Download Math 234 Fall 2015 Assignment 18 - Solutions

Math 234 Fall 2015
Assignment 18 - Solutions - 1 Due Nov 3
Read Section 3.5, and 3.6
Show your work in detail.
1. Page 138:
Practice Problems: 5, 9, 11, 19, 21, 25
4. with given that y h  c 1 cos x  c 2 sin x
y ′′  y  sec x tan x with given that y h  c 1 cos x  c 2 sin x
Let y p  u 1 cos x  u 2 sin x.
i. Solve u ′1 and u ′2 :
sin x
u ′1
− sin x cos x
u ′2
cos x
u ′1
u ′2

0

sec x tan x
−1
cos x

−
sin x
0
− sin x cos x
sin 2 x
cos 2 x
tan x


sec x tan x
cos x − sin x
0
sin x cos x
sec x tan x
− tan 2 x
tanx
ii. Solve u 1 and u 2 :
u 1   − tan 2 xdx  − sec 2 x − 1dx  −tanx − x  x − tanx
sinx
u 2   tanxdx   cos x dx  − ln|cosx|
y p  x − tanx cos x − ln|cosx| sin x
The general solution of the differential equation:
y  c 1 cos x  c 2 sin xx − tanx cos x − ln|cosx| sin x
10. with given that y h  c 1 e 3x  c 2 e −3x (Hint: 9x
 9xe −3x . 
e 3x
y ′′ − 9y  9xe −3x
Let y p  u 1 e 3x  u 2 e −3x .
i. Solve u ′1 and u ′2 :
e 3x
u ′1
e −3x
u ′2
3e 3x −3e −3x
u ′1
u ′2

e 3x
e −3x
0

9xe −3x
−1
0
3e 3x −3e −3x
ii. Solve u 1 and u 2 :
u 1   32 xe −6x dx  − 14 xe −6x −
u 2   − 32 xdx  − 34 x 2
9xe −3x
1
24

3
2e 3x
xe −3x
− 32 x

e −6x
1 −6x 3x
y p  − 14 xe −6x − 24
e e  − 34 x 2 e −3x  − 241 e −3x 6x  1  18x 2 
The general solution of the differential equation :
1 −3x
y  c 1 e 3x  c 2 e −3x − 24
e 6x  1  18x 2 
1
3
2
xe −6x
− 32 x
12. with given that y h  c 1 e x  c 2 xe x
x
y ′′ − 2y ′  y  e 2
1x
Let y p  u 1 e x  u 2 xe x .
i. Solve u ′1 and u ′2 :
e x xe x
u ′1
e x 1  xe x
u ′2
u ′1
u ′2

0

ex
1  x2
−1
e x xe x
−
0
ex
1  x2
e x 1  xe x

x
x 1
1
x2  1
2
ii. Solve u 1 and u 2 :
u 1   − 2 x dx  − 12 lnx 2  1
x 1
u 2   2 1 dx  tan −1 x
x 1
y p  − 12 lnx 2  1e x  x tan −1 xe x
The general solution of the differential equation :
y  c 1 e x  c 2 xe x − 12 lnx 2  1e x  x tan −1 xe x
26. with given that y h  c 1  c 2 cos2x  c 3 sin2x
Let y p  u 1  u 2 cos2x  u 3 sin2x.
i. Solve u ′1 , u ′2 and u ′3 :
1 cos2x
u ′1
sin2x
0 −2 sin2x 2 cos2x
u ′2
0 −4 cos2x −4 sin2x
u ′3
0

0
sec2x
Use Cramer’s Rule:
1 cos2x
det
sin2x
0 −2 sin2x 2 cos2x
 18 sin 2 2x  8 cos 2 2x  8
0 −4 cos2x −4 sin2x
detA 1   det
0
cos2x
sin2x
0
−2 sin2x 2 cos2x
 sec2x2 cos 2 2x  2 sin 2 2x  2
sec2x −4 cos2x −4 sin2x
detA 2   det
1 0
sin2x
0 0
2 cos2x
0 sec2x −4 sin2x
2
 10 − 2 cos2x sec2x  −2
1 cos2x
detA 3   det
0
0 −2 sin2x 0
 1−2 sin2x sec2x − 0  −2 tan2x
0 −4 cos2x sec2x
detA 1 
2 sec2x

 1 sec2x
u ′1 
4
8
detA
detA 2 
u ′2 
 −2  −1
8
4
detA
detA 3 
u ′3 
 − 1 tan2x
4
detA
ii. Solve u 1 , u 2 and u 3 :
u 1   14 sec2xdx  18 ln|sec2x  tan2x|
u 2   − 1 dx  − 14 x
4
u 3   − 1 tan2xdx  − 14 − 12  ln|cos2x|  18 ln|cos2x|
4
1
y p  8 ln|sec2x  tan2x| − 14 x cos2x  18 ln|cos2x| sin2x
The general solution of the differential equation is
y  c 1  c 2 cos2x  c 3 sin2x  18 ln|sec2x  tan2x| − 14 x cos2x 
3
1
8
ln|cos2x| sin2x