Differential equations 2nd edition polking solutions manual

) )
)
2
Differential Equations 2nd Edition Polking SOLUTIONS MANUAL
Full download:
https://testbanklive.com/download/differential-equations-2nd-edition-polking-
solutions-manual/
Chapter 2. First-Order Equations
Section 2.1. Differential Equations and Solutions
1. </J(t, y, y') = t2
y' + (1 + t)y = 0 must be solved for 4. y'(t) + y(t) = (2 - ce-1
+ (2t - 2 + ce-1
= 2t
y', We get
y' =
(1 + t)y
t2
2. </J(t, y, y') = ty' - 2y - t2
must be solved for y'.
We get
f 2y + t2
y=--.
t 5. If y(t) = (4/5) cost+ (8/5) sin t + ce-<1
12
t, then
3. y'(t) = --Cte-<1!2
)t
soy'= --ty.
12
and -ty(t) = -tce-<1
12)12,

)
1
y
(
t
)
'
+
(
l
/
2
)
y
(
t
)
= [-(4/5) sin t + (8/5)
cost - (C /2)e-<1
12
t]
+ (l/2)[(4/5) cost+
(8/5) sin t + ce-(1/2
)
]
= 2cost.

© 2006 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
+
+
.
,
6. If y(t) = 4/(1 + ce-4
t), then
, 16C eC-4t)
y = (1 + C eHt))2
y(4- y) = 1 + �e-4t x [4- 1 + �e-•• J
16(1 + ce-4t) - 16
=------
(1 C eHt))2
16C eC-4t)
(1 + C eHt))2
7. For y(t) = 0, y'(t) = 0 and y(t)(4 - y(t)) =
0(4- 0) = 0.
2.1 Differential Equations and Solutions 13
9. (a) If t2
- 4y2
= c2
, then
d d
-(t2
- 4y2) = -t2
dt dt
2t - 8yy' = 0
t -4yy' = 0.
(b) If y(t) = ±Jt2 - C2 /2, then y' =
±t /(2Jt2 - C2), and
t - 4yy'
8. (a) If t2
+ y2
= C2
then = t - 4[±Jt2 2 2 2
,
d d
-(t2 y2) = -c2
dt dt
2t + 2yy' = 0
t + yy' = 0
=t-t
=0.
- C /2][±t/(2Jt - C )]
(b) If y(t) ±,Jez - t2, then y'
-::r-t/,Jc2 - t2, and
(c) The interval of existence is either -oo < t <
-C or C < t < oo.
t + yy' = t + [±Jc2 - t2][-::r-t;Jc2 - t2J (d)
=t-t
=0.
(c) For JC2 - t2 to be defined we must have
t2
::::; C2 In order that y' (t) = -::r-t/,JC2 - t2,
(d)
we must restrict the domain further. Hence the
interval of existence is -C < t < C.
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

.
t
14 Chapter 2 First-Order Equations
interval (11/6, oo). the interval (-co, oo).
2 3 4 5
11. See Exercise 6.
(-oo, ln(5)/4).
The interval of existence is
13. y(t) = ! t2
+2-. Theintervalofexistenceis (0, oo).
3 3t
-2 -1
1 2 3
14. We need e-1
= y(l) = e-1
(1 + C/(1)) = (1 +
C)e-1
Hence C = 0, and our solution is y(t) =
12. y(t) = (4/17) cost+ (1/17) sin t - (21/17)e-4
on
ie:', This function is defined and differentiable on
the whole real line. Hence the interval of existence

)
,
,
1 4 2
is the whole real line. Butthis equals (1/2)(t+2) only if t � -2, as shown
in the following figure.
0 y
>,. ·-1
·-2
·-3 .._ ._ ......._ .....__
0 2 4
(-2, 0)
15. y(t) = 2/(-l+e-21
13
. Theintervalofexistenceis
(- ln(3)/2, oo).
-1
(0,-3) The second solution proposed by Maple, y(t) =
(1/4) (t-2)2
satisfies theinitialcondition, as y(O) =
(1/4)(0 - 2)2
= 1. But
1
y'(t) = 2(t - 2),
16. The initial value problem is
y' = Jy, y(O) = 1. and
The first solution proposed by Maple, y(t) =
(1/4)(t + 2)2
(1/4)(0 + 2)2
satisfies the initial condition, y(O) =
= 1. Next,
y'(t) = 2(t + 2),
/y0 = J.1.<t -2)2
= .1.11 -21.
and
But this agrees with (1/2)(t - 2) only if t � 2, as

I I /
-2 -1 0 2
y
shown in the following figure. 18.
y'=l-t
I I /
>, 0
I /
(0, 1)
-1
(2, 0)
19.
y ' = t tan(y/2)
<, - -: /
Note that this graph does not pass through (0, 1).
Hence, y(t) = (1/4)(t - 2)2
is not a solution of the
>, 0 - - - - -
initial value problem. -1
/ -: <,
-2 -1 0 2
17. 20.
y'=y+t y ' = f y/(1 + y2)
1.5
/ I I I -: - -: I0.5
>, 0
/ I >, 0 - - - -
-0.5
-1
/ -1
<, <, -1.5
-2 -1 0 2 -2 -1 0 2

21. the right-hand side of
y'=-ty
is undefined.
I y+t
y =-•
y-t
5
-·5 >, 0
-2 0 2
22. -5
4
2 25.
>, 0
--2
-5 0 5
23.
-4
0 5 10
y'=t-y+1
6
4
26.
2
>, 0
-2
-4
-·6
-5 0 5
24. Note the difficulty experienced by the solver as it ap-
proaches the line y = t, where the denominator of
-4 2 4

i ·: I - l
x. ,·_;..;
·1··r--,·,
27. 30.
-15
y ' = 3 y + sin(t) y
0
-10 -5
5 I · ·· ·-1- ·I -· ·· ·· v- ,--,-�-
4
l 1: _1_:_1_) ,_:_,_
I � . '' - . .% /
3 t ,··y,·....:.·.;.-:·;,·1·
2 i 1: . : . .'. ,.: <, .;__ .-:. / .: I i -
1 �
I
v

·
>
,
- -,
,
_:
.
:
_
.,,,;·
�
/
/
/
I
-
·/ I
.
-10
-20
I- 0
-1
-2
-3
-4
-5
. , , .., � 0. -/ ,/ - -I·· f ·· f I
� <: <, ·- /. I ·1 r I . I /
, - :;,,; / /, I '., /- I : i I
··, <, ·.......,,,, ··/ ·1 r> I '/ I I -. 1 - 1 - .
'. -, : -..... _.:..,. .--; .r. :.1 ( 1.: ./ ;t .( J .' I - I - .
, ,. - ,,,, /. I · I I I · I I I- I · I I
<; _;_.-_;, -"/ i -,- .-( -, ·,·.-,-'I ·1 1 -. I - i
_;... .,..:.,,, .j. .,; .,.:.,- -i t, l : ;, t. .,.:,- i - .
-30
-5 -4 -3 -2 -1 0
s
2 3 4 5
28. 31.
y
-2 2
29. 32.
-4

0
I:,
/
50 f � �- � -�- � ·�: � -�- �
.
' '
,
33. tion curve.
200
150
·I·# ·I I ·I·# ·I' I ·II·/· I·/·"/··/·/· l·I· I·
I I I I I I 1· I I I I I I 1 I I I I I0
i I I I I I (II I I I I 1 I I I I Ii
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 I I I I I I I
Tl'/ I Tl 1:1 ·1 I'/" I ·1·.1,·,- I'/' t :t: 1·
I I I I I I t, I I I I I I . I I I I I I
I I I I I I f· I I I I I I 1 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 t. I I I I I I I
a. 100 ·,· i
0
i ·,· i ·,: i ·,· I I i ·,; ·,· i 'ii·, i
I I I I I I r I I I I I I '/ I I I I I I
I I I I I I r I I I I A 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· � f� -�- � -�- � -�- �1,,,,,�,, ,,,,,,,,,1
0 5 10 15
34.
Using the solution curve, we estimate that P(IO) �
124. Thus, there are approximately 124 mg of bac•
teria present after 10 days.
36. We must solve the initial value problem
dA
-
dt = -025A
'
A(O) = 400.
·-6 -4 4 6
Using our numerical solver, we input the equation
and initial condition, arriving at the following solu•
tion curve.
500 �,�,�'�'�'�'�'�'�'�'·
� ' ' ' ' ' ,··· ' ' ' ' ' ' ··, ' ' ' ' '�'
400 '
'.,', '�',�,' ',�'', ��'','�','��-
300
.... '
,,,,,,,,,,,�,,,,,,
200 ..;.,, .....
:...·..:..:......:..· · = : · .
:
, .- ...,_;
............... . .
·..... ..........
100 ��-�-��-�-�-�-�- �?-
35. We must solve the initial value problem 0 ---------------------
0 2 4 6
dP
- = 0.44P,
dt
P(O) = 1.5.
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
de
and initial condition, arriving at the following solu- dt = -0.055c, c(O) = 0.10.

where T is the temperature of the rod at t minutes. 0 --------
Note that since the initial temperature is larger than
the surrounding air (20° Celsius), the minus sign in•
0 10
sures that the model implies that the rod is cooling.
' ' ·, '
' ' '
,,,,,�,,�,,,,,,,�,,,,,,,,,,
� �
� �
a.
tion curve.
39.
Use the solution curve to estimate that it takes a little
less than 15 minutes to cool to 100° Celsius.
The rate at which the population is changing with
respect to time is proportional to the product of the
0.1 . ·· v- .,: · •· .,.·, ·· ····'I.. population and the number of critters less than the
'
,.
11.
' '

' ,
·
·
· ' ,;.
"carrying capacity" (100). Thus,
,;:
0.08 v�. 'v�'v' 'v,: 'v' 'v' 'v :,v' �' '� ''i.,·
,. , dP
��'
��,,,,,,�,,,,,,
•
0.06
o
,,,,,
,,,,�,,'��,� �,�,��,,��,,��,,� dt = kP(lOO- P).
0.04
�'
�
''
-�
'
-�
'
-�'
--�
''
,��
,,�
,,
�
,,
��
,
�,�
,
-�
,
-�
, ,
��-
With k = 0.00125 and an initial population of 20
0.02
��-�--�--��.-�--�--�-�.�---�-·-�-�.-. ritters, we must solve the initial value problem
' ��
c
0 --------------------·
() 10 20 30
dP
- = 0.00125P(100- P),
dt
P(O) = 20.
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,
Note that the right-hand 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 -.
·------.
·-------·
.
------•-
.
dt = -k(T - 20). 00 �-,--,_--,� --,,--,-�/·,-,,-,,-,-,�·
,,,,, ,,,,,
With k == 0.085 and an initial temperature of 300°
Celsius, we must solve the initial value problem
,.,. .; .; .,, .,, ,,,. .,;. .; ., ; .,,
00 ��;��;�;:�;:�;:;;;:.
,,,,,,, .,,,, ,,,,,,,,,,,, .,,,,, ,,,,,,
� �
�;;�:�
�;� �;�
;,
;;,,
;;,�
i-
d
T
t -0.085(T - 20), T(O) = 300, � �,
w--,,
-,
��,
�,
-,
� ,-,�
,,
_,,
_, �·
,, ,,
=
,,, ,, ,
�-----�------ -�-- -��
------------·
20 30
tion curve.
Use the solution curve to estimate that there are about
91 critters in the environment at the end of 30 days.
300
'. ·· ·· .,: ·· ·· ··
··, •· ·· ··
' .

· · ' ·,:' ·,· ' :
,,,,,�,,,,,,
.
-�- . �
' ' ' ,. ' ' ' '
50
250
200
t- 150
' v' ·, '
. .. ,; . . .. . .. .
.. v, .v ..
·'

· .v,

.v, ', .v, ', .v

, ', .v, ', .v

..
., .v

, .v

, .,

. v

.
',,,',','�,�,'��,.,,',',',','�.,,',',',',','
,,,,,,� ' ·,·,',·,·,',·.....,",·...:,',"'°,',"<,'
100 ·
,..;,,;
,...;.
,·,•.•.,. ,,· ,�,,��'·
','· .-'·,'·,', ', �
�-----�--;
-,
,� �------
�...::..
0 --------------------
0 10 20 30

1
+
x
Section 2.2. Solutions to Separable Equations .
2.2 Solutions to Separable Equations 21
1. Separate the variables and integrate.
dy
-=xy
dx
dy
-=xdx
y
dy
dx = y(x + 1)
-dy = (x + l)dx
y
1 2 In IYI =
1
x
2
+ x + C
In IYI =
2x + C 2
ly(x)I = ex2;2+c
y(x) = ±ec . ex2;2
= Aex2;2'
Where the constant A = ±ec is arbitrary.
dy
x-=2y
dx
1 2
-dy = -dx
y x
In IyI = 2 ln Ix I + c
IYI = elnx2+c
y(x) = ±ecx2
Letting D = ±ec, y(x) = Dx2.
dy = ex-y
dx
eYdy = eidx
eY =ex+ C
y(x) = In(ex + C)
dy
- = (1 y2)ex
dx
IYI = e(lf2)x2+x+C
y(x) = ±ece(lf2)x
2
+x
Letting D = ±ec, y = De(l/2)
2
+x.
6. Separate the variables and integrate. Note: Factor
the right-hand side.
dy
dx = (ex+ l)(y - 2)
1
--dy =(ex+ l)dx
y-2
In Iy - 21 = ex + x + C
IY - 21 = eex+x+C
y - 2 = ±eceex+x
Letting D = ±ec, y(x) = Deex+x + 2.
7. Separate the variables and integrate,
dy x
-=--,
dx y+2
(y + 2) dy = x dx,
1 1
2y2+2y=2x2+c,
y2 + 4y - (x2 + 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(x
2
+ D)
2
-
1
- dy = e
x
dx
y(x) = -2 ± Jx2 + (D + 4)
-
+y
tan-1
y =ex+ C
y(x) = tan(ex + C)
If we replace D + 4 with another arbitrary constant
E, then y(x) = -2 ± ,Jx2 + E.

(1 -)
+
-1
)
2
1
:: =yV1)
.!_dy = + - dx
y x -1
In IyI = x + In Ix - 11 + c
IYI = ex+ln lx-lJ+C
y(x) = ±ecexelnlx-11
Letting D = ±ec, y(x) = Dex Ix - 11. It is impor•
tant to note that this solution is not differentiable at
11.
12.
(y3 - 2) dy = x
dx
(y3 - 2)dy = xdx
y4 x2
4-2y = T +c.
The solution is given implicitly by the equation
y4
- 8y - 2x2 = A, where we have set A= 4C.
dy 2x(y + 1)
x = 1 and further information (perhaps in the form dx
=
x2 - 1
of an initial condition) is needed to remove the ab•
solute value and determine the interval of existence.
9. First a little algebra.
x2y' = ylny - y'
(x2 + l)y' = ylny
Separate the variables and integrate.
1 1
--dy=--dx
yln y x2 1
13.
dy 2xdx
y + 1 x2 -1
In IY + 11 = In lx
2
- 11 + C
IY + 11 = elnJx2-11+c = ec lx2 -11
y(x) = A(x2
- 1) - 1.
dy y
1 1
-du= --dx,
u x2 + 1
where u = In y and du = -} dy. Hence, ln lul
tan"! x + C. Solve for u:
lul = etan-1 x+c
U = ±ec etan
1
x-
=
dx x
dy dx
=
y x
}.,,IyI = In Ix I + c
ly(x)I = elnlxl+C = ec lxl
y(x) = Ax.
The initial condition y(l) = -2 gives A = -2. The
10.
Let D = ±ec, replace u with In y, and solve for y.
lny = D/an x
tan-!
y(x) = eDe x
14.
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).
dy 2t(I + y2
x dy = y(l + 2x2
)
dx
dt y
-dy= 1 + 2x dx= [
-1 +2xJ dx
ydy = -2t dt
1 + y2
y x x
In IyI = ln IxI + x2
+ c
ly(x)I = elnlxl+x2+c = eclxlex2
2
y(x) = Axex
- ln(l + y2) = -t2
+ C
2
1 + y2 = e-2,2+c = ece-2t2
1 + y2 = Ae-2t2

( )
.
+
+
With y(O) = 1, 1 + 12
= Ae-2 0 2
and A= 2. Thus,
1 + y2 = 2e-2t2
y = ±J2e-2t2 - 1.
We must choose the branch that contains the initial
2
16.
dy = ex+y
dx
e-y dy = ex dx
-e-Y =ex+ C
e-y = -ex - c
condition y(O) = 1. Thus, y = J2e-2t
solution is defined, provided that
2
- 1. This
With y(O) = 1,
-y = ln(-ex - C)
y = -In(-ex - C)
2e-2r - 1 > 0
e-2t2 > !
2
-2t2
>In!
2
t2
< -21n !2
Thus,
1 = -ln(-e0
- C)
-1- C = e-1
C=-l-e-1
y = - In(-ex + e-1
+ 1).
15.
t2
< ln4
ltl < Jin4.
Thus, the interval of existence is (-v1n4, v1n4).
dy sinx
17.
This solution is defined provided that
-ex+ e-1
+ 1 > 0
ex< e-1
+ 1
x < ln(e-1
+ 1).
Thus, the interval of existence is (-oo, ln(e-1
+ 1)).
dx
=
y
dy = 1 + 2
dt y
ydy = sinxdx
1 2
2y = -cosx C1
y2 = -2cosx + C (C = 2C1)
y(x) = ±,Jc - 2cosx
Using the initial condition we notice that we need
the plus sign, and 1 = y(n/2) = ,Jc. Thus C = 1
and the solution is
y(x) = ,vl - 2cosx.
The interval of existence will be the interval con•
taining Jr /2 where 2 cos x < 1. This is n/3 < x <
18.
_!}z_ = dt
1 + y2
tan-1
(y) = t + C
y(t) = tan(t + 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).
dy x
----
dx 1 2y
(1+2y)dy=xdx
2
Sn/3. y + y2 = x /2 + C

+C
-
This last equation can be written as y2
+y - (x2
/2 +
C) = 0. We solve for y using the quadratic formula
20.
dy x
y(x) = [-1 ± J1 + 4(x2
/2 + C1) J /2
= [-1 ± J2x2
+CJ /2 (C = 1 + 4C1)
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 + J2x2 -1
y(x) = 2 .
For the interval ofexistence we need the interval con•
taining -1 where 2x2
- 1 > 0. This is -oo < x <
-1/v'2,.
19. 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.
i
2
0
x
dx
=
y
ydy=-xdx
1 2 1 2
2y = -
2x
y2 = 2C -x2
y = ±J2C-x2
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.
y
/
/
<,
-2 0 2
/
/ -2 <,
/ <,
/ <,
-4
21. 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-

,
ted is the dashed curve in the next figure.
...e -
23. We have N(t) = N0e-M, and
N (t + T1;2) = Noe-A(t+Ti12)
= Noe-At . e-AT1;2
= N(t). e-AT1;2
1
= N(t) ·
2
if e-H1;2 = 1/2, or T1;2 = In 2/A .
8
22.
-2 / 0 2 4
I
I
I -5
24. (a) A = In 2/ T1;2 = 1.5507 x 10- .
(b) We have No = 1000 and N(t) = 100. Hence
100 = 1000 · «". or t = ln 10/A = 1.4849 x
108
years.
25. We have 80 = N(4) = lOoe-4
).,. Hence J... =
dy y2
+ 1
dx y
ydy = dx
y2 + 1
1
2
ln(y2 + 1) = x + C
y2 + l = e2x+2C = Ae2x (A = e2C)
y2 = Ae2x - 1
y(x) = ±JAe2x - 1
This is the general solution. The initial condition
y(l) = 2 gives
2 = +JAe2 -1
4 = Ae2
- 1
A= 5e-2
ln(l00/80)/4 = 0.0558. Then T1;2 = In 2/J... =
12.4251 hours.
26. Using T112 = 6 hours, we have J... = In 2/T1;2 =
0.1155. Then N(9) = lOe-9). = 3.5355kg.
27. Using T1;2 = 8.04 days, we have J... = ln 2/ T112 =
0.0862. Then N(20) = 50oe-2
m. = 89.1537mg.
28. The decay constants are related to the half-lives
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) = x0e-A210t and of 211Rn by y(t) = y0e-A211t.
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(A210-J..211) /4. Thus we need et(A2io-J..2ll) = 16.
From this we find that t = 11.5 hours.
The particular solution is 29. (a) If N = N0e-J..1
then substituting T)., = 1/J...,
y(x) = J5e2x-2 - 1.
The interval of existence requires that
5e2x-Z - 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.
N = Noe-AT,_
= Noe-A(l/A)
= Noe-1
•
Therefore, after a period of one time constant
T)., = 1/A, the material remaining is Noe-1.
Thus, the amount of radioactive substance has
decreased to e-1
of its original value N0•

-
(b) If the half-life is 12 hours, then
!No= Noe-}..(12)
2
e-m
=
1
2
1
T1;2 = In 2/J.. = 7.927days.
7
6.8
a:
6.6
-12>.. = ln-
2
In l
A=--
-12
Hence, the time constant is
.f:
6.4
6.2
6
2 4 6 8 10
Day
1 -12
T).. = - - - � 17.3hr.
J.. In l2
(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-
31. The half-life is related to the decay constant by
ln2
T1;2= -.
J..226
The decay rate is related to the number of atoms
present by
Substituting,
riods ([O, 4T)..]) follows.
1000
Nln2
T1;2 = -R-.
Calculate the number of atoms present in the lg sam•
ple.
800
600
z
400
200
Now,
1 mol 6.02 x 1023
atoms
N = lg x -- x ------
226 g mole
= 266 x 1021 atoms.
(2.66 x 1021
atoms)(ln 2) 10
0
0 20 40 60 80
T1;2 = 3.7 x 1010 atom/s
= 4.99 x 10 s.
In years, T112 � 1582 yr. The dedicated reader might
check this result in the CRC Table.
30. 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 half-life is
32. (a) Because half of the existing 14
C decays every
5730 years, there will come a time when phys•
ical instruments can no longer measure the re•
maining 14
C. After about 10 half-lives (57300

( 2
years), the amount of original material remain•
ing is
No
1)10
� 0.00097N0,
a very small amount.
(b) The decay constant is calculated with
ln2 ln2
).., = - = - � 0.0001245.
T1;2 5568
We can now write
N = Noe-o.0001245t.
The ratio remaining is 0.617 of the current ratio,
so
0.617No = Noe-o.0001245t
e-o.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)e-kt and write
T = 21 + (31 - 21)e-kt = 21 + lOe-kt.
At t = 1, T = 29°C, which can be used to calculate
k.
29 = 21 + lOe-k(l)
k = - ln(0.8)
k � 0.2231
Thus, T = 21 + 10e-0·2231t. To find the time of
dea1h, enter "normal" body temperature, T = 37°C
and solve fort.
37 = 21 + 10e-0.223lt
In 1.6
t=---
-0.2231
t � -2.l hrs
Thus, the murder occurred at approximately 9:54
PM.
34. 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)) y(O) = 40
Notekispositivesince70 > y(t) andy'(t) > O(the
beer is warming up). This equation separates as
dy
--=kdt
70-y
which has solution y = 70 - Ce-kt. From the
initial condition, y(O) = 40, C = 30. Using
y(lO) = 48, we obtain 48 = 70 - 30e-10k 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 - 2oe-l0k
Therefore,k = (-1/JO)ln(l/2) � .0693. We want
to find the time T when y(T) = 40, which gives the
equation
70 - 20e-kT = 40
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.
36. x' = [at+by+c]' = a+by' = a+bf(at+by+c) =
a+ /(x).Fortheequationy' = (y+t)2
weusex =
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.

=
y2
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:
-y
2x -+xy'
or
y' -1
-=-
y x
This separable differential equation is easily solved
to obtain y(x) = C/x, where C is an arbitrary con•
stant.
38. With the notation as in the previous problem, the
equation of the normal line is
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. The area under the curve y = y(t) from Oto xis
fox y(t) dt
which by assumption, equals (1/4)xy(x) (one•
fourth the area of the rectangle). Therefore
fox y(t) dt = (l/4)xy(x).
Differentiating this equation with respect to x and
using the Fundamental Theorem of Calculus for the
y
"
- y = -1
(
"
-x)
left side gives
-- x
y'(x)
The .x-intercept is found to be Xint = yy' + x. Since
Xint - x is given to be ± 1, we obtain
yy' = ±2
y(x) = (1/4) (y(x) + xy'(x)).
This equation separates as
y' 3
with solution y2
= ±4x + C, where C is an arbitrary y
=
x
constant.
39. Let </> be the angle from the radius to the tangent.
which has the solution y(x) = Cx3.
41. Center the football at the origin with equation
z2+x2 +- =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
vz =
+xi - (y/4)j
--;::=====
J4- x2 - y2/4
From geometry, tan e = rdti]dr. Since e = 2</>, we
obtain
where i and j are the unit vectors parallel to the x and
y-axis. Since the path traced by the drop, y = y(x),
must point in the direction of Vz, we must have
dr
d() = r cot</> = r cot(() /2)
dy . z y
- = slope of the gradient = .2 = - .
x � �

)
2.3 Models of Motion 29
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),
(here, y is the independent variable).
42. Let y (t) be the level of the water and let V (t) be the
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 ,rx2
, where x is the radius. Using the
equation of the side of the bowl (y = x2
, we obtain
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
dV 2dy
- =JTX -
dt dt
From Torricelli's law:
dV 2 2 �
- = -na v = -na y2gy
dt
Since dy /dt = C (a negative constant), we obtain
Cnx" = -na2
/2gy
Solving for y, we obtain y = K x4
where K is a
constant.
dV dy
dt = ny(t) dt ·
On the other hand, dV/dt is equal to na2v where v
is the speed of the water exiting the bowl. From the
hint, v = ,J2gy(t). Thus we obtain the following
differential equation:
44. Following the hint, let () be the polar angle and lo•
cate the destroyer at 4 miles along the positive x-axis.
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 x-axis, the destroyer should first
head from x = 4 to x = 1 (the sub would move
ny(t)
dy
=
dV
= -na2 �
from x = 0 to x = 1 in this same time frame under
- - y2gy(t).dt dt
This equation can be separated and solved for y as
3 )2/3y(t) = (C -
2
affgt
Since y(O) = 1, we obtain C = 1. Setting y(t) = 0,
we obtain
2
t=---
3a2vlg
in units of seconds (here g = 32).
43. Let the unknown curve forming the outside of the
bowl be given by y = y(x) (the bowl is then formed
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 traveled r (()) and the
destroyer would have traveled Ji Jr'(t)2 + 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
3(r(8) - 1) = 1·Jr'(t)Z + r(t)2 d&.
Differentiating this equation gives
by revolving this curve around the y-axis). We can 3r' = ...;'(r')2 + r2 or
dr r(())
- = -- r(O) = 1
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
d() ,Jg
with solution r(()) = e8f-v'3.

,
,
•
,
15
Section 2.3. Models of Motion
1. We need gt = c/5. or t = c/5g = 612,240 sec-
onds. The distance traveled will be s = gt2
/2 =
1.84 x 1014
meters.
2. We need O = -9.8t2
/2 + l5t + 100. The answer is
6.3 seconds.
required is t = m ln(2) /r. The distance traveled is
x = 1'v(s)ds
= mg r(e-rs/m - 1) ds
r lo
3. The depth of the well satisfies d = 4.9t2
where tis
= �g [; (1- e-rt/m) _ t]
m�n2]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.
4. In the first 60s the rocket rises to an elevation of
(100- 9.8)t2
/2 = 162, 360m and achieves a veloc-
= �g [;, _
= �:grn-1n2J
9. 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.
ity of v(60) = (100 - 9.8) * 60 = 5412m/s. After 10.
that the velocity is 5412 - 9. 8t. This is zero at the
highest point, reached when t1 = 552.2s. The alti-
tude at that point is 162, 360 + 5412 t1 - 9.8 t[/2 =
1.657 x 106
m. From there to the ground it takes t2s,
total trip takes 60 + 552.2 + 581.5 = 1193.7s.
(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(e-rt/m_
1)/r. Integrating and setting x (0) = 0, we get
x = 1'
where 4.9tf = 1.657 x 106
or ti = 581.5s. The
v(s)ds
5. The distance dropped in time t is 4.9t2 If T is = mg r(e-rs/m - 1) ds
the time taken for the first half of the trip, then r lo
4.9(T + 1)2
= 2 x 4.9T2
or 4.9(T2
- 2T -1) = 0. mg [m -rt/m J
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.
6. (a) v5/2g
= 7 70 -e )-t
Hence v(2) = -12.4938 and x(2)
-14.5025.
(b) The velocity is 80% of its terminal velocity
when 1 - e-rt/m = 0.8. For the values of m =
70 and r = 34.3 this becomes t = 3.2846s.
(b) Both times are equal to vo/g.
11. Without air resistance, v0 = J2 x 13.5g =
(c) VQ.
7. The velocities must be changed to ft/s, so v0 =
16.2665m/s. With air resistance, v0 is defined by
° v-dv----r J. dy
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
.
Hence,
1vo -
v + mg/r - m 1.s ·
- Vo+ (mg/r) ln(vo + mg/r) - (mg/r) ln(mg/r)
8. We have v(t) = ce-rt/m - mg/r. If v(O) = 0 then C = mg/r, and v(t) = mg(e-rt/m - 1)/r. This is

equalto --mg/2r when e-rt/m = 1/2. Thus the time r
= -13.5- or
m
- vo + 491n(vo + 49) - 49ln(49) = -2.7

1vi r
)
= -- =
2
f. vdv = -
R s v «
2 2 (1 1 )
2.3 Models of Motion 31
This is an implicit equation for v0. Solving on a
calculator or a computer yields v0 = 18.l 142m/s.
12. The: impact velocity Vi is defined by
vdv 1°
------ dy
O v+mg/r m so
(c) If y is the maximum height, the corresponding
velocity is v = 0, so from (3.16)
0=vJ-2GMG- R�J
Solving for y we get the result.
vo < J2GM/R, then v5 < 2GM/R,
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
(d) If
and 2GM/R - v5 > 0. Hence by (c) the
object has a finite maximum height and does
not escape. However, when v0 = J2GM/R,
2G M/R - v5 = 0, and there is no maximum
height.
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
v dv = (-g + R(v)/m) dy = (-9.8 - 0.5v2
dy
Hence if y1 is the maximum height we have
d2
x -mgx
m dt2 = -R-
Using the reduction of order technique as given in
the hint, we obtain
1°
-vd
-v
- = -0.5
1Y1 dy. dv -gx
v =
230 v2 + 19.6 0
- --
Hence
Yi = log(v2
+ 19.6)
= 7.9010.
230
1 0
dx R
which can be separated with a solution given by
v = JC - gx2 / R. The constant C can be evalu•
ated from the initial condition, v(x = R) = 0, to
be C = gR. When x = 0 (the center of the Earth),
we obtain v = Jc= .JgR or approximately 4.93
miles per second.
14.
dv dv dy dv
(a) Follows from a= - v-.
dt dy dt dy
16. We will use GM = gR • Once more we use
(b) v d v = (
GM d
)2 y
GM
v dv = - (R + y)2 dy
R + y
J
r
o
sds=-GM1°
(R
dy
a + y)2
v 1y GM
vo o (R + s)2
ds
1
!(v2
- v5) =-GM(_!_ - - -)
2 R R+y
v = v0 - 2G M R - R + Y
v2
= 2GM [_!_- --]
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

-
5
the table. This force is mgx(t) where mis the mass
density of the chain. Newton's law gives
mx"(t) = mgx(t)
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
Using the hint, x"(t) = dv/dt = v(dv/dx) and so
t = -1 ln (10 + ,J%) � .405 seconds
this equation becomes
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).
Since dx/dt = v = Jg(x2 - 4), we can separate
this equation and integrate to obtain
ln(x + Jx2 - 4) = ,Jgt + K
� 2
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 toe-ax. Hence it is given by R(x, v) =
-keaxv, where k is a positive constant. Newton's
second law gives us mx" = -mg - ke-axx', 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
u(t) = ef-a(t)dt = ef ldt = et.
Multiply both sides of our equation by this integrat•
ing factor and check that the left-hand side of the
resulting equation is the derivative of a product.
et (y' + y) = Ze'
We verify that the left-hand side is the derivative of
e-3ty, so when we integrate we get
e-3ty(t) = -
5
e-3t + C.
3
Solving for y, we get
y(t) = - + Ce31
- •
3
(ety)' = 2et
Integrate and solve for y.
ety = 2er + C
y(t) = 2 + ce:'
2. We have a(t) = 3, so u(t) = e:", Multiplying we
see that the equation becomes
e-3ty' - 3e-3t y = 5e-3t.
3. Compare y' + (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
u(x) = ef-a(x)dx = ef2/xdx = e2ln [x] = lxl2 = X2.
Multiply both sides of our equation by the integrating
factor and note that the left-hand side of the resulting

+
+
2
t+l
•
2.4 Linear Equations 33
equation is the derivative of a product.
x
2
(y'+ � y) = cos x
x2
y' + 2xy = cosx
(x2
y)' = cosx
Integrate and solve for x.
x
2
y = sinx + C
sinx + C
y(x) =
x
2
2
4. We have a(t) = -2t, so u(t) = et • Multiplying by
u, the equation becomes
2 2 2
et y' 2tet y = Stet .
We verify that the left-hand side is the derivative of
2
et y, so when we integrate we get
2 5 2
et y(t) =
2
et C.
Solving for y we get the general solution
y(t) =
5
+ Ce
-t2
.
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•
u(t) = ef -a(t)dt = ef -2/(t+I)dt
= e-2lnjt+ll =It+ 11-2 = (t + 1)-2.
(t + 1)-
2
(x' - --x) = 1
6. If we write the equations as x' = (4/t)x + t3
, we
see that a (t) = 4/t. Thus the integrating factor is
u(t) = e" f(4/t)dt = e-4Inr = t-4.
Multiplying by u, the equation becomes
After verifying that the left-hand side is the deriva•
tive of t-4
x, we can integrate and get
t-4x(t) =Int+ C.
Hence the general solution is
x(t) = t
4
ln t + Ct
4
7. Divide both sides by 1 + x and solve for y'.
I 1 COSX
y =-l+xy+ l+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
u(x) = ef-a(x)dx = efl/(l+x)dx = elnjl+xl = ll+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 left-hand side ofthis result is the deriv•
ative of a product, integrate, and solve for y.
((1 + x)y)' = cosx
(1 + x) y = sin x + C
sinx + C
(Ct+ l)-2
x)' = 1
Integrate and solve for x.
yx( =) --•
l+x
(t + l)-2
x = t + C
x(t) = t(t + 1)2
+ C(t + 1)
2
8. Divide by 1 +x3 to put the equation into normal form
I 3X2 2
y = 1 +x3Y +x

)
-
-
-
We see that a(x) = 3x2
/(1 + x3
. Hence the inte•
grating factor is
Multiplying by this we get
1 , 3x2
x2
1 + x3 y - (1 + x3)2 y = 1 + x3 •
We first verify that the left-hand side is the derivative
of (1 + x3
)
1
y. Then we integrate, getting
1 1
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) = ef-a(x)dx = ef-mdx = e-mx.
Multiply both sides of the differential equation by
the integrating factor and check that the resulting
left-hand side is the derivative of a product.
y' - my = c1emx
e-mx(y' - me-mx) = C1
(e-mx Y)' = C1
·-- y(x) = - ln(l + x3
) + C.
1 +x 3 e-mxy = CJX + C2
Solving for y, we get
1 3
y(x) = 3(1 + x3) ln(l + x3) + C(l + x ).
9. Divide both side of this equation by L and solve for
di/dt.
y = (c1x + c2)emx
11. 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) = ef-a(x)dx = ef secxdx
= eln [sec x+tan x] = I secx + tan xi.
-
di
= -
R
i+
E If secx + tanx > 0, then I secx + tan xi= secx +
- •
dt L L
Compare this with i' = a(t)i + f (t) and note that
a (t) = - R/L. Consequently, an integrating factor
is found with
u(t) = ef -a(t)dx = ef R/Ldt = eRt/L
Multiply both sides of our equation by this integrat•
ing factor and note that the resulting left-hand side
is the derivative of a product.
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
(sec x + tan x) (y' + y sec x) = cos x (sec x + tan x),
or
(secx+tanx)y' +(sec2
x+secx tanx)y = 1 +sin x
Again, check that the left-hand side of this equation
is the derivative of a product, then integrate and solve
eRt/L (d-i +-Rz·) = -Ee Rt/L for y.
dt L L
(eRt/Li)' = E eRt/L
L
Integrate and solve for i.
eRt/Li =
E
eRt/L + c
R
i(t) =
E
+ Ce-Rt/L
R
((secx + tanx)y)' = 1 + sinx
(secx + tanx)y = x - cosx + C
x - cosx + c
y=----•
secx + tanx
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) =ef-a(t)dt =ef-n/tdt =e-nlnltl = ltl-n.

13.
Depending on the sign oft and whether n is even or
odd, ltl-n either equals r» or -t-n. In either case,
when we multiply our equation by either of these
integrating factors, we arrive at
t-nx' - nt-n-lx = e',
Note that the left-hand side of this result is the deriv•
ative of a product, integrate, and solve for x.
(t-nx)' = et
t-nx =et+ C
X = t11
et + Ctn
(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) = e-fa(x)dx = efcosxdx = esinx.
Multiply both sides of the differential equation
by the integrating factor and check that the re•
sulting left-hand side is the derivative of a prod•
uct.
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,
y(x) = 1 - Ae-sinx,
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
(a).
14. Compare y' = y + 2xe2
x with y' = a(x)y + f(x)
and note that a (x) = 1. Consequently, an integrating
u(x) = ef -a(x)dx = ef -1 dx = e=.
esinx (y' + Y COS X) = esinx COS X
(esin X y) I =esin X COS X
esinx Y = esinx + C
y(x) = 1 + ce-sinx
(b) Separate the variables and integrate.
dy
- = cosx(l - y)
dx
dy
-- =cosxdx
I-y
- ln 11 - y I = sin x + C.
Take the exponential of each side.
11 - YI = e-sinx-C
l _ y = ±e-Ce-sinx
If we let A = ±e-c, then
e-xy' - e= y = 2xex
(e-xy)' = 2xex
Integration by parts yields
Consequently,
e-xy = 2xex - 2ex + C
y(x) = 2xe2
x - 2e2
x + ce"
The initial condition provides
3 = y(O) = 2(0)e2
CO) - 2e2
CO) + Ce0
= -2 + C.
Consequently, C = 5 andy(x) = 2xe2
x-2e2
x+sex.
15. Solve for y'.
y(x) = 1 - Ae-sinx,
, 3x 6x
y = - x2 + 1 y + x2 + 1

2
+
2
2
)
Compare this with y' = a (x)y + f(x) and note that
a(x) = -3x/(x2 + 1). Consequently, an integrating
u(x) = ef-a(x)dx = ef 3x/(x2
+1)dx
= e<3/2) ln(x +1) = (x2 + 1)3/2.
Consequently, C = -n/4 and
tan-1 t - zr..
y(t) = (1 + t2)24.
= =17. Compare x' + x cost (1/2) sin 2t with x'
(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
(x2 + 1)3/2 y = 2(x2 + 1)3/2 + C
y = 2 + C(x2 + 1)-3/2
The initial condition gives
-1 == y(O) = 2 + C(02
+ 1)-3
12
= 2 + C.
a(t)x + f(t) and note that a(t) = - cost. Con•
u(t) = ef-a(t)dt = efcostdt = esint.
. . 1 .
esmtx' esmt (cos t)x = -esmt sin 2t
2
(esm. tX )' = -1esm. t sin 2t
2
Therefore, C = -3 and y(x) = 2 - 3(x2
+ 1)-3
/ .
16. Solve for y',
I 4t 1
y = - 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
u(t) = ef -a(t)dx = ef 4t/(l+t2
)dt
= e2ln Jl+r J = (1 + t2)2.
(1 + 2 2y' + 4t(l + 2) =
1
-t ) t -
l+t
((1 + t2)2y)' =
l+t
(1 + t2 2
y = tan-1
t + C
The initial condition y(l) = 0 gives
(1 + 12
)2(0) = tan-1
1 + C.
Use sin 2t = 2 sin t cost.
(esin tX)I = esin t Sin t COS f
Let u = sin t and dv = esint cost dt. Then,
Jesin t cos t sin t d t = Ju d v
= uv - Jvdu
= (sin t)esint - Jesint cost dt
= (sin t)esint _ esint + C
Therefore,
esintx = esint sin t - esint + c
x(t) = sin t - 1 + ce-sint
The initial condition gives
1 = x(O) = sin(O) - 1 + Ce-sin(O) = -1 + C.
Consequently, C = 2 andx(t) = sin t-1 +2e-sint.

/
2+2
•
1 1 . C
1
18. Solve xy' + 2y = sinx for y'.
negative infinity as x approaches zero from the right.
, 2 sinx
y =--y+-
x x
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
5 (n/2,0)
u(x) = ef-a(x)dx = ef2/xdx = e2lnlxl = lxl2 = x2.
Multiply both sides ofour equation by the integrating
-251..L.-��--'
0
x
6
19. Solve for y'.
x2
y' + 2xy = x sinx
(x2
y)' = x sinx
y' =
1
--
2x +3
y + (2x + 3)-1 2
Integration by parts yields
fx sinx dx = -x cosx + fcosx dx
= -x cosx + sinx + C.
Consequently,
x2
y = -x cosx + sinx + C,
y = - - cos x + - sm x + - .
x x2 x2
4 4C
O=y(n/2)= .
T( T(
Consequently, C = -1 and y = -(1/x) cosx +
(l/x2
) sinx - I/x2
We cannot extend any interval to include x = 0, as
Compare this with y' = a (x)y + f(x) and note
that a(x) = l/(2x + 3) and f(x) = (2x + 3)-1
12
•
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.
u(x) = ef-a(x)dx = ef-l/(2x+3)dx
= e-(l/2)ln 12x+31 = 12x + 31-1/2.
However, we will assume that x > -3/2 (a do•
main where both a and f are defined), so u (x) =
(2x + 3)-1
12
• Multiply both sides of our equation
by the integrating factor and note that the left-hand
side of the resulting equation is the derivative of a
product.
(2x + 3)-1
12
y' - (2x + 3)-312 y = (2x + 3)-1
((2x + 3)-1
12
y)' = (2x + 3)-1
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)
(2x + 3)-1
12
y =
or
- ln(2x + 3) + C,
2
as the interval of existence. The solution curve is 1 2 1 2
shown in the following figure. Note how it drops to y = -(2x + 3)1
/
2
ln(2x + 3) + C(2x + 3) 1

I 1 COS f
O=y(-l)=C.
Consequently, y = (1/2)(2x +3)1
12
ln(2x +3). The
interval of existence is (-3/2, +oo) and the solution
curve is shown in the following figure.
6
Consequently, C = 2 and
x - cosx + 2
y = sec x + tan x ·
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.
25
-1
-2
x
6
(0, 1)
-5
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.
u(x) = ef-a(x)dx = ef secxdx
= 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:
((sec x + tan x)y)' = 1 + sin x
(secx + tanx)y = x - cosx + C
Use the initial condition, y (0) = 1.
(sec O + tan 0)(1) = 0 - cos(O) + C
-1.57 x 1.57
21. Solve for x.
x =---x+--
l+t 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) = ef-a(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 + t)x' + x = cost,
((1 + t)x)' = cost,
(1 + t)x = sin t + C.
Use the initial condition.
( 1 - i)(0) = sin ( -i)+ C

•
+
+
-
•
- -
Consequently, C = 1 and This is a linear equation for z. The integrating factor
is 1/x, so we have
x=
1 + si
-
n t
--
1 + t
The interval of existence is maximally extended to
(-cX), -1), as shown in the following figure.
0
_!_ [dz _ .:.] = _!_(-x)
x dx x x
[;]' =-1
z
- = -x+C
x
z(x) = x(C - x)
1
Since z = 1/y, our solution is y(x) = .x(C -x)
x 24. In this case n = 2, so we set z = y-1
Then
--1 .
-4
22. Let z = x1-n. Then
..........
-1
dz dz dy
----
dx dydx
= y-2 [y2 - y]
1
= -1+-
y
= -1 +z
This is a linear equation for z. The integrating factor
dz dz dx ndx
dt = dx dt = (l - n)x- dt ·
This motivates multiplying our equation by (1 -
is e=, so we have
e-x [ddxz - zJ = -e-x
n)x-n to produce
dx
(1 - n)x-n- = (1 - n)a(t)xl-n (1 - n)f(t).
dt
Replacing (1 - n)x-ndx/dt with dz/dt and x1-n
[e-xz]' = -e-x
e-xz = e-x + c
z(x) = 1 + Ce":
Since z = 1/y our solution is y(x) = --
1
-
with z produces the desired result.
dz
- = (1 - n)a(t)z (1 - n)f(t)
dt
25. Solve for y',
y' = -
1
y +x3y3
x
1 + cex
23. In this case n = 2, so we set z = y-1
. 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 = 3. Let z = y1
-
3
= y-2 Then
= -y-2 [
xy2
- �y]
=-x+
1
dz = dz dy = _2Y_3dy.
dx dy dx dx
Multiply the equation by -2y-3
•
xy
z
= -x+-.
x
.
-2y
_3dy
=
2
y
-2
-2x
3
dx x

3
+
2
-
2
-
d
1
1
Replace -2y-3(dy/dx) with dz/dx and y-2
with z.
dz 2
-=-z-2x
dx x
This equation is linear with integrating factor
u(x) = ef-a(x)dx = ef-2/xdx = e-2lnlxl = x-2.
Replace z with r:' and solve for y.
r' = � + ce:"
a
P=----•
b/a Ce-at
Multiply by the integrating factor and integrate.
27. (a) Sincey = Yi +z,y2
= yf+2y1z+z2
• Hence
x-2
z' - 2x-3z = -2x
(x-2
z)' = -2x
x-2
z = -x2
+ C
4 2
z' = y' - Y�
= -[l/ll +<PY+ xl + [l/lyf + <PYI + x]
= l/l[yf - y2] + ¢[y1 - y]
= -l/1 [2y1z + z ] - </Jz
z = -x + Cx
= -(2y1 l/1 + </J)z - tz2
Replace z with y-2
and solve for y.
y-2 = -x4 + Cx2
y = ±1/-/Cx2 - x4
26. Compare this with P' = a(t)P + f(t)Pn and note
that this has the form of Bernoulli's equation with
(b) Since y, = 1/tisasolution, we set z = y+l/t.
Then y = z - l/t, and y2 = z2
- 2z/t + l/t2,
so
z' = y' -
1
t2
= -
1
- t
y
+ y
2
n = 2. Let z = p1 2
= r". Then
(2 t2
-
dz = � dP = _p_2dP.
dt dP dt dt
Multiply the equation by - p-2•
_p-2dP = -aP-1 + b
dt
Replace --P-2
(dP/dt) with dz/dt and p-l with z.
dz
- = -az+b
3z
= -- +z.
t
This is a Bernoulli equation with n = 2. Thus
we set w = 1 /z. Differentiating, we get
I 1 I
w =--z
t2
= _!_ [- 3z + z2Jt2 t
3
= -1dt
This equation is linear with integrating factor
u(t) = ef -a(t)dt = ef adt = eat.
-
tz
= 3w -1.
t
3
This is a linear equation, and t-
ing factor.
is an integrat•
eat
dz
+ ae"z = beat
dt i-3 [w' - 3;] = -t-3
- (eatz) = beat
dt [t-3
w] = -t-3
t-3
w(t) =
2
t-2 + C
w(t) =
2t + Ct3

2 2Bt 1
t
1
1 + Bt2
2Bt
1 + Bt2.
t
t
t
t
t
Now it is a matter of unravelling the changes of
variable. First
1 2 2
z(t) = w(t) = t + 2Ct3 = t + Bt3'
Where we have set B = 2C. Then
1 2 1
y(t) = z(t) - = t + Bt3
This looks a little better if we use partial frac•
tions to write
2 1
y(t) = t + Bt3 -
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 + (31-1/k)e-kt.
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.
30. The homogeneous equation, y' = -3y has solution
Yh (t) = e-3t. 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 - 3vyh
= v'yh - 3yp,
and y; = -3yp + 4, we have v' = 4/yh = 4e3
.
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 10-3. Hence the model equation is
N' = N - N2
/1000. This is a Bernoulli equation,
with n=2. Accordingly we set x = 1 /N. Then
Integrating we see that v(t) = 4e31
/3, and
The general solution is
4 3t
x' = -N'/N2
y(t) = yp(t) + Cyh(t)
.
= -
3
+ Ce- .
= -l/N + 10-3
= -x + 10-3
_
Solving this linear equation, we getx(t) = 10-3
[1 +
ce--t]. Hence N(t) = l/x(t) = 1000/[l + ce-t].
At t = 0, N = 20 = 1 / [1 + CJ. Hence C = 49, and
the solution is N(t) = 1000/[1 + 49e-t].
We have N (t) = 0.9 x 1000 = 900 when t = 6.089
days.
31. 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,
and y; = -2yp + 5, we have v' = 5/yh = 5e2
.
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
Integrating we see that v(t) = 5e2 /2, and

1
•
•
2 2
•
4 2
'
'
32. The homogeneous equation, y' = -(2/x)y has so•
lution v«(x) = x-2
• We look for a particular solution
in the form Yp(x) = v(x)yh(x), where vis an un•
known function. Since
y� = v'yh + vy�
= V Yh - 2yp/X,
and y� = -(2/x)yp + 8x, we have v' = 8x/yh =
8x3. Hence v(x) = 2x4, and
Yp(x) = v(x)yh(x) = 2x
2
y(t) = Yp(t) + Cyh(t) = 2x2
+ Cx-
2
33. The homogeneous equation, y' = -y/t, has solu•
tion v» (t) = 1 / t. We look for a particular solution in
function. Since
y� = v'yh + vy�
= v'yh - 2vyh
= v'yh - Yvf t,
and y� = -yvft + 4t, we have v' = 4t/yh = 4t2
Integrating we get v(t) = 4t3
/3, and
Yp(t) = v(t)yh(t) = 3t .
35. The homogeneous equation, y' = -2xy has solution
2
Yh (x) = e-x • We look for a particular solution in
the form Yv(x) = v(x)yh(x), where vis an unknown
function. Since
y; = v'yh + vy�
= v'v» - 2xyp,
and y� = -2xyp + 4x, we have v' = 4x/yh =
4xex . Hence v(x) = 2ex , and
Yp(x) = v(x)yh(x) = 2.
y(t) = Yp(t) + Cyh(t) = 2 + Ce
x2
.
36. The homogeneous equation, y' = 3y has solution
Yh (t) = e": We look for a particular solution in
function. Since
y� = v'yh + vy�
+ 3vyh
= v'yh
4
y(t) = Yv(t) + Cyh(t) =
3t
2
+ C/t.
34. The homogeneous equation, x' = -2x has solution
xh(t) = e:", We look for a particular solution in
the form Xp(t) = v(t)xh(t), where vis an unknown
function. Since
= v'yh + 3yp,
and y� = 3yp + 4, we have v' = 4/yh = 4e-3
t.
Integrating we see that v(t) = -4e3
/3, and
xPI
= vI
xh + vxhI
= v'xh - 2vxh
= v'xh - 2xp,
and x� = -2xp + t, we have v' = tfxh = te",
Integrating, we get v(t) = (t/2 - l/4)e2
, and
1 Since y(O) = 2, we must have 2 = -4/3 + C, or
xp(t) = v(t)xh(t) =
1
[2t - 1].
4 C = 10/3. Thus the solution is
x(t) = Xp(t) + Cxh(t) = 4[2t - 1] + ce:". y(t) = (-4 + 10e3
t) /3.

2
= +
t ,
1
= +
t )
- -
2
37. The homogeneous equation y' = -y/2 has solution
yh(t) = e-t/2. We look for a particular solution of
function. Since
y� = v'yh + vy�
= v'yh - vyh/2
= v'yh - Yp/2,
and y� = -yp/2+t, we have v' = t/yh(t) = tet/2.
Integrating we find that v(t) = (2t - 4)etf2, and
Cyh(x) = x2 - 1 + ce=', Since y(O) = -1, we
have C = 0, and the solution is
y(x) = x2 - 1.
40. The homogeneous equation x'. = (2/ t
2
)x has solu•
tion xh (t) = e-2/i. We look for a particular solution
of the form Xp(t) = v(t)xh(t), where v is an un•
known function. Since
I I I
xP v xh vxh
= v'xh + 2vxh/t
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
y(t) = (2t - 4) + se-t/2
•
38. The homogeneous equation y' = -y has solution
Yh(t) = e:', We look for a particular solution of
function. Since
= v'xh + 2xp/
2
and x� = 2xp/t
2
+ 1/t2
we have v' = 1/(
2
h) =, t x
e2ft /t2. Integrating we find that v(t) = -e2ft/2,
and xP (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)
-2- (-1 + e2(1-l/t))
.
YpI
= v
I
Yh
+ vyh
I
= v'yh - vyh
41. The homogeneous equation x' = -4tx /(1 + t2) has
solution xh(t) = (1 + t2
- • We look fora particular
)
2
= v
I
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.
solution of the form Xp(t) = v(t)xh(t), where vis
an unknown function. Since
I I I
xP v Xh vxh
The general solution is y(t) = Yp(t) + Cyh(t) = = v'xh - 4txp/(1 + 2
,
et /2 + ce:', From y(O) = 1, we compute that 2 2
C == 1 /2, so the solution is
andx� = -4txp/(1 + t ) + t/(1 + t ), so
v
I
=
t
- .
l
- = t(l + t
2
).
39. The homogeneous equation y' = -2xy has solution
1 + t2 Xh(t)
Integrating, we get v(t) = t2/2 + t4
/4. Thus
2
Yh(x) = e-x • We look for a particular solution of xp(t) = v(t)xh (t) =
2t2 + t4
.
the form yP (x) = v (x)Yh (x), where v is an unknown
function. Since
y� = v'yh + vy�
4(1 + t2 )2
4C + 2t
2
+ t
4
= v'yh - 2xvyh
= v'yh - 2xyp,
x(t) = Xp(t) + Cxh(t) =
40
+ t2)2 .
The initial condition x (0) = 1 implies that C = 1,
2 so the solution is
and y� = -2xyp + 2x3, we see that v' = 2x3ex •
Integrating we get v(x) = (x2 - l)ex2, and Yp(x) = 4 + 2t + t
4

x2 - 1. The general solution is y(x) = Yp(x) + y(t) = 4(1 + t2)2 .

+
T;
42. (a) The equation T + kT = 0 is separable, with
solution Th = Ce-kt, C an arbitrary constant.
(b) The equation T' = -k(T - A) is autonomous.
We seek a constant solution (see Section 2.9)
C) provides
wkA
C=
k2 + w2
and
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.
(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
Substituting these results in Tp = C cos wt +
D sin wt provides the particular solution
wkA k2
A .
cos wt sm wt.
k2 +w2 k2
+w2
Hence, the general solution is
kA .
constant solution by setting the right side equal
to zero.
= Fe-kt +
k2
+ co2
[k sm wt - co cos wt] .
-k(Tp -A)= H
H
Tp-A = -
44. (a) If the period of the ambient temperature is 24
hours, then the computation
K w-
2n
-
2n
-
tt
- - -
H
Tp =A+ K
Hence, the general solution is given by the
equation
T =Th+ Tp = Ce-kt + (A+ HK) .
- T - 24 - 12
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.
43. (a) The solution of the homogeneous equation
T' + kT = 0 is Th = Fe-kt, 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 = kA 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
Note the minimum at 6 am, then the maximum
at 6 pm. What we have is an upside-down sine,
with angular frequency n/12, that is shifted up•
-wC + kD = kA and kC +wD = 0. ward 60° F. Thus, the equation for the ambient
(c) Solving these equations simultaneously (for ex•
ample, multiply the first equation by k, the sec•
temperature must be
. ttt
ond by co, then add the equations to eliminate A = 60 - 20 sm .
12

)
.
nt
+ + 2
Therefore, the model, adjusted for this ambient
temperature, becomes
2.4 linear Equations 45
Thus, the general solution is T = Th + Tp, or
Tp = Ce-r/2 + 60
dT = _! (T - 60 + 20sin nt).dt 2 12
(4.2)
+ 120 [n cos -tit
- 6 sin -ttt J .
(b) The homogeneous equation T' + (l/2)T = 0
has solution Th = ce-r/2
, where C is an arbi•
trary constant. Now, consider the inhomoge•
36 + n2 12 12
(4.6)
neous equation
I 1 Jr(
T +
2
T = 30 - lOsin
12
. (4.3)
Now, when the initial condition T (0) = 50
is substituted into equation (4.6), we obtain
C = -10-120n/(36+n2
. Thus, the general
solution becomes
Note that the right hand side consists of a con•
stant and a sinusoid. Let's try a particular so•
lution having the form T = -
(10 + 120n ) e:'12 + 60
36 + n?
Tp = D + E cos
itt
+ F sin
xt
(4.4) + 120 [ n cos -ttt - 6 sin -nt] .
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 = !n + (-!!..E + !F) sin
2 2 12 2 12
+ (!E + !!..p) cos nt2 12 12
36 + it? 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.
(4.5)
Comparing this with the right-hand side of
equation (4.3), we see that
1
2D = 30,
l( 1
75
70 · ...
65
60 · ·.
55
50
- £ + F = -10, and 45
12 2
1 n
2E + 12F = 0.
Clearly, D = 60, and solving the remaining
two equations simultaneously, we obtain
0 6 12 18 24 30 36 42 48 54 60 66 72
Note that the transient part of the solution dies
120n
E=---
36 n2
and
-720
F= .
36 n
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
These values of D, E, and F, when inserted
into equation (4.4), provide the particular solu•
tion
120n nt 720 . nt
T =60+ cos-- sm-.
P 36 + n2 12 36 + n2 12
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.

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 + ce-3t/IOO. Since
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,
S(O) = 0, the constant C = -20 and the solu• .
rate out= 5 gal/m
x(t)
x lb/gal=
1 .
x(t) lb/m .
tion is S(t) = 20(1 - e-3t/IOO).
S(20) = 10(1 - e-0·6) � 9.038lb.
(b) S(t) = 15 when e-3t/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) =
Ae-t/25
• The initial condition implies that A =
x(O) = 50 gal x 2 lb/gal= 100 lb. Hence the
solution is x(t) = lOOe-t/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) = lOOe-t/25 =
20. This comes to t = 25 ln 5 � 40.2359 min.
(c) x(t) = 10oe-tl25 -+ 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
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).
m m
100 20
As there are 2 lb of salt present in the solution
initially, x(O) = 2 and
dx 5 1
dt = 4 - 20x, x(O) = 2·
Multiply by the integrating factor, eU/20)t, and
integrate.
(e(]/20)tx )' = �e(l/20)t
4
e<t/20)tx = 25e(l/20)t + C
x = 25 + ce-<1!20)t
The initial condition x (0) = 2 gives C = -23
and
x(t) = 25 - 23e-Ol20)t.
Thus, the concentration at time t is given by
x(t) 25 - 23e-0/20)t
c(t) = 100 = 100
and the eventual concentration can be found by
taking the limit as t -+ +oo.
25 - 23e-0/20)r 1
lim = - lb/gal
t-Hoo 100 4
Note that this answer is quite reasonable as the
concentration of solution entering the tank is
also 1 /4 lb/gal.

I
-
=
2.5 Mixing Problems 47
(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•
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
tent x(t). However,
x(t)
or x(t) = lOOc(t).
500c = -
c
I
=-
r
500
r
(500c),
c.
c(t) = 100
Consequently, x'(t) = IOOc'(t). Substituting
these into our balance equation gives
I 5 1
x = 4 - 20x,
I 5 1
lOOc =
4 20
(lOOc),
I 1 1
c = 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•
-
500
This equation is separable, with solution c =
Ae-(r/SOO)t. Use the initial concentration, c(O) =
.05 lb/gal, to produce
c = 0.05e-(r/500)t.
The concentration must reach 1 % in one hour (60
min), so c(60) = 0.01 and
0.01 = o.ose-(r/500)(60)'
! = e-(3/25)r,
5
tion approaches 0.25 lb/gal.
c ' = 1 /80 - 1 /20 c
25
r
3
1n5,
r � 13.4 gal/min.
0.3
0.2
0
0.1
O
,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
;;;;;; ;;;;;;;;;;;;;
///// //////////////
//// ///////////////
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
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
x ' =2-2x/V=2--x-.
io i .
0 50 100 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
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
. x(t) r .
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.
rate out= r gal/mmx
Thus,
500
lb/gal=
500
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
dx .
- = rate m - rate out,
dt
dx r
dt = - 500x.
S(t)/ V (t) lb/gal. Hence the differential equation is
dS -3S
dt - 100 - t

3
+
I
•
This equation is linear and homogeneous. It can be
solved by separating variables. The general solution
isS(t)+A(lOO-t)3.SinceS(O) = lOOx0.05 = 5,
we see that A = 5 x 1o-6
, and the solution is
S(t) = 5 x 10-6 x (100 - t)3.
When V(t) = 100 - t = 50 gal,
We are asked to find when this is one-half of
21.5342. This happens when (30 - t)2
= 450
or at t = 8.7868 min.
8. Letx(t) represent the amount of drug in the organ at
time t. The rate at which the drug enters the organ is
3 3
S(t) = 5 x 10-6
x 503 = 0.625 lb. rate in= a cm /s x K g/cm = ate g/s.
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
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,
in during this initial period is 6 gal/min · 0.5
lb/gal = 3 lb/gal. The rate out is 8 gl/min
rate out= b cm3/sx
x(t)
--- g/cm
b
= ---x(t) g/s.
·x/V = 4x/(50+t).Hencethemodelequation
is
Consequently,
Vo+ rt Vo+ rt
x' = 3 - 4x/(50 + 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. 875 x 108
.
Hence the solution is
8
x(t) = 3t + 30 _ 1.875 x 10
5 (50+t)4
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 equa-
tion is
dx b
-=aK----x.
dt Vo+ rt
The integrating factor is
u(t) = efb/(Vo+rt)dt = e(b/r)ln(Vo+rt) = (Vo+rt)b/r.
((Vo+ rtllrx )' = aK(Vo + rt)b/r
(V,O
+ rt)b/rx = 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« a« V.b/r+l
x = --(Vo+ rt) - 0
(Vo+ rt)-b/r,
+r b+r
x = �(Vo+ rt) [1 - vtr+1(v0 + rt)-b/r-1],
x =-
2X b+r
--
30- t
a« [ ( v;0 )b/r+l]
This homogeneous linear equation can solved
by separating variables to find the general solu•
x = --(Vo + rt) 1 - -- - •
b + r 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)=
a« [
1-
( Vo ) (b+r)/r]
.
21.5342 -- --
x(t) =
900
(30 - t)2. b + r Vo+ rt

v
9. (a) The rate at which pollutant enters the lake is
rate in = p knr'/yr.
The rate at which the pollutant leaves the lake
10. Because the factory stops putting pollutant in the
lake, p = 0 and e' + ((r + p)/V) e = p/V be•
comes
50
is found by multiplying the flow rate by the
concentration of pollutant in the lake.
e' + e = 0.
100
rate out= (r + p) knr'/yr x x(t) krrr'/krn.3
v
r+p 3
= lfx(t)km /yr
Consequently,
dx r+ p
-=p---x.
dt
Bute(t) = x(t)/V, so Ve'(t) = x'(t) and
I r+ p
Ve = p - lf(Ve),
c' + r + p c = .!!_.
v v
(b) With r = 50 and p = 2, the equation becomes
f 2 50+2
c = 100 - ----iooc,
c' = 0.02 - 0.52c.
This is linear and solved in the usual manner.
(e0.52te)' = 0.02eo.052t
e0.52te = 0.02 eo.os2t + K
0.52
Note that we carried r = 50 from Exercise 9.
This equation is separable, with general solution
e = K e-<1
12
)t. The initial concentration is 3.5%,
so e(O) = 0.035 produces K = 0.035 and
e(t) = 0.035e-<1!2)t.
The question asks for the time required to lower the
concentration to 2%. That is, when does e(t)
0.02?
0.02 = 0.035e-Ol2>r
_ = ln 0.02
2 0.035
0.035
t= 2 ln--
0.02
t � 1.1 years
11. (a) The concentrations are plotted in the following
figure. In steady-state the concentration varies
periodically.
1
e = -
26
+ Ke-o.s2t 0.04
The initial concentration is zero, so e(O) = 0
produces K = -1 /26 and
c = 1 -1 e-0.52t
26 26
The question asks when the concentration
reaches 2%, or when e(t) = 0.02. Thus,
0.02 = � (1 - e-o.s2t)
26 '
e-o.s2t = 0.48,
In 0.48
t = - 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.

0 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 - e-o.szr]/0.52.
Let y(t) be the amount of pollutant (measured
Feb Jul
t
Jan 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
·
,
.
Thus there is a shift ofphase between the cause 13.
and the effect.
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
(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.
52km
3
/yr x y/100 = 0.52ykm
3
/yr. Hence
the model equation is
y' = 0.52(x - y) = 2[1 - e-o.szt] - 0.52y.
x'=-�
20
The solution with initial value x(O) = 20 is x(t) =
2oe-t/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
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 - e-O.SZt]/0.52 - 2te-
0.5Zt
= x(t) - 2te-o.szr.
After 3 months, when t l/4, we
have x(l/4) = 0.4689km3
and y(l/4) =
0.0298km3
.
(b) If the factory is shut down, then the flow of
, _ � Y e-t/20 Y_ pollutant at the rate of 2 knr'/yr is stopped.
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
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
(t) = _c_ - 20e-tf20 - 400 e-t/20.
y 80 + t 80 + t
Since y(O) = 40, we can compute that C = 65 x
80 = 5200. Hence the solution is
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 e-t/2•
y(t) = 5200 20e-tf20 _ -
40
-
0
e-t/20. The rate into Lake Sad times is x /2 knr'/yr,
80 + t 80 + t
Tank B will contain 250 gal when t = 20. At this
point we have y(20) = 43.1709 lb.
and the rate out is
y/2 knr'/yr. The model equation is y' =
(x -y)/2 = x1e-tf2/2-y/2. this time we

v v
v
+
v
v v '
use the integrating factor u(t) = et/2
and find
the solution
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.
0.2
0.15
>- 0.1
Initially, there is no salt in Tank I, so x (0) = 0 pro•
duces K = -a V and
X = 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.
Salt leaves Tank II at
rate out II = y(t)/ V lb/gal x b gal/min
= (b/V)y(t) lb/min.
0.05
o�----_. .
Consequently,
dy b b
-= -x- -y.
dt
0 5 10
Time in years Substitute the solution found for x.
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.
Salt leaves Tank II at
rate out I= x(t)/ V lb/gal x b gal/min
= (b/ V)x(t) lb/min.
Consequently,
dx b
-=ab--x.
dt
This equation is linear with general solution
x =av+ Ke-(b/V)t_
dy b b
- = - (aV - aVe-(b/V)t) - -y
dt
dy b
- = --y (ab - abe-<b/V)t).
dt
This equation is also linear, with integrating factor
ib/V)t, SO
(e(b/V)ty)' = ab (e(b/V)t _ I},
e(b/V)ty = aVe(b/V)t - abt + L,
y = a V - abte-(b/V)t + Le-(b/V)t.
Initially, there is no salt in Tank II, so y(O) = 0
produces L = -a V and
Y = a V - abte-(b/V)t - a Ve-(b/V)t.

2 2
2 2 P
1
=! x dx
aF ,
ay
Section 2.6. Exact Differential Equations
1. dF = 2ydx + (2x + 2y)dy
2. dF = (2x - 2y)dx + (-x + 2y)dy
3. dF = xdx + ydy
-/x2 + y2
. dF = -xdx - ydy4
(x2 + y2)3/2
5.
dF =
1
(x2
ydx + y3dx - ydx
x +y
+ x3dy + xy2dy + xdy)
6. dF = (1/x + 2xy3)dx + (1/y + 3x2
y2
)dy
7.
2
10. With P = 1 - y sinx and Q = cosx, we see that
aP . aQ
- =-SinX= -,
ay ax
so the equation is exact. We solve by setting
F(x, y) = fP(x, y) dx = f(1- y sinx) dx
= x + y cos x + </> (y).
To find ¢, we differentiate
aF
Q(x, y) = - = cosx + ¢'(y).
ay
Thus¢' = 0, so we can take¢ = 0. Hence the
solution is F(x, y) = x + y cosx = C.
11. With P = 1 +�and Q =_!,we compute
x x
dF = (
x
x + 1/y) dx
+y 1 aQ 1
= ; ::/:: a; = x2 .
+ 2y
--x)dy
(x2 + y2 y2 Hence the equation is not exact.
12. W"th P = x
and Q = y , we com-
S. dF = ydx - xdy + 4x
2
y3dy + 4y5dy
x2 + y2 pute
,)x2 + y2 Jx2 + y2
-2xy aQ
9. With P = 2x + y and Q = x - 6y, we see that
ay
== (x2 + y2)3/2
=
a;'
er aQ
-=1=-
so the equation is exact. To find the solution we
integrate
ay ax
so the equation is exact. We solve by setting
F(x, y) = fP(x, y) dx
F(x, y) = fP(x, y)dx = f(2x + y)dx
= x2
+ xy + ¢ (y).
Jx2 + y2
= Jx2 + y2 + cp(y).
Q(x, y) = ay = x + ¢ (y).
aF y
Qt», y) = - =
,
= ¢ (y).
Hence¢' = -6y, and we can take ¢(y) = -3y2
•
Hence the solution is F(x, y) = x2+xy-3y2 = C.

ay Jx2 + y2
Thus ¢' = 0, so we can take ¢ = 0. Hence the
solution is F(x, y) = Jx2 + y2 = C.

c
2
2
13. Exactx3
+xy - y3
= C
14. Not exact
15. Exact u2/2 + vu - v2/2 = C
16. Exact ln(u2
+ v2) = C
17. Not exact
2.6 Exact Differential Equations 53
34. ln x - In y and 1 are homogeneous of degree zero.
35. x2 - Cx = y2
36.
F(x, y) = -(1/2) In
2 2
18. Exact. F(u, y) = ylnu - 2u = C
19. Exactx sin2t - t2 = C
20. Exact x2y2 + x4
= C
21. Not exact.
22. -x/y + Inx = C
23. x2y2/2 - lnx + Iny = C
-(y + 1)2
24. ----=C
x3
; Y )
+ arctan(y/x) - Inx = C
37. F(x, y) = xy + (3/2)x2
= C.
38. y2x2 -41ny -2Inx = C
X +Cx4
39. y(x) = 1 - 2Cx3
40. y(x) = x ln(C + 2 lnx)
41. (a) First,
dy dy/dt v0sine-w
dx dx/dt Vo COS O
25. x - (1/2) ln(x2
+ y ) = C
However, cos O = X jJx2 + y2 and sin O =
xy2/2 -y
26. µ(x) = l/x . F(x, y) = = C
x
1 y2
y/Jx2 + y2, so
voy
--;::::== -w
dy Jx2 + y2 VoY - wJx2 + y2
27. µ(x) = -. F(x, y) = xy - Inx - - = C.
x 2
28. µ(y) = 1/�· F(x, y) = 2x� + (2/3)y3
12
= C
dx
= VQX
Jx2 + y2
VQX
yx +x2
29. µ(y) = l/y2
• F(x, y) = = C
y
Divide top and bottom by v0 and replace co/v0
with k.
y - w Jx2 + y2
30. F(x, y) = x2y3 = C
dy Vo y -kJx2 + y2
------=
dx x x
31. x + y and x - y are homogeneous of degree one.
32. x2 - xy - y2 and 4xy are homogeneous of degree
two.
33. x - Jx2
+ y2
and -y are homogenous of degree
one.
(b) Write the equation
dy y - kJx2 + y2
=
dx x
in the form
(y - kJx2 + y2) dx - x dy = 0.

Both terms are homogeneous (degree 1), so
make substitutions y = xv and dy = x dv +
vdx.
(xv -k-/x2 + x2v2) dx -x(x dv + v dx) = 0
After cancelling the common factor x and com•
bining terms,
k�dx-xdv=O.
Separate variables and integrate.
kdx dv
-- =0
x v'f+v2
klnx-ln(�+v)=C
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,
elnxk-ln(�+v) = elnak
xk
-----=ak
v + v'f+v2
(�f=v+�.
Finally, recall that y = xv, so
� = H(�)' - Gr']
v = H(�t'-(�Y-1
(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.
y
Solve for v.
(�f-v=�
(�fk - 2v (�f+ v
2
= 1 + v2
o -1 = 2v (�f
�[(�)'-(�r']=v

)
ay
-
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
y
G(x, y) = x 2 +y2 = 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) = y2
/x = C are the
solid curves in the following figure. The orthogonal
family must satisfy
dy = aF;aF = _ 2x.
dx ay ax y
>, 0
-1
-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.
44. ln(y2
+ x2
-2 -1 0 2
x
- (2/3)y3
= C
2
>, 0
·-1
·-2
-2 -1 0 2
x
45. arctan(y/x) - y4
/4 = C
46. Assuming that m =j:. n - 1, divide both sides of
X dy + y dx = xmyn dx
by x" v" to obtain
X dy + y dx m=n d
----=X X
(xy")
d((xy)l-n) -
---- =xm "<dx .
l-n
Thus, because m - n + 1 =j:. 0,
(xy)l-n xm-n+l
--= +c1-n m-n+l
43. (a) The curves are defined by the equation
F(x, y) = x/(x2 + y2) = c. Hence the or•
thogonal family must satisfy
dy 8F/8F 2xy
dx = a7 = x2 _ y2 ·
(m - n + l)(xy)1-n - (1 - n)xm-n+l = C.
47. arctan(y/x) - (l/4)(y2
+ x2
)2 = C
48. x/y - ln(xy + 1) = C

e = =
•
x+y
49. Cxy=-•
x-y
Rearranging,
dy -x ± Jx2 + y2
50. (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
=------
dx y
e = ¢ + a = fJ + fJ = 2{3 and
2 tan fJ
tan tan 2{3 2 .
1 - tan fJ
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 2y'
x 1 - (y')2•
(b) Use the result from part (a) and cross multiply.
y - y(y')2 = 2xy'
0 = y(y')2
+ 2xy' - y
Use the quadratic formula to solve for y'.
' -x±Jx2+y2
y =
±xdx+ydy =dx.
Jx2 + y2
The trick now is to recognize that the left-hand
side equals ±d(Jx2 + y2
). Thus, when we
integrate,
±d()x2 + y2
) = dx
±Jx2 + y2
= x + C.
Square, then solve for y2.
x2
+ y2 = x2
+ 2Cx + C
2
y2 = 2Cx + C2
This, as was somewhat expected, is the equa•
y tion of a parabola.
Section 2.7. Existence and Uniqueness of Solutions
1. The right hand side of the equation is f(t, y) = theorem are not satisfied.
4 + y2
f 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.

,
1
2.7 Existence and Uniqueness of Solutions 57
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),
the following figure.
2 .
>, 0
f is discontinuous there. Consequently there is no -1
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. The equation is linear. The general solution is
y(t) = t sin t + Ct. Several solutions are plotted
-1 -0.5 0 0.5
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
= -y + t cost'
dt 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.
8. The equation is linear. The general solution is
y(t) = t + 2Ct2
• Several solutions are plotted in
9. The y-derivative of the right hand side f(t, y) =
3y213 is 2y-113, which is not continuous at y = 0.
Hence the hypotheses of Theorem 7.16 are not sat•
isfied.
10. They-derivative of the right hand side f(t, y) =
ty1
12
is ty-1
12
/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 + �. The
interval of solution is (,J3, oo). The solver has trou•
ble near ,./3. The point where the difficulty arises is

circled in the following figure. 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.
y
4
14. The exact solution is y(t) 3 -
J4 + 2 ln(t + 2) - 2 ln 2. The interval of existence
is (-2 + 2e-2
, oo). The solver has trouble near
-2 + 2e-2 � -1.7293. The point where the diffi•
culty arises is circled in the following figure.
t
0 1
0
-2
-3 -2 -1 0 2
-4 -1
13. The exact solution is y(t) = -1 +J4 + 2 ln(l - t).
The interval of existence is (-oo, 1 - e-2
). The
solver has trouble near 1 - e-2
� 0.8647. The point
where the difficulty arises is circled in the following
15. The solution is defined implicitly by the equation
y3 /3 + y2
- 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)1
13
� 2.3811. The
points where the difficulty arises are circled in the

following figure. ure.
q
5
y
2 4
3
--+-��-+-==---+�-=--+-�-t-��-.
-a 3 2
0
0 2 3 4
-4
The exact solution is
5 - 5e-t if O < t < 2,
q(t) = { 5(1 - e-�)e2-t, if t � 2
16. The: solution is defined implicitly by the equation Hence q(4) = 5(1 - e-2)e-2 � 0.5851.
2y3
- 15y2 + 2t3
= -81. The solver that trouble
near (t1,0), where t1 = -(81/2)113 � -3.4341, 18. Thecomputedsolutionisshowninthefollowingfig-
and also near (t2, 5), where t: = 221
13
� 2.8020. ure.
The: points where the difficulty arises are circled in
the following figure.
q
3
y 2
6
4
0
2
0 2 3 4
0
-4 0 2
O, if O < t < 2,
q(t) = { 3(1 - e2-1), if t 2: 2
Hence q(4) = 3(1 - e-2
) � 2.5940.
17. The computed solution is shown in the following fig- 19. The computed solution is shown in the following fig-

Htit
,
ure.
q
3
21. (a) If
then
O,
y(t) = { (t - to)3,
if t ::::: to
if t > to,
2
0
0 2 3 4
2(t -1 +e-t), ifO < t < 2,
'(t+) _ . y(t) - y(to)
y O - 1Im
t'..to t - to
. (t - to)3
- 0
= lim -----
t - to
= lim (t - tol
t---+tt
=0.
On the other hand,
'( _) . y(t) - y(to)
y t = 1 m
1 ----
q (t) = { 2(1 + e-2)e2-t, if t :::::: 2
Hence q(4) = 2(1 + e-2)e-2
� 0.3073.
20. The computed solution is shown in the following fig•
ure.
q
0
t/'to t - to
= lim 0-0
t---+to t - to
=0.
Therefore, y' (to) = 0, since both the left and
right-hand derivatives equal zero.
(b) The right hand side of the equation, f(t, y) =
3
3y2
13
is continuous, but aJJay = 2y-1
13
is not
y = 0. Hence the hypothe•
continuous where
ses of Theorem 7.16 are not satisfied.
2
22. (a) If
0
0 2 3 4
3e-2t'
y(t) = { 5/2 + (3 - 5e2/2)e-2t,
then it is easily seen that
if t < I
if t :::::: 1
0, if O < t < 2,
q (t) = { t - 1 - e2-t, if t :::::: 2
' {-6e-2t'
y (t) = (-6 + 5e2)e-2t
t < 1
t > 1.
Hence q(4) = 3 - e-2 � 2.8647.
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
_G
I , [ ( 5e ) 2t]
y m m .
2t (5 ( 5e
t
{
the derivative from the right,
y�(l)
= lim y(t) - y(l)
t-+J+ t - 1
2
1 [ ( 5 + ( 5e ) 2r)
If
then
f(t) = { �:
if t < 1
if t > 1,
t < I
= lim -- - 3 - - e-
Hl + t - 1 2 2
+ (3_ s;}-2)J
= lim (3-¥)e-�-(3-¥)c 23. If
t > 1.
Thus,y = y(t)isasolutionofy' = -2y+f(t)
on any interval not containing t = 1.
O, t<Ot-+1+ t - 1
Note the indeterminantform 0/0, so l'Hopital's
rule applies.
y(t) = 4
t '
then it is easily seen that
t >: 0,
2
y = lim -2 3 - - e"
+ t-+I+ 2
= 5 - 6e-2
, {O, t < 0y (t) = 4t3,
t > 0.
It remains to check the existence of y'(O). First, the
However, the derivative from the left,
y�(l) = lim y(t) - y(l)
left-derivative.
, (0) _ 1·
- 1
y(t) - y(O) _ 1·
- 1
0 - 0 _ O
-
t-+J- t - 1
2
= lim 3e- - 2 + 3 - T)
e-
2)
- t-+O- t - 0 t-+O- t
Secondly,
t4
I (0) - 1· y(t) - y(O)
lim - lim t3
= 0.
t-+ I- t - 1
3e-2t - 3e-2
Y+ - 1m
t-+O+ t - 0 t-+O+ t t-+O+
= lim -----
t-+ 1- t - 1
Thus, y' (0) = 0 and we can write
Again, an indeterminant form 0/0, so we apply
l'Hopital's rule.
y�(l) = lim (-6e-2
t)
t-+J-
Now,
y'
(t) = { 0
4,
t3,
t<O
i : 0.
= -6e-2
• ty'(t) = {O, t<O
(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
and
4t4
,
4y(t) = {O,4
4t ,
t > 0,
t<O
t > 0,
interval containing t = 1.
(c) We have that
I {-6e-
2
, f < 1
y (t) = (-6 + 5e2)e-2r t > 1.
so y = y(t) is a solution of ty' = 4y. Finally,
y(O) = 0. In a similar manner, it is not difficult to
show that
O, t < 0
cv(t) =
{ 5t4
, t >: 0

is also a solution of the initial value problem ty' =
4y, y(O) = 0. At first glance, it would appear that we h
have contradicted uniqueness. However, if ty' = 4y
is written in normal form,
I 4yy=•
t
then
! (4y) = �
ay t t
(-1, 0)
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.
h
(-2, 0)
(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
Taking the limit as /;:,.t --+ 0,
Ad
dh
t = av.
Using v2
= 2gh, v = .J2gii, and
dh a r::;;::;:
dt = -A" 2gh.
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 = fi dt ·

Differential equations 2nd edition polking solutions manual

Differential equations 2nd edition polking solutions manual

Recommended

Recommended

More Related Content

What's hot

What's hot (16)

Similar to Differential equations 2nd edition polking solutions manual

Similar to Differential equations 2nd edition polking solutions manual (20)

Recently uploaded

Recently uploaded (20)

Differential equations 2nd edition polking solutions manual