BOYCE - Soluções Equações Diferenciais (8ª ed)

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Chapter One
Section 1.1
1.
For , the slopes are ! "#$ negative, and hence the solutions decrease. For , the % "#$
slopes are , and hence the solutions increase. The equilibrium solution appears topositive
be , to which all other solutions converge. & ' "#$ !
3.
For , the slopes are ! ( "#$ )*+,tive, and hence the solutions increase. For % ( "#$
, the slopes are , and hence the solutions decrease. All solutions appear tonegative
diverge away from the equilibrium solution . & ' ( "#$ !
5.
For , the slopes are ! ( "-. )*+,tive, and hence the solutions increase. For
 % ( "-. , the slopes are , and hence the solutions decrease. All solutionsnegative
diverge away from
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the equilibrium solution . & ' ( "-. !
6.
For , the slopes are ! ( . )*+,tive, and hence the solutions increase. For , % ( .
the slopes are , and hence the solutions decrease. All solutions diverge awaynegative
from
the equilibrium solution . & ' ( . !
8. For solutions to approach the equilibrium solution , we must haveall & ' .-/ !
 % 0 ! .-/ ! 0 % .-/1 1 for , and for . The required rates are satisfied by the
differential equation . ' . ( / 1
9. For solutions than to diverge from , must be an other increasing & ' . ' . & ! !
function for , and a function for . The simplest differential ! . % .decreasing
equation
whose solutions satisfy these criteria is . ' ( .1
10. For solutions than to diverge from , we must have other & ' "-/ ' "-/ % 0 ! 1
for , and for . The required rates are satisfied by the differential % "-/ ! 0 ! "-/1
equation . ' / ( "1
12.
Note that for and . The two equilibrium solutions are and ' 0 ' 0 ' $ & ' 01 !
 & ' $ ! 0 ! $ ! . Based on the direction field, for ; thus solutions with initial1
values than diverge from the solution . For , the slopes aregreater $ & ' $ 0 % % $ !
negative between, and hence solutions with initial values and all decrease toward the0 $
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solution . For , the slopes are all ; thus solutions with initial & ' 0 % 0 ! positive
values
less than approach the solution .0 & ' 0 !
14.
Observe that for and . The two equilibrium solutions are ' 0 ' 0 ' . & ' 01 !
and . Based on the direction field, for ; thus solutions with initial & ' . ! 0 ! . ! 1
values than diverge from . For , the slopes are alsogreater . & ' . 0 % % . !
positive between, and hence solutions with initial values and all increase toward the0 .
solution
 & ' . % 0 ! . For , the slopes are all ; thus solutions with initialnegative
values than diverge from the solution .less 0 & ' 0 !
16. Let be the total amount of the drug in the patient's body at ! ! !2 3 & in milligrams
any
given time . The drug is administered into the body at a rate of & 45+ $00 ! constant
67-45#
The rate at which the drug the bloodstream is given by Hence theleaves 0#83 & # !
accumulation rate of the drug is described by the differential equation
93
9&
' $00 ( 0#83 67-45 # !
 !:
Based on the direction field, the amount of drug in the bloodstream approaches the
equilibrium level of ".$067 ;,&4,< 2 =>;4*?5+ # !
18. Following the discussion in the text, the differential equation is !2
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6 ' 67 ( @
9@
9&
 .
or equivalently,
9@
9& 6
' 7 ( @ #
 .
 !: A 0 # After a long time, Hence the object attains a given by9@9& terminal velocity
@ ' #
67
B " 
 !C @ ' 67 ' 0#080D E7-+>C # Using the relation , the required is B. drag coefficient
 !9
19.
All solutions appear to approach a linear asymptote . It is easy to !;,&4 +F*)> >G?2F &* "
verify that is a solution. & ' & ( / !
20.
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All solutions approach the equilibrium solution & ' 0 # !
23.
All solutions appear to from the sinusoid ,diverge & ' ( +,<H& I J ( " ! /
. 8#
!
which is also a solution corresponding to the initial value . 0 ' ( $-. !
25.
All solutions appear to converge to . First, the rate of change is small. The & ' 0 !
slopes
eventually increase very rapidly in .magnitude
26.
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The direction field is rather complicated. Nevertheless, the collection of points at which
the slope field is , is given by the implicit equation The graph ofzero ( K ' .& #/ .
these points is shown below:
The of these curves are at , . It follows that for solutions withy-intercepts ' 0 L K#
initial values , all solutions increase without bound. For solutions with initial ! K#
values in the range , the slopes remain , and % ( K 0 % % K# #and negative
hence
these solutions decrease without bound. Solutions with initial conditions in the range
( K % % 0# initially increase. Once the solutions reach the critical value, given by
the equation , the slopes become negative and negative. These ( K ' .&/ . remain
solutions eventually decrease without bound.
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Section 1.2
1 The differential equation can be rewritten as !2
9 
$ ( 
' 9& #
Integrating both sides of this equation results in , or equivalently,( F< $ ( ' & I C$ $ "
$ ( ' C > 0 ' (& . Applying the initial condition results in the specification of ! 0
the constant as . Hence the solution is C ' $ ( & ' $ I ( $ > #0 0 ! ! (&
All solutions appear to converge to the equilibrium solution & ' $ # !
1 Rewrite the differential equation as !C #
9 
"0 ( . 
' 9& #
Integrating both sides of this equation results in , or( F< "0 ( . ' & I C". "$ $
equivalently,
$ ( ' C > 0 ' (.& . Applying the initial condition results in the specification of ! 0
the constant as . Hence the solution is C ' $ ( & ' $ I ( $ > #0 0 ! ! (.&
All solutions appear to converge to the equilibrium solution , but at a rate & ' $ ! faster
than in Problem 1a #
2 The differential equation can be rewritten as !2 #
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9 
 ( $
' 9& #
Integrating both sides of this equation results in , or equivalently,F< ( $ ' & I C$ $ "
 ( $ ' C > 0 ' & . Applying the initial condition results in the specification of ! 0
the constant as . Hence the solution is C ' ( $ & ' $ I ( $ > #0 0 ! ! &
All solutions appear to diverge from the equilibrium solution . & ' $ !
2 Rewrite the differential equation as !: #
9 
. ( $
' 9& #
Integrating both sides of this equation results in , or equivalently,". "F< . ( $ ' & I C$ $
. ( $ ' C > 0 ' .& . Applying the initial condition results in the specification of ! 0
the constant as . Hence the solution is C ' . ( $ & ' .#$ I ( .#$ > #0 0 ! ! .&
All solutions appear to diverge from the equilibrium solution . & ' .#$ !
2 . The differential equation can be rewritten as !C
9 
. ( "0
' 9& #
Integrating both sides of this equation results in , or equivalently,". "F< . ( "0 ' & I C$ $
 ( $ ' C > 0 ' .& . Applying the initial condition results in the specification of ! 0
the constant as . Hence the solution is C ' ( $ & ' $ I ( $ > #0 0 ! ! .&
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All solutionsappear to diverge from the equilibrium solution . & ' $ !
3 . Rewrite the differential equation as !2
9 
: ( 2 
' 9& ,
which is valid for . Integrating both sides results in , or M : -2 F< : ( 2 ' & I C( "
"
2 $ $
equivalently, . Hence the general solution is : ( 2 ' C > & ' : ( C > -2 #(2& (2& ! !
Note that if , then , and is an equilibrium solution. ' :-2 9 -9& ' 0 & ' :-2 !
 !:
 As increases, the equilibrium solution gets closer to , from above. ! !, 2 & ' 0
Furthermore, the convergence rate of all solutions, that is, , also increases.2
 As increases, then the equilibrium solution !,, : & ' :-2 ! also becomes larger. In
this case, the convergence rate remains the same.
 If and both increase , ! !,,, 2 : but constant:-2 ' then the equilibrium solution
 & ' :-2 ! remains the same, but the convergence rate of all solutions increases.
5 . Consider the simpler equation . As in the previous solutions, re- !2 9 -9& ' ( 2 " "
write the equation as
9 
 
' ( 2 9& #
"
"
Integrating both sides results in & ' C > #" ! (2& ! ! !: # & ' & I E Now set , and substitute into the original differential equation. We"
find that
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( 2 I 0 ' ( 2 I E I :" " ! .
That is, , and hence .( 2E I : ' 0 E ' :-2 ! !C & ' C > I :-2 #. The general solution of the differential equation is This is(2&
exactly the form given by Eq. in the text. Invoking an initial condition , ! !"N 0 ' 0
the solution may also be expressed as & ' :-2 I ( :-2 > # ! !0 (2&
6 . The general solution is , that is, . ! ! ! !2 ) & ' O00 I C > ) & ' O00 I ) ( O00 >& &-. -.0
With , the specific solution becomes . This solution is a) ' D$0 ) & ' O00 ( $0>0
-. ! &
decreasing exponential, and hence the time of extinction is equal to the number of
months
it takes, say , for the population to reach . Solving , we find that& O00 ( $0> ' 0=
-.zero &=
& ' . F< O00-$0 ' $#ND= ! .months ! ! !: ) & ' O00 I ) ( O00 > The solution, , is a exponential as long as0 -.& decreasing
) % O00 O00 I ) ( O00 > ' 00 0
-.. Hence has only root, given by ! &= one
& ' . F< #
O00
O00 ( )
=
0
% &
 ! !C :. The answer in part is a general equation relating time of extinction to the value
of
the initial population. Setting , the equation may be written as& ' ".= months
O00
O00 ( )
' >
0
K,
which has solution . Since is the initial population, the appropriate) ' DON#NKO" )0 0
answer is .) ' DOD0 mice
7 . The general solution is . Based on the discussion in the text, time is ! !2 ) & ' ) > &0 5&
measured in . Assuming , the hypothesis can be expressed asmonths month days" ' /0
) > ' .) 5 ' F< .0 0
"5P . Solving for the rate constant, , with units of ! per month .
 !: Q ' Q-/0. days months . The hypothesis is stated mathematically as ) > ' .)0 05N/30
.
It follows that , and hence the rate constant is given by 5Q-/0 ' F< . 5 ' /0 F< . -Q # ! !
The units are understood to be .per month
9 . Assuming , with the positive direction taken as , !2 no air resistance downward
Newton's
Second Law can be expressed as
6 ' 67
9@
9&
in which is the measured in appropriate units. The equation can7 gravitational constant
be
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written as , with solution The object is released with an9@-9& ' 7 @ & ' 7& I @ # ! 0
initial
velocity .@0
 !: 4. Suppose that the object is released from a height of above the ground. Usingunits
the
fact that , in which is the of the object, we obtain@ ' 9R-9& R downward displacement
the
differential equation for the displacement as With the origin placed at9R-9& ' 7& I @ #0
the point of release, direct integration results in . Based on theR & ' 7& -. I @ & ! . 0
chosen
coordinate system, the object reaches the ground when . Let be the timeR & ' 4 & ' S !
that it takes the object to reach the ground. Then . Using the7S -. I @ S ' 4. 0
quadratic
formula to solve for ,S
S ' #
( @ L @ I .74
7
0 0
#
The answer corresponds to the time it takes for the object to fall to the ground.positive
The
negative answer represents a previous instant at which the object could have been
launched
upward , only to ultimately fall downward with speed , !with the same impact speed @0
from a height of above the ground.4 units
 ! ! !C & ' S @ & 2 #. The impact speed is calculated by substituting into in part That is,
@ S ' @ I .74 ! # 0 .
10 , . The general solution of the differential equation is Given that ! !2 T & ' C > #b (5&
T 0 ' "00 C ' "00 ! , the value of the constant is given by . Hence the amount ofmg
thorium-234 present at any time is given by . Furthermore, based on theT & ' "00 > ! (5&
hypothesis, setting results in Solving for the rate constant, we& ' " D.#08 ' "00 > #(5
find that or .5 ' ( F< D.#08-"00 ' #"ONOK 5 ' #0.D.D ! /week /day
 !C S. Let be the time that it takes the isotope to decay to of its originalone-half
amount.
From part , it follows that , in which . Taking the !2 $0 ' "00 > 5 ' #"ONOK(5S /week
natural logarithm of both sides, we find that or S ' /#$0"8 S ' .8#$"weeks s .92 
11. The general solution of the differential equation is ,9T-9& ' ( 5T T & ' T > ! 0 (5&
in which is the initial amount of the substance. Let be the time that it takesT ' T 00 ! "
the substance to decay to of its original amount , . Setting in theone-half T & '0 "
solution,
we have . Taking the natural logarithm of both sides, it follows that0#$T ' T >0 0
(5"
( 5 ' F< 0#$ 5 ' F< . #" " ! or 
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12. The differential equation governing the amount of radium-226 is ,9T-9& ' ( 5T
with solution Using the result in Problem 11, and the fact that theT & ' T 0 > # ! ! (5&
half-life , the decay rate is given by . The" ' "K.0 5 ' F< . -"K.0years per year !
amount of radium-226, after years, is therefore Let be& T & ' T 0 > # S ! ! (0#0008.NDK&
the time that it takes the isotope to decay to of its original amount. Then setting/-8
& ' S ,
and , we obtain Solving for the decayT S ' T 0 T 0 ' T 0 > # ! ! ! !/ /8 8 (0#0008.NDKS
time, it follows that or ( 0#0008.NDK S ' F< /-8 S ' KN.#/K ! years .
13. The solution of the differential equation, with , isT 0 ' 0 !
T & ' UV " ( > # ! !( -UW&
As , the exponential term vanishes, and hence the limiting value is .&XB T ' UVY
14 . The rate of the chemical is . At any ! !2 H0#0"J /00accumulation grams per hour
given time , the of the chemical in the pond is & T & -"0concentration grams per gallon ! K
.
Consequently, the chemical the pond at a rate of .leaves grams per hour ! !/ Z "0 T &(8
Hence, the rate of change of the chemical is given by
9T
9&
' / ( 0#000/T & ! gm/hr .
Since the pond is initially free of the chemical, .T 0 ' 0 !
 !: . The differential equation can be rewritten as
9T
"0000 ( T
' 0#000/ 9& #
Integrating both sides of the equation results in .( F< "0000 ( T ' 0#000/& I U$ $
Taking
the natural logarithm of both sides gives . Since , the"0000 ( T ' C > T 0 ' 0(0#000/& !
value of the constant is . Hence the amount of chemical in the pond at anyC ' "0000
time
is . Note that . SettingT & ' "0000 " ( > " ! !(0#000/& grams year hours' DNK0
& ' DNK0 T DNK0 ' O.NN#NN, the amount of chemical present after is ,one year grams !
that is, .O#.NNNN kilograms
 !C . With the rate now equal to , the governing equation becomesaccumulation zero
9T-9& ' ( 0#000/T & ! Resetting the time variable, we now assign the newgm/hr .
initial value as .T 0 ' O.NN#NN ! grams
 ! ! !9 C T & ' O.NN#NN > #. The solution of the differential equation in Part is (0#000/&
Hence, one year the source is removed, the amount of chemical in thepond isafter
T DNK0 ' KN0#" ! .grams
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 !> &. Letting be the amount of time after the source is removed, we obtain the equation
"0 ' O.NN#NN > # ( 0#000/ & '(0#000/& Taking the natural logarithm of both sides, 
' F< "0-O.NN#NN & ' ..[ NNK ' .#K ! or .hours years
 !=
15 . It is assumed that dye is no longer entering the pool. In fact, the rate at which the !2
dye the pool is leaves kg/min gm per hour.00 P G & -K0000 ' .00 K0-"000 G & -K0' ( !' ( ! !
.
Hence the equation that governs the amount of dye in the pool is
9G
9&
' ( 0#. G !gm/hr .
The initial amount of dye in the pool is .G 0 ' $000 ! grams
 !: . The solution of the governing differential equation, with the specified initial value,
is G & ' $000 > # ! (0#. &
 !C & ' 8. The amount of dye in the pool after four hours is obtained by setting . That is,
G 8 ' $000 > ' ..8K#K8 K0[ 000 ! (0#D . Since size of the pool is , thegrams gallons
concentration grams/gallon of the dye is .0#0/N8
 !9 S. Let be the time that it takes to reduce the concentration level of the dye to
0#0. "[ .00 . At that time, the amount of dye in the pool is . Usinggrams/gallon grams
the answer in part , we have . Taking the natural logarithm of !: $000 > ' ".00(0#. S
both sides of the equation results in the required time .S ' N#"8 hours
 !> 0#. ' .00-"000. Note that . Consider the differential equation
9G 5
9& "000
' ( G .
Here the parameter corresponds to the , measured in .5 flow rate gallons per minute
Using the same initial value, the solution is given by In orderG & ' $000 > # ! (5 &-"000
to determine the appropriate flow rate, set and . (Recall that of& ' 8 G ' ".00 ".00 gm
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dye has a concentration of ). We obtain the equation 0#0. ".00 ' $000 > #gm/gal (5 -.$0
Taking the natural logarithm of both sides of the equation results in the required flow rate
5 ' /$N .gallons per minute
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Section 1.3
1. The differential equation is second order, since the highest derivative in the equation
is of order . The equation is , since the left hand side is a linear function of two linear 
and
its derivatives.
3. The differential equation is , since the highest derivative of the function fourth order 
is of order . The equation is also , since the terms containing the dependentfour linear
variable is linear in and its derivatives. 
4. The differential equation is , since the only derivative is of order . Thefirst order one
dependent variable is , hence the equation is .squared nonlinear
5. The differential equation is . Furthermore, the equation is ,second order nonlinear
since the dependent variable is an argument of the , which is a linear sine function not
function.
7. . Hence & ' > \ & ' & ' > ( ' 0 #" "" " " ! ! !& 1 11 & 11
Also, and . Thus & ' C*+4 & \ & ' +,<4 & & ' C*+4 & ( ' 0 #. " . . ! ! !1 11 11 .
9. . Substituting into the differential equation, we have & ' /& I & \ & ' / I .& ! !. 1
& / I .& ( /& I & ' /& I .& ( /& ( & ' & ! !. . . . . Hence the given function is a solution.
10. and Clearly, is & ' &-/ \ & ' "-/ & ' & ' & ' 0 # &" "" " " " ! ! ! ! ! !1 11 111 1111
a solution. Likewise, , , & ' > I &-/ \ & ' ( > I "-/ & ' >. . . ! ! !(& 1 (& 11 (&
 & ' ( > & ' >111 (& 1111 (&. . ! !, . Substituting into the left hand side of the equation, we
find that . Hence both> I 8 ( > I / > I &-/ ' > ( 8> I /> I & ' &(& (& (& (& (& (& ! !
functions are solutions of the differential equation.
11. and . Substituting into the left & ' & \ & ' & -. & ' ( & -8"
"-. ("-. (/-.
" " ! ! !1 11
hand side of the equation, we have
.& ( & -8 I /& & -. ( & ' ( & -. I / & -. ( &
' 0
.) * ) *(/-. ("-. "-. "-. "-. "-.
Likewise, and . Substituting into the left & ' & \ & ' ( & & ' . &.
(" (. (/
. . ! ! !1 11
hand side of the differential equation, we have .& . & I /& ( & ( & ' 8 & (. ! !(/ (. (" ("
( / & ( & ' 0(" (" . Hence both functions are solutions of the differential equation.
12. and . Substituting into the left hand & ' & \ & ' ( .& & ' K &" (. " "
(/ (8 ! ! !1 11
side of the differential equation, we have & K & I $& ( .& I 8 & ' K & (. ! !(8 (/ (. (.
( "0 & I 8 & ' 0 & ' & F< & \ & ' & ( .& F< &(. (. .
( (/ (/
.. Likewise, and ! !2 1
 & ' ( $ & I K & F< &11.
(8 (8 ! . Substituting into the left hand side of the equation, we have
& ( $ & I K & F< & I $& & ( .& F< & I 8 & F< & ' ( $ & I K & F< & I. ! ! !(8 (8 (/ (/ ( ( (2 2 2
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I $ & ( "0 & F< & I 8 & F< & ' 0 #( ( (2 2 2 Hence both functions are solutions of the
differential equation.
13. and & ' C*+ & F< C*+ & I & +,< & \ & ' ( +,< & F< C*+ & I & C*+ & ! ! ! !1
 & ' ( C*+ & F< C*+ & ( & +,< & I +>C &11 ! ! . Substituting into the left hand side of the
differential equation, we have ! ! !( C*+ & F< C*+ & ( & +,< & I +>C & I C*+ & F< C*+ & I
I & +,< & ' ( C*+ & F< C*+ & ( & +,< & I +>C & I C*+ & F< C*+ & I & +,< & ' +>C & ! ! .
Hence the function is a solution of the differential equation. & !
15. Let . Then , and substitution into the differential equation & ' > & ' 5 > ! !5& 11 . 5&
results in . Since , we obtain the algebraic equation 5 > I . > ' 0 > M 0 5 I . ' 0#. 5& 5& 5& .
The roots of this equation are 5 ' L , . #"[. #
17. and . Substituting into the differential & ' > \ & ' 5 > & ' 5 > ! ! !5& 1 5& 11 . 5&
equation, we have . Since , we obtain the algebraic5 > I 5> ( K > ' 0 > M 0. 5& 5& 5& 5&
equation , that is, . The roots are , .5 I 5 ( K ' 0 5 ( . 5 I / ' 0 5 ' ( / .. ! ! "[.
18. Let . Then , and . Substituting & ' > & ' 5> & ' 5 > & ' 5 > ! ! ! !5& 1 5& 11 . 5& 111 / 5&
the derivatives into the differential equation, we have . Since5 > ( /5 > I .5> ' 0/ 5& . 5& 5&
> M 0 5 ( /5 I .5 ' 0 #5& / ., we obtain the algebraic equation By inspection, it follows
that . Clearly, the roots are , and 5 5 ( " 5 ( . ' 0 5 ' 0 5 ' " 5 ' . # ! ! " . /
20. and . Substituting the derivatives & ' & \ & ' 5 & & ' 5 5 ( " & ! ! ! !5 1 5 11 5(" (.
into the differential equation, we have . After& 5 5 ( " & ( 8& 5 & I 8 & ' 0. 5 5 5' ( ! ! (. ("
some algebra, it follows that . For , we obtain the5 5 ( " & ( 85 & I 8 & ' 0 & M 0 ! 5 5 5
algebraic equation The roots of this equation are and 5 ( $5 I 8 ' 0 # 5 ' " 5 ' 8 #. " .
21. The order of the partial differential equation is , since the highest derivative, intwo
fact each one of the derivatives, is of . The equation is , since the leftsecond order linear
hand side is a linear function of the partial derivatives.
23. The partial differential equation is , since the highest derivative, and infourth order
fact each of the derivatives, is of order . The equation is , since the left handfour linear
side is a linear function of the partial derivatives.
24. The partial differential equation is , since the highest derivative of thesecond order
function is of order . The equation is , due to the product on? R[ ? P ? ! two nonlinear R
the left hand side of the equation.
25. and ? R[ ' C*+ R C*+4 \ ' ( C*+ R C*+4 ' C*+ R C*+4 #" ! ] ? ] ?]R ] . ." ". .
It is evident that Likewise, given , the second] ? ] ?]R ] 
. .. ." "
. .I ' 0 # ? R[ ' F< R I . ! !
derivatives are
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] ? . 8R
]R R I 
' (
R I 
] ? . 8 
] R I 
' (
R I 
. .
. . . . .
. .
. . . . .
2
2!
 !
.
.
Adding the partial derivatives,
] ? ] ? . 8R . 8 
]R ] R I R I 
I ' ( I (
R I R I 
' (
8 8
R I R I 
' 0
. . . .
. . . . . .. . . .
. . . .
2 2
 ! !
 !
. .
. .
.
 !R I 
.
Hence is also a solution of the differential equation.? R[ . !
27. Let . Then the second derivatives are? R[ & ' +,< R +,< 2&" ! # #
] ?
]R
' ( +,< R +,< 2&
] ?
]&
' ( 2 +,< R +,< 2&
.
.
.
.
.
. .
"
"
# # #
# # #
It is easy to see that . Likewise, given , we have2 ' ? R[ & ' +,< R ( 2&. .
] ? ] ?
]R ]&
. .
. .
" " ! !
] ?
]R
' ( +,< R ( 2&
] ?
]&
' ( 2 +,< R ( 2&
.
.
.
.
.
.
.
 !
 !
Clearly, is also a solution of the partial differential equation.? R[ &. !
28. Given the function , the partial derivatives are? R[ & ' -& > ! #! (R -8 &. .$
? ' ( I
-& > -& R >
. & 8 &
? ' ( I
& >
.&
R >
8 & &
RR
(R -8 & . (R -8 &
. 8 .
&
(R -8 &
.
. (R -8 &
. .
# #
# #
#
! !
$ $
! !
$
. . . .
. . . .
$ $
$ $
It follows that .$. ? ' ? ' (RR &
. &(R >
8 & &
# ) *
#
! $
$
. . (R -8 &. .
. .
$
Hence is a solution of the partial differential equation.? R[ & !
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 page 18
29 . !2
 !: # The path of the particle is a circle, therefore are intrinsic to thepolar coordinates
problem. The variable is radial distance and the angle is measured from the vertical.5 %
Newton's Second Law states that In the direction, the equation of+ F a ' 6 tangential
motion may be expressed as , in which the , that is,+^ ' 62% % tangential acceleration
the linear acceleration the path is is in the directionalong positive2 ' Y9 -9& # H 2% %
. .%
of increasing . Since the only force acting in the tangential direction is the component% J
of weight, the equation of motion is
(67 +,< ' 6Y #
9
9&
%
%.
.
HNote that the equation of motion in the radial direction will include the tension in the
rod .J
 !C . Rearranging the terms results in the differential equation
9 7
9& Y
I +,< ' 0 #
.
.
%
%
—————————————————————————— ——CHAPTER 2. 
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 page 18
Chapter Two
Section 2.1
1 ! !
 !" ! Based on the direction field, all solutions seem to converge to a specific increasing
function.
 ! ! !# ! $ % & ' $ % $() * +(, - & - # & ! The integrating factor is , and hence )$ *.$ *)$
It follows that all solutions converge to the function ' $ % $() * +(, !+ !
2 ! !
 !" . All slopes eventually become positive, hence all solutions will increase without
bound.
 ! ! !# ! $ % & ' $ % $ & () - # & ! The integrating factor is , and hence It is *.$ ) .$ .$
evident that all solutions increase at an exponential rate.
3 ! 
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 ! !" ' $ % + !. All solutions seem to converge to the function /
 ! ! !# ! $ % & ' $ % $ & (. - + - # & ! The integrating factor is , and hence It is .$ . *$ *$
clear that all solutions converge to the specific solution .' $ % +/ !
4 . ! 
 !" . Based on the direction field, the solutions eventually become oscillatory.
 ! !# ! $ % $ The integrating factor is , and hence the general solution is 
' $ % - 012 .$ -
)#30 .$ ) #
4$ . $
 ! ! !
in which is an arbitrary constant. As becomes large, all solutions converge to the# $
function ' $ % )012 .$ (. !+ ! !
5 . ! 
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 !" . All slopes eventually become positive, hence all solutions will increase without
bound.
 ! ! !"# ! $ % &56 * .7$ % & ! The integrating factor is The differential equation *.$
can
be written as , that is, Integration of both& ' * .& ' % )& & ' % )& !*.$ 8 *.$ *$ *.$ *$
8 !
sides of the equation results in the general solution It follows that' $ % * )& - # & ! ! $ .$
all solutions will increase exponentially.
6 ! 
 ! !" ! ' $ % / ! All solutions seem to converge to the function /
 ! !# ! $ % $ The integrating factor is , and hence the general solution is .
' $ % * - -
#30 $ 012 .$ #
$ $ $
 ! ! !
. .
in which is an arbitrary constant. As becomes large, all solutions converge to the# $
function ' $ % / !/ !
7 . ! 
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 page 21
 ! !" ! ' $ % / ! All solutions seem to converge to the function /
 ! ! ! !# ! $ % &56 $ ' $ % $ & - # & ! The integrating factor is , and hence It is . . *$ *$. .
clear that all solutions converge to the function .' $ % // !
8 ! 
 ! !" ! ' $ % / ! All solutions seem to converge to the function /
 ! ! ! # $ !# ! $ % ' $ % $ 2 $ - 9 ( ! Since , the general solution is ! !+ - $ + - $. .. .*+
It follows that all solutions converge to the function .' $ % // !
9 ! !
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 page 22
 !" . All slopes eventually become positive, hence all solutions will increase without
bound.
 ! ! % &"# ! $ % &56 7$ % & The integrating factor is . The differential equation can +. $(.
be written as , that is, Integration& ' - & '(. % )$ & (. % )$ & (.!$(. 8 $(. $(. $(.% && '(.$(. 8
of both sides of the equation results in the general solution All' $ % )$ * : - # & ! ! *$(.
solutions approach the specific solution ' $ % )$ * : !/ !
10 . ! 
 !" ' ; /. For , the slopes are all positive, and hence the corresponding solutions
increase
without bound. For , almost all solutions have negative slopes, and hence solutions' < /
tend to decrease without bound.
 !# ! $ First divide both sides of the equation by . From the resulting , thestandard form
integrating factor is . The differential equation can be ! % &"$ % &56 7$ % +($* +$
written as , that is, Integration leads to the general' ($ * '($ % $ & '($ % $ & !8 . *$ *$8 !
solution For , solutions ' $ % * $& - # $ ! # = / ! *$ diverge, as implied by the direction
field. For the case , the specific solution is # % / ' $ % * $& ! *$, which evidently
approaches as .zero $>?
11 . ! 
 !" ! The solutions appear to be oscillatory.
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 page 23
 ! ! ! ! !# ! $ % & ' $ % 012 .$ * . #30 .$ - # & ! The integrating factor is , and hence $ *$
It is evident that all solutions converge to the specific solution ' $ % 012 .$ * ./ ! !
#30 .$ ! .
12 ! !
 !" . All solutions eventually have positive slopes, and hence increase without bound.
 ! !# ! $ % & The integrating factor is . The differential equation can be .$
written as , that is, Integration of both& ' - & '(. % )$ (. & '(. % )$ (.!$(. 8 $(. . $(. .
8% &
sides of the equation results in the general solution It' $ % )$ * +.$ - .4 - # & ! ! . *$(.
follows that all solutions converge to the specific solution .' $ % )$ * +.$ - .4/ ! .
14. The integrating factor is . After multiplying both sides by , the ! !$ % & $.$
equation can be written as Integrating both sides of the equation results% && ' % $ !2$ 8
in the general solution Invoking the specified condition, we' $ % $ & (. - # & ! ! . *.$ *.$
require that . Hence , and the solution to the initial value& (. - # & % / # % * +(.*. *.
problem is ' $ % $ * + & (. ! ! !. *.$
16. The integrating factor is . Multiplying both sides by , ! !% &"$ % &56 7$ % $ $.$ .
the equation can be written as Integrating both sides of the equation ! !$ ' % #30 $ !. 8
results in the general solution Substituting and setting' $ % 012 $ ($ - # $ ! $ % ! ! . *. !
the value equal to zero gives . Hence the specific solution is # % / ' $ ! % 012 $ ($ ! ! .
17. The integrating factor is , and the differential equationcan be written as !$ % &*.$
 Integrating, we obtain Invoking the specified initial% & !& ' % + ! & ' $ % $ - # !* *2 2$ $8
condition results in the solution ' $ % $ - . & ! ! ! .$
19. After writing the equation in standard orm@ , we find that the integrating factor is
 ! !% &"$ % &56 7$ % $ $4$ 4 . Multiplying both sides by , the equation can be written as% & ! !$ ' % $ & ! $ ' $ % * $ - + & - # !4 *$ 4 *$8 Integrating both sides results in Letting
$ % * + and setting the value equal to zero gives Hence the specific solution of# % / !
the initial value problem is ' $ % * $ - $ & ! ! % &*) *4 *$
21 . ! 
—————————————————————————— ——CHAPTER 2. 
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 page 24
The solutions appear to diverge from an apparent oscillatory solution. From the
direction
field, the critical value of the initial condition seems to be . For , the % * + ; * +/
solutions increase without bound. For , solutions decrease without bound. < * +
 !" ! The integrating factor is . The general solution of the differential !$ % &* (.$
equation is . The solution is sinusoidal as long' $ % A012 $ * 4#30 $ (B - # & ! ! ! ! $(.
as . The # % / initial value of this sinusoidal solution is
 %/ ! ! !A012 / * 4#30 / (B % * 4(B !
 ! !# ! " See part .
22 ! !
All solutions appear to eventually initially increase without bound. The solutions 
increase
or decrease, depending on the initial value . The critical value seems to be % * + !/
 !" ! The integrating factor is , and the general solution of the differential !$ % &* (.$
equation is Invoking the initial condition , the' $ % * )& - # & ! ' / % ! !$() $(.
solution
may also be expressed as Differentiating, follows that' $ % * )& - - ) & ! ! !$() $(.
' / % * + - - ) (. % - + (. !8 ! ! ! The critical value is evidently % * + !/
—————————————————————————— ——CHAPTER 2. 
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 page 25
 !# % * +. For , the solution is / ' $ % * )& - . & $ ! !$() $(., which for large is
dominated by the term containing & !$(.
is .' $ % A012 $ * 4#30 $ (B - # & ! ! ! ! $(.
23 ! !
As , solutions increase without bound if , and solutions decrease$> / ' + % ; !4 !
without bound if ' + % < !4 ! !
 ! ! % &"" $ % &56 7$ % $ & !. The integrating factor is The general solution of the $-+$ $
differential equation is . Invoking the specified value ,' $ % $ & - # & ($ ' + % ! !*$ *$
we have . That is, . Hence the solution can also be expressed as+ - # % & # % & * +
' $ % $ & - & * + & ($ ! !*$ *$ . For small values of , the second term is dominant.$
Setting , critical value of the parameter is & * + % / % +(& !/
 !# ; +(& < +(&. For , solutions increase without bound. For , solutions decrease
without bound. When , the solution is % +(& ' $ % $ & / $> / ! *$, which approaches as 
.
24 . ! 
As , solutions increase without bound if , and solutions decrease$> / ' + % ; !4 !
without bound if ' + % < !4 ! !
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 page 26
 ! ! ! !" ' % ' $ % * #30 $ ($. Given the initial condition, , the solution is * (. (4! !.
!
Since , solutions increase without bound if , and solutionslim
$C
.
/
#30 $ % + ; 4(!
decrease without bound if Hence the critical value is < 4( !!.
 % 4( % /!4B.A4D!!!/ !
. .
 ! ! ! !# ! % 4( ' $ % + * #30 $ ($ ' $ % +(. For , the solution is , and . Hence the!.
$C
lim
/
solution is bounded.
25. The integrating factor is Therefore general solution is ! % &"$ % &56 7$ % & !+. $(.
' $ % 4#30 $ - A012 $ (B - # & ! ! # $ ! ! * (.$ Invoking the initial condition, the specific
solution is . Differentiating, it follows that' $ % 4#30 $ - A012 $ * , & (B ! # $ ! ! $(.
' $ % * 4012 $ - A#30 $ - 4!B & (B
' $ % * 4#30 $ * A012 $ * .!.B & (B
8 $
88 $
 ! ! !' (
 ! ! !' (
* (.
* (.
Setting , the first solution is , which gives the location of the ' $ % / $ % +!):4)8 ! + first
stationary point. Since . The' $ < /88 !+ , the first stationary point in a local maximum
coordinates of the point are . !+!):4) E !A.//A
26. The integrating factor is , and the differential equation ! % &"$ % &56 7$ % &.) $. ()
can
be written as The general solution is ! !& ' % & * $ & (. ! ' $ %. () . () . ()$ $ $8 F.+ * :$G(A -
- F.+ * :$G(A - * .+(A# & ' $ % ' &*. () *. ()/
$ $. Imposing the initial condition, we have . ! !
Since the solution is smooth, the desired intersection will be a point of tangency. Taking
the derivative, Setting , the solution' $ % * )(4 * .' * .+(4 & () ! ' $ % /8 $ 8 ! ! !/ *. ()
is Substituting into the solution, the respective at the$ % H2 .+ * A' (, !+ /
)
. # $ ! value
stationary point is . Setting this result equal to ' $ % - H2 ) * H2 .+ * A' ! !+ /) , ,. 4 A zero,
we obtain the required initial value ' % .+ * , & (A % * +!:4) !/
4() !
27. The integrating factor is , and the differential equation can be written as !$ % &$(4
 ! !& ' % ) & - . & #30 .$ !$ $ $8( ( (4 4 4 The general solution is
' $ % +. - A#30 .$ - :4012 .$ (:B - # & ! ! # $ ! ! * ($ 4
Invoking the initial condition, , the specific solution is' / % / !
' $ % +. - A#30 .$ - :4012 .$ * DAA & (:B ! ! ! !' (* ($ 4
As , the exponential term will decay, and the solution will oscillate about an$>?
average
value amplitude of , with an of +. A( :B !)
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 page 27
29. The integrating factor is , and the differential equation can be written !$ % &*) (.$
as The general solution is ! !& ' % )$ & - . & ! ' $ % * .$ * 4() * 4 & -*) (. *) (. * (.$ $ $ $8
- # & ! ' $ % * .$ * 4() * 4 & - ' - +:() & !) (. ) (./
$ $ $ Imposing the initial condition, ! !
As , the term containing will the solution. Its $>? &) (.$ dominate sign will determine
the divergence properties. Hence the critical value of the initial condition is
' % !/ * +:()
The corresponding solution, , will also decrease without' $ % * .$ * 4() * 4 & ! $
bound.
Note on Problems 31-34 :
Let be , and consider the function , in which I $ ' $ % ' $ - I $ ' $ >? ! ! ! ! !given + +
as . Differentiating, . Letting be a $>? ' $ % ' $ - I $ 8 8 8 ! ! !+ constant, it follows
that ' $ - ' $ % ' $ - ' $ - I $ - I $ !8 8 8 ! ! ! ! ! !+ + Note that the hypothesis on the
function will be satisfied, if . That is, Hence' $ ' $ - ' $ % / ' $ % # & !+ + ++ ! ! ! !8 * $
' $ % # & - I $ ' - ' % I $ - I $ ! ! ! ! !* $ 8 8, which is a solution of the equation 
For convenience, choose . % +
31. Here , and we consider the linear equation The integratingI $ % ) ' - ' % ) ! ! 8
factor is , and the differential equation can be written as The ! !$ % & & ' % )& !$ $ $8
general solution is ' $ % ) - # & ! ! *$
33. Consider the linear equation The integratingI $ % ) * $ ! ' - ' % * + - ) * $ ! ! 8
factor is , and the differential equation can be written as ! ! !$ % & & ' % . * $ & !$ $ $8
The general solution is ' $ % ) * $ - # & ! ! *$
34. Consider the linear equation The integratingI $ % 4 * $ ! ' - ' % 4 * .$ * $ ! ! . 8 .
factor is , and the equation can be written as ! ! !$ % & & ' % 4 * .$ * $ & !$ $ . $8
The general solution is ' $ % 4 * $ - # & ! ! . $*
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Section 2.2
2. For , the differential equation may be written as 5 = * + ' 7' % 75 !# $ !5 ( + - 5. )
Integrating both sides, with respect to the appropriate variables, we obtain the relation
' (. % H2 - # ! ' 5 % J H2 - # !. + .) )* * * *+ - 5 + - 5) ) That is, ! +
3. The differential equation may be written as Integrating both' 7' % * 0125 75 !*.
sides of the equation, with respect to theappropriate variables, we obtain the relation
* ' % #30 5 - # ! 9 * #30 5 ' % + 9*+ That is, , in which is an arbitrary constant. !
Solving for the dependent variable, explicitly, .' 5 % +( 9 * #30 5 ! !
5. Write the differential equation as , or #30 .' 7' % #30 5 75 0&# .' 7' % #30 5 75!*. . . .
Integrating both sides of the equation, with respect to the appropriate variables, we obtain
the relation $ 2 .' % 012 5 #30 5 - 5 - # !
7. The differential equation may be written as Integrating ! !' - & 7' % 5 * & 75 !' *5
both sides of the equation, with respect to the appropriate variables, we obtain the
relation
' - . & % 5 - . & - # !. ' . 5*
8. Write the differential equation as Integrating both sides of the !+ - '. 7' % 5 75 !.
equation, we obtain the relation , that is, ' - ' () % 5 () - # )' - ' % 5 - 9!) ) ) )
9 . The differential equation is separable, with Integration ! ! ' 7' % + * .5 75 !*.
yields Substituting and , we find that * ' % 5 * 5 - # ! 5 % / ' % * +(: # % : !*+ .
Hence the specific solution is . The is' % 5 * 5 * :*+ . explicit form
' 5 % +( ! ! !5 * 5 * :.
 !"
 ! ! !# 5 * 5 * : % 5 - . 5 * ). Note that . Hence the solution becomes . singular at
5 % * . 5 % ) ! and 
10 ! ! ) ! ' 5 % * .5 * .5 - 4 !.
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10 !" !
11 Rewrite the differential equation as Integrating both sides ! ! 5 & 75 % * ' 7' !5
of the equation results in Invoking the initial condition, we5 & * & % * ' (. - # !5 5 .
obtain Hence # % * +(. ! ' % .& * .5 & * +!. 5 5 The of the solution isexplicit form
' 5 % ! ! ).& * .5 & * + ' / % +!5 5 The positive sign is chosen, since !
 !" !
 !# ! 5 % * +!D 5 % /!D: ! The function under the radical becomes near and negative
11 Write the differential equation as Integrating both sides of the ! ! K 7K % 7 !*. *+" "
equation results in the relation Imposing the condition , we* K % H2 - # ! K + % .*+ " !
obtain . # % * +(. The of the solution is explicit form K % .( + * . H2 ! ! !" "
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 !" !
 !# ! ; / ! Clearly, the solution makes sense only if Furthermore, the solution becomes"
singular when , that is, H2 % +(. % & !" " )
13 ! ! !) ! ' 5 % * . H2 + - 5 - 4 !.
 !" !
14 . Write the differential equation as Integrating both ! ! ' 7' % 5 + - 5 75 !*) *+(..
sides of the equation, with respect to the appropriate variables, we obtain the relation
* ' (. % + - 5 - # ! # % * )(. !*. ) . Imposing the initial condition, we obtain 
Hence the specific solution can be expressed as The ' % ) * . + - 5 !*. ) . explicit
form positive of the solution is The ' 5 % +( ) * . + - 5 ! ! + ) . sign is chosen to
satisfy the initial condition.
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 !" !
 !# ! The solution becomes singular when . + - 5 % ) 5 % J B(. !) ). . That is, at 
15 ! ! ) ! ' 5 % * +(. - 5 * +B(4 !.
 !" !
16 . ! Rewrite the differential equation as Integrating both4' 7' % 5 5 - + 75 !) . !
sides
of the equation results in Imposing the initial condition, we obtain' % (4 - # !4 !5 - +. .
# % / ! * 4' % / ! Hence the solution may be expressed as The form !5 - +. . 4 explicit
of the solution is The is chosen based on ' 5 % * ! ' / % ! ! !) ! )5 - + (. * +( .. sign
—————————————————————————— ——CHAPTER 2. 
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 !" !
 !# ! 5 L The solution is valid for all .#
17 ! ! ) ! ' 5 % * B(. * 5 * & - +)(4 !) 5
 !" !
 !# 5 ; * +!4B !. The solution is valid for This value is found by estimating the root of
45 * 4& - +) % / !) 5
18 . Write the differential equation as Integrating both ! ! ! ) - 4' 7' % & * & 75 !*5 5
sides of the equation, with respect to the appropriate variables, we obtain the relation
)' - .' % * & - & - # ! ' / % +. 5 *5 ! ! Imposing the initial condition, , we obtain
# % D! )' - .' % * & - & - D! Thus, the solution can be expressed as Now by. 5 *5 !
completing the square on the left hand side, . ' - )(4 % !. * & - & - :B(A !5 *5 .
Hence the explicit form of the solution is ' 5 % * )(4 - :B(+: * #30M 5 ! ! )
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 !" !
 ! * *# :B * +: #30M 5 N / 5 ; .!+ !. Note the , as long as Hence the solution is valid on
the interval .* .!+ < 5 < .!+
19 ! ! ' 5 % * () - 012 ) #30 5 ! ! !! +) *+ .
 !" !
20 ! ! Rewrite the differential equation as Integrating' 7' % K#012 5( + * 5 75 !. .)
both sides of the equation results in Imposing the condition' () % K#012 5 (. - # !) !.
' / % / # % /! ! , we obtain formThe of the solution is explicit ' 5 % ! ! !+) ). K#012 5 .()
—————————————————————————— ——CHAPTER 2. 
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 !" .
 !# * + O 5 O + !. Evidently, the solution is defined for 
22. The differential equation can be written as Integrating both !)' * 4 7' % )5 75 !. .
sides, we obtain Imposing the initial condition, the specific solution' * 4' % 5 - # !) )
is Referring back to the differential equation, we find that as' * 4' % 5 * + ! ' >?) ) 8
' >J.( ) ! 5 % * +!.D: +!B,A !) The respective values of the abscissas are , 
Hence the solution is valid for * +!.D: < 5 < +!B,A !
24. Write the differential equation as Integrating both sides, ! !) - .' 7' % . * & 75 !5
we obtain Based on the specified initial condition, the solution)' - ' % .5 * & - # !. 5
can be written as , it follows that)' - ' % .5 * & - + !. 5 Completing the square
' 5 % * )(. - .5 * & - +)(4 ! .5 * & - +)(4 N / ! ) 5 5 The solution is defined if ,
that is, . In that interval, , for It can* +!B O 5 O . ' % / 5 % H2 . ! !approximately 8
be verified that . In fact, on the interval of definition. Hence' H2 . < / ' 5 < /88 88 ! !
the solution attains a global maximum at 5 % H2 . !
26. The differential equation can be written as Integrating !+ - '. *+7' % . + - 5 75 ! !
both sides of the equation, we obtain Imposing the given K#$ 2' % .5 - 5 - # !.
initial
condition, the specific solution is Therefore, K#$ 2' % .5 - 5 ! ' 5 % $ 2 !. ! !.5 - 5.
Observe that the solution is defined as long as It is easy to* (. < .5 - 5 < (. !! !.
see that Furthermore, for and Hence.5 - 5 N * +! .5 - 5 % (. 5 % * .!: /!: !. . !
the solution is valid on the interval Referring back to the differential* .!: < 5 < /!: !
—————————————————————————— ——CHAPTER 2. 
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equation, the solution is at Since on the entire intervalstationary 5 % * +! ' 5 ; /88 !
of
definition, the solution attains a global minimum at 5 % * + !
28 . Write the differential equation as Integrating ! ! ! ' 4 * ' 7' % $ + - $ 7$ !*+ *+ *+
both sides of the equation, we obtain TakingH2 ' * H2 ' * 4 % 4$ * 4H2 + - $ - # !* * * * * *
the of both sides, it follows that It followsexponential * * ! !'( ' * 4 % 9 & ( + - $ !4$ 4
that as , . That is, $>? '( ' * 4 % + - 4( ' * 4 >? ' $ > 4 !* * * * ! ! !
 ! !" ! ' / % . 9 % + Setting , we obtain that . Based on the initial condition, the solution
may be expressed as Note that , for all'( ' * 4 % * & ( + - $ ! '( ' * 4 < / ! ! !4$ 4
$ N /! ' < 4 $ N /! Hence for all Referring back to the differential equation, it follows
that is always . This means that the solution is . We find' 8 positive monotone increasing
that the root of the equation is near & ( + - $ % ),, $ % .!A44 !4$ !4
 ! !# ! ' $ % 4 Note the is an equilibrium solution. Examiningthe local direction field,
we see that if , then the corresponding solutions converge to . Referring' / ; / ' % 4 !
back to part , we have , for Setting ! ! # $ ! ! '( ' * 4 % ' ( ' * 4 & ( + - $ ' = 4 !/ / /44$
$ % . ' ( ' * 4 % )(& ' . ( ' . * 4 !, we obtain Now since the function/ / ! ! ! ! !. 4
@ ' % '( ' * 4 ' < 4 ' ; 4 ! ! is for and , we need only solve the equationsmonotone
' ( ' * 4 % ' ( ' * 4 % !/ / / /. .
4 4 ! !* ),, )(& 4/+ )(& ! !and The respective solutions
are and ' % )!::.. ' % 4!4/4. !/ /
30 !@ !
—————————————————————————— ——CHAPTER 2. 
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 page 36
31 !#
32 . Observe that Hence the differential equation ! ! % & 5 - )' (.5' % - !. . + ). 5 . 5' '*+
is .homogeneous
 ! !" ' % 5 P P - 5P % 5 - )5 P (.5 P. The substitution results in . The8 . . . .
transformed equation is This equation is , with generalP % + - P (.5P !8 . ! separable
solution In terms of the original dependent variable, the solution isP - + % # 5 !.
5 - ' % # 5 !. . )
 !# !
33 !# !
—————————————————————————— ——CHAPTER 2. 
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 page 37
34 Observe that Hence the ! ! ' ( ! 45 - )' (F.5 - 'G % * . * . - !* ' '5 5 *+
differential equation is .homogeneous
 ! !" ' % 5 P P - 5P % * . * P( . - P. The substitution results in . The transformed8
equation is This equation is , with generalP % * P - BP - 4 (F. - PG5 !8 . ! separable
solution In terms of the original dependent variable, the solution ! * *P-4 P-+ % 9(5 !. )
is ! * *45 - ' 5-' % 9!.
 !# !
35 . !#
36 Divide by to see that the equation is homogeneous. Substituting , we ! ! 5 ' % 5P.
obtain The resulting differential equation is separable.5 P % + - P !8 . !
 ! !" ! + - P 7P % 5 75 ! Write the equation as Integrating both sides of the equation,* *+.
we obtain the general solution In terms of the original* +( + - P % H2 5 - # ! ! * *
dependent variable, the solution is ' % 5 9 * H2 5 * 5 !# $* * *+
—————————————————————————— ——CHAPTER 2. 
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 page 38
 !# !
37 The differential equation can be expressed as . Hence the ! % & ! ' % *8 + ). 5 . 5' '*+
equation is homogeneous. The substitution results in .' % 5P 5 P % + * BP (.P8 . !
Separating variables, we have .P ++*BP 5.7P % 75 !
 !" ! * Integrating both sides of the transformed equation yields +B
H2 % H2 5 - #* *+ * BP. * * ,
that is, In terms of the original dependent variable, the general+ * BP % 9( 5 !. B* *
solution is B' % 5 * 9( 5 !. . )* *
 !# !
38 The differential equation can be expressed as . Hence the ! % & ! ' % *8 ) +. 5 . 5' ' *+
equation is homogeneous. The substitution results in , that' % 5P 5 P % P * + (.P8 . !
is, .P +P *+ 5. 7P % 75 !
 ! * *" ! H2 % H2 5 - # Integrating both sides of the transformed equation yields ,* *P * +.
that is, In terms of the original dependent variable, the general solutionP * + % 9 5 !. * *
is ' % 9 5 5 - 5 !. . .* *
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 page 39
 !# !
—————————————————————————— ——CHAPTER 2. 
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 page 40
Section 2.3
5 . Let be the amount of salt in the tank. Salt enters the tank of water at a rate of ! Q
. + - 012 $ % - 012 $ ! .Q(+// !+ + + +4 . . 4% & It leaves the tank at a rate of 3R(S12 3R(S12
Hence the differential equation governing the amount of salt at any time is
7Q + +
7$ . 4
% - 012 $ * Q(B/ !
The initial amount of salt is The governing ODE is Q % B/ 3R !/ linear, with integrating
factor Write the equation as The ! % & % &$ % & ! & Q % & !$(B/ $(B/ $(B/8 + +. 4- 012 $
specific solution is Q $ % .B - +.!B012 $ * :.B#30 $ - :)+B/ & (.B/+ 3R ! ! ' (*$(B/
 !" !
 !# ! The amount of salt approaches a , which is an oscillation of amplitudesteady state
+(4 .B 3R ! about a level of 
6 . The equation governing the value of the investment is . The value of ! 7T(7$ % K T
the investment, at any time, is given by Setting , the requiredT $ % T & ! T U % .T ! !/ /K$
time is U % H2 . (K ! !
 !" ! K % D % !/D U V ,!, 'K0 ! For the case , %
 ! ! !# ! K % H2 . (U U % A Referring to Part , . Setting , the required interest rate is to
be approximately K % A!:: !%
8 . Based on the solution in , with , the value of the investments ! ! +: T % /Eq. with/
contributions is given by After years, person A hasT $ % .BE /// & * + ! ! !K$ ten
T % .BE /// +!..: % )/E :4/ ! )BW $ $ ! Beginning at age , the investments can now be
analyzed using the equations and T % )/E :4/ & T % .BE /// & * + !W X
!/A$ !/A$ !
After years, the balances are and thirty $ $T % ))DE D)4 T % .B/E BD,!W X
 !" ! K T % )/E :4/ & For an rate , the balances after years are andunspecified thirty W )/K
T % .BE /// & * + !X !)/K
—————————————————————————— ——CHAPTER 2. 
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 !# .
 !7 ! The two balances can be equal.never
11 . Let be the value of the mortgage. The debt accumulates at a rate of , in ! T KT
which is the interest rate. Monthly payments of are equivalent toK % !/, A//annual $
$ per year The differential equation governing the value of the mortgage is,E :// !
7T(7$ % !/, T * ,E :// ! T Given that is the original amount borrowed, the debt is/
T $ % T & * +/:E ::D & * + ! T )/ % / ! ! !/ !/,$ !/,$ Setting , it follows that
T % ,,E B/// $ .
 !" ! )/ .AAE /// The payment, over years, becomes . The interest paid on thistotal $
purchase is .$ +AAE B//
13 . The balance at a rate of , and at a constant rate of ! K T Yincreases $/yr decreases
$ per year . Hence the balance is modeled by the differential equation .7T(7$ % KT * Y
The balance at any time is given by T $ % T & * & * + ! ! !/ K$ K$YK
 ! !" T $ % FT * G& - !. The solution may also be expressed as Note that if the/ Y YK KK$
withdrawal rate is , the balance will remain at a constant level Y % K T T !/ / /
 ! ! , -# Y ; Y T U % / U % H2 !. Assuming that , for / / / + YK Y*Y/
 !7 K % !/A Y % .Y U % A!::. If and , then / / years .
 ! ! !& T $ % / & " & % ! $ % U. Setting and solving for in Part , Now setting K$ K$ YY*KT/
results in Y % KT & ( & * + !/
U UK K !
 ! !@ ! & Y % +.E /// K % !/A U % ./ In part , let , , and . The required investment
becomes .T % ++,E D+B/ $
14 Let The general solution is Based on the ! ! ! Q % * KQ ! Q $ % Q & !8 *K$/
definition of , consider the equation It follows thathalf-life Q (. % Q & !/ /
BD)/* K
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* BD)/ K % H2F+(.G K % +!./,D Z +/, that is, *4 per year.
 ! !" Q $ % Q & !. Hence the amount of carbon-14 is given by / *+!./,DZ+/ $*4
 ! !# ! Q U % Q (B +(B % & ! Given that , we have the equation Solving for/ *+!./,DZ+/ U*4
the , the apparent age of the remains is approximately decay time years .U % +)E )/4!:B
15. Let be the population of mosquitoes at any time . The rate of of the[ $ $ ! increase
mosquito population is The population by . Hence theK[ ! ./E ///decreases per day
equation that models the population is given by . Note that the7[(7$ % K[ * ./E ///
variable represents . The solution is In the$ [ $ % [ & * & * + !days ! !/ K$ K$./E///K
absence of predators, the governing equation is , with solution7[ (7$ % K[+ +
[ $ % [ & ! [ D % .[ .[ % [ & !+ / + / / / ! !K$ DK Based on the data, set , that is, The growth
rate is determined as Therefore the population,K % H2 . (D % !/,,/. ! ! per day
including the by birds, is predation [ $ % . Z +/ & * ./+E ,,D & * + % ! !B !/,,$ !/,,$
% ./+E ,,D!) * +,DD!) & !!/,,$
16 .The is ! ! # $ ' $ % &56 .(+/ - $(+/ * .#30F$G(+/ ! V .!,:). !doubling-time $
 ! ! !" ! 7'(7$ % '(+/ ' $ % ' / & ! The differential equation is , with solution The$(+/
doubling-time is given by $ % +/H2 . V :!,)+B ! !
 ! !# 7'(7$ % /!B - 012F. $G '(B !. Consider the differential equation The equation is!
separable, with Integrating both sides, with respect to the+ +' B7' % /!+ - 012F. $G 7$ !% &!
appropriate variable, we obtain Invoking the initialH2 ' % $ * #30F. $G (+/ - # ! !! ! !
condition, the solution is The is' $ % &56 + - $ * #30F. $G (+/ ! ! # $ !! ! ! doubling-time
$ V :!)A/4 ! " The approaches the value found in part .doubling-time !
 !7 .
17 . The differential equation is , with integrating factor ! ! 7'(7$ % K $ ' * Y linear
 ! ! ! !' ("$ % &56 K $ 7$ ! ' % * Y $ !* Write the equation as Integration of both8
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 page 43
sides yields the general solution . In this problem,' % * Y 7 - ' / ( $' (" ! ! ! $ $ /
the
integrating factor is ! # $ !$ % &56 #30 $ * $ (B !
 ! !" ! ' $ % / $ % $ The population becomes , if , for some . Referring toextinct \ \
part , ! 
we find that ' $ % / ] !\
. # $ !
/
$
+(B
#
\
&56 #30 * (B 7 % B & ' !$ $ $
It can be shown that the integral on the left hand side increases , from monotonically zero
to a limiting value of approximately . Hence extinction can happen B!/A,) only if
B & ' < B!/A,) ' < /!A))) !+(B # #, that is, 
 ! ! !# " ' $ % / ]. Repeating the argument in part , it follows that \
. # $ !
/
$
+(B
#
\
&56 #30 * (B 7 % & ' !
+
Y
$ $ $
Hence extinction can happen , that is, only if & ' (Y < B!/A,) ' < 4!+::D Y !+(B # #
 !7 ' Y. Evidently, is a function of the parameter .# linear
19 . Let be the of carbon monoxide in the room. The rate of of ! ! Q $ volume increase
CO CO leaves the room is The amount of at a rate of ! !!/4 /!+ % /!//4 @$ (S12 !) ! ! !/!+ Q $ (+.// % Q $ (+./// @$ (S12 ! Hence the total rate of change is given by)
the differential equation This equation is and7Q(7$ % /!//4 * Q $ (+./// ! ! linear
separable, with solution Note that Q $ % 4A * 4A &56 * $(+./// Q % / ! ! @$ ! @$ !) )/
Hence the at any time is given by .concentration %5 $ % Q $ (+.// % Q $ (+. ! ! !
 ! ! !" 5 $ % 4 * 4&56 * $(+.///. The of in the room is A levelconcentration CO %!
of corresponds to . Setting , the solution of the equation/!///+. /!/+. 5 % /!/+.% !$
4 * 4&56 * $(+./// % /!/+. V ): ! is .$ minutes
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20 The concentration is It is easy to see ! ! ! ! # $ % Y - [(K - # * Y * [(K & !/ *K$(^
that # $>? % Y - [(K ! !
 ! !" ! # $ % # & U % H2F.G^ (K U % H2F+/G^ (K! . The are and / B/ +/*K$(^ reduction times
 ! ! !# ! U % H2 +/ :B!. (+.E .// % 4)/!AB The , in , are reduction times years T
 U % H2 +/ +BA (4E ,// % D+!4 _ U % H2 +/ +DB (4:/ % :!/B` a ! ! ! !
 U % H2 +/ ./, (+:E /// % +D!:) !b ! !
21 !# !
 
22 . The differential equation for the motion is Given the ! S7P(7$ % * P()/ *SI !
initial condition , the solution is .P / % ./ P $ % * 44!+ - :4!+ &56 * $(4!B ! ! !m/s
Setting , the ball reaches the maximum height at . IntegratingP $ % / $ % +!:A) !+ + sec
P $ 5 $ % )+A!4B * 44!+ $ * .AA!4B &56 * $(4!B ! ! ! !, the position is given by Hence
the is .maximum height m5 $ % 4B!DA !+
 ! !" ! 5 $ % / $ % B!+.A Setting , the ball hits the ground at .. . sec
 !# !
 
23 The differential equation for the motion is , ! ! S7P(7$ % * P *SIupward .
in which . This equation is , with Integrating % +(+).B 7P % * 7$ !separable SP -SI .
—————————————————————————— ——CHAPTER 2. 
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both sides and invoking the initial condition, SettingP $ % 44!+)) $ 2 !4.B * !... $ ! ! !
P $ % / $ % +!,+: P $ ! !+ +, the ball reaches the maximum height at . Integrating , thesec
position is given by Therefore the5 $ % +,A!DB H2 #30 /!... $ * /!4.B - 4A!BD ! ! # $ !
maximum height m is .5 $ % 4A!B: !+
 !" ! S7P(7$ % - P *SI ! The differential equation for the motion is downward .
This equation is also separable, with For convenience, set atSSI* P . 7P % * 7$ ! $ % /
the of the trajectory. The new initial condition becomes . Integrating bothtop P / % / !
sides and invoking the initial condition, we obtain H2 44!+) * P ( 44!+) - P % $(.!.B# $ ! !
!
Solving for the velocity, Integrating , theP $ % 44!+) + * & ( + - & ! P $ ! ! ! !$ $(.!.B (.!.B
position is given by To estimate the5 $ % ,,!., H2 & ( + - & - +A:!. ! ! !, -$ $ .(.!.B (.!.B
duration sec of the downward motion, set , resulting in . Hence the5 $ % / $ % )!.D: !. .
total time sec that the ball remains in the air is .$ - $ % B!+,.1 .
 !# !
 
24 Measure the positive direction of motion . Based on Newton's ! ! .downward nd
law,
the equation of motion is given by
S % !
7P
7$
* /!DB P -SI / < $ < +/
* +. P -SI $ ; +//
 , 
 , 
Note that gravity acts in the direction, and the drag force is . During thepositive resistive
first ten seconds of fall, the initial value problem is , with initial7P(7$ % * P(D!B - ).
velocity This differential equation is separable and linear, with solutionP / % / ! ! fps
P $ % .4/ + * & P +/ % +D:!D ! ! ! !*$(D!B . Hence fps
 ! !" 5 $ % /. Integrating the velocity, with , the distance fallen is given by
5 $ % .4/ $ - +A// & * +A// ! *$(D!B .
Hence .5 +/ % +/D4!B ! ft
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 !# ! $ % / For computational purposes, reset time to . For the remainder of the motion,
the initial value problem is , with specified initial velocity7P(7$ % * ).P(+B - ).
P / % +D:!D ! P $ % +B - +:+!D & ! $>? ! ! The solution is given by As ,fps *). $(+B
P $ > P % +B ! 5 / % +/D4!B ! !c Integrating the velocity, with , the distance fallenfps
after the parachute is open is given by To find the5 $ % +B $ * DB!A & - ++B/!) ! ! *). $(+B
duration of the second part of the motion, estimate the root of the transcendental equation
+B U * DB!A & - ++B/!) % B/// ! U % .B:!: !*). U(+B The result is sec
 !7 !
 
25 . Measure the positive direction of motion . The equation of motion is ! upward
given by . The initial value problem is ,S7P(7$ % * Y P *SI 7P(7$ % * YP(S * I
with . The solution is SettingP / % P P $ % *SI(Y - P -SI(Y & ! ! ! !/ / *Y$(S
P $ % / $ % S(Y H2 SI - Y P (SI ! ! ! # $ !S S, the maximum height is reached at time /
Integrating the velocity, the position of the body is
5 $ % *SI $(Y - I - F+ * & G!
S SP
Y Y
 ! 0 12 3. *Y$(S/
Hence the maximum height reached is
5 % 5 $ % * I H2 !
SP S SI - Y P
Y Y SI
S S
/ / ! 2 3 0 1.
 ! !" ! d + H2 + - % * - * -e Recall that for , % % % % % %+ + +. ) 4. ) 4
26 . ! !" % * Y P -SI & % * I$ !lim lim
YC/ YC/
*SI- Y P -SI &
Y S
$ *Y$(S !/ *Y$(S
/
 ! ' (% &# ! * - -P & % / & % / ! , since lim lim
SC/ SC/
SI SI
Y Y
*Y$(S *Y$(S
/
28 . In terms of displacement, the differential equation is ! SP 7P(75 % * Y P -SI !
This follows from the : . The differential equation ischain rule 7P 7P 75 7P7$ 75 7$ 7$% % P
separable, with
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 page 47
5 P % * * H2 !
SP S I SI * Y P
Y Y SI
 ! 4 4
.
.
The inverse , since both and are monotone increasing. In terms of the givenexists 5 P
parameters, 5 P % * +!.B P * +B!)+ H2 /!/A+: P * + ! ! * *
 ! !" ! 5 +/ % +)!4B Y % /!.4 . The required value is .meters
 ! !# ! P % +/ 5 % +/ In part , set and .m/s meters
29 Let represent the height above the earth's surface.The equation of motion is ! ! 5
given by , in which is the universal gravitational constant. TheS % *f f7P `S7$ g-5 !.
symbols and are the and of the earth, respectively. By the chain rule,` g mass radius
SP % *f
7P `S
75 g - 5 !. .
This equation is separable, with Integrating both sides,P 7P % *f` g - 5 75 ! !*.
and
invoking the initial condition , the solution is P / % .Ig P % .f` g - 5 - ! !) . *+
- .Ig * .f`(g ! I % f`(g From elementary physics, it follows that . Therefore.
P 5 % .I g( g - 5 ! ! !) , -) Note that mi/hr .I % DAE B4B .
 ! ) , -)" ! 75(7$ % .I g( g - 5 We now consider . This equation is also separable,
with By definition of the variable , the initial condition is) )g - 575 % .I g 7$ ! 5
5 / % /! 5 $ % .I g $ - g *g ! ! ! ' (% &) Integrating both sides, we obtain ) .. ) )(. .()
Setting the distance , and solving for , the duration of such a5 U - g % .4/E /// U !
flight would be U V 4, hours .
32 Both equations are linear and separable. The initial conditions are ! ! ! P / % h #30
W
and . The two solutions are and i / % h012W P $ % h #30W & i $ % * I(K - ! ! !*K$
- h012W - I(K & ! ! *K$
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 ! !" ! Integrating the solutions in part , and invoking the initial conditions, the
coordinates are and5 $ % #30W + * & ! !hK *K$
' $ % * I$(K - I - hK 012W - MK (K * 012W - I(K & !
h
K
 ! % & 2 3. . . *K$
 !# !
 !7 ! U )B/ Let be the time that it takes the ball to go horizontally. Then from above,ft
& % h #30W * D/ (h #30W !*U(B ! At the same time, the height of the ball is given by
' U % * +:/U - .:D - +.Bh012W * FA// - Bh 012WG h #30W * D/ (h #30W ! ! # $ !
Hence and must satisfy the inequalityW h
A//H2 - .:D - +.Bh012W * FA// - Bh 012WG h #30W * D/ (h #30W N +/ !
h #30W * D/
h #30W
0 1 # $ !
33 Solving equation , . The answer is ! ! ! # $ ! ! 1 ' 5 % Y * ' ('8 . +(. positive
chosen, since is an function of .' 5increasing
 !" ' % Y 012 $ 7' % .Y 012 $ #30 $ 7$ !. Let . Then Substituting into the equation in. . .
part , we find that ! 
.Y 012 $ #30 $ 7$ #30 $
75 012 $
% !
.
Hence .Y 012 $ 7$ % 75 !. .
 !# ! % .$ Y 012 7 % 75 ! Letting , we further obtain Integrating both sides of the" ". . ."
equation and noting that corresponds to the , we obtain the solutions$ % % /" origin
5 % Y * 012 (. " ' % Y + * #30 (. ! ! ! # $ ! ! !" " " " ". . and from part 
 ! ! !7 '(5 % + * #30 ( * 012 ! 5 % + ' % .. Note that Setting , , the solution of" " "
the equation is . Substitution into either of the ! !+ * #30 ( * 012 % . V +!4/+" " " "
expressions yields Y V .!+,) !
—————————————————————————— ——CHAPTER 2. 
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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 , the initial value problem/ 4 . L / E 4 !
has a unique solution on the interval !/ E 4 !
3. The function is discontinuous at of . Since , the$ 2 $ < <odd multiples ! ! !. . .
)!
initial value problem has a unique solution on the interval % &! !. .)E !
5. and . These functions are discontinuous at6 $ % .$( I $ % )$ ( ! ! ! !4 * $ 4 * $. ..
5 % J. * . E . !. The initial value problem has a unique solution on the interval !
6. The function is defined and continuous on the interval . Therefore theH2 $ / E? !
initial value problem has a unique solution on the interval . !/ E?
7. The function is continuous everywhere on the plane, along the straight@ $ E ' ! except
line The partial derivative has the ' % * .$(B ! j@(j' % * D$( .$ - B' !. same
region of continuity.
9. The function is discontinuous along the coordinate axes, and on the hyperbola@ $ E ' !
$ * ' % +. . . Furthermore,
j@ J+ ' H2 $'
j' ' + * $ - '
% * .
+ * $ - ' !
* *
 !. . . . .
has the points of discontinuity.same
10. is continuous everywhere on the plane. The partial derivative is also@ $ E ' j@(j' !
continuous everywhere.
12. The function is discontinuous along the lines and . The@ $ E ' $ % JY ' % * + ! !
partial derivative has the region of continuity.j@(j' % #3$ $ ( + - ' ! !. same 
14. The equation is separable, with Integrating both sides, the solution7'(' % .$ 7$ !.
is given by For , solutions exist as long as .' $ % ' (F+ * ' $ G! ' ; / $ < +(' ! / / / /. .
For , solutions are defined for .' O / $/ all
15. The equation is separable, with Integrating both sides and invoking7'(' % * 7$ !)
the initial condition, Solutions exist as long as ,' $ % ' ( .' $ - + ! .' $ - + ; / ! )/ / /
that is, . If , solutions exist for . If , then the.' $ ; * + ' ; / $ ; * +(.' ' % // / / /
solution exists for all . If , solutions exist for .' $ % / $ ' < / $ < * +(.' ! / /
16. The function is discontinuous along the straight lines and .@ $ E ' $ % * + ' % / !
The partial derivative is discontinuous along the same lines. The equation isj@(j'
—————————————————————————— ——CHAPTER 2. 
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 page 50
separable, with Integrating and invoking the initial condition, the' 7' % $ 7$( !. !+ - $)
solution is Solutions exist as long as' $ % H2 + - $ - ' ! ! * *' (.) ) ./ +(.
.
)
H2 + - $ - ' N /5 5) ./ ,
that is, For all it can be verified that yields a valid' N * H2 + - $ ! ' F ' % // / /
. ).
) * *
solution, even though Theorem does not guarantee one , solutions exists as long as.!4!. G* * !+ - $ N &56 * )' (. ! $ ; * +) ./ From above, we must have . Hence the inequality
may be written as It follows that the solutions are valid for$ N &56 * )' (. * + !) . !/# $ !&56 * )' (. * + < $ < ?/ +(). .
17.
18.
Based on the direction field, and the differential equation, for , the slopes' < //
eventually become negative, and hence solutions tend to . For , solutions*? ' < //
increase without bound if Otherwise, the slopes become negative, and$ < / !/ eventually
solutions tend to . Furthermore, is an . Note that slopeszero equilibrium solution' % //
are along the curves and .zero ' % / $' % )
19.
—————————————————————————— ——CHAPTER 2. 
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 page 51
For initial conditions satisfying , the respective solutions all tend to . !$ E ' $' < )/ / zero
Solutions with initial conditions the hyperbola eventually tend toabove or below $' % )
J? ' % /. Also, is an ./ equilibrium solution
20.
Solutions with all tend to . Solutions with initial conditions to the$ < / *? $ E '/ / / !
right of the parabola asymptotically approach the parabola as . Integral$ % + - ' $>?.
curves with initial conditions the parabola and also approach the curve.above !' ; //
The slopes for solutions with initial conditions the parabola and are allbelow !' < //
negative. These solutions tend to *?!
21. Define , in which is the Heaviside step function.' $ % $ * # h $ * # h $#
)(. ! ! ! !.)
Note that and ' # % ' / % / ' # - )(. % +!# # #
.() ! ! !% &
 ! ! ! # % + * )(. ! Let .()
 ! !" ! # % . * )(. ! Let .()
 ! ! ! ! ! !# ! ' . % . ' $ < . / < # < . ' . % / Observe that , for , and that for/ # #)(. )(.. .) )
# N .! # N / J' . L * .E . ! So for any , # ! # $
26 Recalling Eq. in Section , ! ! ! )B .!+
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' % 0 I 0 70 - !
+ #
$ $ 
 ! !. ! !
It is evident that and .' $ % ' $ % 0 I 0 70+ . ! ! ! !"+ +$ $ ! ! 
 ! ! ! !% &"" % &56 * 6 $ 7$ ' % * 6 $ % * 6 $ ' !. By definition, . Hence + +$ $8 ! !+ +
That is, ' - 6 $ ' % /!+ +
8 !
 ! ! ! ! ! ! ! !2 3 2 3"# ! ' % * 6 $ 0 I 0 70 - $ I $ % * 6 $ ' - I $ ! . / .8 + +$ $$ ! ! 
That is, ' - 6 $ ' % I $ !. .
8 ! !
30.Since , set . It follows that and 2 % ) P % ' % * .' % * !*. *)7P 7P7$ 7$ 7$ . 7$
7' 7' ')
Substitution into the differential equation yields , which further* * ' % * ''. 7$
7P) & ' )
results in The latter differential equation is linear, and can be written asP - . P % . !8 & '
 ! !& % . ! P $ % . $ & - #& !. $ *. $ *. $8& & &' ' The solution is given by Converting back to
the original dependent variable, ' % JP !*+(.
31. Since , set . It follows that and 2 % ) P % ' % * .' % * !*. *)7P 7P7$ 7$ 7$ . 7$
7' 7' ')
The differential equation is written as , which upon* * #30 $ - U ' % ''. 7$
7P) !( ' )
further substitution is This ODE is linear, with integratingP - . #30 $ - U P % .!8 !(
factor The solution is ( ( ! ! !% &"$ % &56 . #30 $ - U 7$ % &56 * . 012 $ - .U $ !
P $ % .&56 . 012 $ * .U $ &56 * . 012 - .U 7 - # &56 * . 012 $ - .U $ ! ! ! ! !.( ( $ $ $ (
/
$
Converting back to the original dependent variable, ' % JP !*+(.
33. The solution of the initial value problem , is ' - .' % / ' / % + ' $ % & !+ + + +
8 *.$ ! !
Therefore y On the interval the differential equation is ! ! !+ % ' + % & ! +E? E* + *.
' - ' % / ' $ % #& ! ' + % ' + % #& !. . . .
-8 *$ *+, with Therefore Equating the limits ! ! !
' + % ' + # % & ! ! !* - , we require that Hence the global solution of the initial value*+
problem is
' $ % !
& / O $ O +
& $ ; +
 ! / *.$*+*$, ,
Note the discontinuity of the derivative
' $ % !
* .& / < $ < +
* & $ ; +
 ! / *.$*+*$ ,,
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Section 2.5
1.
For , the only equilibrium point is . , hence the equilibrium' N / ' % / @ / % ; //
\ 8 !
solution is .) !$ % / unstable
2.
The equilibrium points are and . , therefore the' % * (" ' % / @ * (" < /\ \ 8 !
equilibrium solution is .) !$ % * (" asymptotically stable
3.
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4.
The only equilibrium point is . , hence the equilibrium solution' % / @ / ; /\ 8 !
) !$ % /
is .unstable
5.
The only equilibrium point is . , hence the equilibrium solution' % / @ / < /\ 8 !
) !$ % /
is . 0'S6$3$1# HH' stable
6.
—————————————————————————— ——CHAPTER 2. 
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 page 55
7 . !"
8.
The only equilibrium point is . Note that , and that for .' % + @ + % / ' < / ' = +\ 8 8 !
As long as , the corresponding solution is . Hence the' = +/ monotone decreasing
equilibrium solution is .) !$ % + semistable
9.
—————————————————————————— ——CHAPTER 2. 
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 page 56
10.
The equilibrium points are , . . The equilibrium solution' % / J+ @ ' % + * )'\ 8 . !
) !$ % / is , and the remaining two are .unstable asymptotically stable
11.
12.
The equilibrium points are , . . The equilibrium solutions' % / J. @ ' % A' * 4'\ 8 ) !
) ) ! !$ % * . $ % - . and are and , respectively. Theunstable asymptotically stable
equilibrium solution is .) !$ % / semistable
—————————————————————————— ——CHAPTER 2. 
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 page 57
13.
The equilibrium points are and . . Both equilibrium' % / + @ ' % .' * :' - 4'\ 8 . ) !
solutions are .semistable
15 . Inverting the Solution , Eq. shows as a function of the population ! ! ! ++ +) $ '
and
the carrying capacity . With ,k ' % k()/
$ % * H2 !
+ +() + * '(k
K '(k + * +()
4 4 !# $ ! !# $ !
Setting ,' % .'/
$ % * H2 !
+ +() + * .()
K .() + * +()
4 4 !# $ ! !# $ !
That is, If , .$ $% H2 4 ! K % /!/.B % BB!4B+K per year years
 ! !" ! +) ' (k % '(k % In Eq. , set and . As a result, we obtain/ * +
U % * H2 !
+ + *
K + *
4 4# $# $
* +
+ *
Given , and , .* + $% /!+ % /!, K % /!/.B % +DB!DAper year years
16 . ! 
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17. Consider the change of variable Differentiating both sides withh % H2 '(k ! !
respect
to , Substitution into the Gompertz equation yields , with$ h % ' ('! h % * Kh8 8 8
solution It follows that That is,h % h & ! H2 '(k % H2 ' (k & !/ /
*K$ *K$ ! !
'
k
% &56 H2 ' (k & !' ( !/ *K$
 ! ! k % A/!B Z +/ ' (k % /!.B K % /!D+ ' . % BD!BA Z +/. Given , and , .: :/ per year
 !" ! $ Solving for ,
$ % * H2 !
+ H2 '(k
K H2 ' (k
0 1 ! !/
Setting , the corresponding time is .' % /!DBk % .!.+ !$ $ years
19 . The rate of of the volume is given by rate of rate of . ! *increase flow in flow out
That is, Since the cross section is , 7^ (7$ % Y * .IM ! 7^ (7$ % W7M(7$!* ) constant
Hence the governing equation is 7M(7$ % Y * .IM (W!% &)*
 ! % &" ! 7M(7$ % / M % ! Setting , the equilibrium height is Furthermore, since& + Y.I .*
@ M < /8 !& , it follows that the equilibrium height is .asymptotically stable
 ! !# " M. Based on the answer in part , the water level will intrinsically tend to approach .&
Therefore the height of the tank must be than ; that is, .greater M M < ^ (W& &
22 . The equilibrium points are at and . Since , the ! ! ' % / ' % + @ ' % * . '\ \ 8 * *
equilibrium solution is and the equilibrium solution is) )% / % +unstable
asymptotically stable.
 ! # $ !" ' + * ' 7' % 7$. The ODE is separable, with . Integrating both sides and*+ *
invoking the initial condition, the solution is
' $ % !
' &
+ * ' - ' &
 ! /
/ /
*
*
$
$
It is evident that independent of and . ! ! !' ' $ % / ' $ % +/
*? ?
lim lim
$C $C
23 . ! ! ' $ % ' & !/ * $+
 ! !" 75(7$ % 5 ' & ! 75(5 % ' & 7$ !. From part , Separating variables, * */ /* $ * $+ +
Integrating both sides, the solution is 5 $ % 5 &56 ' ( + * & ! ! ' (% &/ /* + * $+
 ! ! ! !# $>? ' $ >/ 5 $ >5 &56 ' ( !. As , and Over a period of time, the/ /* + long
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proportion of carriers . Therefore the proportion of the population that escapesvanishes
the epidemic is the proportion of left at that time, susceptibles 5 &56 ' ( !/ / !* +
25 . Note that , and . So if ! ! # $ ! ! ! @ 5 % 5 g *g * 5 @ 5 % g *g * ) 5# #. 8 . ! !g *g < / 5 % / @ / < /# , the only equilibrium point is . , and hence the solution\ 8
) !$ % / is .asymptotically stable
 ! ! !)" g * g ; / 5 % / J g *g ( !. If , there are equilibrium points ,# #three \
Now , and . Hence the solution is ,@ / ; / @ J g *g ( < / % /8 8 ! !% &) # ) unstable
and the solutions are .) % J g *g ( ) !# asymptotically stable
 !# .
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Section 2.6
1. and . Since , the equation is` 5E ' % .5 - ) l 5E ' % .' * . ` % l % / ! ! ' 5
exact. Integrating with respect to , while holding constant, yields ` 5 ' , !5E ' %
% 5 - )5 - M ' % M ' l. 8 ! !. Now , and equating with results in the possible,'
function . Hence , and the solution isM ' % ' * .' 5E ' % 5 - )5 - ' * .' ! !. . .,
defined .implicitly as 5 - )5 - ' * .' % #. .
2. and . Note that , and hence the` 5E ' % .5 - 4' l 5E ' % .5 * .' ` = l ! ! ' 5
differential equation is not exact.
4. First divide both sides by We now have !.5' - . ! ` 5E ' % ' l 5E ' % 5 ! ! and .
Since , the resulting equation is ` % l % /' 5 exact. Integrating with respect to ,` 5
while holding constant, results in ' , ! !5E ' % 5' - M ' . Differentiating with respect
to , . , we find that , and hence ' % 5 - M ' % l M ' % / M ' % /, ,' '
8 8 ! ! !Setting 
is acceptable. Therefore the solution is defined . Note that ifimplicitly as 5' % #
5' - + % / , the equation is trivially satisfied.
6. Write the given equation