Advanced Engineering Mathematics, by Erwin Kreyszig 10th. Ed.
Problem Set 1.3
No. 1
No. 2
y 3 y'
3
y 3dy
x 3dx
y 3 dy
1
4
y3
0
x3
integrate on both sides
x 3 dx c *
y4
1 x4
4
y4
dy
dx
x4
Or x 4
c*
c; c
y4
c
4c *
c is an arbitrary constant.
No. 3
dy
dx
y' sec 2 y
cos 2 ydy
dx
c o 2s ydy
1
2
1y
2
1
cos 2 y
integrate on both sides.
x c*
cos 2 y dy
1 sin 2 y
4
2 y sin 2 y
sec 2 y
x c*
x c*
4 x c; c
4c * c is an arbitrary constant.
No. 4
y' sin 2 x
dy
y
dy
y
dy
sin 2 x
dx
y cos 2 x
os 2 x
dx
sin 2 x
s 2 x dx
sin 2 x
y cos 2 x
integrate on both sides.
c*
1 ln sin 2 x
2
ln y
c*
2 ln y
ln sin 2 x
2c *
ln y 2
ln sin 2 x
2c *
e ln y
2
y2
e
ln sin 2 x
multiply both sides by 2
e 2c *
e 2c* is an arbitrary constant.
c sin 2 x; c
No. 5
yy ' 36 x 0
ydy
y
36 xdx
y2
2
36 x
ydy
18 x 2
Or 36 x 2
dy
dx
y2
c*
y2
36 xdx
6x2
c
c*
c 2c *
c
No. 6
dy
dx
y' e 2 x 1 y 2
dy
y2
e 2 x 1dx
dy
y2
e 2x 1 y 2
integrate on both sides.
e 2 x 1dx c *
1 e2x 1
2
1
y
Or e 2 x 1
2
y
e2x 1
2
y
c*
0 c
c
2c *
2c * is an arbitrary constant.
No. 7
xy' y 2 x 3 sin 2
y'
u
y
x
2 x 2 sin 2
y
x
divide both sides by x
y
x
Set
xu' u 2 x 2 sin 2 u
xu' 2 x 2 sin 2 u
u ' 2 x sin 2 u
y
x
u
y
xu
y' u
xu'
2 x sin 2 u
du
dx
du
sin 2 u
du
sin 2 u
xdx
cot u
x2
c
y
x
x2
c
ot
2 x d x integrate on both sides.
restore
Or x 2
cot
u
y
x
y
x
c
0
No. 8
y'
y 4x
dv
dx
4
Set y
v2
dv
2
v
dv
dx
v2
4x
v y' 4
v' y '
dv
dx
4
4
dx
dv
2
v 4
dx c *
1 arctan 1 v
2
2
arctan 1 v
tan x c
y 4x
dx c *
x c*
2 x c; c
2
1v
2
1 dv
4
1 v2 1
4
v
2c *
restore
2t a n x c
v
y
4x
2 tan x c
No. 9
xy' y 2
y
y2
x
y'
y
x
xu' xu 2
u
du
u2
du
u2
u
divide both sides by x
y2
x
x2
u
y
x
Set
xu' xu 2
y
x
u, y
u' u 2
integrate on both sides
dx
x c
x c
restore
u
y
x
xu, y' u
xu'
x c
y
x
x
y
Or
x c
No. 10
xy' x
y
x
u
Set
xu'
u
y
x
u, y
xu, y ' u
xu'
integrate on both sides
d
x
u
0
xu' 1
1 dx
x
du
c
divide both sides by x
y
y'
x
y
x
c
u
nx
c
restore
y
x
nx
c
y
y
x
u
x ln x
cx
No. 11
xy' y
0
xy'
dy
y
dx
x
ln y
ln x
e
ln y
e
c1
y
ln x
c
x
y
x
dy
dx
c
4
c
c*
c*
ec*
e
ln y
ln 1
e x
c*
This is the general solution.
6
24
The particular solution has y
24 .
x
No. 12
y' 1 4 y 2
dy
1 4y2
dx
dx
x
c*
From it and the initial condition, y
6
dy
y
y
integrate on both sides
dy
dx c *
1 4y2
1 arctan 2 y
2
x c*
arctan 2 y
2 y an x c
From the initial condition y
0 2 c c
0 an
c
The particular solution has 2 y
2 x c, c
2c *
0
2
tan x 2 .
No. 13
y' cosh 2 x
dy
sin 2 y
dy
cosh 2 x
dx
sin 2 y
dx
cosh 2 x
dy
sin 2 y
cot y
integrate on both sides
dx
cosh2 x
2
1 e 2x
c*
c*
2
1 e 2x
cot y
c, c
c*
1
2
Then, insert the initial value, y 0
cot
2
2
1 e0
c
We have P.S. cot y
c
0 1 c, c
2
1 e 2x
2x
1 1 e
dx
cosh2 x
Remark
sin 2 y
dx
4dx
2
x
u
1
u
e x du
u , dx
4du
x
du
u
4du
2
2
u
4du
2
No. 14
dr dt
dr r
2
u
2tr
2tdt
2
2
2
u
4udu
2
e x ex
ex e x
1 e 2x
2
Set e x
1
2x
2
tanh x
dr r
tdt c
t2
ln r
c
Insert the initial value, r 0
ln ro
0 c c
t2
We have P.S. ln r
t2
Or ln r
ro
ro
ln ro
2
e t
r
ro
2
ro e t
r
No. 15
y' 4 x y y d y 4 x d x
ydy
4 xdx c *
1
2
y2
2x2
c*
1
2
y2
2x2
c*
y2
4x2
Insert the initial value, y
32
2
c
We have P.S. y 2
9
2c *
3
6 c
4x 2
c; c
c
25
Or 4 x 2
25
y2
25
No. 16
y'
x
y 2
Set v
x
v2
v' 1
dv
2
v
y 2
v' 1 y '
v' v 2 1
y ' v' 1
dv
dx
v2 1
x c
arctan v
x c
v
an
x y 2 an
c
Insert the initial value, y
c
restore v
tan x Or y
No. 17
y 3x 4 cos 2 y / x , y
Set u
y/ x
y 2
2
0 2 2 tan c 0 t a nc c 0
The particular solution is x y 2
xy'
x
0
y
xu
2
x
tan x
dy
dx
y'
xu x 2 du
dx
x du
u
代入原式
dx
xu 3 x 4 cos 2 u
x 2 du
dx
3 x 2 cos 2 u
du
dx
c 2 udu
x3
t a nu
y
x
Insert the initial value x 1, y
y
x
Restore u
The particular solution is tan
y
x
dx
y 3x 4 cos 2 u
3x 4 cos 2 u
c
x3
tan
0
x du
3x 2 dx
du
cos 2 u
x 2 dx c
xu
t a n0
c
c
c
1
x3 1
No. 18
No. 19
dy
dt
ky
y t is the amount of yeast at time t, k is the reaction constant.
ce kt
yt
Set y 0
yo c
Set y
yo e k
At t
2, y
At t
4
yt
2 yo
yo e 2k
yo e 4 k
y
yo e kt
ek
2
2
yo e k
2 2 yo
4
yo e k
yo
4 yo
16 yo
No. 20
dy
dt
dy
dt
dy
dt
k1 y
k2 y
k1
y t is the amount of yeast at time t, k1 is the birth-rate constant.
k 2 is death-rate constant.
k2 y
ce k1 k 2 t
yt
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
yo
When t = 3,000 years
e kt
e 0.0001213 3000
e 0.3639
0.69496
No. 22
dv
dt
a dv
v
v
adt assume v is the velocity and a is the acceleration.
dt c
at c
at t1, v1 at1 c
at t2, v2 at 2 c
v2 v1 a 2 t1
As v1
103 m / sec, v2
10 4 10 3
10 3
a
10 4 m / sec and t2
1
10 3 sec
9 10 6 m / sec 2
v1 v 2
2
Traveling distance d
t2
t1
10 3 10 4 10 3
2
No. 23
dV
dp
V
p
dp
p
dV
V
ln V
V
c*
ln p c
c
; c
p
ln 1
p
c*
e c*
No. 24
Set y t as the amount of salt in the tank at time t.
5.5m
69.50%
dy
dt
y
400
y
200
2
1 t
200
ce
G.S. y t
The initial condition y
c 100
100
1 t
200
100e
P. S. y t
At t＝1 hrs＝60 mins
60
200
100e
yt
74.08 lb
No. 25
dT
dt
k
ln T TA
T
TA
kt c *
5 oC
Particular sol. T
At t
1, T
TA
ce kt ; c
12
22 ce 0
5
e c*
c
17
7e kt
22
7e k
12
22
t
1 ln
0.5306
21.9 o C ,
T
If
T
k d TA=22 ℃
ce kt
0, T
At t
dT
T TA
TA
k
0.5306
.9 22
17
9.58 m i n
No. 26
y'
Ay ln y
dy
dx
dy
y ln y
Set u
dy
y ln y
Ay ln y
A dx
ln y
du
(1) becomes ln ln y
Then
Ad x
ln y ce Ax
(1)
dy
y ln y
dy
y
Ax c *
c e c*
If A 0 , y declines.
If A 0 , y grows.
If A 0 , y keeps constant.
du
u
ln u
ln ln y
No. 27
Guess: The survived moisture is 1 0.99
0.015625
2
0.01
0 .0 0 7 8 1
2
0.0078125 0.01 0.015625
The time needed is between 60 and 70 mins.
y'
dy
dt
y
y o e kt
1
2
e 10 k
y
y o e 0.06932 t
ky
and at t
ln 1
10, y
1y
2 o
10k
2
1 ln 2
10
k
As the dryer will have lost 99 % of its moisture i.e., y
0.01 e 0.06932 t ln 0.01
t
ln 100
0.06932
0.0 6 9 3
0.01 y o
0.06932t
66.434 min
No.28
Refer to Prob. 27
No. 29
dT
dt
k
dT
T TA
TA
ln T TA
T
kt c *
kd
TA
TA is the ambient temperature.
ce kt ; c
go into the bar.
Assume the ambient temperature is To
190
60 ce kto
110
60 ce
2.6
If t o
130
o 30
e 30 k
50
k
60 and t1 when Jack was arrested
ce kto
ce
e c* c is the water temperature at t=0 when Jack
o 30
(1)
(2)
0.0319
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
At t=0, v=0 thus c=0. v
7 2
t
2
v
v
dt c
7 t2
2
c
7 2
t
2
At t=10, v
d1
10
vdt
t 0
350m / s
10 7 2
t dt
t 02
7
6
t3
10 3
7
6
1167m
The duration between the engine cut out and the velocity decreases to zero is 350
g
d2
350
2
350
g
122500
2g
6250m
Total distance d1 d 2 1167 6250 7417m
No. 31
Set the equation of the straight line with slope of m as y
The intersect point y'
g yx
gm
mx
g m is a constant.
No. 32
Force normal to the slide surface N
Friction F
0.2 3 W
N
2
0.1 3W
Driving force along the slide
Fs
W sin 30
Net force along the slide Fn
Fs
F
In this case, W
3
W
2
W cos 30
1W
2
1W
2
0.1 3W
9.8m / s 2
45 nt , equivalent acceleration g
Acceleration a
1
2
g
0.1 3g
2
0 .1 3 g
2
0 .1 3
If length of the slide is S and time to reach the end is t, S
S 10 m , a
3.20m / s 2 , thus t
2.5 s
9.8
t
vdt
t 0
3.20m / s 2
t
atdt
t 0
1 at 2
2
t
adt
t 0
The velocity at the end v
at
3.2 2.5 8 m / s
No. 33
0.15S
ln S
0.15
dS
0.15Sd
C*, S eC *e0.15
When S = 1000 So, e
1 ln 1000
0.15
0.15
dS
S
0.15
Ce0.15
C*
So e 0.15
=1000
46.0 5 1 746.0 5 1 7 7.3 2 9 3 t6i m e
2