1
Math 316
Spring 04
set VI
Solutions
7.2 p.455 (5*,6,7,9*,10,21*,22,23*,24)
7.3 p.465(9*,10,21*,22,30*,35,36*,37*,38*)
7.2.5 (455) Use Laplace xforms to solve the IVP:
x' ' x  sin 2t; x0  0  x' 0 .
Solution: We have
Lx' ' x  Lsin 2t 
1
2
2
 s2 X  X  2
 X s   2
s  1s 2  4
s 4
s 4
2
this becomes
X s  
2
A
B
 2
 2
2
s  1s  4 s  1 s  4 . We can instantly write the solution as
B
xt   A sin t  sin 2t , where the constants can be found by the method of partial
2
fractions and a trick: set s 2  z :
2
A
B



z  1z  4 z  1 z  4
2
2
1
2
A
| z  1  ; B 
| z  4   
z  4
s  1
3
3
2
1
B 2
X s   2
A 2

2
s 1 s  4
s 1 2 s2  4
giving for the final answer:
1
xt   2 sin t  sin 2t  .
3
2



7.2.9 (455) Use Laplace xforms to solve the IVP:
x' '4 x'3 x  1; x0   0  x' 0 
Solution: We have
1
Lx' '4 x'3x  L1   s 2 X  4 sX  3 X
s

1
1
 X s  
2
s
s s  4 s  3
Since we have the factorization s  4  3  s  1s  3 this becomes
1
A
B
C
. We can instantly write the solution as
X s  
 

ss  1s  3 s s  1 s  3
xt   A  Be t  Ce 3t , where the constants can be found byt the method of partial
fractions:
2
2
1
A
B
C
 


ss  1s  3 s s  1 s  3
1
1
A
| s 0 
s  1s  3
3
1
1
B
| s  1  
ss  3
2
1
1
C
| s  3 
s s  1
6
giving for the final answer:
1 1
1
xt    e  t  e  3`````````` `````````` `````````` `````````` `````````` `````````` `````````` ``````````
3 2
6
.
``````````
``````````
``````````
``````````
``````````
t
 F s 


L
f

d


7.2.21 (456) Apply the formula: 
to find the inverse
s
0

transform of F s  
1
.
s s2 1
2


Solution: We work this problem two ways:
(a) By using a method like in the previous problem, we introduce s 2  z and write
1
1
A
B
1
1

 
 A
| z 0  1; B  | z  1  1 
2
z 1
z
s s  1 z z  1 z z  1
1
1
F s   2  2
 xt   t  sin t
s
s 1
(b) By using the method suggested here: we have
t

1
1
Lsin t  2
 L1  cos t  L  sin d  

2
s 1
0
 s s 1
.
t

1
Lt  sin t  L  1  cos  d   2 2
 f t   t  sin t
0
 s s 1
2






t
 F s 


L
f

d


7.2.23 (456) Apply the formula: 
to find the inverse
s
0

transform of F s  
1
.
s s 2  1
2
Solution: (a) By using a method like in the previous problem, we introduce
and write
s2  z
``````````
``````````
``````````
``````````
```
3
1
1
A
B
1
1

 
 A
| z 0  1; B  | z 1  1 
2
z 1
z
s s  1 z z  1 z z  1
1
1
F s    2  2
 f t   t  sinh t
s
s 1
(b) By using the method suggested here: we have
t

1
1
Lsinh t  2
 Lcosh t  1  L sinh d  

2
s 1
s
s

1
0


2




t

1
Lsinh t  t  L cosh   1d   2 2
 f t   sinh t  t
s
s

1
0



7.3.9 (465) Apply the translation (shift) theorem to find the inverse transform for
3s  5
F s   2
.
s  6s  25
Solution: We have:
s  3  7
3s  5
3s  5
3s  3  14
4
F s   2
 2

3

2
2
2
2
s  6s  25 s  6s  9  16 s  3  4
s  3  4 2 s  32  4 2
7
f t   3e 3t cos 4t  e 3t sin 4t
2
7.3.21 (465) Apply the translation (shift) theorem to find the inverse transform for
F s  
s
s2  3
2
 2s  2

2
.
.
Solution:
2

s  1  2s  1  3
F s  




s 2  2s  22 s 2  2s  1  12 s  12  12 s  12  12
s  12  2s  2  s  12  1 2s  1  1  2s  1  3 
F s  
s  12  12
s  12  12 s  12  1 s  12  12
s  1  3
1
1
F s  
2

2
2
s  1  1 s  12  1
s  12  12
s2  3
s2  3
s2  3
3
1
f t   e t sin t  te t sin t  e t sin t  t cos t   e t 5 sin t  2t sin t  3t cos t 
2
2
where we used the formulas:




s
1
1

 1


L1 

t sin kt and L1 
 3 sin kt  kt cos kt .


2
2
2
2
2
2


 s k 
 2k
 s k 
 2k




4
7.3.30 (465) Solve the IVP x' '4 x'8x  e t ; x0  x' 0  0 .
Solution: We have
1
1
Lx' '4 x'8 x  Le t    s 2 X  4sX  8 X 
 X s  
s 1
s
X s  


1

s  1 s  4s  8

2
1
A
Bs  C
A s 2  4s  8  Bs  C s  1




s  1 s 2  4s  8 s  1 s 2  4s  8
s  1 s 2  4s  8


1   A  B s 2  4 A  B  C s  8 A  C 
There results the system:
A B  0



4A  B  C  0
8A  C  1
Solving, we subtract eq.s (1) and (3) from eq. 2 to find A:
1
3
1
A   C   , B   so that
5
5
5
1 1
1
s
3
1
1 1
1
s
3
1
X s  






2
2
2
5 s  1 5 s  4 s  8 5 s  4s  8 5 s  1 5 s  2  4 5 s  2 2  4
1 1
1 s  2  2 3
1
1 1
1 s  2
1
2
X s  






2
2
2
5 s  1 5 s  2   4 5 s  2   4 5 s  1 5 s  2  4 10 s  2 2  4
1
1
1
xt   e t  e  2t cos 2t  e  2t sin 2t
5
5
10
7.3.36 (465) Solve the IVP
Solution: We have

  
1


1
1
1
 s 4  2 s 2  1 X s  
 X s  

4
s2
s2
s  2  s  2 s 2  1
A
Bs  C Ds  E

 2


2
s  2 s 1
s2 1
L x 4   2 x' ' x  L e 2t 
X s  
x 4   2 x' ' x  e 2t ; x0  x' 0  x' ' 0  x 3 0  0


s  2s 2  12
 
2
As 2  1  Bs  C s  2 s 2  1  Ds  E s  2 
X s  
s  2s 2  12
Multiplying through etc. we arrive at the polynomial equation:
 A  Bs 4   2B  C s 3  2 A  B  2C  Ds 2   2B  C  2D  E s   A  2C  2E   1
resulting in the system
5
A B  0
 2B  C  0
2 A  B  2C  D  0
 2B  C  2D  E  0
A  2C  2 E  1
from which we find:
1
1
2
1
2
A
, B   , C   , D   , E   so that
25
25
25
5
5
1 1
1 s
2 1
1
s
2
1
X s  





2
2
2
25 s  2 25 s  1 25 s  1 5 s 2  1
5 s2 1 2
1 2t 1
2
t
1
xt  
e  cos t  sin t  sin t  sin t  t cos t  
25
25
25
10
5
1
xt  
2e 2t  10t  2 cos t  5t  14sin t
50






7.3.37 (465)
Solve the IVP
x' '4 x'13x  te t ; x0  0, x' 0  2
Solution: We have
Lx' '4 x'13x  Lte t  
X s  
X s  
1
s  1
2
 s 2 X  2  4sX  13 X 
1
s  12

2
1
2s 2  4s  3
A
B
Cs  D





2
2
2
2
2
s  4s  13 s  1 s 2  4 s  13 s  1 s 2  4s  13 s  1 s  1
s  4s  13


2s  4s  3

2
s  1
2
s
2

 4s  13





A
B
Cs  D

 2

2
s  1 s  1 s  4s  13


2s 2  4s  3
Cs  D
s  12
 As  1  B  2
2
s  4s  13
s  4 s  13




Where the last equation was found by multiplying both sides by s  1 . Now set s = -1
to find B:
2s 2  4s  3
1
B 2
| s  1 
10
s  4s  13
and differentiate with respect to s to find A:
2
B 
B 
A



2s  42s 2  4s  3  |   1
d 2s 2  4s  3
4s  4

|


s  1
 s 2  4s  13
ds s 2  4s  13
s 2  4s  132  s1 50


6
X s  
2s 2  4s  3
s  1
2
s
2

 4 s  13


A
B
Cs  D

 2

2
s  1 s  1 s  4s  13
 



2 s 2  4s  3  As  1 s 2  4s  13  B s 2  4s  13  Cs  D s  1 
2
2 s 2  4s  3   A  C s 3  5 A  B  2C  D s 2  17 A  4 B  C  2 D s  13 A  13B  D 
resulting in the system
AC  0
5 A  B  2C  D  2
17 A  4 B  C  2 D  4
13 A  13B  D  3
Taking advantage of the known values of A,B we have
1
A  C  0  C  A 
50
1 1
1 49
5 A  B  2C  D  2  D  2  5 A  B  2C  2   

10 10 25 25
(we can verify that these are correct by substituting in the last two equations, and see that
they hold:
17 20 1 196
17 A  4 B  C  2 D  4   


4
50 50 50 100
13 65 98
13 A  13B  D  3  


3
100 50 50
Finally, we substitute the constants and complete the squares in the quadratic
denominator:
X s   
1 1
1
1
1
s  98



2
2
50 s  1 10 s  1
50 s  4s  13
X s   
1 1
1
1
1
s  98



2
2
50 s  1 10 s  1
50 s  4s  4  9




1 1
1
1
1 s  2  96



2
50 s  1 10 s  1
50 s  22  9
1
t
1
32
xt    e t  e t  e  2t cos 3t  e  2t sin 3t
50
10
50
50
X s   


7
7.3.38 (465) Solve the IVP
Solution: We have
Lx' '6 x'18 x  Lcos 2t 
x' '6x'18x  cos 2t; x0  1, x' 0  1
s
s
 s 2 X  s  1  6sX  1  18 X  2

s 4
s 4
s5
s
s 3  5s 2  5s  20
As  B
Cs  D
X s   2
 2

 2
 2

2
2
2
s  6s  18
s  4 s  6s  18
s  4 s  6s  18
s 4
s  6 s  18
As  B
Cs  D
As  B C s  3  D  3C 
X s   2
 2
 2


s 4
s  6s  9  9
s 4
s  32  9

 


xt   A cos 2t 
2
 





B
D  3C 3t
sin 2t  Ce 3t cos 3t 
e sin 3t
2
3
To do this problem by partial fractions, we proceed as usual:
s 3  5s 2  5s  20
As  B
Cs  D
X s   2
 2
 2

2
s  4s  6s  18 s  4 s  6s  18

 

s 3  5s 2  5s  20   As  B  s 2  6s  18  s 2  4 Cs  D  
s 3  5s 2  5s  20   A  C s 3  6 A  B  D s 2  18 A  6 B  4C s  18B  4 D 
resulting in the system
A  C  1 C  1 A
6A  B  D  5  D  5  6A  B
18 A  6 B  4C  5  18 A  6 B  41  A  5  14 A  6 B  1
18 B  4 D  20  18 B  45  6 A  B   24 A  14 B  0
which gives:
7
12
163
796
D  3C 307
A
,B 
,C 
,D 


170
170
170
170
3
510
so that finally
7
6
163 3t
307 3t
xt  
cos 2t 
sin 2t 
e cos 3t 
e sin 3t
170
170
170
510

