of 0 aBb 1aBb. That is,
# P
81B
-8 œ ( c0 aBb 1aBbd=38
.B
P !
P
" "!!
81B
œ
c0 aBb 1aBbd=38
.B
(
&! !
"!!
#! =38 8#1 81
œ %!
Þ
8# 1 #
Finally, the thermal diffusivity of copper is "Þ"% -7# Î=/- . Therefore the temperature
distribution in the rod is
? aB ß >b œ #!
B %! _ #! =38 8#1 81 "Þ"% 8# 1# >Î"!!!!
81B
# "
/
=38
Þ
&
1 8œ"
8#
"!!
________________________________________________________________________
page 666
—————————————————————————— CHAPTER 10. ——
a- b. > œ & ß "!ß #!ß %! =/- À
> œ "!!ß #!!ß $!!ß &!! =/- À
________________________________________________________________________
page 667
—————————————————————————— CHAPTER 10. ——
a. b. The steady-state temperature of the center of the rod will be 1a&!b œ "!°G .
Using a one-term approximation,
? aB ß >b ¸ "!
)!! %! 1 "Þ"% 1# >Î"!!!!
/
Þ
1#
Numerical investigation shows that "! ?a&! ß >b "" for >
$(&& =/- .
________________________________________________________________________
page 668
—————————————————————————— CHAPTER 10. ——
11a+bÞ The heat conduction problem is formulated as
?BB œ ?> ,
?a! ß >b œ $! ,
?aB ß !b œ 0 aBb ,
! B $! , > ! à
?a$! ß >b œ ! , > ! à
! B $! ,
in which the initial condition is given by 0 aBb œ Ba'! BbÎ$! . Express the solution as
?aB ß >b œ @aBb AaB ß >b , where @aBb œ $! B and A satisfies the heat conduction
problem
ABB œ A> ,
! B $! , > ! à
A a ! ß >b œ ! ,
Aa$! ß >b œ ! , > ! à
AaB ß !b œ 0 aBb @aBb ,
! B $! .
As shown in Section "!Þ& ,
AaB ß >b œ " -8 /8 1 >Î*!! =38
_
# #
8œ"
81B
,
$!
in which the coefficients -8 are the Fourier sine coefficients of 0 aBb @aBb. That is,
# P
81B
.B
( c0 aBb 1aBbd=38
P !
P
" $!
81B
œ
.B
( c0 aBb 1aBbd=38
"& !
$!
#a" -9= 81b 8# 1# a" -9= 81b
œ '!
Þ
8$ 1 $
-8 œ
Therefore the temperature distribution in the rod is
'! _ #a" -9= 81b 8# 1# a" -9= 81b 8# 1# >Î*!!
81B
? aB ß >b œ $! B $ "
/
=38
Þ
$
8
1 8œ"
$!
a,b. > œ & ß "!ß #!ß %! =/- À
________________________________________________________________________
page 669
—————————————————————————— CHAPTER 10. ——
> œ &!ß (&ß "!!ß #!! =/- À
________________________________________________________________________
page 670
—————————————————————————— CHAPTER 10. ——
a- b .
Based on the heat conduction equation, the rate of change of the temperature at any given
point is proportional to the concavity of the graph of ? versus B , that is, ?BB . Evidently,
near > œ '! , the concavity of ?aB ß >b changes.
13a+b. The heat conduction problem is formulated as
?BB œ % ?> ,
?B a! ß >b œ ! ,
?aB ß !b œ 0 aBb ,
! B %! , > ! à
?B a%! ß >b œ ! , > ! à
! B %! ,
in which the initial condition is given by 0 aBb œ Ba'! BbÎ$! .
As shown in the discussion on rods with insulated ends, the solution is given by
? a B ß >b œ
_
-!
81B
# # #
" -8 /8 1 ! >Î"'!! -9=
,
#
%!
8œ"
where -8 are the Fourier cosine coefficients. In this problem,
________________________________________________________________________
page 671
—————————————————————————— CHAPTER 10. ——
# P
-! œ ( 0 aBb.B
P !
" %! Ba'! Bb
œ
.B
(
#! !
$!
œ %!!Î* ,
and for 8
",
# P
81 B
-8 œ ( 0 aBb-9=
.B
P !
P
" %! Ba'! Bb
81B
œ
-9=
.B
(
#! !
$!
%!
"'!a$ -9= 81b
œ
Þ
$8# 1#
Therefore the temperature distribution in the rod is
#!! "'! _ a$ -9= 81b 8# 1# >Î'%!!
81 B
? a B ß >b œ
# "
/
-9=
.
#
*
$1 8 œ "
8
%!
a,b. > œ &!ß "!!ß "&!ß #!! =/- À
________________________________________________________________________
page 672
—————————————————————————— CHAPTER 10. ——
> œ %!ß '!!ß )!!ß "!!! =/- À
a- b. Since
# #
lim /8 1 >Î'%!! -9=
>Ä_
81B
œ!
%!
for each B , it follows that the steady-state temperature is ?_ œ #!!Î* .
________________________________________________________________________
page 673
—————————————————————————— CHAPTER 10. ——
a. b. We first note that
? a%! ß >b œ
#!! "'! _ a "b8 a$ -9= 81b 8# 1# >Î'%!!
# "
/
.
8#
*
$1 8 œ "
For large values of > , an approximation is given by
? a%! ß >b ¸
#!! $#! 1# >Î'%!!
# /
.
*
$1
Numerical investigation shows that ##Þ## ?a%! ß >b #$Þ## for >
"&&! =/- .
16a+b. The heat conduction problem is formulated as
?BB œ ?> ,
?a! ß >b œ ! ,
?aB ß !b œ 0 aBb ,
! B $! , > ! à
?B a$! ß >b œ ! , > ! à
! B $! ,
in which the initial condition is given by 0 aBb œ $! B. Based on the results of Prob.
"&,
the solution is given by
? aB ß >b œ " -8 /a#8"b 1 >Î$'!! =38
_
8œ"
in which
# #
81B
,
'!
# P
a#8 "b1B
.B
( 0 aBb=38
P !
'!
" $!
a#8 "b1B
œ
.B
( a$! Bb=38
"& !
'!
# -9= 81 a#8 "b1
œ "#!
Þ
a#8 "b# 1#
-8 œ
________________________________________________________________________
page 674
—————————————————————————— CHAPTER 10. ——
Therefore the solution of the heat conduction problem is
? aB ß >b œ "#! "
_
8œ"
# -9= 81 a#8 "b1
a#8
"b # 1 #
/a#8"b 1 >Î$'!! =38
# #
81 B
.
'!
a,b. > œ "!ß #!ß $!ß %! =/- À
> œ %!ß '!ß )!ß "!! =/- À
________________________________________________________________________
page 675
—————————————————————————— CHAPTER 10. ——
> œ "!!ß "&!ß #!!ß #&! =/- À
________________________________________________________________________
page 676
—————————————————————————— CHAPTER 10. ——
a- b .
The location of B2 moves from B œ ! to B œ $! .
a. b .
17a+b. The heat conduction problem is formulated as
?BB œ ?> ,
! B $! , > ! à
?a! ß >b œ %! ,
?B a$! ß >b œ ! , > ! à
?aB ß !b œ $! B ,
! B $! ,
The steady-state temperature satisfies the boundary value problem
@ ww œ ! , @a!b œ %! and @ w a$!b œ ! .
It easy to see we must have @aBb œ %! . Express the solution as
?aB ß >b œ %! AaB ß >b ,
________________________________________________________________________
page 677
—————————————————————————— CHAPTER 10. ——
in which A satisfies the heat conduction problem
ABB œ A> ,
! B $! , > ! à
A a ! ß >b œ ! ,
AB a$! ß >b œ ! , > ! à
AaB ß !b œ "! B ,
! B $! .
As shown in Prob. "&, the solution is given by
A aB ß >b œ " -8 /a#8"b 1 >Î$'!! =38
_
# #
8œ"
in which
81B
,
'!
a#8 "b1B
# P
.B
( 0 aBb=38
P !
'!
a#8 "b1B
" $!
œ
.B
( a "! Bb=38
"& !
'!
' -9= 81 a#8 "b1
œ %!
Þ
a#8 "b# 1#
-8 œ
Therefore the solution of the original heat conduction problem is
? aB ß >b œ %! %! "
_
8œ"
' -9= 81 a#8 "b1
a#8
"b # 1 #
/a#8"b 1 >Î$'!! =38
# #
81 B
.
'!
a,b. > œ "!ß $!ß &!ß (! =/- À
________________________________________________________________________
page 678
—————————————————————————— CHAPTER 10. ——
> œ "!!ß #!!ß $!!ß %!! =/- À
a- b. Observe the concavity of the curves. Note also that the temperature at the insulated
end tends to the value of the fixed temperature at the boundary B œ ! .
18. Setting - œ .# , the general solution of the ODE \ ww .# \ œ ! is
\ aBb œ 5" /3.B 5 # / 3.B Þ
The boundary conditions C w a!b œ C w aPb œ ! lead to the system of equations
. 5" /
3.P
. 5" . 5 # œ !
.5 # / 3.P œ ! Þ
a‡b
If . œ ! , then the solution of the ODE is \ œ EB F . The boundary conditions
require that \ œ F .
If . Á ! , then the system algebraic equations has a nontrivial solution if and only if the
coefficient matrix is singular. Set the determinant equal to zero to obtain
/ 3.P / 3.P œ ! .
________________________________________________________________________
page 679
—————————————————————————— CHAPTER 10. ——
Let . œ / 35 . Then 3.P œ 3/ P 5 P , and the previous equation can be written
as
/ 5 P / 3/ P / 5 P / 3 / P œ ! .
Using Euler's relation, /3/ P œ -9= / P 3 =38 / P , we obtain
/5P a-9= / 3 =38 / b /5P a-9= / 3 =38 / b œ ! .
Equating the real and imaginary parts of the equation,
ˆ/5P /5P ‰-9= / P œ !
ˆ/5P /5P ‰=38 / P œ ! Þ
Based on the second equation, / P œ 81 , 8 − ˆ . Since -9= 8P Á ! , it follows that
/5P œ /5P , or /#5P œ " . Hence 5 œ ! , and . œ 81ÎP , 8 − ˆ .
Note that if 5 Á !, then the last two equations have no solution. It follows that the
system
of equations a‡b has no nontrivial solutions.
20a+b. Consider solutions of the form ?aB ß >b œ \ aBbX a>b. Substitution into the partial
differential equation results in
!# \ ww X œ X w .
Divide both sides of the differential equation by the product \X to obtain
\ ww
Xw
œ # .
\
! X
Since both sides of the resulting equation are functions of different variables, each must
be equal to a constant, say - . We obtain the ordinary differential equations
\ ww -\ œ ! and X w -!# X œ ! .
Invoking the first boundary condition,
?a! ß >b œ \ a!bX a>b œ ! .
At the other boundary,
?B aP ß >b # ?aP ß >b œ c\ w aPb # \ aPbdX a>b œ ! .
Since these conditions are valid for all > ! , it follows that
\ a!b œ ! and \ w aPb # \ aPb œ ! .
________________________________________________________________________
page 680
—————————————————————————— CHAPTER 10. ——
a,bÞ We consider the boundary value problem
a‡b
\ ww -\ œ ! , ! B P à
\ a!b œ ! , \ w aPb # \ aPb œ ! .
Assume that - is real, with - œ .# . The general solution of the ODE is
\ aBb œ -" -9=2a.Bb -# =382a.Bb .
The first boundary condition requires that -" œ !Þ Imposing the second boundary
condition,
-# .-9=2a.Pb # -# =382a.Pb œ ! .
If -# Á ! , then .-9=2a.Pb # =382a.Pb œ ! , which can also be written as
a. # b/.P a. # b/.P œ ! .
If # œ . , then it follows that -9=2a.Pb œ =382a.Pb, and hence . œ !. If # Á .,
then /.P œ /.P again implies that . œ ! . For the case . œ ! , the general solution is
\ aBb œ EB F . Imposing the boundary conditions, we have F œ ! and
E # EP œ ! .
If # œ "ÎP , then \ aBb œ EB is a solution of a‡b. Otherwise E œ ! .
a- b. Let - œ .# , with . ! . The general solution of a‡b is
\ aBb œ -" -9=a.Bb -# =38a.Bb .
The first boundary condition requires that -" œ !Þ From the second boundary condition,
-# .-9=a.Pb # -# =38a.Pb œ ! .
For a nontrivial solution, we must have
.-9=a.Pb # =38a.Pb œ ! .
a. b. The last equation can also be written as
>+8 .P œ
.
Þ
#
The eigenvalues - obtained from the solutions of a‡‡b, which are infinite in number.
In the graph below, we assume #P œ " .
a‡‡b
________________________________________________________________________
page 681
—————————————————————————— CHAPTER 10. ——
For #P œ " À
Denote the nonzero solutions of a‡‡b by ." , .# , .$ ,â .
a/b. We can in principle calculate the eigenvalues -8 œ .#8 . Hence the associated
eigenfunctions are \8 œ =38 .8 B . Furthermore, the solutions of the temporal equations
are X8 œ /B:a !# .#8 >bÞ The fundamental solutions of the heat conduction problem
are given as
?8 aB ß >b œ /! .8 > =38 .8 B ,
# #
which lead to the general solution
? aB ß >b œ " -8 /! .8 > =38 .8 B .
_
# #
8œ"
________________________________________________________________________
page 682
—————————————————————————— CHAPTER 10. ——
Section 10.7
2a+b. The initial velocity is zero. Therefore the solution, as given by Eq. a#!b, is
?aB ß >b œ " -8 =38
_
8œ"
81 B
81 + >
-9=
,
P
P
in which the coefficients are the Fourier sine coefficients of 0 aBb. That is,
# P
81B
-8 œ ( 0 aBb=38
.B
P !
P
PÎ%
$PÎ%
P
#
%B
81 B
81 B
%P %B
81 B
œ –(
=38
.B (
=38
.B (
=38
.B—
P !
P
P
P
P
P
PÎ%
$PÎ%
œ)
=38 81Î% =38 $81Î%
Þ
8# 1 #
Therefore the displacement of the string is given by
) _
81
$81
81B
81+ >
?aB ß >b œ # " ”=38
=38
-9=
.
•=38
1 8œ"
%
%
P
P
a,b. With + œ " and P œ "! ,
) _
81
$81
81B
81>
?aB ß >b œ # " ”=38
=38
-9=
.
•=38
1 8œ"
%
%
"!
"!
________________________________________________________________________
page 683
—————————————————————————— CHAPTER 10. ——
a- b .
________________________________________________________________________
page 684
—————————————————————————— CHAPTER 10. ——
a. b .
3a+b. The initial velocity is zero. As given by Eq. a#!b, the solution is
?aB ß >b œ " -8 =38
_
8œ"
81 B
81 + >
,
-9=
P
P
in which the coefficients are the Fourier sine coefficients of 0 aBb. That is,
________________________________________________________________________
page 685
—————————————————————————— CHAPTER 10. ——
# P
81B
-8 œ ( 0 aBb=38
.B
P !
P
# P )BaP Bb#
81 B
œ (
=38
.B
$
P !
P
P
# -9= 81
œ $#
Þ
8$ 1 $
Therefore the displacement of the string is given by
$# _ # -9= 81
81 B
81 + >
?aB ß >b œ $ "
=38
-9=
.
$
1 8œ"
8
P
P
a,b. With + œ " and P œ "! ,
$# _ # -9= 81
81 B
81 >
?aB ß >b œ $ "
=38
-9=
.
$
1 8œ"
8
"!
"!
________________________________________________________________________
page 686
—————————————————————————— CHAPTER 10. ——
a- b .
________________________________________________________________________
page 687
—————————————————————————— CHAPTER 10. ——
a. b .
4a+b. As given by Eq. a#!b, the solution is
?aB ß >b œ " -8 =38
_
8œ"
81 B
81 + >
-9=
,
P
P
in which the coefficients are the Fourier sine coefficients of 0 aBb. That is,
________________________________________________________________________
page 688
—————————————————————————— CHAPTER 10. ——
# P
81B
-8 œ ( 0 aBb=38
.B
P !
P
# PÎ#"
81B
œ (
=38
.B
P PÎ#"
P
=38 8#1 =38 8P1
œ%
Þ
81
Therefore the displacement of the string is given by
% _ "
81
81
81B
81+ >
?aB ß >b œ " ’=38
=38 “=38
-9=
.
1 8œ"8
#
P
P
P
a,b. With + œ " and P œ "! ,
% _ "
81
81
81B
81>
?aB ß >b œ " ’=38
=38 “=38
-9=
.
1 8œ"8
#
"!
"!
"!
________________________________________________________________________
page 689
—————————————————————————— CHAPTER 10. ——
a- b .
________________________________________________________________________
page 690
—————————————————————————— CHAPTER 10. ——
a. b .
5a+b. The initial displacement is zero. Therefore the solution, as given by Eq. a$%b, is
?aB ß >b œ " 5 8 =38
_
8œ"
81 B
81 + >
=38
,
P
P
in which the coefficients are the Fourier sine coefficients of ?> aB ß !b œ 1aBb. It follows
that
________________________________________________________________________
page 691
—————————————————————————— CHAPTER 10. ——
P
#
81 B
58 œ
.B
( 1aBb=38
81 + !
P
œ
PÎ#
P
#
#B
81 B
#aP Bb
81 B
=38
.B
=38
.B—
(
(
–
81 + !
P
P
P
P
PÎ#
œ )P
=38 81Î#
Þ
8$ 1 $ +
Therefore the displacement of the string is given by
?aB ß >b œ
)P _ "
81
81 B
81 + >
"
=38
=38
=38
.
+ 1 $ 8 œ " 8$
#
P
P
a,b. With + œ " and P œ "! ,
)! _ "
81
81B
81>
?aB ß >b œ $ " $ =38
=38
=38
.
1 8œ"8
#
"!
"!
________________________________________________________________________
page 692
—————————————————————————— CHAPTER 10. ——
a- b .
________________________________________________________________________
page 693
—————————————————————————— CHAPTER 10. ——
a. b .
7a+b. The initial displacement is zero. As given by Eq. a$%b, the solution is
?aB ß >b œ " 5 8 =38
_
8œ"
81B
81 + >
=38
,
P
P
in which the coefficients are the Fourier sine coefficients of ?> aB ß !b œ 1aBb. It follows
________________________________________________________________________
page 694
—————————————————————————— CHAPTER 10. ——
that
P
#
81B
.B
( 1aBb=38
81 + !
P
P
#
)BaP Bb#
81 B
œ
=38
.B
(
$
81 + !
P
P
# -9= 81
œ $#P
Þ
8% 1% +
58 œ
Therefore the displacement of the string is given by
?aB ß >b œ
$#P _ # -9= 81
81 B
81 + >
"
=38
=38
.
%
%
+1 8 œ "
8
P
P
a,b. With + œ " and P œ "! ,
?aB ß >b œ
$#! _ # -9= 81
81 B
81 >
"
=38
=38
.
1% 8 œ "
8%
"!
"!
________________________________________________________________________
page 695
—————————————————————————— CHAPTER 10. ——
a- b .
________________________________________________________________________
page 696
—————————————————————————— CHAPTER 10. ——
a. b .
8a+b. As given by Eq. a$%b, the solution is
?aB ß >b œ " 5 8 =38
_
8œ"
81 B
81 + >
=38
,
P
P
in which the coefficients are the Fourier sine coefficients of ?> aB ß !b œ 1aBb. It follows
that
________________________________________________________________________
page 697
—————————————————————————— CHAPTER 10. ——
P
#
81B
58 œ
.B
( 1aBb=38
81+ !
P
PÎ#"
#
81B
œ
=38
.B
(
81+ PÎ#"
P
=38 8#1 =38 8P1
œ %P
Þ
8# 1 # +
Therefore the displacement of the string is given by
%P _ "
81
81
81 B
81 + >
" # ’=38
?aB ß >b œ
=38 “=38
=38
.
#
+1 8 œ " 8
#
P
P
P
a,b. With + œ " and P œ "! ,
%! _ "
81
81
81B
81>
?aB ß >b œ # " # ’=38
=38 “=38
=38
.
1 8œ"8
#
"!
"!
"!
________________________________________________________________________
page 698
—————————————————————————— CHAPTER 10. ——
a- b .
________________________________________________________________________
page 699
—————————————————————————— CHAPTER 10. ——
a. b .
11a+b. As shown in Prob. * , the solution is
?aB ß >b œ " -8 =38
_
8œ"
a#8 "b1B
a#8 "b1+ >
-9=
,
#P
#P
in which the coefficients are the Fourier sine coefficients of 0 aBb. It follows
that
________________________________________________________________________
page 700
—————————————————————————— CHAPTER 10. ——
a#8 "b1B
# P
-8 œ ( 0 aBb=38
.B
P !
#P
a#8 "b1B
# P )BaP Bb#
œ (
=38
.B
$
P !
P
#P
$-9= 81 a#8 "b1
œ &"#
Þ
a#8 "b% 1%
Therefore the displacement of the string is given by
?aB ß >b œ
&"# _ $-9= 81 a#8 "b1
a#8 "b1B
a#8 "b1+ >
"
=38
-9=
.
%
%
1 8œ"
#P
#P
a#8 "b
Note that the period is X œ %PÎ+ .
a,b. With + œ " and P œ "! ,
&"# _ $-9= 81 a#8 "b1
a#8 "b1B
a#8 "b1>
?aB ß >b œ % "
=38
-9=
.
%
1 8œ"
#!
#!
a#8 "b
________________________________________________________________________
page 701
—————————————————————————— CHAPTER 10. ——
a- b .
________________________________________________________________________
page 702
—————————————————————————— CHAPTER 10. ——
a. b .
12. The wave equation is given by
+#
` #?
` #?
œ
.
`B#
`>#
Setting = œ BÎP , we have
`?
`? .=
" `?
œ
œ
.
`B
`= .B
P `=
It follows that
________________________________________________________________________
page 703
—————————————————————————— CHAPTER 10. ——
` #?
" ` #?
.
œ
`B#
P# `=#
Likewise, with 7 œ +>ÎP ,
`?
+ `?
` #?
+# ` # ?
œ
and
œ
.
`>
P `7
`>#
P# ` 7 #
Substitution into the original equation results in
` #?
` #?
œ
.
`=#
`7 #
15. The given specifications are P œ & ft , X œ &! lb , and weight per unit length
# œ !Þ!#' lb/ft . It follows that 3 œ # Î$#Þ# œ )!Þ(& ‚ "!& slugs/ft .
a+b. The transverse waves propagate with a speed of + œ ÈX Î3 œ #%) ft/sec .
a,b. The natural frequencies are =8 œ 81+ÎP œ %*Þ) 18 rad/sec Þ
a- b. The new wave speed is + œ ÈaX ?X bÎ3 . For a string with fixed ends, the
natural modes are proportional to the functions
81B
Q8 aBb œ =38
,
P
which are independent of + .
19. The solution of the wave equation
+# @BB œ @>>
in an infinite one-dimensional medium subject to the initial conditions
@aB ß !b œ 0 aBb , @> aB ß !b œ ! , _ B _
is given by
@ aB ß >b œ
"
c0 aB +>b 0 aB +>bd.
#
The solution of the wave equation
+# ABB œ A>> ,
on the same domain, subject to the initial conditions
AaB ß !b œ ! , A> aB ß !b œ 1aBb , _ B _
is given by
________________________________________________________________________
page 704
—————————————————————————— CHAPTER 10. ——
A aB ß >b œ
" B+>
1 a 0 b. 0 .
(
#+ B+>
Let ?aB ß >b œ @aB ß >b AaB ß >b . Since the PDE is linear, it is easy to see that ?aB ß >b
is a solution of the wave equation +# ?BB œ ?>> . Furthermore, we have
?aB ß !b œ @aB ß !b AaB ß !b œ 0 aBb
and
?> aB ß !b œ @> aB ß !b A> aB ß !b œ 1aBb .
Hence ?aB ß >b is a solution of the general wave propagation problem.
20. The solution of the specified wave propagation problem is
?aB ß >b œ " -8 =38
_
8œ"
81 B
81 + >
-9=
.
P
P
Using a standard trigonometric identity,
=38
81 B
81 + >
"
81 B 81 + >
81 B 81 + >
-9=
œ ”=38Œ
=38Œ
•
P
P
#
P
P
P
P
"
81
81
œ ’=38 aB +>b =38 aB +>b“Þ
#
P
P
We can therefore also write the solution as
" _
81
81
?aB ß >b œ " -8 ’=38 aB +>b =38 aB +>b“ .
# 8œ"
P
P
Assuming that the series can be split up,
_
" _
81
81
?aB ß >b œ – " -8 =38 aB +>b " -8 =38 aB +>b— .
# 8œ"
P
P
8œ"
Comparing the solution to the one given by Eq. a#)b, we can infer that
2aBb œ " -8 =38
_
8œ"
81B
.
P
21. Let 2a0b be a #P-periodic function defined by
2 a0 b œ œ
0 a0 b , ! Ÿ 0 Ÿ P à
0 a 0b , P Ÿ 0 Ÿ ! .
Set ?aB ß >b œ "# c2aB +>b 2 aB +>bd. Assuming the appropriate differentiability
________________________________________________________________________
page 705
—————————————————————————— CHAPTER 10. ——
conditions on 2 ,
`?
"
œ c2 w aB +>b 2 w aB +>bd
`B
#
and
` #?
" ww
œ
c2 aB +>b 2 ww aB +>bd.
#
`B
#
Likewise,
` #?
+# ww
œ
c2 aB +>b 2 ww aB +>bd.
#
`>
#
It follows immediately that
+#
Let >
` #?
` #?
œ
.
`B#
`>#
! . Checking the first boundary condition,
?a! ß >b œ
"
"
c2a +>b 2 a+>bd œ c 2 a+>b 2 a+>bd œ ! .
#
#
Checking the other boundary condition,
?aP ß >b œ
"
c2aP +>b 2 aP +>bd
#
"
œ c 2a+> Pb 2 a+> PbdÞ
#
Since 2 is #P-periodic, 2a+> Pb œ 2 a+> P #Pb. Therefore ?aP ß >b œ ! .
Furthermore, for ! Ÿ B Ÿ P ,
?aB ß !b œ
"
c2aBb 2aBbd œ 2aBb œ 0 aBbÞ
#
Hence ?aB ß >b is a solution of the problem.
23. Assuming that we can differentiate term-by-term,
_
`?
-8 8
81 B
81 + >
œ 1+ "
=38
=38
`>
P
P
P
8œ"
and
_
`?
-8 8
81 B
81 + >
œ 1"
-9=
-9=
Þ
`B
P
P
P
8œ"
Formally,
________________________________________________________________________
page 706
—————————————————————————— CHAPTER 10. ——
_
#
`? #
81 B
81 + >
# # " -8 8
=38#
Š
‹ =38#
Œ œ 1 +
`>
P
P
P
8œ"
1# +# " J87 aBß >b
_
8Á7
and
_
#
`?
81 B
81 + >
# " -8 8
œ
1
-9=#
Š
‹ -9=#
Œ
`B
P
P
P
8œ"
#
1# " K87 aBß >b ,
_
8Á7
in which J87 aBß >b and K87 aBß >b contain products of the natural modes and their
derivatives. Based on the orthogonality of the natural modes,
_
#
`?
81 + >
# # P " -8 8
.B
œ
1
+
Š
‹ =38#
( Œ
`>
# 8œ" P
P
!
P
#
and
_
#
`?
81 + >
# P " -8 8
Þ
Š
‹ -9=#
( Œ .B œ 1
`B
# 8œ" P
P
!
P
#
Recall that +# œ X Î3 . It follows that
( –3 Œ
P
!
_
#
`?
`?
81 + >
# X P " -8 8
X
.B
œ
1
Š
‹ =38#
Œ —
`>
`B
# 8œ" P
P
#
#
1#
X P _ -8 8 #
81 + >
"Š
.
‹ -9=#
# 8œ" P
P
Therefore,
_
"
`?
"
`?
# X " # #
3
X
.B
œ
1
8- .
( – Œ
Œ —
#
`>
#
`B
%P 8 œ " 8
!
P
#
#
________________________________________________________________________
page 707
—————————————————————————— CHAPTER 11. ——
Chapter Eleven
Section 11.1
1. Since the right hand sides of the ODE and the boundary conditions are all zero, the
boundary value problem is homogeneous.
3. The right hand side of the ODE is nonzero. Therefore the boundary value problem is
nonhomogeneous.
6. The ODE can also be written as
C ww -ˆ" B# ‰C œ ! .
Although the second boundary condition has a more general form, the boundary value
problem is homogeneous.
7. First assume that - œ ! . The general solution of the ODE is CaBb œ -" B -# . The
boundary condition at B œ ! requires that -# œ ! . Imposing the second condition,
- " a 1 "b - # œ ! .
It follows that -" œ -# œ ! . Hence there are no nontrivial solutions.
Suppose that - œ .# . In this case, the general solution of the ODE is
CaBb œ -" -9=2 .B -# =382 .B .
The first boundary condition requires that -" œ ! . Imposing the second condition,
-" a-9=2 .1 .=382 .1b -# a=382 .1 .-9=2 .1b œ ! .
The two boundary conditions result in
-# a>+82 .1 .b œ ! .
Since the only solution of the equation >+82 .1 . œ ! is . œ ! , we have -# œ ! .
Hence there are no nontrivial solutions.
Let - œ .# , with . ! . Then the general solution of the ODE is
CaBb œ -" -9= .B -# =38 .B .
Imposing the boundary conditions, we obtain -" œ ! and
-" a-9= .1 .=38 .1b -# a=38 .1 .-9= .1b œ ! .
For a nontrivial solution of the ODE, we require that =38 .1 .-9= .1 œ ! . Note that
-9= .1 œ ! Ê =38 .1 œ ! ,
which is false. It follows that >+8 .1 œ . . From a plot of 1 >+8 1. and 1. ,
________________________________________________________________________
page 720
—————————————————————————— CHAPTER 11. ——
we find that there is a sequence of solutions, ." ¸ !Þ()(' , .# ¸ "Þ'("' , â ; For large
values of 8,
1
1 .8 ¸ a#8 "b .
#
Therefore the eigenfunctions are 98 aBb œ =38 .8 B , with corresponding eigenvalues
-" ¸ !Þ'#!% , -# ¸ #Þ(*%$ , â Þ
Asymptotically,
a#8 "b#
-8 ¸
.
%
8. With - œ ! , the general solution of the ODE is CaBb œ -" B -# . Imposing the two
boundary conditions, -" œ ! and #-" -# œ ! . It follows that -" œ -# œ ! . Hence
there are no nontrivial solutions.
Setting - œ .# , the general solution of the ODE is
CaBb œ -" -9=2 .B -# =382 .B .
The first boundary condition requires that -# œ ! . Imposing the second condition,
-" a-9=2 . .=382 .b -# a=382 . .-9=2 .b œ ! .
The two boundary conditions result in
-" a" . >+82 .b œ ! .
Since . >+82 .
! , it follows that -" œ ! , and there are no nontrivial solutions.
Let - œ .# , with . ! . Then the general solution of the ODE is
CaBb œ -" -9= .B -# =38 .B .
Imposing the boundary conditions, we obtain -# œ ! and
________________________________________________________________________
page 721
—————————————————————————— CHAPTER 11. ——
-" a-9= . .=38 .b -# a=38 . .-9= .b œ ! .
For a nontrivial solution of the ODE, we require that -9= . .=38 . œ ! . First note
that
-9= . œ ! Ê . œ ! or =38 . œ ! .
Therefore we find that " . >+8 . œ ! . From a plot of . >+8 . , there is a sequence of
solutions, ." ¸ !Þ)'!$ , .# ¸ $Þ%#&' , â ; For large 8,
. 8 ¸ a8 "b 1 .
Therefore the eigenfunctions are 98 aBb œ -9= .8 B , with corresponding eigenvalues
-" ¸ !Þ(%!# , -# ¸ ""Þ($%* , â Þ
Asymptotically,
- 8 ¸ a8 "b # 1 # .
12. First note that T aBb œ " , UaBb œ #B and V aBb œ - . Based on Prob. "" , the
integrating factor is a solution of the ODE
. w aBb œ #B .aBb .
The differential equation is first order linear, with solution .aBb œ - /B:a B# b. It then
follows that the Hermite equation can be written as
’/B C w “ - /B C œ ! .
#
w
#
14. For the Laguerre equation, T aBb œ B , UaBb œ " B and V aBb œ - . Using the
result of Prob. "", the integrating factor is a solution of the ODE
B . w aBb œ B .aBb .
The general solution of . w aBb œ .aBb is .aBb œ - /B . Therefore the Laguerre
equation can be written as
________________________________________________________________________
page 722
—————————————————————————— CHAPTER 11. ——
cB /B C w d - /B C œ ! .
w
15. For the Chebyshev equation, T aBb œ " B# , UaBb œ B and V aBb œ !# . The
integrating factor is a solution of the ODE
ˆ" B# ‰. w aBb œ B .aBb .
The differential equation is separable, with
..
B
œ
.
.
" B#
The general solution of the resulting ODE is
.aBb œ
.
Èk" B# k
Recall that the Chebyshev equation is typically defined for kBk Ÿ " . Therefore it can also
be written as
’È" B# C w “
w
!#
È" B#
C œ !.
16. We consider solutions of the form ?aB ß >b œ \ aBbX a>b . Substitution into the PDE
results in
\X ww - \X w 5 \X œ !# \ ww X .
Dividing both sides of the equation by \X , we obtain
\X ww
\X w
\ ww X
,
5 œ !#
\X
\X
\X
that is,
ww
X ww
Xw
#\
œ!
5.
X
X
\
Since both sides of the resulting equation are functions of different variables, each must
be
equal to a constant, say - . Therefore we obtain two ordinary differential equations
!# \ ww a- 5 b\ œ !
and X ww -X w -X œ ! .
17a+b. Setting C œ =aBb? , we have C w œ = w ? = ?w and C ww œ = ww ? #= w ? w = ?ww .
Substitution into the given ODE results in
= ww ? #= w ? w = ?ww #a= w ? = ?w b a" -b=? œ ! .
________________________________________________________________________
page 723
—————————————————————————— CHAPTER 11. ——
Collecting the various terms,
= ?ww a#= w #=b? w c= ww #= w a" -b=d? œ ! .
The second term on the left vanishes as long as = w œ = .
a,b. With =aBb œ /B , the transformed differential equation can be written as
?ww -? œ ! .
Since the boundary conditions are homogeneous, we also have ?a!b œ ?a"b œ ! . It now
follows that the eigenfunctions are ?8 œ =38 È-8 B , with corresponding eigenvalues
- 8 œ 8# 1 # .
Therefore the eigenfunctions for the original problem are 98 aBb œ /B =38 81B , with
corresponding eigenvalues
" -8 œ " 8# 1 # .
a- b. The given equation is a second order constant coefficient differential equation. The
characteristic equation is
with roots <"ß# œ " „ È - .
<# #< a" -b œ ! ,
If - œ ! , then the general solution is C œ -" /B -# B/B . Imposing the two boundary
conditions, we find that -" œ -# œ ! , and hence there are no nontrivial solutions. If
- ! , then the general solution is
C œ -" /B:Š" È - ‹B -# /B:Š" È - ‹B .
It again follows that -" œ -# œ ! , and hence there are no nontrivial solutions.
Therefore - ! , and the general solution is
C œ -" /B -9= È- B -# /B =38 È- B .
Invoking the boundary conditions, we have -" œ ! and -# /=38È- œ ! . For a
nontrivial
solution, È- œ 81 .
19. First write the differential equation as
C ww a" -bC w -C œ ! ,
which is a second order constant coefficient differential equation. The characteristic
equation is
________________________________________________________________________
page 724
—————————————————————————— CHAPTER 11. ——
< # a" - b< - œ ! ,
with roots <" œ " and <# œ - . For - Á " , the general solution is
C œ -" /B -# /-B .
Imposing the boundary conditions, we require that -" -# œ ! and -" /" -# /- œ ! .
For a nontrivial solution, it follows that /" œ /- , and hence - œ " , which is contrary
to the assumption.
If - œ " , then the general solution is
C œ -" /B -# B/B .
The boundary conditions require that -" œ ! and -" /" -# /" œ ! . Hence there are
no nontrivial solutions.
21. Suppose that - œ ! . In that case the general solution is C œ -" B -# . The
boundary
conditions require that -" #-# œ ! and -" -# œ ! . We find that -" œ -# œ ! , and
hence there are no nontrivial solutions.
a+b. Let - œ .# , with . ! . Then the general solution of the ODE is
CaBb œ -" -9= .B -# =38 .B .
The boundary conditions require that
#-" . -# œ ! and -" -9= . -# =38 . œ ! .
These equations have a nonzero solution only if
#=38 . . -9= . œ ! ,
which can also be written as
#>+8 . . œ ! .
Based on the graph, the positive roots of the determinantal equation are
________________________________________________________________________
page 725
—————————————————————————— CHAPTER 11. ——
1
." ¸ %Þ#(%) , .# ¸ (Þ&*'& ,â ; for large 8, .8 ¸ a#8 "b .
#
Therefore the eigenvalues are
-" ¸ ")Þ#($) , -# ¸ &(Þ(!(& ,â ; for large 8, -8 ¸ a#8 "b#
a,b. Setting - œ .# ! , the general solution of the ODE is
1#
.
%
CaBb œ -" -9=2 .B -# =382 .B .
Imposing the boundary conditions, we obtain the equations
#-" . -# œ ! and -" -9=2 . -# =382 . œ ! .
These equations have a nonzero solution only if
#=382 . . -9=2 . œ ! .
The latter equation is satisfied only for . œ ! and . œ „ "Þ*"&! . Hence the only
negative eigenvalue is -" œ $Þ''($ .
24. Based on the physical problem, - œ 7=# ÎIM ! . Let - œ .% . The characteristic
equation is <% .% œ ! , with roots <"ß# œ „ .3 , <$ œ . and <% œ . . Hence the
general solution is
CaBb œ -" -9=2 .B -# =382 .B -$ -9= .B -% =38 .B .
a+b. Simply supported on both ends À Ca!b œ C ww a!b œ ! à C aPb œ C ww aPb œ ! Þ
Invoking the boundary conditions, we obtain the system of equations
-" - $ œ !
-" - $ œ !
-" -9=2 .P -# =382 .P -$ -9= .P -% =38 .P œ !
#
-" . -9=2 .P -# .# =382 .P -$ .# -9= .P -% .# =38 .P œ ! .
The determinantal equation is
.% =382 .P =38 .P œ ! .
The nonzero roots are .8 œ 81ÎP , 8 œ "ß #ß â . The first two equations result in
-" œ -$ œ ! . The last two equations,
-# =382 81 -% =38 81 œ !
-# =382 81 -% =38 81 œ ! ,
imply that -# œ ! . Therefore the eigenfunctions are 98 œ =38 .8 B , with corresponding
eigenvalues -8 œ 8% 1% ÎP% .
________________________________________________________________________
page 726
—————————————————————————— CHAPTER 11. ——
a,b. Simply supported À Ca!b œ C ww a!b œ ! à clamped À C aPb œ C w aPb œ ! Þ
Invoking the boundary conditions, we obtain the system of equations
-" - $ œ !
-" - $ œ !
-" -9=2 .P -# =382 .P -$ -9= .P -% =38 .P œ !
-" .=382 .P -# .-9=2 .P -$ .=38 .P -% .-9= .P œ ! .
The determinantal equation is
#.$ =382 .P -9= .P #.$ -9=2 .P =38 .P œ ! .
Based on numerical analysis, ." ¸ $Þ*#''ÎP and .# ¸ (Þ!')'ÎP .
The first two equations result in -" œ -$ œ ! . The last two equations,
-# =382 .8 P -% =38 .8 P œ !
-# -9=2 .8 P -% -9= .8 P œ ! ,
imply that
-# œ
=38 .8 P
-% .
=382 .8 P
Therefore the eigenfunctions are
98 œ
=38 .8 P
=382 .8 B =38 .8 B ,
=382 .8 P
with corresponding eigenvalues -8 œ .%8 .
a- b. Clamped À Ca!b œ C w a!b œ ! à free À C ww aPb œ C www aPb œ ! Þ
Invoking the boundary conditions, we obtain the system of equations
________________________________________________________________________
page 727
—————————————————————————— CHAPTER 11. ——
-" - $ œ !
. -# . - % œ !
#
#
#
-" . -9=2 .P -# . =382 .P -$ . -9= .P -% .# =38 .P œ !
-" .$ =382 .P -# .$ -9=2 .P -$ .$ =38 .P -% .$ -9= .P œ ! .
The determinantal equation is
" -9=2 .P -9= .P œ ! .
The first two nonzero roots are ." ¸ "Þ)(&"ÎP and .# ¸ %Þ'*%"ÎP . With -$ œ -"
and -% œ -# , the system of equations reduce to
-" a-9=2 .8 P -9= .8 Pb -# a=382 .8 P =38 .8 Pb œ !
-" a=382 .8 P =38 .8 Pb -# a-9=2 .8 P -9= .8 Pb œ ! .
Let E8 œ a-9=2 .8 P -9= .8 PbÎa=382 .8 P =38 .8 Pb . The eigenfunctions are
given by
98 aBb œ -9=2 .8 B -9= .8 B E8 a=38 .8 B =382 .8 Bb ,
with corresponding eigenvalues -8 œ .%8 .
25a+b. Assume that the solution has the form ?aB ß >b œ \ aBbX a>b . Substitution into
the
PDE results in
I ww
\ X œ \X ww .
3
Dividing both sides of the equation by \X , we obtain
I \ ww X
\X ww
œ
,
3 \X
\X
that is,
________________________________________________________________________
page 728
—————————————————————————— CHAPTER 11. ——
\ ww
3 X ww
œ
.
\
I X
Since both sides of the resulting equation are functions of different variables, each must
be
equal to a constant, say - . Therefore we obtain two ordinary differential equations
\ ww -\ œ !
and X ww -
I
X œ !.
3
a,b. Given that ?a! ß >b œ \ a!bX a>b for > ! , it follows that \ a!b œ ! . The second
boundary condition can be expressed as
IE\ w aPbX a>b 7\ aPbX ww a>b œ ! , > ! .
From the result in Part a+b,
IE\ w aPbX a>b -7
I
\ aPbX a>b œ ! , > ! .
3
Since the condition is to be satisfied for all > ! , we arrive at the boundary condition
7
\ w aPb \ aPb œ ! .
3E
a- b. If - œ ! , the general solution of the spatial equation is
\ aBb œ -" B -# .
The boundary condition require that -" œ -# œ !. Hence there are no nontrivial
solutions.
If - œ .# ! , then the general solution is
\ aBb œ -" -9=2 .B -# =382 .B .
The first boundary condition implies that -" œ !. The second boundary condition
requires
that
7
-# -9=2 .P -# .
=382 .P œ ! .
3E
The solution is nontrivial only if
3E
.
7
! , there are no nontrivial solutions.
. >+82 .P œ
Since . >+82 .P
Let - œ .# ! . The general solution of the spatial equation is
\ aBb œ -" -9= .B -# =38 .B .
________________________________________________________________________
page 729
—————————————————————————— CHAPTER 11. ——
The first boundary condition implies that -" œ !. The second boundary condition
requires
that
7
-# -9= .P -# .
=38 .P œ ! .
3E
For a nontrivial solution, it is necessary that
7
-9= .P .
=38 .P œ ! ,
3E
or
>+8 .P œ
For the case a7Î3EPb œ !Þ& ,
3E
.
7.
we find that ." P ¸ "Þ!('* and .# P ¸ $Þ'%$' . Therefore the eigenfunctions are given
by 98 aBb œ =38 .8 B . The corresponding eigenvalues are solutions of
-9= È-8 P
PÈ
-8 =38 È-8 P œ ! .
#
The first two eigenvalues are approximated as -" ¸ "Þ"&*(ÎP# and -# ¸ "$Þ#('ÎP# .
________________________________________________________________________
page 730
—————————————————————————— CHAPTER 11. ——
Section 11.2
2. Based on the boundary conditions, - ! . The general solution of the ODE is
CaBb œ -" -9=È-B -# =38È-B .
The boundary condition C w a!b œ ! requires that -# œ ! . Imposing the second boundary
condition, we find that -" -9=È- œ ! . So for a nontrivial solution, È- œ a#8 "b1Î# ,
8 œ "ß #ß â Þ Therefore the eigenfunctions are given by
98 aBb œ 58 -9=
a#8 "b1B
.
#
In this problem, <aBb œ " , and the normalization condition is
58# (
!
"
a#8 "b1B
”-9=
• .B œ " .
#
#
It follows that 58# œ # . Therefore the normalized eigenfunctions are
98 aBb œ È# -9=
a#8 "b1B
, 8 œ "ß #ß â Þ
#
3. Based on the boundary conditions, - !. For - œ !, the eigenfunction is
9! aBb œ 5 ! Þ
Set 5! œ " . With - ! , the general solution of the ODE is
CaBb œ -" -9=È-B -# =38È-B .
Invoking the boundary conditions, we require that -# œ ! and -" È- =38È- œ ! .
Since
- ! , the eigenvalues are -8 œ 8# 1# , 8 œ "ß #ß â , with corresponding eigenfunctions
98 aBb œ 58 -9= 81B .
The normalization condition is
58# ( -9=# 81B .B œ " .
"
!
It follows that 58# œ # . Therefore the normalized eigenfunctions are
9! aBb œ " , and 98 aBb œ È# -9= 81B , 8 œ "ß #ß â Þ
4. From Prob. ) in Section ""Þ" , the eigenfunctions are 98 aBb œ 58 -9= È-8 B , in
which
-9=È-8 È-8 =38È-8 œ ! . The normalization condition is
________________________________________________________________________
page 731
—————————————————————————— CHAPTER 11. ——
58# (
!
"
-9=# È-8 B .B œ " .
First note that
-9= È-8 =38 È-8 È-8
.
#È-8
#
( -9= È-8 B .B œ
"
!
Based on the determinantal equation,
-9= È-8 =38 È-8 È-8
" =38# È-8
œ
#
#È-8
$ -9= #È-8
œ
.
%
Therefore
58# œ
%
$ -9= #È-8
and the normalized eigenfunctions are given by
98 aBb œ
# -9= È-8 B
É$ -9= #È-8
.
6. As shown in Prob. ", the normalized eigenfunctions are
98 aBb œ È# =38
a#8 "b1B
, 8 œ "ß #ß âÞ
#
Based on Eq. a$%b, with <aBb œ " , the coefficients in the eigenfunction expansion are
given by
-7 œ ( 0 aBb97 aBb.B
"
!
œ È#( =38
"
#È#
!
œ
a#7 "b1B
.B
#
a#7 "b1
.
Therefore we obtain the formal expansion
"œ
#È# _
"
a#8 "b1B
"
=38
.
1 8 œ " #8 "
#
________________________________________________________________________
page 732
—————————————————————————— CHAPTER 11. ——
8. We consider the normalized eigenfunctions
98 aBb œ È# =38
a#8 "b1B
, 8 œ "ß #ß âÞ
#
Based on Eq. a$%b, with <aBb œ " , the coefficients in the eigenfunction expansion are
given by
-7 œ ( 0 aBb97 aBb.B
"
!
œ È #(
œ
"Î#
=38
a#7 "b1B
.B
#
#È#
a#7 "b1
”" -9=
•.
a#7 "b1
%
!
Therefore we obtain the formal expansion
0 aBb œ
#È# _
a#8 "b1
a#8 "b1B
" ”" -9=
.
•=38
1 8œ"
%
#
9. The normalized eigenfunctions are
98 aBb œ È# =38
a#8 "b1B
, 8 œ "ß #ß âÞ
#
Based on Eq. a$%b, with <aBb œ " , the coefficients in the eigenfunction expansion are
given by
-7 œ ( 0 aBb97 aBb.B
"
"
a#7 "b1B
a#7 "b1B
.B È#( =38
.B
#
#
!
"Î#
)
71
71
œ
# # ’=38 # -9= # “ .
a#7 "b 1
!
œ È #(
"Î#
#B =38
Therefore the formal expansion of the given function is
0 aBb œ
) _ =38 8#1 -9= 8#1
a#8 "b1B
"
=38
.
#
1# 8 œ " a#8 "b
#
11. From Prob. %, the normalized eigenfunctions are given by
98 aBb œ
# -9= È-8 B
É$ -9= #È-8
,
________________________________________________________________________
page 733
—————————————————————————— CHAPTER 11. ——
in which the eigenvalues satisfy -9=È-8 È-8 =38È-8 œ !. Based on Eq. a$%b, the
coefficients in the eigenfunction expansion are given by
-7 œ ( 0 aBb97 aBb.B
"
!
œ
œ
#
É$ -9= #È-7
( B -9= È-7 B.B
"
!
È# ˆ# -9= È-7 "‰
in which !7 œ É" =38# È-7 Þ
-7 !7
,
12. The normalized eigenfunctions are given by
98 aBb œ
# -9= È-8 B
É$ -9= #È-8
,
in which the eigenvalues satisfy -9=È-8 È-8 =38È-8 œ !. Based on Eq. a$%b, the
coefficients in the eigenfunction expansion are given by
-7 œ ( 0 aBb97 aBb.B
"
!
œ
œ
#
É$ -9= #È-7
( a" Bb-9= È-7 B.B
"
!
È# ˆ" -9= È-7 ‰
-7 !7
,
in which !7 œ É" =38# È-7 Þ
13. We consider the normalized eigenfunctions
98 aBb œ
# -9= È-8 B
É$ -9= #È-8
,
in which the eigenvalues satisfy -9=È-8 È-8 =38È-8 œ !. The coefficients in the
eigenfunction expansion are given by
________________________________________________________________________
page 734
—————————————————————————— CHAPTER 11. ——
-8 œ ( 0 aBb98 aBb.B
"
!
œ
œ
#
É$ -9= #È-8
(
È# =38ˆÈ-8 Î#‰
,
È -8 !8
"Î#
!
-9= È-8 B.B
in which !8 œ É" =38# È-8 Þ
15. The differential equation can be written as
ˆ" B# ‰C w ‘ w C œ ! ,
with :aBb œ " B# and ; aBb œ " . The boundary conditions are homogeneous and
separated. Hence the BVP is self-adjoint.
16. Since the boundary conditions are not separated, the inner product is computed:
Given ? and @, sufficiently smooth and satisfying the boundary conditions,
aPc?dß @b œ ( c? ww @ ?@d.B
"
!
œ ? w @ ¹ ( c? w @ w ?@d.B
"
"
!
!
œ c? w @ ?@ w d ¹ a?ß Pc@dbÞ
"
!
Based on the given boundary conditions,
? w a"b@a"b ? w a!b@a!b œ ?a!b@a"b #?a"b@a!b
?a"b@ w a"b ?a!b@ w a!b œ ?a"b@a!b #?a!b@a"b Þ
Since
c? w @ ?@ w d ¹ œ ?a"b@a!b ?a!b@a"b ,
"
!
the BVP is not self-adjoint.
18. The differential equation can be written as
cC w d œ - C ,
w
with :aBb œ " , ; aBb œ ! , and <aBb œ " . The boundary conditions are homogeneous
and separated. Hence the BVP is self-adjoint.
________________________________________________________________________
page 735
—————————————————————————— CHAPTER 11. ——
19. If +# œ ! , then
? w a"b@a"b ?a"b@ w a"b œ
and since ?a!b œ @a!b œ ! ,
,# w
,#
? a"b@ w a"b ? w a"b@ w a"b œ ! ,
,"
,"
? w a!b@a!b ?a!b@ w a!b œ ! .
If ,# œ ! , then ?a"b œ @a"b œ ! implies that
? w a"b@a"b ?a"b@ w a"b œ ! .
Furthermore,
? w a!b@a!b ?a!b@ w a!b œ
+# w
+#
? a!b@ w a!b ? w a!b@ w a!b œ ! .
+"
+"
Clearly, the results are also true if +# œ ,# œ ! .
20. Suppose that 9" aBb and 9# aBb are linearly independent eigenfunctions associated
with an eigenvalue - . The Wronskian is given by
[ a9" , 9# baBb œ 9" aBb9#w aBb 9# aBb9"w aBb.
Each of the eigenfunctions satisfies the boundary condition +" Ca!b +# C w a!b œ ! . If
either +" œ ! or +# œ ! , then clearly [ a9" , 9# ba!b œ ! . On the other hand, if +# is
not equal to zero, then
[ a9" , 9# ba!b œ 9" a!b9#w a!b 9# a!b9"w a!b
+"
+"
œ 9" a!b9# a!b 9# a!b9" a!b
+#
+#
œ !.
By Theorem $Þ$Þ# , [ a9" , 9# baBb œ ! for all ! Ÿ B Ÿ " . Based on Theorem $Þ$Þ$ ,
9" aBb and 9# aBb must be linearly dependent. Hence - must be a simple eigenvalue.
22. We consider the operator
PcCd œ c:aBbC w d ; aBbC
w
on the interval ! B " , together with the boundary conditions
+" Ca!b +# C w a!b œ ! , ," C a"b ,# C w a"b œ ! .
Let ? œ 9 3< and @ œ 0 3( . If ? and @ both satisfy the boundary conditions, then
the real and imaginary parts also satisfy the same boundary conditions. Using the inner
product
a? ß @b œ ( ?aBb@aBb.B ,
"
!
________________________________________________________________________
page 736
—————————————————————————— CHAPTER 11. ——
w
aPc?d ß @b œ ( c:aBb? w d @ ; aBb?@‘.B
"
!
w
œ ( ˜ c:aBba9 w 3< w bd @ ; aBb?@™.B
"
!
œ :aBba9 w 3< w b@¹ ( e:aBba9 w 3< w b @ w ; aBb?@f.B Þ
"
"
!
!
Integrating by parts, again,
w
w
w
w
w w
( e:aBba9 3< b@ f.B œ a9 3<b:aBb@ ¹ ( ˜c:aBb@ d ?™.B .
"
"
"
!
!
!
Collecting the boundary terms,
:aBbca9 w 3< w b@ a9 3<b@ w d¹ œ :aBbca9 w 3< w ba0 3(b a9 3<ba0 w 3( w bd¹
"
"
!
!
.
The real part is given by
:aBbca9 w 0 < w (b a90 w <( w bd¹ œ :aBbca9 w 0 90 w b a< w ( <( w bd¹
"
!
"
!
œ :aBbc9 0 90 d¹ :aBbc< ( <( w d¹
w
w
"
!
w
"
!
Þ
Since 9 , < , 0 and ( satisfy the boundary conditions, it follows that
:aBbca9 w 0 < w (b a90 w <( w bd¹ œ ! .
"
!
Similarly, the imaginary part also vanishes. That is,
:aBbca< w 0 <0 w b a9 w ( 9( w bd¹ œ ! .
"
!
Therefore
w
aPc?d ß @b œ ( ˜ c:aBb@ w d ? ; aBb?@™.B
"
œ a P c @ d ß ?b
œ a ? ß P c@ d b .
!
The result follows from the fact that a? ß Pc@db œ a? ß Pc@db .
24. Based on the physical problem, - œ T ÎIM ! . Let - œ .# . The characteristic
equation is <% .# <# œ ! , with roots <"ß# œ ! , <$ œ .3 and <% œ .3 . Hence the
general solution is
CaBb œ -" -# B -$ -9= .B -% =38 .B .
________________________________________________________________________
page 737
—————————————————————————— CHAPTER 11. ——
a+b. Simply supported on both ends À Ca!b œ C ww a!b œ ! à C aPb œ C ww aPb œ ! Þ
Invoking the boundary conditions, we obtain the system of equations
-" - $ œ !
-$ œ !
-$ -9= .P -% =38 .P œ !
-" -# P -$ -9= .P -% =38 .P œ ! .
The determinantal equation is
=38 .P œ ! .
The nonzero roots are .8 œ 81ÎP , 8 œ "ß #ß â . Therefore the eigenfunctions are
98 œ =38 .8 B , with corresponding eigenvalues -8 œ 8# 1# ÎP# . Hence the smallest
eigenvalue is -" œ 1# ÎP# .
a,b. Simply supported À Ca!b œ C ww a!b œ ! à clamped À C aPb œ C w aPb œ ! Þ
Invoking the boundary conditions, we obtain the system of equations
-" - $ œ !
-$ œ !
-# -$ .=38 .P -% .-9= .P œ !
-" -# P -$ -9= .P -% =38 .P œ ! .
The determinantal equation is
.P -9= .P =38 .P œ ! .
It follows that the eigenfunctions are given by
98 aBb œ =38È-8 B ŠÈ-8 -9=È-8 P‹B ,
and the eigenvalues satisfy the equation PÈ-8 -9= È-8 P =38 È-8 P œ ! .
The smallest eigenvalue is estimated as -" ¸ a%Þ%*$%b# ÎP# Þ
a- b. Clamped À Ca!b œ C w a!b œ ! à clamped À C aPb œ C w aPb œ ! Þ
Invoking the boundary conditions, we obtain the system of equations
-" - $ œ !
-# .- % œ !
-" -# P -$ -9= .P -% =38 .P œ !
-# -$ .=38 .P -% .-9= .P œ ! .
The determinantal equation is
# #-9= .P œ .P =38 .P Þ
It follows that the eigenfunctions are given by
________________________________________________________________________
page 738
—————————————————————————— CHAPTER 11. ——
98 aBb œ " -9=È-8 B ,
and the eigenvalues satisfy the equation # #-9= È-8 P œ È-8 P =38 È-8 P Þ
The smallest eigenvalue is -" œ a#1b# ÎP# Þ
26. As shown is Prob. #& , the general solution is
CaBb œ -" -# B -$ -9= .B -% =38 .B .
Imposing the boundary conditions, we obtain the system of equations
-# œ !
-" - $ œ !
-# .- % œ !
-$ -9= .P -% =38 .P œ ! .
For a nontrivial solution, it is necessary that
-9= .P œ ! .
We find that -# œ -% œ ! , and hence the eigenfunctions are given by
98 aBb œ " -9=È-8 B .
The corresponding eigenvalues are -8 œ a#8 "b# 1# Î%P# , 8 œ "ß #ß â . The smallest
eigenvalue is -" œ 1# Î%P# Þ
________________________________________________________________________
page 739
—————————————————————————— CHAPTER 11. ——
Section 11.3
4. The eigensystem of the associated homogeneous problem is given in Prob. "" of
Section ""Þ# . The normalized eigenfunctions are
98 aBb œ
È# -9= È-8 B
É" =38# È-8
,
in which the eigenvalues satisfy -9=È-8 È-8 =38È-8 œ !. Rewrite the given
differential equation as C ww œ #C B . Since . œ # Á -8 , the formal solution of
the nonhomogeneous problem is
CaBb œ "
-8
98 aBb,
- #
8œ" 8
_
in which
-8 œ ( 0 aBb98 aBb.B
"
!
œ
œ
È#
É" =38# È-8
( B -9= È-8 B .B
"
È#ˆ# -9= È-8 "‰
!
-8 É" =38# È-8
.
Therefore we obtain the formal expansion
CaBb œ # "
_
È#ˆ# -9= È-8 "‰ -9= È-8 B
8œ"
-8 a-8 #bˆ" =38# È-8 ‰
.
5. The solution follows that in Prob. " , except that the coefficients are given by
-8 œ ( 0 aBb98 aBb.B
"
!
œ È #(
œ%
"Î#
#B =38 81B .B È#(
È# =38a81Î#b
!
8# 1 #
Therefore the formal solution is
"
"Î#
a# #Bb =38 81B .B
.
=38a81Î#b =38 81B
.
8# 1# a8# 1# #b
8œ"
CaBb œ ) "
_
________________________________________________________________________
page 740
—————————————————————————— CHAPTER 11. ——
6. The differential equation can be written as C ww œ .C 0 aBb. Note that ; aBb œ !
and <aBb œ " . As shown in Prob. " in Section ""Þ# , the normalized eigenfunctions are
98 aBb œ È# =38
a#8 "bB
,
#
with associated eigenvalues -8 œ a#8 "b# 1# Î% . Based on Theorem ""Þ$Þ" , the
formal solution is given by
-8
a#8 "bB
=38
,
a-8 .b
#
8œ"
CaBb œ È# "
_
as long as . Á -8 Þ The coefficients in the series expansion are computed as
-8 œ È#( 0 aBb=38
"
!
a#8 "bB
.B .
#
7. As shown in Prob. " in Section ""Þ# , the normalized eigenfunctions are
98 aBb œ È# -9=
a#8 "bB
,
#
with associated eigenvalues -8 œ a#8 "b# 1# Î% . Based on Theorem ""Þ$Þ" , the
formal solution is given by
-8
a#8 "bB
-9=
,
a-8 .b
#
8œ"
CaBb œ È# "
_
as long as . Á -8 Þ The coefficients in the series expansion are computed as
a#8 "bB
-8 œ È#( 0 aBb-9=
.B .
#
!
"
9. The normalized eigenfunctions are
98 aBb œ
È# -9= È-8 B
É" =38# È-8
.
The eigenvalues satisfy -9=È-8 È-8 =38È-8 œ !. Based on Theorem ""Þ$Þ" , the
formal solution is given by
CaBb œ È# "
_
8 œ " a-8
-8 -9= È-8 B
.bÉ" =38# È-8
,
as long as . Á -8 Þ The coefficients in the series expansion are computed as
________________________________________________________________________
page 741
—————————————————————————— CHAPTER 11. ——
-8 œ
È#
É" =38# È-8
( 0 aBb-9= È-8 B .B .
"
!
13. The differential equation can be written as C ww œ 1# C -9= 1B + . Note that
. œ 1# and 0 aBb œ -9= 1B + . Furthermore, . œ 1# is an eigenvalue corresponding
to the eigenfunction 9" aBb œ È# =38 1B . A solution exists only if 0 aBb and 9" aBb are
orthogonal. Since
( a-9= 1B +b=38 1B .B œ #+Î1 ,
"
!
there exists a solution as long as + œ ! . In that case, the ODE is
C ww 1# C œ -9= 1B .
The complementary solution is C- aBb œ -" -9= 1B -# =38 1B . A particular solution is
] aBb œ EB -9= 1B FB =38 1B . Using the method of undetermined coefficients, we
find that E œ ! and F œ "Î#1 . Therefore the general solution is
B
CaBb œ -" -9= 1B -# =38 1B
=38 1B .
#1
The boundary conditions require that -" œ ! . Hence the solution of the boundary value
problem is
B
CaBb œ -# =38 1B
=38 1B .
#1
15. Let CaBb œ 9" aBb 9# aBb . It follows that PcCd œ Pc9" d Pc9# d œ 0 aBb. Also,
+" Ca!b +# C w a!b œ +" 9" a!b +" 9# a!b +# 9"w a!b +# 9#w a!b
œ +" 9" a!b +# 9"w a!b +" 9# a!b +# 9#w a!b
œ !.
Similarly, the boundary condition at B œ " is satisfied as well.
16. The complementary solution is C- aBb œ -" -9= 1B -# =38 1B . A particular solution
is ] aBb œ E FB . Using the method of undetermined coefficients, we find that E œ !
and F œ " . Therefore the general solution is
CaBb œ -" -9= 1B -# =38 1B B .
Imposing the boundary conditions, we find that -" œ " . Therefore the solution of the
BVP is
CaBb œ -9= 1B -# =38 1B B .
Now attempt to solve the problem as shown in Prob. "& . Let BVP-1 be given by
________________________________________________________________________
page 742
—————————————————————————— CHAPTER 11. ——
? ww 1# ? œ 1# B ,
?a!b œ ! , ?a"b œ ! Þ
The general solution of the ODE is
?aBb œ -" -9= 1B -# =38 1B B .
The boundary conditions require that -" œ ! and -" " œ ! . We find that BVP-1 has
no solution. Let BVP-2 be given by
@ ww 1# @ œ ! ,
@ a !b œ " , @ a " b œ ! Þ
The general solution of the ODE is @aBb œ -" -9= 1B -# =38 1B . Imposing the
boundary conditions, we obtain -" œ " and -" œ ! . Thus BVP-2 has no solution.
17. Setting CaBb œ ?aBb @aBb, substitution results in
? ww @ ww :aBbc? w @ w d ; aBbc? @d œ ? ww :aBb? w ; aBb?
@ ww :aBb@ w ; aBb@ Þ
Since the left hand side of the equation is zero,
? ww :aBb? w ; aBb? œ c@ ww :aBb@ w ; aBb@dÞ
Furthermore, ?a!b œ Ca!b @a!b œ ! and ?a"b œ Ca"b @a"b œ ! . The simplest
function having the assumed properties is @aBb œ a, +bB + . In this case,
1aBb œ a+ ,b:aBb a+ ,bB ; aBb + ; aBb Þ
20. The associated homogeneous PDE is ?> œ ?BB , ! B " , with
?B a! ß >b œ !, ?B a"ß >b ?a"ß >b œ ! and ?aB ß !b œ " B.
Applying the method of separation of variables, we obtain the eigenvalue problem
\ ww -\ œ ! , with boundary conditions \ w a!b œ ! and \ w a"b \ a"b œ !. It was
shown in Prob. % , in Section ""Þ# , that the normalized eigenfunctions are
98 aBb œ
where -9=È-8 È-8 =38È-8 œ !.
È# -9= È-8 B
É" =38# È-8
,
We assume a solution of the form
?aB ß >b œ ",8 a>b98 aBb .
_
8 œ"
________________________________________________________________________
page 743
—————————————————————————— CHAPTER 11. ——
Substitution into the given PDE results in
",8w a>b98 aBb œ ",8 a>b98ww aBb />
_
_
8 œ"
8 œ"
œ "-8 ,8 a>b98 aBb /> ,
_
8 œ"
that is,
"c,8w a>b -8 ,8 a>bd98 aBb œ /> Þ
_
8 œ"
We now note that
È# =38È-8
"œ"
_
8 œ" È-8 É"
=38# È-8
98 aBb .
Therefore
œ ""8 /> 98 aBb ,
_
/
>
8 œ"
in which "8 œ È# =38È-8 ΔÈ-8 É" =38# È-8 • . Combining these results,
",8w a>b -8 ,8 a>b "8 /> ‘98 aBb œ ! Þ
_
8 œ"
Since the resulting equation is valid for ! B " , it follows that
,8w a>b -8 ,8 a>b œ "8 /> , 8 œ "ß #ß â .
Prior to solving the sequence of ODEs, we establish the initial conditions. These are
obtained from the expansion
?aB ß !b œ " B œ "!8 98 aBb ,
_
8 œ"
in which !8 œ È# ˆ" -9=È-8 ‰Î”-8 É" =38# È-8 •Þ That is, ,8 a!b œ !8 .
Therefore the solutions of the first order ODEs are
,8 a>b œ
"8 ˆ/> /-8 > ‰
!8 /-8 > , 8 œ "ß #ß â .
a - 8 "b
Hence the solution of the boundary value problem is
________________________________________________________________________
page 744
—————————————————————————— CHAPTER 11. ——
?aB ß >b œ "–
_
8 œ"
"8 ˆ/> /-8 > ‰
!8 /-8 > —98 aBb Þ
a - 8 "b
21. Based on the boundary conditions, the normalized eigenfunctions are given by
98 aBb œ È# =38 81B ,
with associated eigenvalues -8 œ 8# 1# Þ We now assume a solution of the form
?aB ß >b œ ",8 a>b98 aBb .
_
8 œ"
Substitution into the given PDE results in
",8w a>b98 aBb œ ",8 a>b98ww aBb " k" #Bk
_
_
8 œ"
8 œ"
œ "-8 ,8 a>b98 aBb " k" #Bk,
_
8 œ"
that is,
"c,8w a>b -8 ,8 a>bd98 aBb œ " k" #BkÞ
_
8 œ"
It was shown in Prob. & that
" k" #Bk œ "%
_
8 œ"
È# =38a81Î#b
8# 1#
98 aBb .
Substituting on the right hand side and collecting terms, we obtain
"–,8w a>b -8 ,8 a>b %
_
8 œ"
È# =38a81Î#b
8# 1 #
—98 aBb œ !Þ
Since the resulting equation is valid for ! B " , it follows that
,8w a>b 8# 1# ,8 a>b œ %
È# =38a81Î#b
8# 1#
, 8 œ "ß #ß â .
Based on the given initial condition, we also have ,8 a!b œ ! , for 8 œ "ß #ß â . The
solutions of the first order ODEs are
,8 a>b œ %
È# =38a81Î#b
8% 1%
Š" /8 1 > ‹, 8 œ "ß #ß â .
# #
Hence the solution of the boundary value problem is
________________________________________________________________________
page 745
—————————————————————————— CHAPTER 11. ——
) _ =38a81Î#b
# #
?aB ß >b œ % "
Š" /8 1 > ‹=38 81B Þ
%
1 8 œ"
8
23a+b. Let ?aB ß >b be a solution of the boundary value problem and @aBb be a solution
of the related BVP. Substituting for ?aB ß >b œ AaB ß >b @aBb, we have
<aBb?> œ <aBbA>
and
c:aBb?B dB ; aBb? J aBb œ c:aBbAB dB ; aBbA c:aBb@ w d ; aBb@ J aBb
œ c:aBbAB dB ; aBbA J aBb J aBb
œ c:aBbAB dB ; aBbA .
w
Hence AaB ß >b is a solution of the homogeneous PDE
<aBbA> œ c:aBbAB dB ; aBbA .
The required boundary conditions are
Aa! ß >b œ ?a! ß >b @a!b œ ! ,
Aa" ß >b œ ?a" ß >b @a"b œ ! .
The associated initial condition is AaB ß !b œ ?aB ß !b @aBb œ 0 aBb @aBb.
a,b. Let @aBb be a solution of the ODE
c:aBb@ w d ; aBb@ œ J aBb ,
w
and satisfying the boundary conditions @ w a!b 2" @a!b œ X" , @ w a"b 2# @a"b œ X# .
If AaB ß >b œ ?aB ß >b @aBb, then it is easy to show the A satisfies the PDE and initial
condition given in Part a+b. Furthermore,
AB a! ß >b 2" Aa! ß >b œ ?B a! ß >b @ w a!b 2" ?a! ß >b 2" @a!b
œ ?B a! ß >b 2" ?a! ß >b @ w a!b 2" @a!b
œ !.
Similarly, the other boundary condition is also homogeneous.
________________________________________________________________________
page 746
—————————————————————————— CHAPTER 11. ——
25. In this problem, J aBb œ 1# -9= 1B . First find a solution of the boundary value
problem
@ ww œ 1# -9= 1B , @ w a!b œ ! , @a"b œ " .
The general solution is @aBb œ EB F -9= 1B . Imposing the initial conditions, the
solution of the related BVP is @aBb œ -9= 1B . Now let AaB ß >b œ ?aB ß >b -9= 1B .
It follows that AaB ß >b satisfies the homogeneous boundary value problem, and the initial
condition AaB ß !b œ -9=a$1BÎ#b -9= 1B a -9= 1Bb œ -9=a$1BÎ#b .
We now seek solutions of the homogeneous problem of the form
AaB ß >b œ ",8 a>b98 aBb ,
_
8 œ"
in which 98 aBb œ È# -9= a#8 "b1BÎ# are the normalized eigenfunctions of the
homogeneous problem and -8 œ a#8 "b# 1# Î% , with 8 œ "ß #ß â . Substitution into
the PDE for A, we have
",8w a>b98 aBb œ ",8 a>b98ww aBb
_
_
8 œ"
8 œ"
œ "-8 ,8 a>b98 aBb .
_
8 œ"
Since the latter equation is valid for ! B " , it follows that
,8w a>b -8 ,8 a>b œ ! , 8 œ "ß #ß â ,
with ,8 a>b œ ,8 a!b/B:a -8 >b. Hence
AaB ß >b œ ",8 a!b/B:a -8 >b98 aBb .
_
8 œ"
Imposing the initial condition, we require that
È# ",8 a!b-9= a#8 "b1B œ -9= $1B .
#
#
8 œ"
_
It is evident that all of the coefficients are zero, except for ,# a!b œ "ÎÈ# . Therefore
AaB ß >b œ /B:ˆ *1# >Î%‰-9=
$1B
,
#
and the solution of the original BVP is
?aB ß >b œ /B:ˆ *1# >Î%‰-9=
$1B
-9= 1B .
#
________________________________________________________________________
page 747
—————————————————————————— CHAPTER 11. ——
26a+b. Let ?aB ß >b œ \ aBbX a>b. Substituting into the homogeneous form of a3b,
<aBb\X ww œ c:aBb\ w d X ; aBb\X .
w
Now divide both sides of the resulting equation by \X to obtain
X ww
c:aBb\ w d w
; aBb
œ
œ -.
X
<aBb\
<aBb
It follows that
c:aBb\ w d ; aBb\ œ -<aBb\ .
w
Since the boundary conditions a33b are valid for all > ! , we also have
\ w a!b 2" \ a!b œ ! , \ w a"b 2# \ a"b œ ! .
a,b. Let -8 and 98 aBb denote the eigenvalues and eigenfunctions of the BVP in Part a+b.
Assume a solution, of the PDE a3b, of the form
?aB ß >b œ ",8 a>b98 aBb Þ
_
8 œ"
Substituting into a3b,
w
<aBb ",8ww a>b98 œ ",8 a>b˜c:aBb98w d ; aBb98 ™ J aB ß >b
_
_
8 œ"
8 œ"
_
œ ",8 a>bc -8 <aBb98 d J aB ß >b .
8 œ"
Rearranging the terms,
<aBb "c,8ww a>b -8 ,8 a>bd98 œ J aB ß >b ,
_
8 œ"
or
"c,8ww a>b -8 ,8 a>bd98 œ
_
8 œ"
J a B ß >b
.
<aBb
Now expand the right hand side in terms of the eigenfunctions. That is, write
_
J a B ß >b
œ "#8 a>b98 aBb ,
<aBb
8 œ"
in which
________________________________________________________________________
page 748
—————————————————————————— CHAPTER 11. ——
#8 a>b œ ( <aBb
"
!
J a B ß >b
98 aBb.B
<aBb
œ ( J aB ß >b98 aBb.B , 8 œ "ß #ß â .
"
!
Combining these results, we have
"c,8ww a>b -8 ,8 a>b #8 a>bd98 œ ! .
_
It follows that
8 œ"
,8ww a>b -8 ,8 a>b œ #8 a>b , 8 œ "ß #ß â .
In order to solve this sequence of ODEs, we require initial conditions ,8 a!b and ,8w a!b Þ
Note that
?aB ß !b œ ",8 a!b98 aBb and ?> aB ß !b œ ",8w a!b98 aBb .
_
_
8 œ"
8 œ"
Based on the given initial conditions,
0 aBb œ ",8 a!b98 aBb and 1aBb œ ",8w a!b98 aBb .
_
_
8 œ"
8 œ"
Hence ,8 a!b œ !8 and ,8w a!b œ "8 , the expansion coefficients for 0 aBb and 1aBb in
terms of the eigenfunctions, 98 aBb .
27a+b. Since the eigenvectors are orthogonal, they form a basis. Given any vector b ,
b œ ",3 0 a3b Þ
8
3œ"
Taking the inner product, with 0 a4b , of both sides of the equation, we have
ˆb ß 0 a4b ‰ œ ,4 ˆ0 a4b ß 0 a4b ‰Þ
a,b. Consider solutions of the form
x œ "+3 0 a3b Þ
8
3œ"
Substituting into Eq. a3b, and using the above form of b ,
"+3 A0 a3b ". +3 0 a3b œ ",3 0 a3b Þ
8
8
8
3œ"
3œ"
3œ"
It follows that
________________________________________________________________________
page 749
—————————————————————————— CHAPTER 11. ——
"c+3 -3 . +3 ,3 d0 a3b œ 0 Þ
8
3œ"
Since the eigenvectors are linearly independent,
+3 -3 . +3 ,3 œ ! , for 3 œ "ß #ß âß 8 .
That is,
+3 œ ,3 Îa-3 .b , 3 œ "ß #ß âß 8 .
Assuming that the eigenvectors are normalized, the solution is given by
ab ß 0 a3b b a3b
0 ,
.
3
3œ"
xœ"
8
as long as . is not equal to one of the eigenvalues.
29. First write the ODE as C ww C œ 0 aBb . A fundamental set of solutions of the
homogeneous equation is given by
C" œ -9= B and C# œ =38 B .
The Wronskian is equal to [ c-9= B ß =38 Bd œ " . Applying the method of variation of
parameters, a particular solution is
] aBb œ C" aBb?" aBb C# aBb?# aBb ,
in which
?" aBb œ ( =38a=b0 a=b.= and ?# aBb œ ( -9=a=b0 a=b.= .
B
B
!
!
Therefore the general solution is
C œ 9aBb œ -" -9= B -# =38 B -9= B( =38a=b0 a=b.= =38 B( -9=a=b0 a=b.=.
B
B
!
!
Imposing the boundary conditions, we must have -" œ ! and
-# =38 " -9= "( =38a=b0 a=b.= =38 "( -9=a=b0 a=b.= œ ! .
"
!
"
!
It follows that
-# œ
"
"
( =38a" =b0 a=b.= ,
=38 " !
and
________________________________________________________________________
page 750
—————————————————————————— CHAPTER 11. ——
B
=38 B "
9aBb œ
( =38a" =b0 a=b.= ( =38aB =b0 a=b.=.
=38 " !
!
Using standard identities,
=38 B † =38a" =b =38 " † =38aB =b œ =38 = † =38a" Bb .
Therefore
=38 B † =38a" =b
=38 = † =38a" Bb
=38aB =b œ
.
=38 "
=38 "
Splitting up the first integral, we obtain
9aBb œ (
B
!
"
=38 = † =38a" Bb
=38 B † =38a" =b
0 a=b.= (
0 a=b.=
=38 "
=38 "
B
œ ( KaB ß =b0 a=b.= ,
"
!
in which
K aB ß =b œ
=38 =†=38a"Bb
=38 "
=38 B†=38a"=b
=38 "
ß
!Ÿ=ŸB
ß
B Ÿ = Ÿ "Þ
31. The general solution of the homogeneous problem is
C œ -" -# B .
By inspection, it is easy to see that C" aBb œ " satisfies the BC C w a!b œ ! and that the
function C# aBb œ " B satisfies the BC C a"b œ ! . The Wronskian of these solutions is
[ cC" ß C# d œ " . Based on Prob. $! , with :aBb œ " , the Green's function is given by
K aB ß =b œ œ
a" Bb ß
a " =b ß
!Ÿ=ŸB
B Ÿ = Ÿ "Þ
Therefore the solution of the given BVP is
9aBb œ ( a" Bb0 a=b.= ( a" =b0 a=b.= .
B
!
"
B
32. The general solution of the homogeneous problem is
C œ -" -# B .
We find that C" aBb œ B satisfies the BC Ca!b œ ! . Imposing the boundary condition
________________________________________________________________________
page 751
—————————————————————————— CHAPTER 11. ——
Ca"b C w a"b œ ! , we must have -" #-# œ ! . Hence choose C# aBb œ # B . The
Wronskian of these solutions is [ cC" ß C# d œ # . Based on Prob. $! , with :aBb œ " , the
Green's function is given by
K aB ß =b œ œ
=aB #bÎ# ß
Ba= #bÎ# ß
!Ÿ=ŸB
B Ÿ = Ÿ "Þ
Therefore the solution of the given BVP is
" B
" "
9aBb œ ( =aB #b0 a=b.= ( Ba= #b0 a=b.= .
# !
# B
34. The general solution of the homogeneous problem is
C œ -" -# B .
By inspection, it is easy to see that C" aBb œ B satisfies the BC Ca!b œ ! and that the
function C# aBb œ " satisfies the BC C w a"b œ ! . The Wronskian of these solutions is
[ cC" ß C# d œ " . Based on Prob. $! , with :aBb œ " , the Green's function is given by
K aB ß =b œ œ
=ß
Bß
!Ÿ=ŸB
B Ÿ = Ÿ "Þ
Therefore the solution of the given BVP is
9aBb œ ( =0 a=b.= ( B0 a=b.= .
B
"
!
B
35a+b. We proceed to show that if the expression given by Eq. a3@b is substituted into
the
integral of Eq. a333b, then the result is the solution of the nonhomogeneous problem. As
long as we can interchange the summation and integration,
C œ 9aBb œ ( KaB ß = ß .b0 a=b.=
"
93 aBb "
( 0 a=b93 a=b.= .
- . !
8œ" 3
!
_
œ"
Note that
( 0 a=b93 a=b.= œ -3 .
"
!
Therefore
-3 93 aBb
,
.
3
8œ"
C œ 9aBb œ "
_
________________________________________________________________________
page 752
—————————————————————————— CHAPTER 11. ——
as given by Eq. a"$b in the text. It is assumed that the eigenfunctions are normalized and
-3 Á . .
a,b. For any fixed value of B, KaB ß = ß .b is a function of = and the parameter . . With
appropriate assumptions on K , we can write the eigenfunction expansion
KaB ß = ß .b œ "+3 aB ß .b93 a=b .
_
3œ"
Since the eigenfunctions are orthonormal with respect to <aBb,
( KaB ß = ß .b<a=b93 a=b.= œ +3 aB ß .b .
"
!
Now let
C3 aBb œ ( KaB ß = ß .b<a=b93 a=b.= .
"
!
Based on the association 0 aBb œ <aBb93 aBb , it is evident that
PcC3 d œ . <aBbC3 aBb <aBb93 aBb .
In order to evaluate the left hand side, we consider the eigenfunction expansion
C3 aBb œ " ,35 95 aBb .
_
5œ"
It follows that
PcC3 d œ " ,35 Pc95 d
_
5œ"
_
œ " ,35 -5 <aBb95 aBb .
5œ"
Therefore
<aBb " ,35 -5 95 aBb œ . <aBb" ,35 95 aBb <aBb93 aBb ,
and since <aBb Á ! ,
_
_
5œ"
5œ"
" ,35 -5 95 aBb œ . " ,35 95 aBb 93 aBb .
_
_
5œ"
5œ"
Rearranging the terms, we find that
________________________________________________________________________
page 753
—————————————————————————— CHAPTER 11. ——
93 aBb œ " ,35 a-5 .b95 aBb .
_
5œ"
Since the eigenfunctions are linearly independent, ,35 a-5 .b œ $35 , and thus
C3 aBb œ "
"
$35
95 aBb œ
93 aBb .
- .
-3 .
5œ" 5
_
We conclude that
+ 3 aB ß . b œ
which verifies that
"
93 aBb ,
-3 .
93 aBb93 a=b
.
-3 .
3œ"
K aB ß = ß .b œ "
_
36. First note that . # CÎ.=# œ ! for = Á B . On the interval ! = B , the solution
of the ODE is C" a=b œ -" -# = . Given that C a!b œ ! , we have C" a=b œ -# = . On the
interval B = " , the solution is C# a=b œ ." .# = . Imposing the condition C a"b œ ! ,
we have C# a=b œ ." a" =b. Assuming continuity of the solution, at = œ B ,
-# B œ ." a" Bb ,
which gives -# œ ." a" BbÎB . Next, integrate both sides of the given ODE over an
infinitesimal interval containing = œ B À
(
B
B
It follows that
B
.# C
.=
œ
$ a= Bb.= œ " .
(
.=#
B
C w aB b C w aB b œ " ,
and hence -# a ." b œ " . Solving for the two coefficients, we obtain -# œ " B and
." œ B . Therefore the solution of the BVP is given by
C a =b œ œ
=a" Bb ß
Ba" =b ß
!Ÿ=ŸB
B Ÿ = Ÿ ",
which is identical to the Green's function in Prob. #).
________________________________________________________________________
page 754
—————————————————————————— CHAPTER 11. ——
Section 11.4
1. Let 98 aBb œ N! ˆÈ-8 B‰ be the eigenfunctions of the singular problem
aB C w b œ -BC , ! B " ,
C ß C w bounded as B p ! , C a"b œ ! .
w
Let 9aBb be a solution of the given BVP, and set
9aBb œ " ,8 98 aBb .
a‡b
_
8œ!
Then
aB 9 w b œ . B9 0 aBb
0 aBb
œ . B9 B
.
B
w
Substituting a‡b, we obtain
" ,8 -8 B 98 aBb œ .B " ,8 98 aBb B " -8 98 aBb ,
_
_
_
8œ!
8œ!
8œ!
in which the -8 are the expansion coefficients of 0 aBbÎB for B ! . That is,
"
"
0 aBb
98 aBb.B
B
(
#
m98 aBbm !
B
"
"
œ
( 0 aBb98 aBb.B .
m98 aBbm# !
-8 œ
It follows that if B Á ! ,
" c-8 ,8 a-8 .bd98 aBb œ ! .
_
8œ!
As long as . Á -8 , linear independence of the eigenfunctions implies that
-8
, 8 œ "ß #ß â .
,8 œ
-8 .
Therefore a formal solution is given by
-8
N! ŠÈ-8 B‹,
.
8œ! 8
9aBb œ "
_
in which È-8 are the positive roots of N! aBb œ ! .
________________________________________________________________________
page 755
—————————————————————————— CHAPTER 11. ——
3a+b. Setting > œ È- B , it follows that
.C
.C
.#C
.#C
œ Èand
œ
.
.B
.>
.B#
.>#
The given ODE can be expressed as
È-
.
> È .C
5 # È
œ È- > C ,
.> È.>
>
or
.
.C
5#
Œ>
œ >CÞ
.>
.>
>
An equivalent form is given by
>#
.C
.C
>
ˆ># 5 # ‰C œ ! ,
.>
.>
which is known as a Bessel equation of order 5 . A bounded solution is N5 a>b .
a,b. N5 ŠÈ- B‹ satisfies the boundary condition at B œ ! . Imposing the other
boundary
condition, it is necessary that N5 ŠÈ- ‹ œ ! . Therefore the eigenvalues are given by
-8 , 8 œ "ß #â , where È-8 are the positive zeroes of N5 aBb . The eigenfunctions of
the BVP are 98 aBb œ N5 ˆÈ-8 B‰Þ
a- b. The BVP is a singular Sturm-Liouville problem with
PcCd œ aB C w b
w
5#
C and <aBb œ " .
B
We note that
-8 ( B 98 aBb97 aBb.B œ ( Pc98 d 97 aBb.B
"
"
!
!
œ ( 98 aBbPc97 d.B
"
!
œ -7 ( B 98 aBb97 aBb.B Þ
"
!
Therefore
a-8 -7 b( B 98 aBb97 aBb.B œ ! Þ
"
!
________________________________________________________________________
page 756
—————————————————————————— CHAPTER 11. ——
So for 8 Á 7 , we have -8 Á -7 and
( B 98 aBb97 aBb.B œ ! Þ
"
!
a. b. Consider the expansion
0 aBb œ " +8 98 aBb Þ
_
8œ!
Multiplying both sides of equation by B 94 aBb and integrating from ! to " , and using the
orthogonality of the eigenfunction,
( B 0 aBb94 aBb.B œ " +8 ( B 94 aBb98 aBb.B
_
"
!
"
!
8œ!
œ +4 ( B 94 aBb94 aBb.B Þ
"
!
Therefore
+4 œ ( B 0 aBb94 aBb.BÎ( Bc94 aBbd# .B , 4 œ "ß #ß â Þ
"
"
!
!
a/b. Let 9aBb be a solution of the given BVP, and set
9aBb œ " ,8 98 aBb ,
a‡b
_
where 98 aBb œ N5 ˆÈ-8 B‰. Then
Substituting a‡b, we obtain
8œ!
Pc9d œ . B9 0 aBb
0 aBb
œ . B9 B
.
B
" ,8 -8 B 98 aBb œ .B " ,8 98 aBb B " -8 98 aBb ,
_
_
_
8œ!
8œ!
8œ!
in which the -8 are the expansion coefficients of 0 aBbÎB for B ! . That is,
"
"
0 aBb
B
98 aBb.B
(
#
m98 aBbm !
B
"
"
œ
( 0 aBbN5 ŠÈ-8 B‹.B .
mN5 ˆÈ-8 B‰m# !
-8 œ
________________________________________________________________________
page 757
—————————————————————————— CHAPTER 11. ——
It follows that if B Á ! ,
" c-8 ,8 a-8 .bdN5 ŠÈ-8 B‹ œ ! .
_
8œ!
As long as . Á -8 , linear independence of the eigenfunctions implies that
-8
, 8 œ "ß #ß â .
,8 œ
-8 .
Therefore a formal solution is given by
-8
N5 ŠÈ-8 B‹.
.
8œ! 8
9aBb œ "
_
5a+b. Setting - œ !# in Prob. "& of Section ""Þ" , the Chebyshev equation can also be
written as
’È" B# C w “ œ
w
Note that
C.
È" B#
:aBb œ È" B# , ; aBb œ ! , and <aBb œ "ÎÈ" B# ,
hence both boundary points are singular.
a,b. Observe that :a" &b œ È#& &# and :a " &b œ È#& &# Þ It follows
that if ?aBb and @aBb satisfy the boundary conditions a333b, then
lim :a" &bc? w a" &b@a" &b ?a" &b@ w a" &bd œ !
& Ä !
and
lim :a " &bc? w a " &b@a " &b ?a " &b@ w a " &bd œ ! .
& Ä !
Therefore Eq. a"(b is satisfied and the boundary value problem is self-adjoint.
a- b. For 8 Á ! ,
8# (
"
X! aBb X8 aBb
.B œ ( X! aBb PcX8 d.B
" È" B#
"
"
œ ( PcX! d X8 aBb.B
"
"
since PcX! d œ ! † X! œ ! . Otherwise,
œ !,
________________________________________________________________________
page 758
—————————————————————————— CHAPTER 11. ——
"
X8 aBb X7 aBb
8 (
.B œ ( PcX8 d X7 aBb.B
" È" B#
"
#
"
œ ( X8 aBbPcX7 d.B
"
"
œ 7# (
Therefore
X8 aBb X7 aBb
.B œ ! Þ
" È" B#
ˆ8# 7# ‰(
So for 8 Á 7 ,
X8 aBb X7 aBb
.B Þ
" È" B#
"
"
X8 aBb X7 aBb
.B œ ! Þ
" È" B#
(
"
________________________________________________________________________
page 759
—————————————————————————— CHAPTER 11. ——
Section 11.5
3. The equations relating to this problem are given by Eqs. a#b to a"(b in the text. Based
on the boundary conditions, the eigenfunctions are 98 aBb œ N! a-8 <b and the associated
eigenvalues -" ß -# ß â are the positive zeroes of N! a-b . The general solution has the
form
?a< ß >b œ " c-8 N! a-8 <b -9= -8 +> 58 N! a-8 <b =38 -8 +>d .
_
8œ"
The initial conditions require that
?a< ß !b œ " -8 N! a-8 <b œ 0 a<b
_
8œ"
and
?> a< ß !b œ " +-8 58 N! a-8 <b œ 1a<b .
_
8œ"
The coefficients -8 and 58 are obtained from the respective eigenfunction expansions.
That is,
-8 œ
"
"
<0 a<bN! a-8 <b.<
(
mN! a-8 <bm# !
and
"
"
58 œ
( <1a<bN! a-8 <b.< ,
+-8 mN! a-8 <bm# !
in which
mN! a-8 <bm# œ ( <cN! a-8 <bd# .<
"
!
for 8 œ "ß #ß â .
8. A more general equation was considered in Prob. #$ of Section "!Þ& . Assuming a
solution of the form ?a< ß >b œ V a<bX a>b, substitution into the PDE results in
!# ”V ww X
" w
V X • œ VX w .
<
Dividing both sides of the equation by the factor VX , we obtain
________________________________________________________________________
page 760
—————————————————————————— CHAPTER 11. ——
V ww
" Vw
Xw
œ # .
V
< V
! X
Since both sides of the resulting differential equation depend on different variables, each
side must be equal to a constant, say -# . That is,
V ww
" Vw
Xw
œ # œ -# .
V
< V
! X
It follows that X w !# -# X œ ! , and
V ww
" Vw
œ -# ,
V
< V
which can be written as <# V ww < V w -# <# V œ ! . Introducing the variable 0 œ -< ,
the last equation can be expressed as 0# V ww 0 V w 0# V œ ! , which is the Bessel
equation of order zero.
The temporal equation has solutions which are multiples of X a>b œ /B:a !# -# >b. The
general solution of the Bessel equation is
V a<b œ ," N! a-8 <b ,# ]! a-8 <b .
Since the steady state temperature will be zero, all solutions must be bounded, and hence
we set ,# œ ! . Furthermore, the boundary condition ?a" ß >b œ ! requires that V a"b œ !
and hence N! a-b œ ! . It follows that the eigenfunctions are 98 aBb œ N! a-8 <b , with the
associated eigenvalues -" ß -# ß â , which are the positive zeroes of N! a-b . Therefore
the fundamental solutions of the PDE are ?8 a< ß >b œ N! a-8 <b/B:a !# -8# >b, and the
general solution has the form
?a< ß >b œ " -8 N! a-8 <b/B:ˆ !# -8# >‰ .
_
8œ"
The initial condition requires that
?a< ß !b œ " -8 N! a-8 <b œ 0 a<b.
_
8œ"
The coefficients in the general solution are obtained from the eigenfunction expansion of
0 a<b. That is,
"
"
-8 œ
( <0 a<bN! a-8 <b.< ,
mN! a-8 <bm# !
in which
mN! a-8 <bm# œ ( <cN! a-8 <bd# .<
"
!
a8 œ "ß #ß â b.
________________________________________________________________________
page 761
—————————————————————————— CHAPTER 11. ——
Section 11.6
1. The sine expansion of 0 aBb œ " , on ! B " , is given by
0 aBb œ # "
_
" -9= 71
=38 71B ,
71
7œ"
with partial sums
W8 aBb œ # "
8
" -9= 71
=38 71B .
71
7œ"
The mean square error in this problem is
V8 œ ( k" W8 aBbk# .B .
"
!
Several values are shown in the Table :
8
V8
&
!Þ!'(
"!
!Þ!%
"&
!Þ!#'
#!
!Þ!#
Further numerical calculation shows that V8 !Þ!# for 8
#" .
3a+b. The sine expansion of 0 aBb œ Ba" Bb , on ! B " , is given by
0 aBb œ # "
_
" -9= 71
=38 71B ,
71
7œ"
with partial sums
W8 aBb œ % "
8
" -9= 71
=38 71B .
$ 1$
7
7œ"
a,ß - b. The mean square error in this problem is
V8 œ ( k Ba" Bb W8 aBbk# .B .
"
!
________________________________________________________________________
page 762
—————————————————————————— CHAPTER 11. ——
We find that V" œ !Þ!!!!%) . The graphs of 0 aBb and W" aBb are plotted below :
6a+b. The function is bounded on intervals not containing B œ ! , so for & ! ,
"Î#
.B œ # #È& Þ
( 0 aBb.B œ ( B
"
"
&
&
Hence the improper integral is evaluated as
"Î#
.B œ # Þ
( 0 aBb.B œ lim ( B
"
"
&Ä!
!
&
On the other hand, 0 # aBb œ "ÎB for B Á ! , and
"
( 0 aBb.B œ ( B .B œ 68È& Þ
"
&
"
#
&
Therefore the improper integral does not exist.
a,b. Since 0 # aBb ´ " , it is evident that the Riemann integral of 0 # aBb exists. Let
________________________________________________________________________
page 763
—————————————————————————— CHAPTER 11. ——
TR œ e! œ B" ß B# ß âß BR " œ "f
be a partition of c! ß "d into equal subintervals. We can always choose a rational point,
03 , in each of the subintervals so that the Riemann sum
V a0" ß 0# ß âß 0R b œ " 0 a08 b
R
8œ"
"
œ ".
R
Likewise, can always choose an irrational point, (3 , in each of the subintervals so that
the Riemann sum
V a(" ß (# ß âß (R b œ " 0 a(8 b
R
8œ"
It follows that 0 aBb is not Riemann integrable.
"
œ ".
R
8. With T! aBb œ " and T" aBb œ B, the normalization conditions are satisfied. Using
the usual inner product on c " ß "d,
( T! aBbT" aBb.B œ !
"
"
and hence the polynomials are also orthogonalÞ Let T# aBb œ +# B# +" B +! . The
normalization condition requires that +# +" +! œ " . For orthogonality, we need
#
#
( ˆ+# B +" B +! ‰.B œ ! and ( Bˆ+# B +" B +! ‰.B œ ! .
"
"
"
"
It follows that +# œ $Î# , +" œ ! and +! œ "Î# . Hence T# aBb œ a$B# "bÎ# .
Now let T$ aBb œ +$ B$ +# B# +" B +! . The coefficients must be chosen so that
+$ +# +" +! œ " and the orthogonality conditions
( T3 aBbT4 aBb.B œ ! a3 Á 4b
"
"
are satisfied. Solution of the resulting algebraic equations leads to +$ œ &Î# , +# œ ! ,
+1 œ $Î# and +! œ ! . Therefore T$ aBb œ a&B$ $BbÎ# .
11. The implied sequence of coefficients is +8 œ " , 8 " . Since the limit of these
coefficients is not zero, the series cannot be an eigenfunction expansion.
13. Consider the eigenfunction expansion
0 aBb œ "+3 93 aBbÞ
_
3œ"
Formally,
________________________________________________________________________
page 764
—————————————————————————— CHAPTER 11. ——
0 aBb œ "+3# 93# aBb # "+3 +4 93 aBb94 aBb Þ
_
#
3œ"
3Á4
Integrating term-by-term,
#
#
( <aBb0 aBb.B œ "( +3 <aBb93 aBb.B # "( +3 +4 <aBb93 aBb94 aBb.B
"
_
#
!
œ
"
"
3œ" !
_
"
"+3# ( 93# aBb.B ,
!
3œ"
3Á4
!
since the eigenfunctions are orthogonal. Assuming that they are also normalized,
#
( <aBb0 aBb.B œ "+3 .
"
!
_
#
3œ"
________________________________________________________________________
page 765
