Advanced Engineering Mathematics, by Erwin Kreyszig 10th. Ed.
Problem Set 1.3
No. 1
No. 2
y 3 y'x 3  0
y 3 dy   x 3 dx
y3
dy
dx
 x3
integrate on both sides
3
3
 y dy    x dx  c *
1 y4   1 x4  c *
4
4
y 4   x 4  c; c  4c *
Or x 4  y 4  c
c is an arbitrary constant.
No. 3
dy
 sec 2 y  1
dx
cos 2 y
y' sec 2 y
cos 2 ydy  dx
integrate on both sides.
2
 c o s ydy   dx  c *
 2 1  cos 2 y dy   dx  c *
1
1
2
y  1 sin 2 y  x  c *
4
2 y  sin 2 y  4 x  c; c  4c * c is an arbitrary constant.
No. 4
dy
sin 2x  y cos 2x
dx
y ' sin 2x  y cos 2x
dy
  cos 2x dx
y
sin 2x

integrate on both sides.
dy
   cos 2x dx  c *
y
sin 2x
ln y  1 ln sin 2x  c *
multiply both sides by 2
2
2 ln y  ln sin 2x  2c *
ln y 2  ln sin 2x  2c *
e ln y
2
e
ln sin 2x
 e 2c*


y 2  c sin 2x; c  e 2c * is an arbitrary constant.
No. 5
dy
 36 x
dx
yy '36 x  0
y
ydy  36 xdx
 ydy  36  xdx  c *
y2
 18 x 2  c *
2
y 2  36 x 2  c
c  2c *
Or 36 x 2  y 2  c
No. 6
dy
dx
y'  e 2 x1 y 2
dy
y2

 e 2 x 1dx
 e 2 x 1 y 2
integrate on both sides.
dy
  e 2 x 1dx  c *
2
y
 1  1 e 2 x 1  c *
y
 2  e 2 x 1  2c *
2
y
Or e 2 x 1  2  c  0 c 2c * is an arbitrary constant.
y
No. 7
xy'  y  2 x 3 sin 2
y' 
y
x
 2 x 2 sin 2
y
x
divide both sides by x
y
x
Set
u  xu'  u  2 x 2 sin 2 u
xu'  2 x 2 sin 2 u
u '  2 x sin 2 u
y
u
x
y  xu
y '  u  xu'
du
 2 x sin 2 u
dx

du
sin 2 u
du
sin 2 u
  2 xdx  c
 cot u  x 2  c
 cot
y
x
 2 x d x integrate on both sides.
u
restore
Or x 2  cot
 x2  c
y
x
y
c 0
x
No. 8
y'   y  4 x 2
dv
 4  v2
dx
Set y  4 x  v y '4  v' y '  dv  4
dx
dv
 v2  4
dx
dv
 dx
2
v 4
dv
 2   dx  c *
v 4
1 arctan 1 v
2
2

1 dv
4
1 v 2 1
4
  dx  c *
 x  c*
arctan 1 v  2 x  c; c  2c *
2
1v
2
 tan 2 x  c 
v  2 t a n2x  c
restore
v  y  4x
 y  4x  2 tan 2x  c
No. 9
xy'  y 2  y
y' 
divide both sides by x
y2
y
y2
y
 
x
x
x
2
x
x
u  xu'  xu 2  u
du
u2

 dx
Set
xu'  xu 2
 xc
 u, y  xu, y'  u  xu'
u'  u 2
integrate on both sides
du
  dx  c
u2
1
u
y
x
restore
u
y
x
1
y
x
 xc
x
 xc
y
No. 10
xy'  x  y
y
divide both sides by x
y
x
y'  1 
x x c 0
Or
Set
y
x
 u, y  xu, y'  u  xu'
u  xu'  1  u
xu' 1
du  1 dx
integrate on both sides
x
 du  
dx
x
c
u  ln x  c
restore
y
 ln x  c
x
y  x ln x  cx
u
y
x
No. 11
xy' y  0

xy'   y
x
dy
 y
dx
dy
  dx
y
x
dy
   dx  c *
y
x
ln y   ln x  c *
e
ln y
e
yc1
x
 ln x
 c*
c  e c*
e
ln y
ln 1
 e x  c*
This is the general solution.
From it and the initial condition, y4  6
6  c c  24
4
The particular solution has y  24 .
x
No. 12
y'  1  4 y 2
dy
1 4 y 2
 dx
integrate on both sides

dy
1 4 y 2
  dx  c *
 x  c*
1 arctan 2 y
2
arctan 2 y  2 x  c, c  2c *
2 y  tan 2x  c
From the initial condition y1  0
0  tan 2  c  0  2  c c  2
The particular solution has 2 y  tan 2x  2 .
No. 13
dy
cosh 2 x  sin 2 y
dx
y' cosh 2 x  sin 2 y
dy
 dx
2
sin y
cosh 2 x

integrate on both sides
dy
  dx  c *
2
sin y
cosh 2 x
 cot y 
cot y 
2  c *
1 e 2 x
2  c, c  c *
1 e 2 x
Then, insert the initial value, y 0  1 
2
cot  
2
2
1 e 0
 c  1 c
We have P.S. cot y 
Remark 
0  1  c, c  1
2
1 e 2 x
 1  1 e
2x
1 e 2 x
dx
dx

2
cosh x
 e x e x


2





2

Set e x  u, dx  e  x du  du
u

4du
u  u  1 
 u

2

4udu
 u 2 1


2
No. 14
dr dt  2tr
dr r  2tdt

4du
 2 
u  u 1 
 u 


2

4du
 u 2 1


u
2 
2
 u 2 1  e 2 x 1

 

2
x x
 e  e   tanh x
e x e x
4dx
 e x  e  x 


2
 dr r   2tdt  c
ln r  t 2  c
Insert the initial value, r 0  ro
ln ro  0  c  c
We have P.S. ln r  t 2  ln ro
Or ln r  t 2
ro
r
ro
 e t
2
r  ro e  t
2
No. 15
y'  4 x y y d y 4 x d x
 ydy   4 xdx  c *
1 y 2  2 x 2  c *
2
1 y 2  2 x 2  c *
2
y 2  4 x 2  c; c  2c *
Insert the initial value, y2  3
3 2  4  2 2  c
9  16  c
We have P.S. y 2  4 x 2  25
c  25
Or 4 x 2  y 2  25
No. 16
y'  x  y  22
Set v  x  y  2
v '1  v 2

v'  1  y '
v'  v 2  1
y '  v'1
dv
 v2 1
dx
dv
  dx  c
2
v 1
arctan v  x  c
v  t a nx  c 
x  y  2  tan x  c
Insert the initial value, y0  2
restore v  x  y  2
0  2  2  tan c 0  t a nc c  0
The particular solution is x  y  2  tan x Or y  2  x  tan x
No. 17
xy'  y  3x 4 cos 2  y / x , y1  0
Set u  y / x
y  xu
y' 
dy
 u  x du
dx
dx
代入原式
xu  x 2 du  xu  3 x 4 cos 2 u


x u  x du  y  3x 4 cos 2 u
dx
x 2 du  3 x 4 cos 2 u
dx
dx
du
 3 x 2 cos 2 u
dx
du
 3x 2 dx
2
cos u
t a nu  x 3  c
2
2
 sec udu   3x dx  c
y
y
t a n  x3  c
x
x
Insert the initial value x  1, y  0
t a n0  1  c
Restore u 
The particular solution is tan
y
x
c  1
 x3  1
No. 18
No. 19
dy
 ky
dt
y t  is the amount of yeast at time t, k is the reaction constant.
yt   ce kt
yt   yo e kt
Set y 0   y o  c
k
Set y1  yo e  2 yo
 
4
y4  yo e 4k  yo e k   yo 24  16 yo
At t  2, y2  yo e
At t  4
ek  2
2k
2
 yo e k  22 yo  4 yo
No. 20
dy
 k1 y
dt
dy
 k 2 y
dt
y t  is the amount of yeast at time t, k1 is the birth-rate constant.
k 2 is death-rate constant.
dy
 k1  k 2  y
dt
yt   ce k1 k 2 t
i 
If k1  k 2 y t  always increasing
ii 
If k1  k 2 y t  is increasing until diminish.
iii 
If k1  k 2 y t  keeps constant.
No. 21
Refer to Example 4 in this section
y  y o e kt , k  0.0001213
y
 e kt  e  0.0001213 3000  e  0.3639  0.69496  69.50%
yo
When t = 3,000 years
No. 22
dv
a
dt
dv  adt assume v is the velocity and a is the acceleration.
 dv   adt  c
v  at  c
at t1, v1  at1  c
at t2, v2  at 2  c
v2  v1  at 2  t1 
As v1  103 m / sec, v2  10 4 m / sec and t2  t1  103 sec
4
3
a  10 10  9  10 6 m / sec 2
10  3
4
v v
 3
Traveling distance d   1 2 t 2  t1    10 10 10  3  5.5m
2
 2 


No. 23
dV
dp

 V
p
dp
dV
   c *
V
p
ln V   ln p  c  ln 1  c *
p
V  c ; c  e c*
p
No. 24
Set y t  as the amount of salt in the tank at time t.
dy
dt

y
400
2  
y
200
 1 t
G.S. yt   ce 200
The initial condition y0  100
c  100
 1 t
P. S. yt   100e 200
At t＝1 hrs＝60 mins
 60
y t   100e 200  74.08 lb 
No. 25
dT
dt
 k T  T A 
dT
T T A
 k d t TA=22 ℃
T  T A  ce kt ; c  e c *
ln T  TA  kt  c *
T  T A  ce kt
5  22  ce 0
At t  0 , T  5 o C
c  17
Particular sol. T  22  17e  kt
At t  1 , T  12
T  21.9 o C ,
If
12  22  17e  k
t
1
0.5306
ln
k  0.5306
21.9  22 
17
 9.58 m i n
No. 26
y'   Ay ln y
dy
 A d x
y ln y
dy
  Ay ln y
dx

dy
  A dx
y ln y
Set u  ln y
du 
(1)
dy
y

dy
  du  ln u  ln ln y
y ln y
u
(1) becomes ln ln y   Ax  c *
Then
ln y  ce  Ax
c  e c*
If A  0 , y declines.
If A  0 , y grows.
If A  0 , y keeps constant.
No. 27
Guess: The survived moisture is 1  0.99  0.01
12 6  0.015625 12 7  0.0 0 7 8 1 2 5
0.0078125  0.01  0.015625
The time needed is between 60 and 70 mins.
dy
 ky
dt
y' 
y  y o e  kt
1  e 10 k
2
and at t  10, y  1 y o
2
ln 1  10k
2
k  1 ln 2  0.0 6 9 3 2
10
y  y o e 0.06932 t
As the dryer will have lost 99 % of its moisture i.e., y  0.01yo
0.01  e 0.06932 t ln 0.01  0.06932t
t  ln 100  66.434 min
0.06932
No.28
Refer to Prob. 27
No. 29
dT
dt
 k T  T A 
dT
T T A
 k d t TA is the ambient temperature.
T  T A  ce kt ; c  e c * c is the water temperature at t=0 when Jack
ln T  TA  kt  c *
go into the bar.
Assume the ambient temperature is To  60 and t1 when Jack was arrested
190  60  ce  kto
130  ce  kto
110  60  ce  k t o  30 
50  ce  k t o  30  (2)
1
2  
2.6  e 30 k
(1)
k  0.0319
If t o  30 min as claimed by Jack
From (1) c  338 o F It is impossible for water temperature above the boiled temperature 212
℉.
These results do not give Jack an alibi.
No. 30
dv
dt
 a  g  7t
dv  7t d t
 dv   7tdt  c
v  7 t2  c
2
At t=0, v=0 thus c=0. v  7 t 2
2
At t=10, v  7 t 2  350m / s
2
10
10
d1  t  0 vdt  t  0 7 t 2 dt  7  t 3  7  10 3  1167m
2
6
6
The duration between the engine cut out and the velocity decreases to zero is 350
g
d 2  350  350  122500  6250m
2
g
2g
Total distance d1  d 2  1167  6250  7417m
No. 31
Set the equation of the straight line with slope of m as y  mx
 
The intersect point y'  g y x  g m
g m  is a constant.
No. 32
Force normal to the slide surface N  W cos 30  3 W
2
Friction F   N  0.2 3 W  0.1 3W
2
Driving force along the slide
Fs  W sin 30  1 W
2
Net force along the slide Fn  Fs  F  1 W  0.1 3W
2
In this case, W  45 nt , equivalent acceleration g  9.8m / s 2
2
 2

Acceleration a  1 g  0.1 3 g  1  0.1 3 g  1  0.1 3  9.8  3.20m / s 2
2
t
t
If length of the slide is S and time to reach the end is t, S  t  0 vdt  t  0 atdt  1 at 2
2
S  10 m , a  3.20m / s 2 , thus t  2.5 s
t
The velocity at the end v  t  0 adt  at  3.2  2.5  8 m / s
No. 33
S  0.15S
dS  0.15Sd

dS  0.15 d  C *

S
ln S  0.15  C*, S eC *e 0.15  Ce0.15  So e 0.15
When S = 1000 So, e
0.15
=1000
  1 ln 1000  46.0 5 1 746.0 5 1 7 7.3 2 9 3 t6i m e s
0.15
2