Solutions Manual for Differential Equations 2nd Edition by Polking IBSN 9780134689586 Full clear download( no error formatting) at: http://downloadlink.org/p/solutionsmanualfordifferentialequations2ndeditionbypolkingibsn9780134689586/
Chapter 2. FirstOrder Equations
Section 2.1. Differential Equations and Solutions 1. </J(t, y, y') = t2y' + (1 + t)y = 0 must be solved for
4. y'(t) + y(t) = (2  ce1 ) + (2t  2 + ce1) = 2t
y', We get
y' =
2. </J(t, y, y') We get
= ty'
(1
+ t)y t2
 2y  t2 must be solved for y'.
f
2y
+ t2
y=.
t
5. If y(t) = (4/5) cost+ (8/5) sin t + ce<112 )t, then
12 3. y'(t) = soy'= ty.
2 Cte<1!2)t
and ty(t)
=
tce<112)12,
y ( t ) '
+ ( l / 2 ) y ( t )
= [(4/5) sin t + (8/5) cost  (C /2)e<112 t] + (l/2)[(4/5) cost+ (8/5) sin t + ce(1/2) ]
= 2cost.
)
1
2.1 6. If y(t) = 4/(1
(a) If t2

4y2 = c2, then
16C eC4t)
,
d
+ C eHt))2
y = (1 y(4 y)
9.
+ ce4t), then
= 1 + �e4t x [4 1 +
�e••
13
Differential Equations and Solutions
d
(t2  4y2) = t2 dt dt 2t  8yy' = 0 t 4yy' = 0.
J
+ ce  16 + C eHt))2 4t)
16(1
=
(1
16C eC4t)
=
7. For y(t) = 0, y'(t) 0(4 0) = 0. 8.
(a) If t2
+ y2
(b) If y(t) = ±Jt2  C2 /2, 2 ±t /(2Jt  C2), and
+ C eHt))2
(1
0 and y(t)(4  y(t))
t  4yy' = t  4[±Jt2
,
=tt
d (t2
dt 2t
::rt/,Jc2
t
 t2,
=
=
= C2 then
(b) If y(t)
then y'
d + y2) = c2 dt + 2yy' = 0 t + yy' = 0
±,Jez  t2, and
then
2 
C /2][±t/(2Jt
2
2

C
)]
=0.
y'
+ yy' = t + [±J c2  t2][::rt ;J c2  t2J
(c) The interval of existence is either oo < t < C or C < t < oo.
(d)
=tt
=0. (c) For J C2  t2 to be defined we must have t2 ::::; C2. In order that y' (t) = ::rt / ,JC2  t2, we must restrict the domain further. Hence the interval of existence is C < t < C. (d)
10. If y(t) = 3/(6t  11), then y' = 3 · 6/(6t 11)2 = 18/(6t 11)2• On the other hand, 2y2 = 2[3/(6t  11)]2 = 18/(6t  11)2 , so we have a solution to the differential equation. Since y(2) = 3/(12  11) = 3, we have a solution to the ini• tial value problem. The interval of existence is the interval containing 2 where 6t  11 # 0. This is the
14
Chapter 2
FirstOrder Equations the interval (  co, oo).
interval (11/6, oo).
2
3
11. See Exercise 6. (oo, ln(5)/4).
2
4
5
The interval of existence is
13. y(t)
= ! t2+2. 3
3t
Theintervalofexistenceis (0, oo).
1
1
2
3
14. We need e1 = y(l) = e1(1 + C/(1)) = (1 C)e1. Hence C = 0, and our solution is y(t) 12. y(t) = (4/17) cost+ (1/17) sin t  (21/17)e4 t on
+ =
ie:', This function is defined and differentiable on the whole real line. Hence the interval of existence
2.1
Differential Equations and Solutions
Butthis equals (1/2)(t+2) only if t in the following figure.
is the whole real line.
0 >,.
�
15
2, as shown
y
·1 ·2 ·3 .._ ._ 0
......._ 2
.....__
4
(2, 0)
15. y(t) = 2/(l+e2113 ). Theintervalofexistenceis ( ln(3)/2, oo).
1 The second solution proposed by Maple, y(t) = (1/4) (t2)2, satisfies theinitialcondition, as y(O) =
(0,3)
(1/4)(0  2)2 = 1. But
1
y'(t) = 2(t  2),
16. The initial value problem is y' =
Jy,
y(O) = 1.
The first solution proposed by Maple, y(t) = (1/4)(t + 2)2 , satisfies the initial condition, y(O) = (1/4)(0 + 2)2 = 1. Next, 1 y'(t) = 2(t
+ 2),
and
/y0 =
J.41.<t 2)
2
=
.1.11 21. 2
and But this agrees with (1/2)(t  2) only if t � 2, as
16
Chapter 2
FirstOrder Equations 18.
shown in the following figure.
y'=lt
I I I / I I
y >,
0 1
(0, 1)
2
\
/
\
\
/
2
0
1
\
(2, 0)
19. y ' = t tan(y/2)
\ >,
: /
<,


 

/ :
<,
\
0 
Note that this graph does not pass through (0, 1). Hence, y(t) = (1/4)(t  2)2 is not a solution of the initial value problem.
1
2
17.
0
1
2
20. y'=y+t
/
\ >,
0
1
\ \ \ \ 2
1
y ' = f y/(1 + y2)
I I / I
1.5
0.5 >,
0
I 
:

:
I
  
0.5
\ 0
/ 2
1 1.5
\ 2
<, 1
<, 0
\ 2
2.1
Differential Equations and Solutions
the righthand side of
21.
y+t yt
I
y'=ty
y =• is undefined.
5
·5
>,
2
0
2
0
22.
5 5
5
0
4 2 >,
25.
0
2
4 5
0
10
23. y'=ty+1
6
26.
4 2 >,
0
2
4 ·6 5
24.
0
5
Note the difficulty experienced by the solver as it approaches the line y = t, where the denominator of
4
2
4
17
18 Chapter 2 FirstOrder Equations 27.
30.
y ' = 3 y + sin(t)
y
5 4 3
0
15
10
5
2 1
10
0
I
1
2
3
4
20
5
I
l I t
\· ·\· ·\·
,_:_,_ \ \ ''  . .% / ,··y,·....:.·.;.:·;,·1·
·1··r,·,
x. \ ,·_;..;
1: \. : \. .'. ,.: I \ ·\,,._� /I ·/ v \ > , _:: ., ,;· / / i ·: I 
i
<, .;__ .:. / .:
I
�
. \
\,
� \
,
..,
�
0. /
,/ 
<: <, ·/. I ·1 ,  :;,,; / /, I '.,
'\. , : ..... _.:..,. .; .r. :.1
·1 (
, ,.  ,,,, /.
I · I
I
_;... .,..:.,,, .j. .,; .,.:.,
i
\
v ,,� 
\· ·\· ·1 ·I
1: _1_:_1_) � \ .\ \
\ ··,
<, ·.......,,,, ··/
<; _;_._;, "/
r> I '/
5 4 3 2 1
0
r
1.: ./ ;t .( I · I
I
i , .( , ·,·.,'I
30
I·· f ·· f I .I / I : i I I . 1 J .' I I I · I ·1 1 . I 
t,
l : ;,
y
29.
32.
4
I / I 1  .
I  . I
2 3 4 5
s
2
I
l .
i t. .,.:, i  .
31.
28.
i
2
2.1
Differential Equations and Solutions
19
tion curve.
33.
200
·I·# ·I I I I i I I I I I I I
f
150
a. 100
Tl'/ I I I I
I I I I
·,· i
I I I I ) I
0
I I I I
/
I ·I·# ·I' I ·II·/· I I I 1· I I I I I I I (II I I I I I I. I I I I I I I I. I I I I I Tl 1:1 ·1 I'/" I I I t, I I I I I I I f· I I I I I I I I· I I I I I I I /· I I I I i ·,: i ·,· I I
i ·,·
I I I I I I
r r
I I I I I I
I I I
I I I
I I I
I·/·"/··/·/· I I 1 I I I I 1 I I I I I I I I I I I I I I ·1·.1,·, I'/' I I . I I I I I 1 I I I I I I I I I I t. I I I I 0 0
I:,
i
l·I· I· I I I I Ii I I I I I I t :t: 1· I I I I I I I I I I I I
·,; ·,· i 'ii·, i
I I I '/ I I I I I I I A I '/ I I I I I I I 'I I·, I I I I I )
� f� � � �  � � � 50 f1,,,,,�,, � � � � � ·�: � � �·i·,,,,,,,,,1 I
0
10
5
15
Using the solution curve, we estimate that P(IO) � 124. Thus, there are approximately 124 mg of bac• teria present after 10 days. 34.
36. We must solve the initial value problem dA
 = 025A dt
.
A(O) = 400.
'
Using our numerical solver, we input the equation and initial condition, arriving at the following solu• tion curve. ·6
4
4
6 500 �,�,�'�'�'�'�'�'�'�'·
'� ' ' ' ' ' ,·· ' ' ' ' ' ' ··, ' ' ' ' '�''
400 300 200
'' ' ' ' ' ''
....
'
.....
..;.,,
,
:...·..:..:......:..·
:
,
·
=
:
35. We must solve the initial value problem
dt
P(O) = 1.5.
.
...,_;
...............
.
.
..........
� � � � � � � � �  � 
� ?
0 0
dP  = 0.44P,
·
.
·.....
100
''' ''
� ,�, ,� , �� ,� ,� �' .,,,,,,,,,,,,,� ,,,,,,
2
4
6
Use the solution curve to estimate A(4) � 150. Thus, there are approximately 150 mg of material remaining after 4 days. 37. We must solve the initial value problem
Using our numerical solver, we input the equation
de
and initial condition, arriving at the following solu
dt = 0.055c,
c(O) = 0.10.
20
Chapter 2 FirstOrder Equations Using our numerical solver, we input the equation and initial condition, arriving at the following solu• tion curve.
0.1
\. \ ·\· \ v \ .,: \ ·\ \ •\· \ .,.·, ·\· \ ' 11. \ \
,.
\ \
·
' ' ' ,·
\ \
\ \ ·
' \
\
\
\
'
,;: \
·\·\·\·'I..
\ \ ,;.
' ·, '
Use the solution curve to estimate that it takes a little less than 15 minutes to cool to 100° Celsius. 39. The rate at which the population is changing with
respect to time is proportional to the product of the population and the number of critters less than the "carrying capacity" (100). Thus,
'
�. 'v�'v ' 'v ,: 'v ' 'v ' 'v \ :,v ' \� ' ''� '\ '\'i.,· 0.08 v
,. ,, ,,, ,,� �,,,, ,,,,� , ,,, ,,, ,, ,
dP
��,,,,,,�,,,,,, •
���'
0.06 ,,,,,
�� �� � '��� , � ,� ,� � ,, ,, ,, ,,,, ,, �
o
dt
= kP(lOO P).
�
�  � , ,, , �, �
' ' ' '' ,, , , , ,, 0.04 '� '' �, � � �� � � �� �� ��
0.02
������.����.��·��.. '
With k
= 0.00125 and an initial
population of 20
ritters, we must solve the initial value problem
���
c dP dt
0 · () 10 20 30
Use the solution curve to estimate that it takes a little more than 29 days for the concentration level to dip below 0.02.
38. The rate at which the rod cools is proportional to the difference between the temperature of the rod and the surrounding air (20° Celsius). Thus,
 = 0.00125P(100 P),
Note that the righthand side of this equation is posi• tive if the number of critters is less than the carrying capacity (100). Thus, we have growth. Usingournu• merical solver, we input the equation and the initial condition, arriving at the following solution curve.
100
dT
0.085(T  20),
T(O)
.
.
= 300,
where T is the temperature of the rod at t minutes. Note that since the initial temperature is larger than the surrounding air (20° Celsius), the minus sign in• sures that the model implies that the rod is cooling. Using our numerical solver, we input the equation and initial condition, arriving at the following solu• tion curve.
'.
\ ·\· \ ·\· \
. ,:
\ ·\· \ ·\· \ ·\·
··,
•\· \ ·\· \ ·\·
.
00 ��;�� �;�;:�;:�;:;;;:. � � ,,,,,,, .,,,, ,,,,,, ,,,,,, ,,,,,,�.,,,,, �
a.
�
�
�;;�:� �;� �;� ;,;; ;; i,,,�
� w�� �� , ,_,_, �· �,
,,
=
300
.
00 � ,,_,� ,,,�/·,,,, �· ,,, ,,,,, ,,,,, ,.,. .; .; .,, .,, ,,,. .,;. .; ., ; .,,
With k == 0.085 and an initial temperature of 300° Celsius, we must solve the initial value problem T
··· •
    
dt = k(T  20).
dt
P(O) = 20.
'.
,,,
, ,,
,� , ,
,, ,,, ,, ,
�� � ��
0 0
10 · 20 30
Use the solution curve to estimate that there are about 91 critters in the environment at the end of 30 days.
\
250 �.. \\ '\
200 t
\
\
\
\ \
\ \
\
\
\
\
\
\
\
\
\
�
\ \ \ \ v' \ \ \ \ \ \ ·, \ \ \ \ \ '\ .,; \ \. \ .\ \ .\. \ \. \ .\. \ \. \ \ \ v, \ \ \ \ \ \ .v \ \ \ \ \ ..
\. \ .\. \
' ' .v,' '\, .v,,. '\, .v, '\,' .v,' '\, .v...,\ .v, \' .v, \' .,. v . ,',,',','� ,',',',','�., ,' ,',',',',' ,� ,'��,. ,,,,,,�,,,,,, ·'\ · \ .v,
\
\
\
\
\
\
\
\
150 ·,,,,,,� · ' ·,:' ·,· ' : ' ·,·,',·,·,',·. . .,",·...:,',"'°,',"<,' . ,,,,,,� ,,�;, �' ''''' ', � · ,· .·,·,,  ...::.. 100 ·. ;, ;...;.·,•.•.. 50
,· ��,� ��
0 10 0 20 30
2.2
Solutions to Separable Equations
21
Section 2.2. Solutions to Separable Equations . 1. Separate the variables and integrate.
5. Separate the variables and integrate.
dy
dy
dx = y(x
=xy dx
dy
1
dy = (x
=xdx
y
y
1
2
ly(x)I y(x)
1
+C
+x +C
2 IYI = e(lf2)x2+x+C
= ex2;2+c
2 y(x) = ±ec e(lf2)x +x
= ±ec . ex2;2 = Aex2;2'
Where the constant A = ±ec is arbitrary. 2.
+ l)dx
In IYI = x2
2
In IYI = x
+ 1)
2
Letting D = ±ec, y = De(l/2) x +x. 6.
Separate the variables and integrate. Note: Factor the righthand side.
Separate the variables and integrate.
dy
dx = (ex+ l)(y  2)
dy x=2y dx
1
1 dy =(ex+ l)dx y2 In I y  21 = ex + x + C
2
dy = dx y x
In I y I = 2 ln Ix I
I YI
y  2 = ±eceex+x
= elnx2+c
y(x) = ±ecx2 Letting D
IY  21 = eex+x+C
+c Letting D
= ±ec, y(x) = Dx2.
= ±ec, y(x) = Deex+x + 2.
7. Separate the variables and integrate,
dy
(y
dy = exy dx eYdy = ei dx
1
y2
+ C)
4. Separate the variables and integrate.
dy
 = (1
dx
1
+
dx y+2 2) dy = x dx,
1
2y2+2y=2x2+c,
eY =ex+ C y(x) = In(ex
x
=,
3. Separate the variables and integrate.
+ y2)ex
+ 4y
+ D)
= 0,
With D replacing 2C in the last step. We can use the
quadratic formula to solve for y. y(x) = 4
± )16 + 4(x2 + D) 2
y(x)
x
  dy = e dx 
 (x2
= 2 ± Jx2 + (D + 4)
+y
tan1 y =ex+ C y(x) = tan(ex
+ C)
If we replace D + 4 with another arbitrary constant E, then y(x) = 2 ± ,Jx2 + E.
22 Chapter 2 FirstOrder Equations 8. Separate the variables and integrate.
:: =yV1) (1 + )
.!_dy = y
x 1
In I y I = x
+ In Ix
11. (y3  2) dy = x dx (y3  2)dy = xdx
dx
y4
11 + c

IYI = ex+ln lxlJ+C
T +c.
The solution is given implicitly by the equation y4  8y  2x2 = A, where we have set A= 4C.
y(x) = ±ecexelnlx11
Letting D = ±ec, y(x) = Dex Ix  11. It is impor• tant to note that this solution is not differentiable at
12. dy
x = 1 and further information (perhaps in the form
dx
of an initial condition) is needed to remove the ab• solute value and determine the interval of existence.
+ 1)
2x(y
=
dy
x2

1
2xdx
+ 1 x2 1 2 In IY + 11 = In lx  11 + C IY + 11 = elnJx211+c = ec lx2 11 y
9. First a little algebra. x2y' = ylny  y' (x2
x2
42y =
+ l)y' = ylny
y(x) = A(x2  1)  1.
Separate the variables and integrate. 1
13.
1
dy=dx yln y x2 + 1
1
dy
1
dx dy
du= dx, u x2 + 1
where u tan"! x
= In y
y
and du = } dy. Hence, ln lul =
etan1
x+c
y(x) = Ax.
The initial condition y(l) = 2 gives A = 2. The solution is y(x) = 2x. The solution is defined for all x, but the differential equation is not defined at x = 0 so the interval of existence is (0, oo).
Let D = ±ec, replace u with In y, and solve for y.
lny = D/an y(x) =
1
x
tan!
eDe
x
ly(x)I = elnlxl+C = ec lxl
= ±ec etan 1 x
U
=
x dx
}., I y I = In Ix I + c
+ C. Solve for u: lul
y
=
x
14.
10.
dy
x dy = y(l dx
dy = 1 
IyI
+ 2x
= ln
dt
2
dx=
x
y
In
+ 2x2) [
1 +2x 
x
Ix I + x + c 2
ly(x)I = elnlxl+x2+c = eclxlex2 y(x) = Axex
2
2t(I
J dx
1
1
 ln(l 2
+ y2) y
ydy = 2t dt + y2
+ y2) 1 + y2 1 + y2
= t2
+C
= e2,2+c = ec e2t2 = Ae2t2
2.2
= 1, 1 + 12 = Ae2(0 )2 and A= 2. Thus,
With y(O)
1
+ y2
Solutions to Separable Equations
16. dy = ex+y dx ey dy = ex dx
= 2e2t2
y = ±J2e2t2  1.
eY =ex+ C
c
ey = ex We must choose the branch that contains the initial 2
y = ln(ex  C) y = In(ex  C)
condition y(O) = 1. Thus, y = J2e2t  1. This solution is defined, provided that With y(O) = 1,
2
2e2r
23
1 > 0

e2t2 >
1 = ln(e0  C)
!
1 C = e1
2
C=le1.
2t2 >In! 2
2
t < 21n
Thus,
!2
This solution is defined provided that
t2 < ln4
ltl
<
+ e1 + 1).
y =  In ( ex
ex+ e1
Jin4.
+1
> 0
+1 ln(e1 + 1).
ex< e1 Thus, the interval of existence is
(v1n4, v1n4).
x <
Thus, the interval of existence is (oo, ln(e1 + 1)).
15. 17. dy dx
sinx =
dy = 1 dt
y
ydy = sinxdx 1
2y
2
y2
+ C1 = 2cosx + C
_!}z_ 1
= cosx
y(x) =
(C
For the initial condition we have 1 = y(O) = tan C, so C = n/4 and the solution is y(t) = tan(t + Jr /4). Since the tangent is continuous on the interval (n/2, n/2), the solution y(t) = tan(t + n/4) is continuous on the interval (3n /4, n /4).
and the solution is
,vl  2cosx.
The interval of existence will be the interval con• taining Jr /2 where 2 cos x < 1. This is n /3 < x <
+C
y(t) = tan(t + C)
= 2C1)
Using the initial condition we notice that we need the plus sign, and 1 = y(n /2) = ,Jc. Thus C = 1
Sn/3.
= dt
tan1(y) = t
±,Jc  2cosx
y(x) =
+ y2
+ y2
18. dy x dx 1 + 2y (1+2y)dy=xdx
2
y
+ y2
= x
/2 + C
24 Chapter 2
FirstOrder Equations
This last equation can be written as y2 + y  (x2 /2 + C) = 0. We solve for y using the quadratic formula
dy x = dx y ydy=xdx
+ 4(x2 /2 + C1) J /2
y(x) = [1 ± J1 =
20.
[1 ± J2x2 +CJ /2
(C =
1+
4C1)
1
2y
For the initial condition y(1) = 0 we need to take the plus sign in order to counter the 1. Then the initial condition becomes O = [1 + J2 + C]/2, which means that C = 1. Thus the solution is 1
y(x) =
+ J2x2 1 2
y2
1
=  x
2
2
+C
= 2C x2
y =
±J2Cx2
With y(O) = 2, we choose the positive branch and 2 = J2c leads to C = 2 and the solution y = J4  x2 with interval of existence (2, 2). This solution is plotted as a solid curve in the fol• lowing figure. With y(O) = 2, we choose the neg• ative branch and 2 = J'fc leads to C = 2 and the solution y  J4  x2 with interval of existence (2, 2). This solution is shown as a dashed curve in the figure.
.
For the interval of existence we need the interval con• taining 1 where 2x2  1 > 0. This is oo < x <
1/v'2,.
19.
2
Withy(O) = 1,wegetthesolutiony(x) = Jl +x2, with interval of existence (oo, oo). This solu• tion is plotted with the solid curve in the follow• ing figure. With y(O) = 1, we get the solu• tion y(x) = JT+x2, with interval of existence (oo, oo). This solution is plotted with the dashed curve in the following figure.
y
i
/
2
0
x

/
<,
2
0 /
/
2
/
2 <,
21.
<, <,
/
4
With y(O) = 3 the solution is y(t) = 2 + «'. on ( oo, oo). This solution is plotted is the solid curve in the next figure. With y(O) = 1 the solution is y(t) = 2  e:', on (oo, oo). This solution is plot
2.2 23. We have N(t)
ted is the dashed curve in the next figure.
N (t
Solutions to Separable Equations
25
= N0eM, and + T1;2)
= NoeA(t+Ti12) = NoeAt . eAT1;2
= N(t). eAT1;2 = N(t) ·
1
2
if eH1;2 = 1/2, or T1;2 = In 2/A .
...e 2
I
/
8
0
24.
4
2
= In 2/ T1;2 = 1.5507 x
5
Hence J...
=
=
In 2/J...
=
26. Using T112 = 6 hours, we have J... = In 2/T1;2 0.1155. Then N(9) = lOe9). = 3.5355kg.
=
27. Using T1;2 = 8.04 days, we have J... = ln 2/ T112 0.0862. Then N(20) = 50oe2m. = 89.1537mg.
=
25. We have 80
2
+1
= dx
+ 1) = x + C y2 + l = e2x+2C = Ae2x
ln(y2
y2
N(4)
=
lOoe4).,. T1;2
y
dx ydy
=
ln(l00/80)/4 = 0.0558. Then 12.4251 hours.
+1
y2
dy
1
.
108 years.
22.
y2
10
(b) We have No = 1000 and N(t) = 100. Hence 100 = 1000 · «". or t = ln 10/A = 1.4849 x
I
I
(a) A
=
y(x) =
Ae2x 
±J
(A
= e2C)
1
Ae2x 
1
This is the general solution. The initial condition y(l) = 2 gives
2=
+J Ae2 1
4 = Ae2

1
2
28. The decay constants are related to the halflives by A210 = ln(2)/2.42 = 0.2864 and J...211 = ln(2)/15 = 0.0462. The amount of P'Rn is given by x(t) = x0eA210t and of 211 Rn by y(t) = y0eA211t. The initial condition is that y(O)/x(O) = Yo/xo = 0.2/0.8, = 1/4, so 4x0 = 4yo. We are looking for a time t when 0.8/0.2 = 4 = y(t)/x(t) = et(A210J..211) /4. Thus we need et(A2ioJ..2ll) = 16. From this we find that t = 11.5 hours.
A= 5e
,
The particular solution is
y(x) =
J 5e2x2  1.
The interval of existence requires that 5e2xZ

1 > 0
2x  2 > ln(l/5) x > 1  ln(5)/2 � 0.1953. Thus the interval of existence is 1  ln(5) /2 < x < 00.
29.
(a) If N
= N0eJ..
1
then substituting T).,
N
= 1/J...,
= NoeAT,_ =
NoeA(l/A)
= Noe1• Therefore, after a period of one time constant T)., = 1/A, the material remaining is Noe1. Thus, the amount of radioactive substance has decreased to e1 of its original value N0•
26
Chapter 2 FirstOrder Equations (b) If the halflife is 12 hours, then
T1;2 = In 2/J.. = 7.927days.
7
!No= Noe}..(12) 2
6.8
e m = 1 2 12>..
1
a:
= ln
.f:
2
In l A=12
6.6
6.4 6.2 6
2
4
6
8
10
Day
Hence, the time constant is
31. The halflife is related to the decay constant by
1 12 T). =    � 17.3hr. J.. In l2
ln2
T1;2= .
J..226
(c) If 1000 mg of the substance is present ini• tially, then the amount of substance remain• ing as a function of time is given by N = 1000e(tln(l/2))/I2. The graph over four time pe
The decay rate is related to the number of atoms present by
Substituting, Nln2 T1;2 = R.
riods ([O, 4T). ]) follows.
Calculate the number of atoms present in the lg sam• ple. 1000
1 mol 6.02 x 1023 atoms 226 g mole 21 = 266 x 10 atoms.
N = lg x  x 
800 600
z 400
Now,
200 0
30.
(2.66 x 1021 atoms)(ln 2) T1;2 0
20
40
60
80
The data is plotted on the following figure. The line is drawn using the slope found by linear regression. It has slope A = 0.0869. Hence the halflife is
=
3.7 x 10 10 atom/s
10
= 4 .99 x
10 s.
In years, T112 � 1582 yr. The dedicated reader might check this result in the CRC Table. 32. (a) Because half of the existing 14C decays every 5730 years, there will come a time when phys• ical instruments can no longer measure the re• maining 14C. After about 10 halflives (57300
2.2
years), the amount of original material remain• ing is No
1)10 (2
� 0.00097N0,
ln2 5568
).., =  =  � 0.0001245.
We can now write N = Noeo.0001245t.
The ratio remaining is 0.617 of the current ratio, so 0.617 No = Noeo.0001245t eo.0001245t = 0.617 0.0001245t
= In 0.617
In 0.617 t=0.0001245 Thus, the charcoal is approximately 3879 years old. 33. Let t = 0 correspond to midnight. Thus, T (0) = 31 °C. Because the temperature of the surrounding medium is A = 21 °C, we can use T = A + (To A)ekt and write T = 21 + (31  21)ekt = 21 + lOekt.
At t = 1, T k.
Solutions to Separable Equations
= 29°C, which can be used to calculate
27
Let y(t) be the temperature of the beer at time t minutes after being placed into the room. From New• ton's law of cooling, we obtain y'(t) = k(70  y(t))
a very small amount. (b) The decay constant is calculated with ln2 T1;2
34.
y(O) = 40
Notekispositivesince70 > y(t) andy'(t) > O(the beer is warming up). This equation separates as dy
=kdt 70y which has solution y = 70  Cekt. From the initial condition, y(O) = 40, C = 30. Using y(lO) = 48, we obtain 48 = 70  30e10k or k = (1/10) ln(ll/15) or k = .0310. When t = 25, we obtain y(25) = 70 30e·598 � 56.18°. 35. The same differential equation and solution hold as in the previous problem: y(t) = 70 
ce:"
We let t = 0 correspond to when the beer was dis• covered, so y(O) = 50. This means C = 20. We also have y(lO) = 60 or 60 = 70  2oel0k Therefore,k = (1/JO)ln(l/2) � .0693. We want to find the time T when y(T) = 40, which gives the equation 70  20ekT = 40
29 = 21 + lOek(l) k =  ln(0.8) k � 0.2231 Thus, T = 21 + 10e0·2231t. To find the time of dea1h, enter "normal" body temperature, T = 37°C and solve fort.
Since we know k, we can solve this equation for T to obtain T = (1/k)ln(3/2) � 5.85
or about 5 .85 minutes before the beer was discovered on the counter.
37 = 21 + 10e0.223lt In 1.6 t=0.2231 t � 2.l hrs Thus, the murder occurred at approximately 9:54 PM.
36. x' = [at+by+c]' = a+by' = a+bf (at+by+c) = a+ /(x).Fortheequationy' = (y+t)2weusex = t+y. Thenx' = l+y' = l+(y+t)2 = l+x2.Solv• ing this separable equation in the usual way we get thegeneralsolutionx(t) = tan(t+C).Intermsofthe unknown y, we get y(t) = x(t)t = tan(t+C)t.
28 Chapter 2
FirstOrder Equations
37. The tangent line at the point (x, y) is y  y y'(x)(x  x) (the variables for the tangent line have the hats). The P intercept is.q., = y/y' +x. Since (x, y) bisects the tangent line, we have Xint = 2x. Therefore: 2x
or
= y'
y +x y'
which can be separated as dr / r = cot(() /2)d(). This can be solved for r as r(()) = C sin2(e /2), where C is a constant. 40.
from Oto xis
fox y(t) dt
1
=y x
which by assumption, equals (1/4)xy(x) (one• fourth the area of the rectangle). Therefore
This separable differential equation is easily solved to obtain y(x) = C /x, where C is an arbitrary con• stant.
fox y(t) dt = (l/4)xy(x).
38. With the notation as in the previous problem, the equation of the normal line is
" y y
= y(t)
The area under the curve y
Differentiating this equation with respect to x and using the Fundamental Theorem of Calculus for the left side gives
1 " = ( x) x y'(x)
The .xintercept is found to be Xint = yy' Xint  x is given to be ± 1, we obtain
+ x.
y(x) = (1/4) (y(x)
Since
+ xy' (x)).
This equation separates as
yy' = ±2
y'
with solution y2 = ±4x + C, where C is an arbitrary constant.
y
3 =
x
which has the solution y(x) 39. Let </> be the angle from the radius to the tangent. 41.
= Cx3.
Center the football at the origin with equation
z2+x2
y2
+4
=4.
The top half of the football is the graph of the func• tion
z = .J4  x2

y2 /4.
The (x, y )components of the path of a rain drop form a curve in the (x, y) plane which must always point in the direction of the gradient of the function z (the path of steepest descent). The gradient of z is given by
+xi  (y/4)j
v z = ;::===== J4 x2  y2/4
From geometry, tan obtain
e = r dti]dr.
Since e
dr d() = r cot</> = r cot(() /2)
= 2</>, we
where i and j are the unit vectors parallel to the x and yaxis. Since the path traced by the drop, y = y(x ), must point in the direction of V z, we must have dy . z y  = slope of the gradient = .2 =  . x � �
2.3 This differential equation can be separated and solved as x = Cy4• C can be solved from the initial position of the drop (xo, Yo) to be C = xo/Yci· The final answer is given by inserting x = Cy4 into the expression for z: (x, y, z) = (Cy4, y,
J4  C2y8

y2/4),
dV dt
dt
dy = ny(t) dt ·
On the other hand, dV /dt is equal to where v is the speed of the water exiting the bowl. From the hint, v = ,J2gy(t). Thus we obtain the following differential equation:
dy
=
dt
dV
= na2
�
dt
= (C 
Since dy /dt = C (a negative constant), we obtain
Cn x" = na2 /2gy Solving for y, we obtain y = K x4 where K is a constant.
cate the destroyer at 4 miles along the positive xaxis. The destroyer wants to follow a path so that its arc length is always three times that of the sub. To ac• count for the possibility that the sub heads straight along the positive xaxis, the destroyer should first head from x = 4 to x = 1 (the sub would move from x = 0 to x = 1 in this same time frame under
y2gy(t).

This equation can be separated and solved for y as
y(t)
3
2
a\ffgt
)2/3
Since y(O) = 1, we obtain C = 1. Setting y(t) = 0, we obtain 2
this scenario). Now the destroyer must circle around the sub along a polar coordinate path r = r(()). We have r(O) = 1. If the destroyer intersects the sub at ( (), r (())), then the sub will have traveled2 r (()) and the destroyer would have traveled Jr'(t) + r(t)2 d() (arc length along the curve r = r(())). Since the speed of the destroyer is three times that of the sub, we obtain
Ji
t=
3a2vlg
in units of seconds (here g 43.
dt
44. Following the hint, let () be the polar angle and lo• na2v
ny(t)
2dy =JTX 
dV 2 2 �  = na v = na y2gy dt
42. Let y ( t) be the level of the water and let V ( t) be the
dV
29
From Torricelli's law:
(here, y is the independent variable). volume of the water in the bowl at time t. On the one hand, we have dV /dt = cross sectional area of the water level x dy / dt. The cross sectional area of the water level is ,r x2, where x is the radius. Using the equation of the side of the bowl (y = x2 ), we obtain
Models of Motion
change of volume, d V /dt is the cross sectional area multiplied by the rate of change in height, dy /dt. The cross sectional area is nx2 = nx(y)2• Thus
3(r(8)  1) =
= 32).
Let the unknown curve forming the outside of the bowl be given by y = y(x) (the bowl is then formed by revolving this curve around the yaxis). We can also write this equation as x = x (y) (reversing the roles of the independent and dependent variables). As in the analysis of the last problem, the rate of
1·
Jr'(t)Z
+ r(t)2 d&.
Differentiating this equation gives
3r'
= ...;'(r')2 + r2
or
with solution r(()) = e8fv'3.
dr
r(())
d()
,Jg
 =  r(O)
=1
30
Chapter 2
FirstOrder Equations
Section 2.3. Models of Motion 1. We need gt = c/5. or t = c/5g = 612,240 seconds. The distance traveled will be s = gt2 /2 = 1.84 x 1014 meters. 2. We need O = 9.8t2 /2 6.3 seconds.
+ l5t + 100.
3. The depth of the well satisfies d
required is t = m ln(2) / r. The distance traveled is x
The answer is
= mg
total trip takes 60 + 552.2 where 4.9tf
=
ti
= 581.5s.
�g [; (1 ert/m) _ t]
=
� g [;, _
�:g rn1n2J
10.
(a) First, the terminal velocity gives us 20 = mg/r, or r = mg/20 = 70 x 9.8/20 = 34.3. Next, we have v(t) = ce:"!"  mg/r. Since v(O) = 0, C = mg/r, and v(t) = mg(ert/m_ 1) / r. Integrating and setting x (0) = 0, we get
1'
x =
v(s)ds
The
= mg
4.9(T + 1)2
= 2 x 4.9T2 , or 4.9(T2  2T 1) = 0. Solving we find that T = 1 + v'2 = 2.4142s. So the body fell 2 x 4.9T2 = 57 .12m, and it took T + 1 = 3.4142s.
mg [m
J
rt/m
7 70 e v(2) = 12.4938
)t
Hence and x(2) 14.5025. (b) The velocity is 80% of its terminal velocity when 1  ert/m = 0.8. For the values of m = 70 and r = 34.3 this becomes t = 3.2846s. 11.
Without air resistance, v0
J2 x
=
13.5g
=
16.2665m/s. With air resistance, v0 is defined by
VQ.
7. The velocities must be changed to ft/s, so v0
=
60 mi/h = 60 x 5280 /3600 = 88ft/s, and v = 30 mi/h == 44ft/s. Then a = ( v2  v5) /2(x  xo) = 5.8ft/s2.
1
°
vdvr
vo 
Hence,
v
+ mg/r
= cert/m
 mg/r. If v(O)
= 0 then
C

m
J.
15
1.s
dy ·
+ mg/r)  (mg/r) ln(mg/r) = mg/r, and v(t) = mg(ert/m  1)/r. This is
 Vo+ (mg/r) ln(vo 8. We have v(t)
1) ds
=
(a) v5/2g
(c)
r (ers/m 
r lo
the time taken for the first half of the trip, then
(b) Both times are equal to vo / g.
m�n2]
The resistance force has the form R = r v. When v = 0.2, R = 1 so r = 5. The terminal velocity is Vtenn = mg/r = 0.196m/s.
5. The distance dropped in time t is 4.9t2 • If T is
6.
1) ds
9.
+ 581.5 = 1193.7s.
1.657 x 106 , or
r (ers/m 
=
=
4. In the first 60s the rocket rises to an elevation of (100 9.8)t2 /2 = 162, 360m and achieves a velocity of v(60) = (100  9.8) * 60 = 5412m/s. After that the velocity is 5412  9. 8t. This is zero at the highest point, reached when t1 = 552.2s. The altitude at that point is 162, 360 + 5412 t1  9.8 t[ /2 = 1.657 x 106m. From there to the ground it takes t2s,
v(s)ds
r lo
= 4.9t2 , where tis
the amount of time it takes the stone to hit the water. It also satisfies d = 340s, where s = 8  t is the amount of time it takes for the noise of the splash to reach the ear. Thus we must solve the quadratic equation 4.9t2 = 340(8  t). The solution is t = 7.2438sec. The depth is d = 340(8  t) = 257.lm.
1'
=
equalto mg /2r when ert/m = 1/2. Thus the time
r or m 491n(vo + 49)  49ln(49)
= 13.5
 vo
+
= 2.7
2.3 Models of Motion 31 (c) If y is the maximum height, the corresponding velocity is v = 0, so from (3.16)
This is an implicit equation for v0. Solving on a calculator or a computer yields v0 = 18.l 142m/s. 12. The: impact velocity
1
Vi
is defined by
r 1° vdv dy
vi
v+mg/r
O
Solving for y we get the result.
so
m
vo < J2GM/ R, then v5 < 2GM/ R, (d) If and 2GM/R  v5 > 0. Hence by (c) the object has a finite maximum height and does not escape. However, when v0 = J2G M/ R, 2G M/ R  v5 = 0, and there is no maximum height.
From which we get Vi  (mg/r) ln(vi + mg/r) + (mg/r) ln(mg/r) r = 50, or m Vi  19.6ln(vi + 19.6) + 19.6ln(19.6) = 25 This is an implicit equation for Vj. Solving on a cal• culator or a computer yields vi = 17 .340lm/s. 13. Following the lead of Exercise 11, we find that
15.
Let x(t) be the distance from the mass to the center of the Earth. The force of gravity is kx (proportional to the distance from the center of the Earth). Since the force of gravity at the surface (when x = R) is mg, we must have k = mg/ R. Newton's law becomes d2x
dy = (9.8  0.5v2 ) dy
+ R(v)/m)
v dv = (g
v vd  = 0.5 1Y1 2 230 v + 19.6 0
1
mgx
m dt2 = RUsing the reduction of order technique as given in the hint, we obtain
Hence if y1 is the maximum height we have °
dv gx v =  
dy.
dx
230 2
+ 19.6) 1
0
= 7.9010.
14.
dv (a) Follows from a= dt
dy dv = dv = v. dy dt dy
16. We will use GM = g R 2• Once more we use
v dv =
(b)
GM ( R
GM v dv =  (R
R
which can be separated with a solution given by v = JC  g x2 / R. The constant C can be evalu• ated from the initial condition, v(x = R) = 0, to be C = g R. When x = 0 (the center of the Earth), we obtain v = Jc= .JgR or approximately 4.93 miles per second.
Hence Yi = log(v
+ y)2
d )2
f.
v
!(v2 2
vdv = 
+y

Jo
1y o
GM (R
+ s)2
2
2

2G M
a
v2 = 2GM
ds
1 v5) =GM(_!_   ) R R+y v = v0
y
dy rsds=GM1°
vo
R�J
0=vJ2GMG
(1R  +1 ) R
Y
dy (R
+ y)2
[_!_R s v] «
2agR a+R 17. The force acting on the chain is the force of grav• ity applied to the piece of the chain that hangs off
32 Chapter 2 FirstOrder Equations where K is a constant. From the initial condition that x = 2 when t = 0, we obtain K = ln 2. Inserting x = 10 and solving for t, we obtain
the table. This force is mgx(t) where mis the mass density of the chain. Newton's law gives mx"(t) = mgx(t)
Using the hint, x"(t) = dv/dt = v(dv/dx) and so
1 ln (10 t = 
this equation becomes
�
+
,J%)
� .405 seconds
2
dv
v =gx dx
Separating this equation and integrating gives v2 = gx2 + C. Since v = 0 when x = 2 (initial velocity is zero), we obtain C = 4g. Therefore
v
= )g(x2 4).
J
Since dx/dt = v = g(x2  4), we can separate this equation and integrate to obtain ln(x
+ J x2

4) = ,Jgt
+K
18.
v = g  (k/m)v, v = velocity. Velocity on im• pact is 58.86 meters/per second downward and 90.9 seconds until he hits the ground.
19.
Let x be the height of the parachuter and let v be his velocity. The resistance force is proportional to v and toeax. Hence it is given by R(x, v) = keaxv, where k is a positive constant. Newton's second law gives us mx" = mg  keax x', or mx" + keaxx' +mg= 0.
Section 2.4. Linear Equations 1. Compare y' = y + 2 with y' = a(t)y + f (t) and note that a(t) = 1. Consequently, an integrating factor is found with
We verify that the lefthand side is the derivative of e3ty, so when we integrate we get 5 3
u(t) = efa(t)dt = ef ldt = et.
Multiply both sides of our equation by this integrat• ing factor and check that the lefthand side of the resulting equation is the derivative of a product. et (y'
e3t y(t) = e3t
Solving for y, we get
+ y) = Ze'
y(t) = 
y(t) = 2
+C
+ ce:'
2. We have a(t) = 3, so u(t) = e:", Multiplying we see that the equation becomes e3t y'  3e3t y = 5e3t.

+ Ce31
•
+
(2/x) y = (cosx)/x2 with y' = a(x)x + f(x) and note that a(x) = 2/x. Con• sequently, an integrating factor is found with
3. Compare y'
ety = 2er
5 3
(ety)' = 2et
Integrate and solve for y.
+ C.
u(x)
= efa(x)dx = ef2/xdx = e2ln [x] = lxl2 = X2.
Multiply both sides of our equation by the integrating factor and note that the lefthand side of the resulting
2.4
2
(y' + � y)
33
3
6. If we write the equations as x' = (4/t)x + t , we see that a (t) = 4 / t. Thus the integrating factor is
equation is the derivative of a product. x
Linear Equations
= cos x
u(t) = e" f(4/t)dt =
= t4.
e4Inr
2
x y' + 2xy = cosx
Multiplying by u, the equation becomes
(x2y)' = cosx
Integrate and solve for x. After verifying that the lefthand side is the deriva• tive of t4x, we can integrate and get
2
x y = sinx + C y(x) =
sinx + C
x
2
t4x(t) =Int+ C.
4. We have a(t) = 2t, so u(t) = et u, the equation becomes
2 •
Multiplying by
Hence the general solution is 4
2
et y'
+ 2tet 2 y
x(t) = t ln t
2
We verify that the lefthand side is the derivative of 2 et y, so when we integrate we get 2
et y(t) =
5
2
et
2
+ C.
5
2 + Ce
t2
.
=It+
112
= (t +
1)2.
Multiply both sides of our equation by the integrating factor and note that the lefthand side of the resulting equation is the derivative of a product. (t
+ 1)2
and solve for y'.
1
I
COSX
u(x) = efa(x)dx = efl/(l+x)dx =
u(t) = ef a(t)dt = ef 2/(t+I)dt e2lnjt+ll
+x
Compare this result with y' = a(x)y + f (x) and note that a(x) = 1/(1 + x). Consequently, an integrating factor is found with
5. Compare x'  2x/(t + 1) = (t + 1)2 with x' = a(t)x + f (t) and note that a(t) = 2/(t + 1). Con• sequently, an integrating factor is found with
=
7. Divide both sides by 1
y =l+xy+ l+x
Solving for y we get the general solution y(t) =
+ Ct4•
= Stet .
elnjl+xl
If 1 + x > 0, then 11 + x I = 1 + x. If 1 + x < 0, then 11 +xi= (1 +x). Ineithercase,ifwemul• tiply both sides of our equation by either integrating factor, we arrive at (1 + x)y' + y = cosx.
Check that the lefthand side of this result is the deriv• ative of a product, integrate, and solve for y. ((1
+ x)y)' = cosx
(x'  x) =
1
(1 + x) y = sin x + C
(Ct+ l)2x)' =
1
sinx + C yx ( = ) • l+x
t+l
= ll+xl.
Integrate and solve for x. 8. Divide by 1 + x3 to put the equation into normal form (t
+ l)2x
=t
+C 2
x(t) = t(t + 1)2 + C(t + 1)
I
3X2
y = 1 +x3Y +x
2
34
Chapter 2
FirstOrder Equations
We see that a(x) = 3x2 /(1 + x3 ). Hence the inte• grating factor is
10. Compare y' =my+ c1emx with y' = a(x)y + f(x) and note that a(x) = m. Consequently, an integrat• ing factor is found with u(x) = efa(x)dx = efmdx = emx. Multiply both sides of the differential equation by the integrating factor and check that the resulting lefthand side is the derivative of a product.
Multiplying by this we get
1
1 , 3x2 x2 + x3 y  (1 + x3)2 y = 1 + x3
y'  my = c1emx
•
emx(y'  memx) =
C1
(emx Y )' =
C1
We first verify that the lefthand side is the derivative of (1 + x3)1 y. Then we integrate, getting
1
Integrate and solve for y.
1
·
1 +x
y(x) =  ln(l + x3) + C. 3
emxy =
CJX
+ C2
y = (c1x
+ c2)emx
Solving for y, we get
11. 1 3 y(x) = 3(1 + x3) ln(l + x3) + C(l + x ).
Compare y' = cosx  y secx with y' = a(x)y + f (x) and note that a(x) =  secx. Consequently, an integrating factor is found with u(x)
9. Divide both side of this equation by L and solve for di/dt. di
 =dt
R
= efa(x)dx = ef secxdx =
E
eln [sec x+tan x]
= I secx + tan xi.
If secx + tanx > 0, then I secx
i+ • L L
+ tan xi= secx +
tanx. If secx + tanx < 0, then I secx + tan xi =  (secx + tanx). In either case, when we multiply both sides of the differential equation by this inte• grating factor, we arrive at
Compare this with i' = a(t)i + f (t) and note that a (t) =  R / L. Consequently, an integrating factor is found with
( sec x + tan x) (y' + y sec x) = cos x ( sec x + tan x),
u(t) = ef a(t)dx = ef R/Ldt = eRt/L
or Multiply both sides of our equation by this integrat• ing factor and note that the resulting lefthand side is the derivative of a product.
e Rt/L (d i dt
R z·) +L
(secx+tanx)y' +(sec2 x+secx tanx)y = 1 +sin x Again, check that the lefthand side of this equation is the derivative of a product, then integrate and solve for y.
E e Rt/L = L
((secx + tanx)y)' = 1 + sinx
(eRt/Li)' = E eRt/L L
(secx + tanx)y = x  cosx + C x  cosx + c y=• secx + tanx
Integrate and solve for i. E eRt/Li = eRt/L + c R E i(t) =  + CeRt/L R
12.
Compare x'  (n/t)x = e' t" with x' = a(t)x + f (t) and note thata(t) = n/t. Consequently, an integrat• ing factor is found with u(t) =efa(t)dt =efn/tdt =enlnltl = ltln.
2.4 Linear Equations 35 where A is any real number, except zero. How• ever, when we separated the variables above by dividing by y  1, this was a valid operation only if y :/ 1. This hints at another solution. Note that y = 1 easily checks in the original equation. Consequently,
Depending on the sign oft and whether n is even or odd, ltln either equals r» or tn. In either case, when we multiply our equation by either of these integrating factors, we arrive at
= e',
tnx'  ntnlx
Note that the lefthand side of this result is the deriv• ative of a product, integrate, and solve for x. (tnx)'
y(x) = 1  Aesinx,
= et
where A is any real number. Note that this will produce the same solutions a y = 1 +Ce sin x, C any real number, the solution found in part
tnx =et+ C X
13.
= t et + Ctn 11
(a).
(a) Compare y' + y cos x = cos x with y' = a(x)y + f (x) and note that a(x) =  cosx. Consequently, an integrating factor is found with u(x) = efa(x)dx = efcosxdx = esinx. Multiply both sides of the differential equation by the integrating factor and check that the re• sulting lefthand side is the derivative of a prod• uct. esinx (y'
+ Y COS X) y) I
(esin X
= esinx
COS X
= esin
COS X
X
y(x) = 1
Compare y' = y + 2xe2x with y' = a(x)y + f(x) and note that a (x) = 1. Consequently, an integrating factor is found with u(x)
= ef a(x)dx = ef 1 dx = e=.
Multiply both sides of our equation by the integrating factor and note that the lefthand side of the resulting equation is the derivative of a product. exy'  e= y
= 2xex
(exy)' = 2xex Integration by parts yields
Integrate and solve for y. esinx Y = esinx
14.
+C
+ cesinx
(b) Separate the variables and integrate. dy  = cosx(l  y)
dx
Consequently,
y(x) = 2xe2x
dy
+C  2e2x + ce"
exy = 2xex  2ex
 =cosxdx Iy
 ln
11  y I
= sin x
+ C.
Take the exponential of each side.
11  YI
=
esinxC
The initial condition provides 3 = y(O) = 2(0)e2CO) Consequently, C

2e2CO)
+ Ce0
= 2
+ C.
= 5 andy(x) = 2xe2x2e2x+sex.
l _ y = ±eCesinx
If we let A
= ±ec, then y(x) = 1  Aesinx,
15. Solve for y'.
, 3x y =  x2 + 1 y
6x
+ x2 + 1
© 2006 P e a rs on Educa tion, Inc., Uppe r S a ddle Rive r, NJ . All rights re s e rve d. This ma te ria l is prote cte d unde r a ll copyright la ws a s the y curre ntly e xis t. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
36 Chapter 2 FirstOrder Equations Compare this with y' = a (x) y + f (x) and note that a(x) = 3x/(x2 + 1). Consequently, an integrating factor is found with
Consequently, C = n/4 and tan1 t y(t) =
= efa(x)dx = ef 3x/(x +1)dx = e<3/2) ln(x2+1) = (x2 + 1)3/2. 2
u(x)
17. Compare x' Multiply both sides of our equation by the integrating factor and note that the lefthand side of the resulting equation is the derivative of a product. (x2
+ 1)3/2 y' + 3x(x2 + l)1;2 y = 6x(x2 + 1)1;2 ( (x2 + 1)3/2 y )' = 6x(x2 + 1) 1/2
Integrate and solve for y. (x2
+ 1)3/2 y
u(t) = efa(t)dt = efcostdt = esint.
Multiply both sides of our equation by the integrating factor and note that the lefthand side of the resulting equation is the derivative of a product.
+ 1)3/2 + C = 2 + C(x2 + 1)3/2
= 3 and y(x) = 2  3(x2 + 1)3/ 2.
.
esmt x'
Use sin 2t
.
+ esmt (cos t)x =
I
1
4t
= 2 sin t cost.
J
esin t
cos t sin t d t =
Multiply both sides of our equation by the integrating factor and note that the lefthand side of the resulting equation is the derivative of a product. 2t 2 ) y'
+ 4t(l + ((1
2t ) =
1
+
t2 )2y
The initial condition y(l) = 0 gives
+ 12)2(0)
= tan11
+ C.
t COS f
J
vdu
= (sin t)esint 
esint cost dt
+C
Therefore,
x(t) = sin t  1
+C
J
= (sin t)esint _ esint
esint x = esint sin t  esint

l+t
l+t = tan1 t
Sin
u dv
+ t2)2y)' =
(1
(1
J
= uv 
= ef a(t)dx = ef 4t/(l+t2)dt = e2ln Jl+r J = (1 + t2)2. 2
+
.
Let u = sin t and dv = esint cost dt. Then,
=  1 + t2 y + 1 + t2
Compare this with y' = a(t)y + f (t) and note that a(t) = 4t /(1 + t2). Consequently, an integrating factor is found with u(t)
1
esmt sin 2t
2 . )' 1 . ( esmt X = esmt sin 2t 2
(esin t X) I = esin t
16. Solve for y', y
=
= 2 + C(02 + 1)312 = 2 + C.
1 == y(O)
(1
(1/2) sin 2t with x'
+ f (t) and note that a(t) =  cost. Conâ€˘ sequently, an integrating factor is found with
The initial condition gives
Therefore, C
=
zr..
+ t2)24.
a(t)x
= 2(x2
y
+ x cost
(1
+c
+ cesint
The initial condition gives 1 = x(O) = sin(O)  1 Consequently, C
+ Cesin(O)
= 1
+ C.
= 2 andx(t) = sin t1 +2esint.
2.4 18. Solve xy'
+ 2y
37
negative infinity as x approaches zero from the right.
= sinx for y'.
,
Linear Equations
2
sinx
x
x
y =y+
5
(n/2,0)
Compare this with y' = a(x)y + f (x) and note that a(x) = 2/x and f(x) = (sinx)/x. Itis important to note that neither a nor f is continuous at x = 0, a fact that will heavily influence our interval of ex• istence. An integrating factor is found with u(x) = efa(x)dx = ef2/xdx = e2lnlxl = lxl2 = x2.
Multiply both sides of our equation by the integrating factor and note that the lefthand side of the resulting equation is the derivative of a product.
x2y'
x sinx dx
0
6
x
19. Solve for y'.
1
+ 2xy = x sinx (x2y)' = x sinx
y' =  y 2x +3
+ (2x + 3)1/2
Compare this with y' = a (x) y + f (x) and note that a(x) = l/(2x + 3) and f(x) = (2x + 3)112• It is important to note that a is continuous every• where except x = 3 /2, but f is continuous only on ( 3 /2, +oo), facts that will heavily influence our interval of existence. An integrating factor is found with
Integration by parts yields
f
251..L.� � � � � � � '
f
= x cosx + cosx dx = x cosx + sinx + C.
Consequently,
u(x) = efa(x)dx = efl/(2x+3)dx
x2y = x cosx 1
y =   cos x x
= e(l/2)ln 12x+31 = 12x
+ sinx + C, 1 x
.
C x
However, we will assume that x >  3 /2 (a do• main where both a and f are defined), so u (x) = (2x + 3)112• Multiply both sides of our equation by the integrating factor and note that the lefthand side of the resulting equation is the derivative of a product.
+ 2 sm x + 2 .
The initial condition provides O=y(n/2)=
4 4C . +2 2 T( T(
Consequently, C = 1 and y (l/x2) sinx  I/x2 •
=
(1/x) cosx
We cannot extend any interval to include x
+ 311/2.
+
= 0, as
our solution is undefined there. The initial condition y(1r/2) = 0 forces the solution through a point with x == n/2, a fact which causes us to select (0, +oo) as the interval of existence. The solution curve is shown in the following figure. Note how it drops to
(2x
+ 3)112 y' 
+ 3)312 y = (2x + 3)1 + 3)112 y )' = (2x + 3)1
(2x
( (2x
Integrate and solve for y. (2x
+ 3)112y
1 =  ln(2x
2
+ 3) + C,
or 1
y = (2x 2
+ 3)1/2 ln(2x + 3) + C(2x + 3)112
38
Chapter 2
FirstOrder Equations
Consequently, C
The initial condition provides
= 2 and
x  cosx + 2 y = sec x + tan x ·
O=y(l)=C.
Consequently, y = (1/2)(2x +3)112 ln(2x +3). The interval of existence is (3/2, +oo) and the solution curve is shown in the following figure.
The initial condition forces the graph to pass through (0, 1), but a(x) has nearby discontinuities at x = n /2 and x = n /2. Consequently, the interval of existence is maximally extended to ( rr /2, n /2), as shown in the following figure.
6 25
(0, 1)
1
2
x
6
20. Compare y' = cosx  y secx with y' = a(x)y + f (x) and note that a(x) =  secx and f (x) = cos x. Although f is continuous everywhere, a has discontinuities at x = n /2 + kn, k an integer.
5 1.57 21. Solve for x.
1 l+t
I
= elnlsecx+tanxl = I secx + tan x], If secx + tanx > 0, the I secx + tan xi = secx + tanx. If secx + tanx < 0, the I secx + tan xi = (secx + tanx). Multiplying our equation by ei• ther integrating factor produces the same result. (secx + tanx)y' + (secx tanx + sec2 x)y = 1 + sinx, From which follows:
Use the initial condition, y (0) = 1. (sec O + tan 0)(1) = 0  cos(O) + C
l+t
Compare this result with x' = a(t)x + f (t) and note that a(t) = 1/(1 + t) and f (t) = cost /(1 + t), neither of which are continuous at t = 1. An inte• grating factor is found with u(t) = efa(t)dt = ef 1/(t+l) = eln ll+tl =
11 + ti.
However, the initial condition dictates that our so• lution pass through the point (n /2, 0). Because of the discontinuity at t = 1, our solution must remain to the left of t = 1. Consequently, with t < 1, u (t) =  (1 + t). However, multiplying our equation by u(t) produces (1
( ( sec x + tan x) y)' = 1 + sin x (secx + tanx)y = x  cosx + C
COS f
x =x+
An integrating factor is found with u(x) = efa(x)dx = ef secxdx
1.57
x
+ t)x' + x = cost, ((1 + t)x)' = cost, (1 + t)x = sin t + C.
Use the initial condition. (1 
i)
(0) = sin (
i)
+C
2.4 Linear Equations 39 This is a linear equation for z. The integrating factor is 1/x, so we have
Consequently, C = 1 and 1
+
sin t
x=
+t
1
_!_ [dz _ .:.] = _!_(x)
x
The interval of existence is maximally extended to (cX), 1), as shown in the following figure.
dx
x
x
[;]' =1
z
 = x+C x z(x) = x(C  x)
0 Since z
1
= 1/y, our solution is y(x) = x(C x) .
24. In this case n
x
= 2, so we set z = y1 • Then dz dz dy dx dydx = y2 [y2  y]
1 .
..........
4
1
1 y
= 1+= 1 +z
22. Let z = x1n. Then dz dt
This is a linear equation for z. The integrating factor
dz dx dt
= dx
ndx  n)x dt ·
= (l
is e=, so we have
J
z  z = e x e x [d dx
This motivates multiplying our equation by (1 
[exz]' exz
n)xn to produce dx dt
(1  n)xn = (1  n)a(t)xln
+ (1  n)f (t).
Replacing (1  n)xndx/dt with dz/dt and x1n
=
=
ex ex + c
z(x) = 1
Since
z=
+ Ce":
1/y our solution is y(x)
with z produces the desired result. dz dt
+ (1  n)f(t)
 = (1  n)a(t)z
23. In this case
1
= 1 + cex
25. Solve for y', 1 x
y' =  y +x3y3 n =
2, so we set z = y1. Then
dz
dz dy
Compare this with y' = a(x)y + f (x)yn and note • that this has the form of Bernoulli's equation with
dx
dy dx
n
y]
= y 2 [xy 2  �
=x+
1
= 3. Let
z=
y13 = y2 Then
dz dx
= dz
= _2Y_3dy. dx
Multiply the equation by 2y3
• xy =
dy dy dx
z
x+. x
.
2y
_3dy 2 2 3  = y 2x dx x
40
Chapter 2
FirstOrder Equations
Replace 2y3(dy /dx) with dz/dx and y2 with z.
Replace
z with r:'
dz 2 =z2x 3 dx x This equation is linear with integrating factor u(x)
(x2z)' = 2x x2z = x2 + C 4
z= Replace
2
+ Cx
x
(a) Sincey
= Yi +z,y2 = yf+2y1z+z2•
= (2y1 l/1
z with y and solve for y.
y =
Hence
+ </J)z  tz2
(b) Since y, = 1/tisasolution, we set z = y+l/t. Then y = z  l/t, and y2 = z2  2z/t + l/t2, so
+ Cx2
±1// Cx2

x4
1
Compare this with P' = a(t)P + f(t)Pn and note that this has the form of Bernoulli's equation with
= 2.
27.
2
y2 = x4
n
+ Ceat
b/a
z' = y'  Y� = [l/ll +<PY+ xl + [l/lyf + <PYI + x] = l/l[yf  y2] + ¢[y1  y] 2 = l/1 [2y1z + z ]  </Jz
x2z'  2x3z = 2x
26.
P=•
= efa(x)dx = ef2/xdx = e2lnlxl = x2.
Multiply by the integrating factor and integrate.
+ ce:"
= � a
r'
y.
and solve for
Let z
= p12 = r ".
_p2dP = aP1 dt
pl
with z.
= ef a(t)dt = ef adt = eat.
Multiply by the integrating factor and integrate.
d dt
 (eat z) = beat
1
I
This equation is linear with integrating factor
+ ae" z = beat
(2
w =z t2
dz  = az+b dt
dz dt
y

t +y
2
t2
3z 2 t This is a Bernoulli equation with n = 2. Thus we set w = 1 / z. Differentiating, we get
+b
Replace P2(dP /dt) with dz/dt and
1
=  +z.
Multiply the equation by  p2•
eat 
=
Then
dz = � dP = _p_2dP. dt dP dt dt
u(t)
z' = y'  t2
I
= _ !_2 [ 3z + z2J t
t
3 =
1 tz
= 3w 1.
t This is a linear equation, and t 3 is an integrat• ing factor.
3;]
= t3
[t3w]
= t3
i3 [ w' 
t3w(t) = w(t) =
1
2 1
t2
+C
2t + Ct
3
2.4
1
2
2
+ 2Ct3
= t
We need to compute the time to when T(to) = 37, using k � 0.2231 from Exercise 35 of Section 2. This is a nonlinear equation, but using a calculator or a computer we can find that to� 0.8022. Since t = 0 corresponds to midnight, this means that the time of death is approximately 11: 12 PM.
+ Bt3'
Where we have set B = 2C. Then 1
y(t) = z(t) 
2
1
t = t + Bt3
This looks a little better if we use partial frac• tions to write 2
y(t) = t
=
+ Bt3
30.
1

2Bt
t
1
+
1
2Bt 1 + Bt2.
1
y;
Bt2
= v'yh  3yp,
and y; = 3yp + 4, we have v' = 4/yh = 4e3 t. Integrating we see that v(t) = 4e31 /3, and
The general solution is 4 y(t) = yp(t)
x' = N'/N2
= l/N + 103 = x + 103_ Solving this linear equation, we getx(t) = 103[1 + cet]. Hence N(t) = l/x(t) = 1000/[l + cet]. At t = 0, N = 20 = 1 / [1 + CJ. Hence C = 49, and the solution is N(t) = 1000/[1 + 49et].
= 0.9 x
1000
31.
+ Cyh(t).
= 
3
3t
+ Ce
.
The homogeneous equation, y' = 2y has solution Yh(t) = e:", We look for a particular solution in the form Yp(t) = v(t)yh(t), where vis an unknown function. Since y; = v'yh
+ vy�
= v'yh  2vyh = v'yh  2yp,
= 900 when t = 6.089
29. Newton's law of cooling says the rate of change of temperature is equal to k times the difference between the current temperature and the ambient temperature. In this case the ambient temperature is decreasing from 0°C, and at a constant rate of l°C per hour. Hence the model equation is T' = k(T + t), where we are taking t = 0 to be mid• night. This is a linear equation. The solution is
= v'yh + vy� = v'yh  3vyh
28. The model is N' = kN(lOOO  N), where the pro• portionality constant k is yet to be determined. Since we know that when N = 100, the rate of infec• tion is 90/day, we have k · 100 · 900 = 90, we find that k = 1 x 103. Hence the model equation is N' = N  N2 /1000. This is a Bernoulli equation, with n=2. Accordingly we set x = 1 / N. Then
We have N (t) days.
The homogeneous equation, y' = 3y has solution Yh (t) = e3t. We look for a particular solution in the form yp(t) = v(t)yh(t), where vis an unknown function. Since
t
2
41
T(t) = t + 1/ k + ce:", Since T(O) = 31, the constant evaluates to C = 31  1/ k. The solution is T(t) = t + l/k + (311/k)ekt.
Now it is a matter of unravelling the changes of variable. First z(t) = w(t) = t
Linear Equations
and y; = 2yp
+ 5, we have
v' = 5/yh = 5e2 t. t
Integrating we see that v(t) = 5e2 /2, and
The general solution is
42 Chapter 2 FirstOrder Equations 32. The homogeneous equation, y' = (2/x)y has so• lution v« (x) = x2• We look for a particular solution in the form Yp(x) = v(x)yh(x), where vis an un• known function. Since
+ vy�
y� = v'yh
=
1
V
+ vy�
y; = v'yh
Yh  2yp/X,
and y� = (2/x)yp + 8x, we have v' 8x3. Hence v(x) = 2x4, and Yp(x)
35. The homogeneous equation, y' = 2xy has solution 2 Yh (x) = ex • We look for a particular solution in the form Yv(x) = v(x)yh(x), where vis an unknown function. Since
= v' v»
= 8x/yh = and y�
= v(x)yh(x) = 2x2•
=
2xyp
+ 4x,
2
 2xyp, we have v'
=
4x/yh
=
2
4xex . Hence v(x) = 2ex , and
The general solution is y(t)
33.
= Yp(t) + Cyh(t) = 2x2 + Cx2•
Yp(x)
The homogeneous equation, y' = y / t, has solu• tion v» (t) = 1 / t. We look for a particular solution in the form Yp(t) = v(t)yh(t), where vis an unknown function. Since
The general solution is y(t) = Yp(t)
= v'yh  2vyh
36.
= v'yh  Yvf t, and y� = yvft + 4t, we have v' = 4t/yh Integrating we get v(t) = 4t3 /3, and
= 4t2•
The homogeneous equation, y' = 3 y has solution Yh (t) = e": We look for a particular solution in the form Yp(t) = v(t)yh(t), where vis an unknown
function. Since
= v(t)yh(t) = 34 t 2 .
y� = v'yh
y(t) = Yv(t)
+ vy� + 3vyh
= v'yh
The general solution is
34.
+ Cyh(t) = 2 + Cex2 .
= v'yh + vy�
y�
Yp(t)
= v(x)yh(x) = 2.
= v'yh 4
+ 3yp,
2
+ Cyh(t) = 3t + C/t.
The homogeneous equation, x' = 2x has solution
and y�
=
3yp
+ 4,
we have v'
=
4/yh = 4e3t.
Integrating we see that v(t) = 4e3 ' /3, and
xh(t) = e:", We look for a particular solution in the form Xp(t) = v(t)xh(t), where vis an unknown
function. Since I xPI = v xh
= v'xh
+ vxh I
The general solution is
 2vxh
= v'xh  2xp, and x� = 2xp + t, we have v' = tfxh = te", Integrating, we get v(t) = (t /2  l/4)e2 ', and
xp(t)
= v(t)xh(t)
=
= 2,
1
Since y(O)
[2t  1]. 4
C = 10/3. Thus the solution is
we must have 2
= 4/3 + C,
1 x(t) = Xp(t)
+ Cxh(t)
= 4[2t  1]
+ ce:".
y(t) =
(4 + 10e3t) /3.
or
2.4 37. The homogeneous equation y' = y /2 has solution yh(t) = et/2. We look for a particular solution of the form Yp(t) = v(t)yh(t), where vis an unknown function. Since y� = v'yh
= v'yh  vyh/2 = v'yh  Yp/2,
Cyh(x) = x2  1 + ce=', Since y(O) = 1, we have C = 0, and the solution is y(x) = x2  1.
yp(t) = v(t)yh(t) = (2t  4). The general solution is y(t) = Yp(t) + Cyh(t) = (2t 4) + c«:": From y(O) = 1 we compute that C = 5, so the solution is
+ set/2•
38. The homogeneous equation y' = y has solution Yh(t) = e:', We look for a particular solution of the form Yp(t) = v(t)yh(t), where vis an unknown function. Since
+ vyh
= v'yh
 vyh
I
I
+ vxh 2 = v'xh + 2vxh/t
I
and y� = yp/2+t, we have v' = t/yh(t) = tet/2. Integrating we find that v(t) = (2t  4)etf2, and
YpI = v Yh
I
= v Yh  Yp, and y� = =s» + et, we see that v' = et /Yh = e", Integrating we get v(t) = e2t /2, and yp(t) = et /2.
+ ce:', From y(O) C == 1 /2, so the solution is
=
+ 2xp/t 2,
= v'xh
and x� = 2xp/t2 + 1/t2, we have v' = 1/(t 2xh) = e2ft /t2. Integrating we find that v(t) = e2ft /2, and x P (t) = 1 /2. The general solution is x (t) = xp(t) + Cxh(t) = 1/2 + c«?", Since x(l) = 0, we find that C = e2 /2, and the solution is x(t)
1
+ e2(1l/t)) .
2 (1
)
et /2
I
xP = v xh
41. The homogeneous equation x' = 4tx /(1 + t2) has solution xh(t) = (1 + t2  • We look fora particular
I
The general solution is y(t) = Yp(t)
+ Cyh(t)
solution of the form Xp(t) = v(t)xh(t), where vis an unknown function. Since I
xP
= v Xh + vxh I
I
= v'xh  4txp/(1
=
1, we compute that
2
andx� = 4txp/(1
1
39. The homogeneous equation y' = 2xy has solution 2 •
We look for a particular solution of
the form yP (x) = v (x) Yh (x), where v is an unknown function. Since y� = v'yh
= v'yh
l
+ t2
Integrating, we get v(t)
2
2t2
+ t4
.
4(1
+ t2)2
4C
+ 2t2 + t4 + t 2 )2 .
The general solution is
+ Cxh(t)
=
40
The initial condition x (0) = 1 implies that C = 1, so the solution is
+
and y� = 2xyp 2x3, we see that v' = 2x3ex • Integrating we get v(x) = (x2  l)ex2, and Yp(x) =
+ t 2 ).
= t2 /2 + t4 /4. Thus
xp(t) = v(t)xh (t) =
x(t) = Xp(t)
= v'yh  2xyp,
so
Xh(t)
+ vy�  2xvyh
t ) 2 ,
+ t2 ) + t/(1 + t2 ),
t
I
+
v =  .  = t(l
Yh(x) = ex
43
40. The homogeneous equation x'. = (2/ t2)x has solu• tion xh (t) = e2/i. We look for a particular solution of the form Xp(t) = v(t)xh(t), where v is an un• known function. Since
+ vy�
y(t) = (2t  4)
Linear Equations
4
+ 2t2 + t4
x2  1. The general solution is y(x) = Yp(x)
+
y(t) =
4(1
+ t2)2
.
44 Chapter 2 FirstOrder Equations 42.
(a) The equation T + kT = 0 is separable, with solution Th = Cekt, C an arbitrary constant.
C) provides
C=
(b) The equation T' = k(T  A) is autonomous. We seek a constant solution (see Section 2.9) that makes the right side equal to zero. Hence, Tp == A is a particular solution of the inhomo• geneous equation. (c) The general solution is T = Th + Tp ce:" + A, with Can arbitrary constant.
wk A k2 + w2
Substituting these results in Tp = C cos wt D sin wt provides the particular solution
=
wkA cos wt +w 2
k2
(d) Again, the solution of the homogeneous equa• tion T' + kT = 0 is Th = ce:", with Can arbitrary constant. The inhomogeneous equa• tion T' = k(T A)+ His also autonomous (the right side is independent oft). We seek a
+
+
k2 A . sm wt. k +w 2 2
Hence, the general solution is
kA = Fekt
constant solution by setting the right side equal to zero. 44.
k(Tp A)= H
and
H
+
k 2 + co2
. [k sm wt  co cos wt] .
(a) If the period of the ambient temperature is 24 hours, then the computation
TpA = 2n 2n tt   w T  24  12
K
H Tp =A+ K Hence, the general solution is given by the equation
T =Th+ Tp = Ce kt
43.
+ ( A+
H) K
.
gives the angular frequency. Because the sinu• soid has a maximum of 80° F and a minimum of 40° F, the amplitude will be half of the dif• ference, or 20. A sketch of the ambient tem• perature follows.
(a) The solution of the homogeneous equation T' + kT = 0 is Th = Fekt, with Fan ar• bitrary constant. (b) We guess that Tp = C cos wt + D sin wt is a particular solution. Substituting Tp and T; = =Cco sin wt + Dec cos wt in the left side of T' + kT = k A sin wt, then gathering coeffi• cients of cos wt and sin wt, we obtain
T; +kTp
= ( we +kD)
sin wt+(kC +wD) cos wt. (4.1) Comparing this with the right side of + kTp = kA sin wt, we see that
T;
wC + kD = kA
and
kC +wD
= 0.
(c) Solving these equations simultaneously (for ex• ample, multiply the first equation by k, the sec• ond by co, then add the equations to eliminate
Note the minimum at 6 am, then the maximum at 6 pm. What we have is an upsidedown sine, with angular frequency n / 12, that is shifted up• ward 60° F. Thus, the equation for the ambient temperature must be .
A = 60  20 sm
tt t
12
.
2.4 Therefore, the model, adjusted for this ambient temperature, becomes
dTdt = _!2 (T  60 + 20sin nt). 12
Thus, the general solution is
Tp = Cer/2 + 60 120
+
36 +
(b) The homogeneous equation T' + (l/2)T = 0 has solution Th = cer/2, where C is an arbi• trary constant. Now, consider the inhomoge•
1
I
T +
2
T = 30  lOsin
Jr(
12
.
(4.3)
Note that the right hand side consists of a con• stant and a sinusoid. Let's try a particular so• lution having the form
Tp = D + E cos
itt
+ F sin
xt
.
2
=
!n + (!!..E + !F) sin 2
12
+
(!E + !!..p) cos 2
2
12
nt12
nt12 (4.5)
Comparing this with the righthand side of equation (4.3), we see that 1
2 
£+
12 1
2
and
120
+
36 + it?
+ 60
nt]
[ n cos tt t  6 sin  . 12
12 (4.7)
(c) The plot of the ambient temperature is shown as a dashed curve in the following figure. The temperature T inside the cabin is shown as a solid curve.
75 70
· ...
65 · ·.
45
n
0 6 12 18 24 30 36 42 48 54 60 66 72
Clearly, D = 60, and solving the remaining two equations simultaneously, we obtain 120n 36 + n
120n ) 2 ( 10 + 36 + n? e:' 1
50
F = 10,
2E + 12F = 0.
E=2
J
[ n cos tit ttt .  6 sin 12 12 (4.6)
55
1
l(
T = 
60
D = 30,
n2
Now, when the initial condition T (0) = 50 is substituted into equation (4.6), we obtain C = 10120n /(36+n2) . Thus, the general solution becomes
(4.4)
12 12 Substitute this guess and its derivative into the left hand side of equation (4.4) and collect co• efficients to get
T' + !T
T = Th + Tp, or
(4.2)
neous equation
45
linear Equations
and
F=
720 36 + n2
.
These values of D, E, and F, when inserted into equation (4.4), provide the particular solu• tion 120n nt 720 . nt T =60+ cossm. 2 2 P 36 + n 12 36 + n 12
Note that the transient part of the solution dies out quickly. Indeed, because of the factor of e:'!", the time constant (See Section 2.2, Exer• cise ??) is Tc = 2 hr. Thus, in about four time constants, or 8 hours, this part of the temper• ature solution is negligible. Finally, note how the temperature in the cabin reacts to and trails the ambient temperature outside, which makes sense.
46 Chapter 2 FirstOrder Equations
Section 2.5. Mixing Problems 1.
(a) Let S(t) denote the amount of sugar in the tank, measured in pounds. The rate in is 3 gal/min x 0.2 lb/gal = 0.6 lb/min. The rate out is 3 gal/min x S/lOOlb/gal = 3S/100lb/min. Hence dS .  = rate m  rate out dt
= 0.6  3S/100 This linear equation can be solved using the integrating factor u(t) = e3t/IOO to get the gen• eral solution S(t) = 20 + ce3t/IOO. Since S(O) = 0, the constant C = 20 and the solu• tion is S(t) = 20(1  e3t/IOO). S(20) = 10(1  e0·6) � 9.038lb. (b) S(t) = 15 when e3t/IOO = 1  15/20 1/4 . Taking logarithms this translates to t (100ln4)/3 � 46.2098m. (c) Ast + oo S(t) + 20.
2.
(a) Let x(t) represent the number of pounds of sugar in the tank at time t. The rate in is 0, and the rate out is 2 gal/min· x/50 lb/gal = x /25 lb/min. Hence the model equation is x' == x /25. The general solution is x(t) = Aet/25• The initial condition implies that A = x(O) = 50 gal x 2 lb/gal= 100 lb. Hence the solution is x(t) = lOOet/25. After 10 minutes we have x ( 10) = 67 .032 lb of sugar in the tank. (b) We have to find t such that x(t) = lOOet/25 = 20. This comes to t = 25 ln 5 � 40.2359 min.
(c) x(t) = 10oetl25 + 0 as t + oo. 3.
(a) Let x (t) represent the number of pounds of salt in the tank at time t. The rate at which the salt in the tank is changing with respect to time is equal to the rate at which salt enters the tank minus the rate at which salt leaves the tank, i.e., dx .  = rate m  rate out. dt
Solution enters the tank at 5 gal/min, but the concentration of this solution is 1/4 lb/gal. Consequently, rate in= 5 gal/minx �lb/gal= � lb/min. Solution leaves the tank at 5 gal/min, but at what concentration? Assuming perfect mixing, the concentration of salt in the solution is found by dividing the amount of salt by the volume of solution, c(t) = x(t)/100. Consequently,
.
m
lb/gal=
. x(t) lb/m .
20
100
As there are 2 lb of salt present in the solution initially, x(O) = 2 and
5
dx dt =
4
1 20x,
x(O) = 2·
Multiply by the integrating factor, eU/20)t, and integrate. (e(]/20)t x
)' = � e(l/20)t
4 e<t/20)t x
=
25e(l/20)t
x = 25
+C
+ ce<1!20)t
The initial condition x (0) = 2 gives C = 23 and x(t) = 25  23eOl20)t. Thus, the concentration at time t is given by 25  23e0/20)t
x(t)
c(t) = 100 =
100
and the eventual concentration can be found by taking the limit as t + +oo. lim
tHoo
In order that the units match in this equation, dx/dt, the rate In, and the rate Out must each be measured in pounds per minute (lb/min).
1
x(t)
rate out= 5 gal/m x
25  23e0/20)r 100
1 =  lb/gal 4
Note that this answer is quite reasonable as the concentration of solution entering the tank is also 1 /4 lb/gal.
m
2.5 (b) We found it convenient to manipulate our original differential equation before using our solver. The key idea is simple: we want to sketch the concentration c(t), not the salt con• tent x(t). However,
500c
x(t) c(t)
=
I
=
I
x(t) = lOOc(t).
or
Mixing Problems
47
Let c(t) represent the concentration at time t. Thus, c(t) = x(t)/500, or 500c(t) = x(t) and 500c'(t) = x' (t). Substitute these into the rate equation to pro• duce
c =
r 500
r
(500c), c.
100 500 This equation is separable, with solution c Ae(r/SOO)t. Use the initial concentration, c(O) = .05 lb/gal, to produce
Consequently, x'(t) = IOOc'(t). Substituting these into our balance equation gives
5
I
=
1
4  20x, 5 1 = 4 20 (lOOc),
x = lOOc
I
I
=
c
1
c
= 0.05e(r/500)t.
The concentration must reach 1 % in one hour (60 min), so c(60) = 0.01 and
1
80  20c,
with c(O) = x(0)/100 = 2/100 = 0.02. The numerical solution of this ODE is presented in the following figure. Note how the concentra•
0.01 =
o.ose(r/500)(60)'
! = e(3/25)r, 5
tion approaches 0.25 lb/gal.
r
= 25 1n5, 3
r � 13.4 gal/min.
c ' = 1 /80  1 /20 c ,,,,,,,,,,,,,,,,,,,, 0.3 ,,,,,,,,,,,,,,,,,,,,
0.2 0
0.1 O
;;;;;; ///// ////
5. The volume is increasing at the rate of 2 gal/min, so
the volume at time t is V (t) = 20 + 2t. The tank is full when V (t) = 50, or when t = 15 min. If x(t) is the amount of salt in the tank at time t, then the concentration is x (t) / V (t). The rate in is 4 gal/min · 0.5 lb/gal = 2 lb/min. The rate out is 2 gal/min x / V lb/gal. Hence the model equation is
;;;;;;;;;;;;;
////////////// ///////////////
1111/////II///I/I/// Ill lllllllll/111111 lllllllllllllll/1/II II lllllllllllll/111 IIIIIIIIIIIIIJIIJIII I IIIIIIIIIIIIIIIIII 11/III/II I llllllllll Ill II/Ill II Ill/I/I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I
50
0
x ' =22x/V=2x. io i .
100
This linear equation can be solved using the integrat• ing factor u (t) = 10 + t, giving the general solution x(t) = 10 + t + C/(10 + t). The initial condition x (0) = 0 enables us to compute that C = 100, so the solution is x(t) = 10 + t  100/(10 + t). At t = 15, when the tank is full, we have x(15) = 21 lb.
4. Let x(t) represent the amount of salt in the solution at time t. Let r represent the rate (gal/min) that water enters (and leaves) the tank. Consequently, the rate at which salt enters the tank is O gal/min, but the . rate out= r gal/mmx
x(t) 500
r
lb/gal=
500
Thus, dx .  = rate m  rate out, dt dx r dt =  500x.
.
x(t) lb/mm.
6.
The volume in the tank is decreasing at 1 gal/min, so the volume is V (t) = 100  t. There is no sugar coming in, and the rate out is 3 gal/min x S(t)/ V (t) lb/gal. Hence the differential equation is
dS
3S
dt  100  t
48 Chapter 2 FirstOrder Equations This equation is linear and homogeneous. It can be solved by separating variables. The general solution isS(t)+A(lOOt)3.SinceS(O) = lOOx0.05 = 5, we see that A = 5 x 1 o6, and the solution is S(t) = 5 x 106 x (100  t)3. When V(t) = 100  t = 50 gal, S(t) = 5 x 106 x 503 = 0.625 lb.
7.
(a) The volume of liquid in the tank is increasing by 2 gal/min. Hence the volume is V (t) = 100 + 2t gal. Let x(t) be the amount of pol• lutant in the tank, measured in lbs. The rate in during this initial period is 6 gal/min · 0.5 lb/gal = 3 lb/gal. The rate out is 8 gl/min ·x/V = 4x/(50+t).Hencethemodelequation is x' = 3  4x/(50
8. Letx(t) represent the amount of drug in the organ at time t. The rate at which the drug enters the organ is rate in= a cm3 /s x
x(t) = 3t + 30 _ 1.875 x 10
8
(50+t)4
Vo+ rt
Vo+ rt
dx b =aKx. dt Vo+ rt
The integrating factor is u(t) = efb/(Vo+rt)dt = e(b/r)ln(Vo+rt) = (Vo+rt)b/r.
Multiply by the integrating factor and integrate. ((Vo+ rtllr x )' = aK(Vo (V,O
+ rt)b/r
+ rt)b/r x =
aK (V,0 + rt)h/r+ l + L r(b/r + 1) aK x =(Vo+ rt)+ L(Vo + rt)b/r b+r
No drug in the system initially gives x (0) = 0 and L = aKVtr+I /(b + r). Consequently, a«
x = (Vo+ rt) +r x = � (Vo+ rt)
2X
b+r a«
This homogeneous linear equation can solved by separating variables to find the general solu•
b
Consequently,
30 t
I
x =
x(t)
rate out= b cm3 /sx  g/cm3 = x(t) g/s.
After 10 minutes the tank contains x(lO) = 21.5324 lb of salt. (b) Now the volume is decreasing at the rate of 4 gal/min from the initial volume of 120 gal. Hence if we start with t = 0 at the 10 minute mark, the volume is V (t) = 120 4t gal. Now the rate in is 0, and the rate out is 8 gal/min ·x/V = 2x/(30  t). Hence the model equation is
g/cm3 = ate g/s.
K
The rate at which the drug leaves the organ equals the rate at which fluid leaves the organ, multiplied by the concentration of the drug in the fluid at that time. Hence,
+ t).
This linear equation can be solved using the in• tegrating factor u(t) = (50 + r)". The general solution is x(t) = 3(50 + t)/5 + C(50 + t)4. The initial condition x (0) = 0 allows us to compute the constant to be C = 1. 87 5 x 108. Hence the solution is
5
We are asked to find when this is onehalf of 21.5342. This happens when (30  t)2 = 450 or at t = 8.7868 min.
+ rt)
0
(Vo+ rt)b/r,
b+r
[1  vtr+1(v + rt)b/r0
[
x = (Vo
b+r
a« V.b/r+l
v; 0
( 1 
1],
)b/r+l]
 
Vo+ rt
•
tion x(t) = A(30  t)2 • At t = 0 we have
The concentration is found by dividing x(t) by
x(O) = 21.5342, from which we find that A = 21.5342/900, and the solution is
V (t) = Vo +rt. Consequently,
c(t)=
21.5342 x(t) =
900
(30  t)2.
a« [ ( Vo ) (b+r)/r] 1. b+r Vo+ rt
9.
2.5 Mixing Problems 49 10. Because the factory stops putting pollutant in the lake, p = 0 and e' + ((r + p)/V) e = p/V be• comes
(a) The rate at which pollutant enters the lake is rate in = p knr' /yr. The rate at which the pollutant leaves the lake
50
is found by multiplying the flow rate by the concentration of pollutant in the lake.
+ p) knr' /yr x
rate out= (r
r+p
e'
+
100
e = 0.
Note that we carried r = 50 from Exercise 9. This equation is separable, with general solution e = K e<112)t. The initial concentration is 3.5%, so e(O) = 0.035 produces K = 0.035 and
x(t) krrr' /krn.3
v
3
= lfx(t)km /yr
Consequently,
dx
r+ p
dt
v
e(t) = 0.035e<1!2)t.
=px.
The question asks for the time required to lower the concentration to 2%. That is, when does e(t) 0.02?
Bute(t) = x(t)/V, so Ve'(t) = x'(t) and
r+ p
I
Ve = p  lf(Ve),
c'
0.02 = 0.035eOl2>r
+ r + p c = .!!_.
v
v
_\ = ln 0.02
(b) With r = 50 and p = 2, the equation becomes 50+2
2
f
2
c = 100 
ioo c,
c' = 0.02  0.52c.
0.035 0.035 t= 2 ln0.02 t � 1.1 years
This is linear and solved in the usual manner. 11.
(e0.52t e )' = 0.02eo.052t e0.52t e = 0.02 eo.os2t 0.52 1
e= 26
+K
+ Keo.s2t
0.04
The initial concentration is zero, so e(O) produces K = 1 /26 and
=
0
1 e0.52t c= 1 26 26 The question asks when the concentration reaches 2%, or when e(t) = 0.02. Thus, 0.02 = � (1 26 eo.s2t = 0.48,
 eo.s2t)
In 0.48 t
(a) The concentrations are plotted in the following figure. In steadystate the concentration varies periodically.
=  0.52 '
t � 1.41 years.
0
0.035 30
40
50
60
' (b) The following figure shows one year of the os• cillation, and indicates that the maximum con• centration occurs early in February. This is four months after the time of the minimum flow.
50 Chapter 2 FirstOrder Equations Thus there is a shift of phase between the cause and the effect.
13.
(a) Let x(t) be the amount of pollutant (measured in km") in Lake Happy Times. The rate in for Lake Happy Times is 2 krrr' /yr. The rate out is 52km3 /yr x x/100 = 0.52x knr' /yr. Hence the model equation is x' = 2 0.52x. This linear equation can be solved using the integrating factor u(t) = e0·52t. With the ini• tial condition x (0) = 0 we find the solution x(t) = 2[1  eo.szr]/0.52.
0
Feb
Jul t
Jan
12. For Tank A we have a constant volume of 100 gal. Let x(t) denote the amount of salt in Tank A. The rate into Tank A is 0, and the rate out is 5 galls x x/lOOlb/gal = x/20lb/s. Hence the model equa• tion is
x'=�
20 The solution with initial value x(O) = 20 is x(t) = 2oet/20. The volume of solution in Tank B is increasing at 2.5 galls. Hence the volume at time t is 200 + 2.5t. Let y(t) denote the amount of salt in Tank B. Then the rate into Tank B is the same as the rate out of Tank A, x /20. The rate out of Tank B is 2.5gal/s x y/(200+2.5t)lb/gal = y/(80+t)lb/s. Hence the model equation is
,_ � Y y  20
80 + t 
80 + t
This linear equation can be solved using the inte• grating factor u (t) = 80 + t. The general solution is (t) =
y
_c_  20etf20 80 + t
400 et/20. 80 + t
Since y(O) = 40, we can compute that C 80 = 5200. Hence the solution is y(t)
=
5200 80 + t
= 65
y' = 0.52(x  y) = 2[1  eo.szt]  0.52y.
This linear equation can also be · solved using the integrating factor u(t) = e0 52t. With the initial condition y(O) = 0 we find the solution y(t)
= 2[1
 eO.SZt]/0.52  2te0.5Zt
= x(t)  2teo.szr. After 3 months, when t l/4, we have x(l/4) = 0.4689km 3 and y(l/4) = 0.0298km3. (b) If the factory is shut down, then the flow of pollutant at the rate of 2 knr' /yr is stopped.
Y_
et/20
Let y(t) be the amount of pollutant (measured in knr') in Lake Sad Times. The rate into lake Sad Times is the same as the rate out of Lake Happy Times, or 0.52x knr' /yr. The rate out is 3 3 52km /yr x y/100 = 0.52ykm /yr. Hence the model equation is
x
400 20etf20 _ et/20. 80 + t
Tank B will contain 250 gal when t = 20. At this point we have y(20) = 43.1709 lb.
This means that the flow between the lakes and that out of Lake Sad Times will be reduced to 50 knr' /yr in order to maintain the volumes. We will start time over at this point and we have the initial conditions x(O) = x1 = 0.4689km3 , and y(O) = y1 = 0.0298km3. Now there is no flow of pollutant into Lake Happy Times, and the rate out is x /2 krrr' /yr. Hence the model equation is x' = x/2. The solution is x(t) = x1 et/2• The rate into Lake Sad times is x /2 knr' /yr, and the rate out is y /2 knr' /yr. The model equation is y' = (x y)/2 = x1etf2/2y/2. this time we
2
use the integrating factor u(t) = et/ and find the solution
2.5 Mixing Problems 51 Initially, there is no salt in Tank I, so x (0) = 0 pro• duces K = a V and X
The plot of the solution over 10 years is shown in the following figure. It is perhaps a little sur• prising to see that the level of pollution in Lake Sad Times continues to rise for some time after the factory is closed.
= aV  aVe(b/V)t.
Let y (t) represent the amount of salt in Tank II at time t. Salt enters Tank II at the same rate as it leaves Tank I. Consequently, rate in II= (b/V)x(t) lb/min.
0.2
Salt leaves Tank II at
0.15
rate out II = y(t) / V lb/gal x b gal/min = (b/V)y(t) lb/min.
> 0.1
Consequently, 0.05
o� _. 0
dy b b = x y. dt v v
.
5
10
Time in years
Using a computer or a calculator, we find that y(t) = yif2 when t = 10.18 yrs. 14. Let x (t) represent the amount of salt in Tank I at time t. The rate at which salt enters Tank I is rate In I= a lb/gal x b gal/min= ab lb/min.
Substitute the solution found for x. dy b b  =  (aV  aVe(b/V)t)  y dt v v ' dy b  = y + (ab  abe<b/V)t). dt v
This equation is also linear, with integrating factor ib/V)t,
SO
Salt leaves Tank II at rate out I= x(t)/ V lb/gal x b gal/min = (b/ V)x(t) lb/min.
(e(b/V)ty)' = ab (e(b/V)t _
dx dt
b
=abx.
v
+ L,  abte(b/V)t + Le(b/V)t.
e(b/V)ty = aVe(b/V)t  abt y = aV
Consequently,
I},
Initially, there is no salt in Tank II, so y(O) = 0 produces L = a V and
This equation is linear with general solution x =av+ Ke(b/V)t_
Y = a V  abte(b/V)t  a Ve(b/V)t.
52
Chapter 2
FirstOrder Equations
Section 2.6. Exact Differential Equations 1. dF = 2ydx
+ (2x + 2y)dy
2. dF = (2x  2y)dx 3. dF
10. With P = 1  y sinx and Q = cosx, we see that
+ (x + 2y)dy
aP ay
.
aQ ax
 =SinX= ,
= xdx + ydy /x2 + y2
so the equation is exact. We solve by setting F(x, y) =
4. dF = xdx  ydy (x2 + y2)3/2
f
P(x, y) dx =
f
(1 y sinx) dx
= x + y cos x + </> (y). 5. dF =
To find ¢, we differentiate
1 x 2 +y 2
(x2ydx
+ y3dx
 ydx
aF Q(x, y) =  = cosx + ¢'(y). ay
+ x3dy + xy2dy + xdy)
Thus¢' = 0, so we can take¢ = 0. Hence the solution is F(x, y) = x + y cosx = C.
6. dF = (1/x + 2xy3)dx + (1/y + 3x2y2)dy 7.
11. With P = 1 +�and Q =_!,we compute x x
2 dF = (
x
x 2 +y 2
+ 1/y) dx
2y
+
x)

( x2 + y2
aQ 1 ayP = ;1 ::/:: a; = x2 .
dy
y2
Hence the equation is not exact. 12 .
S. dF
=
ydx  xdy
1 W"th P =
2
+ 4x y3dy + 4y5dy x2 + y2
9. With P = 2x + y and Q = x  6y, we see that
er
x
,)x2
+ y2
pute
ay
y
and Q = Jx2
, we com
+ y2
2xy
aQ
== (x2 + y2)3/2 =
a;'
so the equation is exact. To find the solution we integrate
aQ
=1=ay
ax
so the equation is exact. We solve by setting
F(x, y)
=
f
= x
2
P(x, y)dx
=
f
(2x
+ y)dx
+ xy + ¢ (y).
To find ¢, we differentiate
Q(x, y) =
aF , ay = x + ¢ (y).
F(x, y) =
f
=!
P(x, y) dx
x
dx
+ y2 = Jx2 + y2 + cp(y). Jx2
To find ¢, we differentiate aF Qt», y) =  =
y
,
= ¢ (y).
Hence¢' = 6y, and we can take ¢(y) = 3y2• Hence the solution is F(x, y) = x2+xy3y2 = C.
ay
Jx2
+ y2
Thus ¢' = 0, so we can take ¢ = 0. Hence the solution is F(x, y) = Jx2 + y2 = C.
2.6 13. Exactx3 +xy  y3
Exact Differential Equations
34. ln x  In y and 1 are homogeneous of degree zero.
=C
14. Not exact
35. x2  Cx = y2
15. Exact u2 /2
+ vu
 v2 /2
16. Exact ln(u + v2) 2
=C
36.
=C F(x, y)
= (1/2) In
17. Not exact
= ylnu
18. Exact. F(u, y)
19. Exactx sin2t  t2 20.
53
Exact x2y2
 2u
c
=C
2
+ arctan(y/x)
=C 37.
+ x4 = C
F(x, y)
2 ;
Y
)
 Inx = C
= xy + (3/2)x2 = C.
38. y2x2 41ny 2Inx = C 21. Not exact. 39. 22. x/y
+ Inx = C
23. x2y2/2  lnx 24.
(y
+ Iny =
40. y(x) = x ln(C
C
+ 1)2
41.
=C x3
25. x  (1/2) ln(x
2
+y ) =C
26. µ(x) = l/x . F(x, y) =
28. µ(y)
xy2/2 y x
29. µ(y)
= C
y2 F(x, y)
= 1/ � · = l/y2•
(a) First, dy
dy/dt
dx
dx/dt
= xy  Inx  2 = C.
F(x, y)
yx +x2
y
=C
+ y2, so
dy
voy ;::::== w Jx2 + y2
dx
=
F(x, y) = x2y3 = C
Jx2
+ y and x
 y are homogeneous of degree one.
dy
+ y2
and y are homogenous of degree
Jx
COS
+ y2
2
O and sin O =
VoY  wJx2
+ y2
VQX
+ y2
+ y2
Vo
=
y kJx2
x
x
(b) Write the equation
32. x2  xy  y2 and 4xy are homogeneous of degree two. 33. x  Jx2 one.
Vo
Divide top and bottom by v0 and replace co/ v0 with k.
dx
31. x
v0sinew
VQX
y  w Jx2
30.
Xj
y/Jx2
= 2x� + (2/3)y312 = C
F(x, y) =
=
However, cos O
1
= . x
+ 2 lnx)
2
2
27. µ(x)
+Cx4 y(x) = 1  2Cx3 X
dy y  kJx2 = dx x
+ y2
in the form (y  kJx2
+ y2) dx  x dy =
0.
+ y2
54
Chapter 2 FirstOrder Equations Both terms are homogeneous (degree 1), so make substitutions y = xv and dy = x dv +
Finally, recall that y = xv, so
vdx. (xv k/x2
+ x2v2) dx x(x dv + v dx) = 0
After cancelling the common factor x and com•
� =
H(�)'  Gr']
v=
H(�t' (�Y1
bining terms, k� dxxdv=O.
Separate variables and integrate. kdx
dv
x
v'f+v2

=0
klnxln(� +v)=C
(c) The following three graphs show the cases where a = 1, and k = 1/2, 1, 3/2. When 0 < k < 1, the wind speed is less than that of the goose and the goose flies home. When k = 1 the two speeds are equal, and try as he might, the goose can't get home. Instead he approaches a point due north of the nest. When k > 1 the wind speed is greater, so the goose loses ground and keeps getting further from the nest.
Note the initial condition (x, y) = (a, 0). Be• cause y = xv, v must also equal zero at this
point. Thus, (x, v) = (a, 0) and klna  ln(/I=o2 + 0) = C C = klna.
Therefore, klnx ln(�
+ v)
= klna.
Taking the exponential of both sides, elnxkln(� +v)
= elnak
xk
y
=ak
v
+ v'f+v2
(�f
=v+� .
Solve for v.
(�fv=�
(�fk  2v (�f + v
2
o
= 1
+ v2
1 = 2v
�[(�)'(�r']=v
(�f
2.6
Exact Differential Equations
55
(b) The differential equation is homogeneous. Solving in the usual way we find that the or• thogonal family is defined implicitly by
=
G(x, y)
y x2
+y 2
= C.
The original curves are the solid curves in the following figure, and the orthogonal family is dashed. 2 /
42.
The hyperbolas with F (x, y) = y / x = C are the solid curves in the following figure. The orthogonal family must satisfy

'\
2
>,
0 1
dy = aF;aF = _ 2x. dx ay ax y
\
2
The: solution to this separable equation is found to be given implicitly by G(x, y) = 2x2+ y2 = C. These curves are the dashed ellipses in the accompanying figure. They do appear to be orthogonal.
2
1
2
0
x
+ x2 )
44.
ln(y2
45.
arctan(y/x)  y4/4 = C
46.
Assuming that m =j:. n  1, divide both sides of

X
>,
/
<,
2
(2/3)y3 = C
dy
+ y dx = xmyn dx
by x" v" to obtain
0
X
dy + y dx m=n d X (xy") d((xy)ln)  =xm "<dx . ln
=X
·1
·2
2
1
0
2
Thus, because m  n
x 43.
8F/8F
= ay a7 =
2xy x2 _ y2 ·
=j:. 0,
(xy)ln xmn+l = +c 1n mn+l 1 (m  n + l)(xy) n  (1  n)xmn+l = C.
(a) The curves are defined by the equation F(x, y) = x/(x2 + y2) = c. Hence the or• thogonal family must satisfy dy dx
+1
47.
arctan(y/x)  (l/4)(y2
48.
x/y  ln(xy
+ 1)
= C
+ x2)2
= C
56
Chapter 2
FirstOrder Equations
x+y 49. Cxy=• xy 50.
Rearranging, dy x ± Jx2 + y2 =dx y
(a) An exterior angle of a triangle equals the sum of its two remote interior angles, so e = ¢+a. We're given that a= fJ and e and fJ are corre• sponding angles on the same side of a transver• sal cutting parallel lines, so ¢ = fJ. Thus,
becomes ±xdx+ydy =dx. Jx2 + y2
e = ¢ + a = fJ + fJ = 2{3 and tan e
= tan 2{3 =
2 tan fJ 1  tan
2
fJ
.
The trick now is to recognize that the lefthand side equals ±d(Jx2 + y2). Thus, when we integrate,
However, tan e = y / x and tan fJ equals the slope of the tangent line to y = y(x) at the point (x, y); i.e., tan fJ = y'. Thus, y x
±d()x2
2y' 1  (y')2•
+
±Jx2
(b) Use the result from part (a) and cross multiply.
y2) = dx
+ y2
=x
+ C.
Square, then solve for y2.
y  y(y')2 = 2xy' 0
x2
= y(y')2 + 2xy'  y
Use the quadratic formula to solve for y'. ' x±Jx2+y2 y =
+ y2 = x2 + 2Cx + C2 y2 = 2Cx + C2
This, as was somewhat expected, is the equa• tion of a parabola.
y
Section 2.7. Existence and Uniqueness of Solutions 1. The right hand side of the equation is 4
+ y2 • f
f (t, y) =
theorem are not satisfied.
is continuous in the whole plane. Its
partial derivative aJ/8y = 2y is also continuous on the whole plane. Hence the hypotheses are satisfied and the theorem guarantees a unique solution. 2. The right hand side of the equation is f ( t, y) = ,Jy. f is defined only where y ::::: 0, and it is continu• ous there. However, aJJay = 1/(2,Jy), which is only continuous for y > 0. Our initial condition is at Yo = 0, and to = 4. There is no rectangle con• taining (to, y0) where both f and aJ/ay are defined and continuous. Consequently the hypotheses of the
3. The right hand side of the equation is f (t, y) = t tan:" y, which is continuous in the whole plane. aJ/ay = t/(1 + y2) is also continuous in the whole plane. Hence the hypotheses are satisfied and the theorem guarantees a unique solution. 4.
The right hand side of the equation is f (s, cv) = co sin co + s, which is continuous in the whole plane. aJ;acv = sin co + co cos co is also continuous in the whole plane. Hence the hypotheses are satisfied and the theorem guarantees a unique solution.
5. The right hand side of the equation is f (t, x) = t/(x + 1), which is continuous in the whole plane, except where x = 1. aJ;ax = t/(x + 1)2 is also continuous in the whole plane, except where x == 1. Hence the hypotheses are satisfied in a rectangle containing the initial point (0, 0), so the theorem guarantees a unique solution. 6. The: right hand side of the equation is f (x, y) y / x + 2, which is continuous in the whole plane, except where x = 0. Since the initial point is (0, 1),
f is discontinuous there. Consequently there is no rectangle containing this point in which f is continu• ous. The hypotheses are not satisfied, so the theorem does not guarantee a unique solution.
2.7 Existence and Uniqueness of Solutions the following figure.
2
>,
57
.
0
.. 
1
.·
.·.

2"''''''
1
0.5
0
0.5
7. The equation is linear. The general solution is y(t) = t sin t + Ct. Several solutions are plotted in the following figure.
Since the general solution is y(t) = t
+ 2Ct2 , every
solution satisfies y(O) = 0. There is no solution with y(O) = 2. If we put the equation into normal form dy
2y  t
dt
we see that the right hand side f(t, y) = (2y  t)/t fails to be continuous at t = 0. Consequently the hy• potheses of the existence theorem are not satisfied.
Since every solution satisfies y(O) = 0, there is no solution with y(O) = 3. If we put the equation into normal form y dt
1
= y + t cost' t
we see that the right hand side f ( t, y) fails to be continuous at t = 0. Consequently the hypotheses of the existence theorem are not satisfied.
9. The yderivative of the right hand side f (t, y) = 3y213 is 2y113, which is not continuous at y = 0. Hence the hypotheses of Theorem 7.16 are not sat• isfied.
10. Theyderivative of the right hand side f (t, y) = ty112 is ty112 /2 which is not continuous at y = 0. Hence the hypotheses of Theorem 7.16 are not sat• isfied.
11. The exact solution is y(t) = 1 8. The equation is linear. The general solution is y(t) = t + 2Ct2• Several solutions are plotted in
+�.
The
interval of solution is ( ,J3, oo). The solver has trou• ble near ,./3. The point where the difficulty arises is
58
Chapter 2
FirstOrder Equations
figure.
circled in the following figure.
y
4 2 0
0
� t
2
1
2
2
2
4
4
12. Theexactsolutionisy(t) = l+Jt2  4t  1. The interval of existence is (oo, 2  ,JS). The solver has trouble near 2  ,Js � 0.2361. The point where the difficulty arises is circled in the following figure.
14. The
exact
solution
is
3
y(t)

J4 + 2 ln(t + 2)  2 ln 2. The interval of existence is (2 + 2e2, oo). The solver has trouble near 2 + 2e2 � 1.7293. The point where the diffi• culty arises is circled in the following figure.
y 4
0
1
t 0
2
3
2
0
2
1
4
13. The exact solution is y(t) = 1 + J4 + 2 ln(l  t). The interval of existence is (oo, 1  e2). The solver has trouble near 1  e2 � 0.8647. The point where the difficulty arises is circled in the following
1
15.
The solution is defined implicitly by the equation  3y = 2t3 /3. The solver has trouble near (t1, 1), where t1 = (5/2)113 � 1.3572, and also near (t2, 3), where tz = (27 /2)113 � 2.3811. The points where the difficulty arises are circled in the y3 /3 + y2
2.7
Existence and Uniqueness of Solutions
59
ure.
following figure.
q
5 y
4
2
3 +� � +==+� =+� t� � .
a
3
2
0
4
3
2
0
4
The exact solution is 5  5et q(t) = { 5(1  e�)e2t,
16. The: solution is defined implicitly by the equation 2y3  15y2 + 2t3 = 81. The solver that trouble near (t1,0), where t1 = (81/2)113 � 3.4341, and also near (t2, 5), where t: = 22113 � 2.8020. The: points where the difficulty arises are circled in the following figure.
if O < t < 2, if t � 2
Hence q(4) = 5(1  e2)e2 � 0.5851.
18. Thecomputedsolutionisshowninthefollowingfigure.
q 3
y
2
6 4 0
2
0
4
0
2
3
2
0
The exact solution is O, q(t) = { 3(1  e21),
Hence q(4) = 3(1  e2) 17. The computed solution is shown in the following fig
4
�
if O < t < 2, if t 2: 2
2.5940.
19. The computed solution is shown in the following fig
60
Chapter 2
FirstOrder Equations 21.
ure.
(a) If
O,
if t ::::: to if t > to,
y(t) = { (t  to)3, q
3
then
y
2
'(t+) _ 1. y(t)  y(to) O  Im t'\..to t  to (t  to)3  0
.
= lim Htit t  to = lim (t  tol t+tt
0
=0.
4
3
2
0
On the other hand,
The exact solution is
2(t 1 +et), q (t) = { 2(1
'( _)
ifO < t < 2,
+ e2)e2t,
y t 0
if t :::::: 2
.
y(t)  y(to)
1

= 1m t/'to
t  to
= lim 00
Hence q(4) 20.
= 2(1 + e2)e2
�
0.3073.
t+to t 
to
=0.
The computed solution is shown in the following fig• ure.
Therefore, y' (to) = 0, since both the left and righthand derivatives equal zero. (b) The right hand side of the equation,
q
f (t, y) =
3y213 , is continuous, but aJJay = 2y113 is not
3
y = 0. Hence the hypothe• continuous where ses of Theorem 7 .16 are not satisfied.
2
22.
(a) If
3e2t' y(t) = { 5/2
+ (3  5e2 /2)e2t,
if t < I if t :::::: 1
0 2
0
3
4
The exact solution is
then it is easily seen that
'
y (t) =
0, q (t) = { t  1  e2t,
Hence q(4) = 3  e2
�
if O < t < 2, if t : :::: 2
2.8647.
{6e2t' (6 + 5e2)e2t
t < 1 t > 1.
It remains to find the derivative at t = 1. Re• member, because of the "cusp" at t = 1, we suspect that this derivative will not exist. First,
2.7
= lim y(t)  y(l)
Hl +
then
1
t 
+(
1 [ (5 = lim t  1
2
2
5e ) 2 3   e2
r)
t < I t > 1.
_ G + (3 _ s;}2) J =
if t < 1 if t > 1,
f(t) = { �:
y�(l)
t+J+
lim (3¥)e� (3¥)
Thus,y = y(t)isasolutionofy' = 2y+ f(t) on any interval not containing t = 1. 2
c
23.
If
t 1
t+1+
2 , [ 2 ( 3  5e ) e" y I = lim + t+I+ 2
,
y (t)
=
{O,4t
0
3,
t < t > 0.
leftderivative. y
y�(l) = lim y(t)  y(l)
t1

, (0) _ 1· y(t)  y(O) _ 1· 00 _O  1m  1m  . 
(52 + (
2
2
3
5e ) ) T e
t 0
t+O

Thus, y' (0)
Again, an indeterminant form 0/0, so we apply l'Hopital's rule. y�(l) = lim
t 0
t+O+
3e2
= lim t+ 1t  1
(6e2t)
lim t+O+ t
= 6e2•
y ' (t)
=
{ 0,
t<O i : 0.
'
=
{O,
t<O
ty (t)
(b) The derivative from the left doesn't equal the derivative from the right. The function y = y(t) is not differentiable at t = 1 and cannot be a solution of the differential equation on any
4t3,
t
> 0,
and
4y(t)
= {O, 4 4t ,
t<O t > 0,
= y(t) is a solution of ty' = 4y. Finally, y(O) = 0. In a similar manner, it is not difficult to
interval containing t = 1.
so y
(c) We have that
show that I
y (t) =
{6e(6
+ 5e2)e2r
t+O+
= 0 and we can write
4t 4 ,
t,
lim t3 = 0.
Now,
t+J
2
t
t4 (0)  1· y(t)  y(O) Y+  1m
t 1
t+ I
t+O
Secondly,
I
3e2t
>: 0,
It remains to check the existence of y'(O). First, the
However, the derivative from the left,
2t
t<O t
then it is easily seen that
2t]
= 5  6e2
= lim 3e
O, 4 {t '
y(t) =
Note the indeterminantform 0/0, so l'Hopital's rule applies.
t+J
61
Existence and Uniqueness of Solutions
If
the derivative from the right,
f
< 1
t > 1.
cv(t) =
O, { 5t 4 ,
t < 0 t >: 0
62
Chapter 2
FirstOrder Equations
is also a solution of the initial value problem ty' = 4 y, y(O) = 0. At first glance, it would appear that we have contradicted uniqueness. However, if ty' = 4 y is written in normal form,
h
4y
I
y=•
t
(1, 0)
then
!
ay
(4y) = � t
t
is not continuous on any rectangular region contain• ing the vertical axis (where t = 0), so the hypotheses of the Uniqueness Theorem are not satisfied. There is no contradiction of uniqueness.
24.
(a) The point here is the fact that you don't know the moment the water completely drained. Here are two possibilities.
(b) Let A represent the cross area of the drum and h the height of the water in the drum. Then t:,.h represents the change in height and A t:,.h the volume of water that has left the drum. A particle of water leaving the drain at speed v travels a distance v /;:,.t in a time Sr, Because a is the cross section of the drain, the volume of water leaving the drain in time /;:,.t is av !;:,.t. Because the water leaving the drum in time /;:,.t must exit the drain, At:,.h = av S: !;:,.h
A =av. St
h
Taking the limit as
/;:,.t +
0,
dh
Adt = av. Using v2 = 2gh, v = .J2gii, and dh dt =
a r::;;::;:
A" 2gh.
(2, 0)
The minus sign is present because the drum is draining.
(c) If we let co = ah ands= f3t, then by the chain rule dco dco dh dt a dh ds = dh . dt . ds = dt ·
fi
2.7 Multiply both sides of our equation by a/ {3,
(:!..) (�) /fih. f3
� dh = /3 dt A
Replace (a/ f3)(dh/dt) with dcalds and h with co]«.
!:=mmfi!dw ds
= _!_
(:!..) /2gav'w.
/3 A
Let ho represent the height of a full tank. This motivates the selection of a = 1 / ho and co = h/ ho, as co = 0 when the tank is empty and co = 1 when the tank is full. Thus, dw ds
= _ !_ /3
(:!..) {iiv'w, V ho
A
which motivates the selection of
which upon substitution, gives us dw =vw. ds (d) Separate the variables and integrate.
+C
1 w1;2 = (C  s) 2 1 co =
4
2
(C  s)
However, as evidenced in part (a), we only want the left half of this parabola. After the drum empties, it remains empty for all time. Thus, for any C < so, 1 w(s) =
2
4(C  s) '
{ 0,
63
is a solution of w' = fo, w(so) = 0. Finally, this emergence of multiple solutions does not contradict uniqueness, because in
�cvw) = aw
1_
2,jw
is not continuous on any rectangle containing the horizontal axis (defined by to = 0). 25. The equation x' = f (t, x) satisfies the hypotheses of the uniqueness theorem. Notice that x1 (0) = x2(0) = 0. If they were both solutions x' = f(t, x) near t = 0, then by the uniqueness theorem they would have to be equal everywhere. Since they are not, they cannot both be solutions of the differential equation. 26. The equation x' = f (t, x) satisfies the hypotheses of the uniqueness theorem. Notice that XI (rr /2) = x2(n /2) = 0. If they were both solutions x' = f (t, x) near t = n /2, then by the uniqueness theo• rem they would have to be equal everywhere. Since they are not, they cannot both be solutions of the differential equation. 27. Notice that xj (t) = 0 is a solution to the same differ• ential equation with initial value x1 (0) = 0 < 1 = x(O). The right hand side of the differential equa• tion, f (t, x) = x cos2 t and aflax = cos2 tare both continuous on the whole plane. Consequently the uniqueness theorem applies, so the solution curves for x and x1 cannot cross. Hence we must have x(t) > XI (t) = 0 for all t. 28. Notice that YI (t) = 3 is a solution to the same dif• ferential equation with initial value Yt (1) = 3 > 1 = y(l). The right hand side of the differential
w112 dco = ds 2w112 = s
Existence and Uniqueness of Solutions
s<C
s :=:: C,
equation, f (t, y) = (y  3)ecos(ty) and aflay = ecos(ty)[]  t(y  3) sin(ty)] are both continuous on the whole plane. Consequently the uniqueness theo• rem applies, so the solution curves for y and y1 can• not cross. Hence we must have y(t) < YI (t) = 3 for all t. 29. Notice that the right hand side of the equation is f (t, y) = (y2  1 )ety and f is continuous on the whole plane. Its partial derivative af!ay = 2yety + t(y2  l)ety is also continuous on the whole plane. Thus the hypotheses of the uniqueness theo• rem are satisfied. By direct substitution we discover
64 Chapter 2 FirstOrder Equations that YI (t) = 1 and Y2(t) = 1 are both solutions to the differential equation. If y is a solution and satisfies y(l) = 0, then YI (1) < y(l) < y2(1). By the uniqueness theorem we must have y1 (t) < y(t) < Y2(t) for all t for which y is defined. Hence 1 < y(t) < 1 for all t for which y is defined. 30. Notice that x1 (t) = 0 and x2(t) = 1 are solu• tions to the same differential equation with initial values x1 (0) = 0 < 1/2 = x(O) < 1 = x2(0). The right hand side of the differential equation, f(t, x) = (x3  x)/(1 + t2x2), and aJ (3x2  1)(1 = ax
+ t2 x 2 )
31.
Notice that x.fr) = t2isasolutiontothesamediffer• ential equation with initial value x1 (0) = 0 < 1 = x(O). The right hand side of the differential equa• tion, f (t, x) = x  t2 + 2t and aJJax = 1 are both continuous on the whole plane. Consequently the uniqueness theorem applies, so the solution curves for x and x1 cannot cross. Hence we must have t2 = XI (t) < x(t) for all t.
32.
Notice that y1 (t) = cost is a solution to the same dif• ferential equation with initial value y1 (0) = 1 < 2 = y(O). The right hand side of the differential equa• tion, f(t, y) = y2  cos2 t  sin t and aflay = 2y are both continuous on the whole plane. Conse• quently the uniqueness theorem applies, so the so• lution curves for y and y1 cannot cross. Hence we must have y(t) > y1 (t) =cost for all t.
 2t2x(x3  x)
are both continuous on the whole plane. Conse• quently the uniqueness theorem applies, so the solu• tion curves for x, x1, and x2 cannot cross. Hence we must have O = x1 (t) < x(t) < x2(t) = 1 for all t.
Section 2.8. Dependence of Solutions on Initial Conditions 1. x(O)
= 0.8009
13. Ten! :)
2. x(O)
= .9084
14.
3. x(O) = 0.9596 4. x(O)
= 0.9826
5. x(O)
= 0.7275
6. x(O)
= 0.72897
7. x(O)
= 0.7290106
8. x(O)
= 0.729011125
9. x(O)
= 3.2314
10. x(O)
= 3.23208
11. x(O)
= 3.2320923
12. x(O) = 3.23092999999
1esint  (1/lO)eltl S y(t) S 1 esint + (1/lO)eltl
2.9 The three middle curves are the solutions to the differential equation corresponding to the initial conditions x(O) = .1, 0 .1, and the outside curves are the graphs of ei. and en. Note how the solutions of the differential equation remain inside the graphs of ei, and en.
17.
esint _
o.or«
:'.:S y(t) :'.:S l _ esint
65
(a) The right hand side of the equation is f ( t, x) = 2x + sint, and of/ax = 2. Hence M = max(lof/oxl) = 2, and Theorem 7.15 pre• dicts that ly(t)  x(t)I :::s ly(O)  x(O)leMltl = ly(O)  x(O)le21tl. (b) The equation is linear, and we find that x(t) = [2 sin t  cos t]/5, and y(t) = [2 sin t 
15. The only adjustment from the previous exercise is that we now want lxo  Yol < 0.01. This leads to 1
Autonomous Equations and Stability
cos t]/5  e21 /10. Hence x(t)  y(t) = e21 /10 = lx(O)  y(O)le2 t
+ O.Oleltl
:'.:S ly(O)  x(O)le21r1.
and this image.
(c) Since e2r = e21tl for t < 0, we see that the
x 4
maximum predicted error is achieved for all t < 0. 18. The right hand side of the equation is f (t, x) = x2  t, and aflax = 2x. On the rectangle R we have [x] :'.:S 2, so M = max 1aJ/oxl = max 12xl = 4. Thus the bound predicted by Theorem 7 .15 is
The maximum predicted error is where [r] = 1, and it is 40.9486. the two solutions are plotted in the following figure. 4 16.
(a) Therighthandsideoftheequationisf(t,x) = (x  1) cost. Thus of/ox = cost, and max loflax I = max I cos ti = 1. Hence Theo• rem 7.15 predicts that lx(t)  y(t)I :::S lx(O) y(O)leltl. (b) The equation is separable and linear, and the solutions are x(t) = 1  esint and y(t) = 1 9esint /10. Hencetheseparationisx(t)y(t) = esint /10. Since sin t :::S [r], we see that lx(t)y(t)I = esint /10 :'.:S e1t1/10 = lx(O)y(O)le1tl. (c) Since sin t < [r] except at t = 0, we have lx(t)  y(t)I < eltl/10, except at t = 0.
1.l
1
The actual bound is about 2, which is much less than 41, the theoretical bound.
66
Chapter 2
FirstOrder Equations
Section 2.9. Autonomous Equations and Stability explicit dependence of the righthand side of this differential equation on the independent variable t causes the equation to be nonautonomous.
1. Note that P' = 0.05P  1000 is autonomous, having form P' = f (P). Solving the equation 0 = f(P) = 0.05P  1000, we find the equilib• rium point P = 20000. Thus, P(t) = 20000 is an unstable equilibrium solution, as shown in the fol• lowing figure. 4.
x
10
4
Note that P' = 0.13P(l  P /200) is autonomous, having form P' = f (P). Solve the equation f (P) = 0 to find the equilibrium points.
4 �������������������� //II/IIIIIIII//I/III IIIIIIII/I/////IIIII
0.13P
3 ;�;;;;;;��;�;;�;�;;�
(1  z:o) = 0
,,,,,,,,,,,,,,,,,,,,
0.
2

,,,,,,,,,,,,,,,,,,,,
P
''''''''''''''''''''
=0
or
P = 200
,, ,, ,, ',, ', ', '' '' ', ', ', ', ', ', ', ', ', ', '' ' ,,'' ,, ,, ,,', ',, ',', ', '' ', ','' ', ', '' ',', ' '' ' '' ''''._0 ......''..' '' ' ' ' ' _'....._'''' ' '50 0 100
Thus, P(t) = 0 and P(t) = 200 are equilibrium solutions, as shown in the following figure. P = 0 unstable and P = 200 is asymptotically stable.
2. Note that y' = 1  2y + y2 is autonomous, having form y' = f (y). Solve the equation f (y) = 0 to find the equilibrium points. 12y
+ y2 = 0
300
,,,,,,,,,,,,,,,,,,,,
,,,,,,,,,,,,,,,,,,,,
y = 1.
200 1=,. .,..,. .==...==,..,==....==,. .,.. .,==.. ,==,..,==....==,. .,.
             
Thus, y(t) = 1 is an unstable equilibrium solution, as shown in the following figure.
;;
o,
100
;;;
;;;
;;;;;;;
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;;;;;;;;;;;;;;;;;;;; ;;;;;;;;;;;;;;;;;;;;
=======
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0 ,,,,,,,,,,,,,,,,,,,,
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I I I I
5
10
0
3. Note that x' = t2

x2 is not autonomous, having
form x' = f(t, x), where f(t, x) = t2

x2 • The
5. The equation is autonomous. The point q = 2 is an unstable equilibrium point, as the following figure shows. In addition every solution of sin q = 0 is an equilibrium point. These are the points kn, where k is any integer, positive of negative. The stability of the equilibrium points alternates between asymptotic
2.9
Autonomous Equations and Stability
= 2 is unstable.
is asymptotically stable and y
stable and unstable, as is seen in the figure.
67
q
y I
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6. The equation is not autonomous because of the cost term. 7. Note that the graph off (y) intercepts the yaxis at y = 3. Consequently, y = 3 is an equilibrium point (f (3) = 0) and y(t) = 3 is an equilibrium solution, shown in the following figure. The solution y = 3 is unstable.
9. Since f (y) has zeros at y = 1 and y = 1, these are equilibrium points. Correspondingly, y(t) = 1 and y(t) = 1 are equilibrium solutions, and are plot• ted in the following figure. Both are unstable.
y y / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / /
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•• t
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8. Note that the graph of f (y) intercepts the yaxis at y = 0 and y = 2. Consequently, y = 0 and y = 2 are equilibrium points(! (0) = 0 and f (2) = 0) and y(t) = 0 and y(t) = 2 are equilibrium solutions, shown in the following figure. The solution y = 0
10.
Since f (y) has zeros at y = 2, y = 1/2, y = 1, and at y = 2, all four are equilibrium points. Cor• respondingly, y(t) = 2, y(t) = 1/2, y(t) = 1, and y(t) = 2 are equilibrium solutions, and are plot• ted in the following figure. y =  2 and y = 1 are
68 Chapter 2 FirstOrder Equations asymptotically stable and the other two are unstable.
in the following figure.
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� , ; , > , ; , > > > , , ; > > > > , ,
Because the differential equation y' = f (y) is au• tonomous, the slope at any point (t, x) in the direc• tion field does not depend on t, only on y, as shown in the following figure.
\ \ \ \ <,
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ , , , , , , , , <, ,

/
/ / / / / / / / / /
I
I I I I I I I I I I
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I I I I I I I I I I
There are two equilibrium points. The smaller of them is unstable and the other is asymptotically sta• ble.
13. The key thing to note is the fact that v' and f (y) are equal. Consequently, the value of f (y) is the slope of the direction line positioned at (t, y).
• At y = 3, f (y) = 0 and the slope is zero. Thus y = 3 is an equilibrium point. This is shown in the following figure. • To the right of y = 3, note that the graph off dips below the yaxis. Therefore, as y increases beyond 3, the slope becomes increasingly nega• tive. This is also shown in the following figure.
The equilibrium point is asymptotically stable.
12. Because the differential equation y' = f (y) is au• tonomous, the slope at any point (t, x) in the direc• tion field does not depend on t , only on y, as shown
• To the left of y = 3, note that the graph off rise above the yaxis. Therefore, as y decreases be• low 3, the slope becomes increasingly positive. This is also shown in the following figure. In particular, this means that the equilibrium point y = 3 is asymptotically stable.
2.9
y
\
\ \
Slope is more negative as y increases.
, Slope is zero at y = 3.
3 /
I Slope is more positive as y decreases. Finally, because the equation y' = f (y) is autonomous, the slope of a direction line positioned at (t, y) depends only on y and not on t. Consequently, the rest of the direction field is easily completed, as shown in the next figure.
\ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\ , /
I
\ \ \ \ \ \ \ \ \ , , , , , , , , , ,
69
• Between y = 0 and y = 4, the graph of f lies above the yaxis. Consequently, f (y) is positive for O < y < 4. Moreover, the graph of f has a maximum about halfway between y = 0 and y = 4. Consequently, the slope of the direction field lines will be positive between y = 0 and y = 4, with a maximum positive slope occurring about halfway between y = 0 and y = 4. This is shown in the following figure.
• To the left of y = 0, note that the graph of f falls below the yaxis. Furthermore, as y de• creases below O, f (y) (the slope of the direction line at (t, y)) becomes increasingly negative. This is also shown in the next figure.
From these considerations we see that the equilib• rium point y = 0 is unstable, and y = 4 is asymp• totically stable.
/ / / / / / / / / /
I I I I I I I I I I y
I I 4
14.
Autonomous Equations and Stability
The key thing to note is the fact that y' and f (y) are equal. Consequently, the value of f (y) is the slope of the direction line positioned at (t, y). • At y = 0 and y = 4, f (y) = 0, so the slope of the direction lines at these yvalues is zero. These points are the equilibrium points. This is shown in the following figure. •• To the right of y = 4, note that the graph of
f dips below the yaxis. Furthermore, as yin• creases beyond 4, /(y) (the slope of the direc• tion line at (t, y)) becomes increasingly nega• tive. This is also shown in the following figure.
I I \
I I I
Slope is more negative as y increases beyond y = 4. Slope is zero at y = 4. Slope is most positive midway between y = 0 and y = 4 Slope is zero at y = 0. t Slope is more negative as y decreases below y
= 0.
Finally, because the equation y' = f (y) is au• tonomous, the slope of a direction line positioned at (t, y) depends only on y and not on t. Consequently, the rest of the direction field is easily completed, as
70
Chapter 2 FirstOrder Equations shown in the next figure.
I I
I
stable equilibrium solution, y(t) = 2.
YI I \ I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I
15. (i) In this case, .f (y) = 2 y, whose graph is shown in the following figure.
j(y)=2y
16. (i) In this case, .f (y) = 2 y 7, whose graph is shown in the next figure. j(y)=2y7
(ii) The phase line is easily captured from the previâ€˘ ous figure, and is shown in the following figure. (ii) The phase line is easily captured from this figure, and is shown in next figure. 2  y
(iii) The phase line in the second figure indicates that solutions increase if y < 2 and decrease if y > 2. This allows us to easily construct the phase portrait shown in the ty plane in the next figure. Note the
4Âˇ 7
y
2
(iii) The phase line in the second figure indicates that solutions decrease if y < 7 /2 and increase if
y > 7 /2. This allows us to easily construct the phase
portrait shown in the ty plane in the next figure. Note the unstable equilibrium solution, y(t) = 7 /2.
2.9 Autonomous Equations and Stability 71 (iii) The phase line in the second figure indicates that solutions increase if y < 1, decrease for 1 < y < 4, and increase if y > 4. This allows us to easily construct the phase portrait shown in the ty plane in the next figure. Note the unstable equilib• rium solution, y(t) = 4, and the stable equilibrium solution, y(t) = 1.
17. (i) In this case, f (y) = (y + l)(y4), whose graph is shown in the next figure. 18. (i) In this case, f (y) = 6 + y  y2 factors as f (y) = (2 + y)(3 y), whose graph is shown in the next figure.
f(y)=6+yy2
(ii) The phase line is easily captured from the previ• ous figure, and is shown in the next figure.
.
.
1
..
4•
..
.. y
(ii) The phase line is easily captured from the previ
72 Chapter 2 FirstOrder Equations ous figure, and is shown in the next figure.
figure.
f(y)=9yy3
(iii) The phase line in the second figure indicates that solutions decrease if y <  2, increase for 2 < y < 3, and decrease if y > 3. This allows us to easily construct the phase portrait shown in the ty plane in the next figure. Note the unstable equilib• rium solution, y(t) = 2, and the stable equilibrium solution, y(t) = 3.
(ii) The phase line is easily captured from the previ• ous figure, and is shown in the next figure.
�t•3�0·�3•� Y
19. (i) In this case, f (y) = 9y  y3 factors as f (y) = y(y + 3)(y  3), whose graph is shown in the next
(iii) The phase line in the second figure indicates that solutions increase if y < 3, decrease for  3 < y < 0, increase if O < y < 3, and decrease for y > 3. This allows us to easily construct the phase portrait shown in the ty plane in the next figure. Note the stable equilibrium solution, y(t) = 3, the un• stable equilibrium solution, y(t) = 0, and the stable
2.9 equilibrium solution, y(t)
Autonomous Equations and Stability
73
3 < y < 1, decrease if 1 < y < 3, and increase for y > 3. This allows us to easily con• struct the phase portrait shown in the ty plane in the next figure. Note the unstable equilibrium so• lution, y(t) = 3, the stable equilibrium solution, y(t) = 1, and the unstable equilibrium solution,
= 3.
y
y(t) = 3.
Jy
(0,3)
�
\
,
::: t
(0,1)
= (y + l)(y  9) factors as f(y) = (y+l)(y3)(y+3), whose graph is shown
20. (i) ln this case,
f (y)
2
(0,3) ""'\
in the next figure.
,
21. Due to the periodic nature of this equation, we sketch only a few regions. You can easily use the periodicity to produce more regions. (i) In this case, f (y) = sin y, whose graph is shown in the next figure. f(y)=siny
(ii) The phase line is easily captured from the previ• ous figure, and is shown in the next figure .
. ..
3
1
.
•
3
..
.. y
(iii) The phase line in the second figure indicates that solutions decrease if y < 3, increase for
74
Chapter 2 FirstOrder Equations (ii) The phase line is easily captured from the previ• ous figure, and is shown in the next figure.
. ..
•
0
.
.
..
1(
•
in the next figure.
.. .. y
2rr
f(y)=cos 2y
(iii) The phase line in the second figure indicates that solutions decrease if TC < y < 0, increase for 0 < y < TC, decrease if tt < y < Zn , and increase for 2n < y < 3n. This allows us to easily con• struct the phase portrait shown in the ty plane in the next figure. Note the unstable equilibrium solution, y(t) = 0, the stable equilibrium solution, y(t) = n , and the unstable equilibrium solution, y(t) = Zn:
y
(ii) The phase line is easily captured from the previ• ous figure, and is shown in the next figure .
.
.
.
n
4
22. Due to the periodic nature of this equation, we sketch only a few regions. You can easily use the periodicity to produce more regions.
(i) In this case,
f (y)
= cos 2 y, whose graph is shown
.
•
37f
..
.. y
4
(iii) The phase line in the second figure indicates that solutions increase if TC/ 4 < y < TC/ 4, de• crease for TC/ 4 < y < 3n /4, and increase if 3n / 4 < y < 5n / 4. This allows us to easily con• struct the phase portrait shown in the ty plane in the next figure. Note the stable equilibrium solution, y(t) = TC /4, and the unstable equilibrium solution,
2.9
Autonomous Equations and Stability
75
y(t) = 3n/4. f(y)=6y
23. The: equation is linear, so multiply by the integrating factor and integrate.
y
(et y)' = 6e1
+c y(t) = 6 + c« ' ly = 6et
(0,6)
The: initial condition y(O) = 2 produces C = 4 and y(t) = 6  4et. Now, et approaches zero as t �  +oo, so
lim y(t)
t++oo
= t+oo lim (6  4et) = 6.
Compare y' = f (y) with v' = 6  y. Then f (y) = 6  y, whose graph is shown in the first fig• ure below. The phase line on the yaxis in this figure shows that y = 6 is a stable equilibrium point, so a trajectory with initial condition y(O) = 2 should approach the stable equilibrium solution y(t) = 6, as shown in the second figure. This agrees nicely with the analytical solution.
24. Writing the equation as y' = 5  2y, we see that the right hand side is f (y) = 5  2 y. The graph of f is in the next figure. We have also indicated the direction of the solutions on the yaxis, which shows that y = 5 /2 is an asymptotically stable equi• librium point. Thus any solution curve will approach
76
Chapter 2 FirstOrder Equations y = 5 /2 as t increases.
as t increases. f (y) =
(1
+ y)(5  y)
f(y) = 52y
To find the exact solution, we separate variables and use partial fractions to get 1 +1 J 61 [ l+y 5y
dy=dt
Integrate,
1
The exact solution can be found since the equations is separable (and linear). With some work we find that it is y(t) = 5(1 
e2t]/2.
1
+ y I  In 15  y I 6 6 In 11 + YI  In 15  YI In 11
+ c, = 6t + 6C, =t
1 In I Y + 1 = 6t y5
Clearly the solution
has the indicated limiting behavior. y
+1
y5
=
+ 6C
Ae6r,
where A = ±e6c. Using the initial condition y(O) 2 we see that A= 1, so y
+1
=e
=
6t
y5
25. The equation has the form y' = f (y), where f (y) = (1 + y)(5  y). The graph off is in the next figure. We have also indicated the direction of the solutions on the yaxis. This shows that y = 1 is an unsta• ble equilibrium point, and y = 5 is an asymptotically stable equilibrium point. Therefore, a solution start• ing with y(O) = 2 will increase and approach y = 5
Solving for y, we find that
5e6t
y (t)

1
5  e61
= 1 + e6t = 1 + e6t .
From this we see that y(t) � 5 as t � oo, agreeing with what we discovered earlier.
2.9 26.
Autonomous Equations and Stability
77
versus y.
Separating variables, dy  = (3
+ y)(l
dt
 y)
f (y) =
(3
+ y)(l
 y)
dy
=dt. (3
+ y)(l  y)
A partial fraction decomposition allows us to con• tinue.
1 [ 1 +1  ] 4 3+y In 13
1y
dy
se
+ y I  In 11  y I
dt
= 4t
+C
+ YI =4t+C In 31y
1
3 1
+y
I
1y
= ec e4t
3 + y = Ae4t
1y
Note the equilibrium points at y = 3 and y = 1. Moreover, note that between 3 and 2, solutions in• crease to the stable point at y = 1. Thus,
with y(O) = 2,
3
+ 2 = Ae4(0)
12
lim y(t)
=}A= 5
t+OO
27.
+y
= 5e4t
3
+y
= 5e4 t
1y
3
+ 5e4t
+ 5ye4t
= y(5e4t  1)
28.
3 + 5e4t y = 5e4t  1 · Multiply top and bottom by e4t.
3e4t y
=
+5
5  e4t
29.
Thus,
. o+ s 1 1my==l. 50
tH)O
Using qualitative analysis, plot the graph of the right• hand side of dy  = (3
dt
+ y)(l
 y)
We have the equation x' = f (x) = 4  x2. The equilibrium points are at x = ±2, where f (x) = 0. We have f'(x) = 2x. Since f'(2) = 4 > 0, x = 2 is unstable. Since !'(2) = 4 < 0, x = 2 is asymptotically stable.
and 3
= 1.
30.
Wehavetheequationx' = f(x) =x(xl)(x+2). The equilibrium points are at x = 0, 1, and 2, where f(x) = 0. We have f'(x) = 3x2 + 2x  2. Since f'(O) = 2 < 0, x = 0 is asymptotically stable. Because f' ( 1) = 3 > 0, x = 1 is unstable. Finally, because f'(2) = 2 > 0, x = 2 is also unstable. (a) f(x) = x2, f(x) = x3, or f(x) = x4. (b) f(x) = x3, f(x) = x5, or f(x) = x1. Notice that we are measuring the displacement as positive below the plane. First divide through by m to get dv k  =gv. dt m Note that this equation is autonomous, having form v' = f ( v). The graph of f is a line, with slope
78 Chapter 2 FirstOrder Equations k/m and intercept g, as shown in the following figure.
Let c(t) represent the concentration of salt in the solution at time t. Thus, c(t) = x(t)/100 and lOOc' = x',
lOOc' = 6 
1 (lOOc) 50
I 6 1 c =c 100 50
Let f (c) = 6/100  (1/50)c. Setting f (c) = 0 produces the equilibrium point c = 3, as shown in the following figure.
f (c)= 180!oc The phase line on the vaxis in this figure shows that v = (mg)/ k is a stable equilibrium point. Our skyâ€˘ diver starts from rest, so the solution trajectory with v(O) = 0 should approach the stable equilibrium soâ€˘ lution, v(t) = (mg)/ k. Consequently, the terminal velocity is (mg)/ k. 31.
Let x(t) represent the amount of salt in the tank at time t. The rate at which solution enters the tank is given by Rate In= 2 gal/minx 3 lb/gal= 6 lb/min. The rate at which solution leaves the tank is Rate Out= 2 gal/minx
:: lb/gal= ]_x lb/min. 100 50
Consequently, dx 1 =6x. dt 50
The phase line on the caxis in this figure shows that c = 3 is a stable equilibrium point so a trajectory with initial condition c(O) = 0 (the initial concenâ€˘ tration of salt is zero) should approach the stable equilibrium solution c(t) = 3.
Solutions Manual for Differential Equations 2nd Edition by Polking IBSN 9780134689586 Full clear download( no error formatting) at: http://downloadlink.org/p/solutionsmanualfordifferentialequations2ndeditionbypolkingibsn9780134689586/ differential equations 2nd edition pdf polking, differential equations, 2nd ed., prentice hall. pdf differential equations polking pdf download polking differential equations pdf differential equations polking boggess arnold pdf differential equations with boundary value problems (2nd edition) pdf differential equations with boundary value problems polking pdf polking differential equations solutions
Solutions Manual for Differential Equations 2nd Edition by Polking IBSN 9780134689586 Full clear download( no error formatting) at: http://d...
Published on Dec 17, 2017
Solutions Manual for Differential Equations 2nd Edition by Polking IBSN 9780134689586 Full clear download( no error formatting) at: http://d...