16Spring Hwk 07 Soln. PDF

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MATH 3301 (Differential Equations)- Lamar University- Professor Dawkins Homework solution set #7 Spring 2016, MATH 3301 (Differential Equations)- Lamar University- Professor Dawkins Homework solution set #7 Spring 2016...

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Math 3301

Homework Set 7 – Solutions

1. (2 pts) F ( s ) =

4. (2 pts) H ( s ) =

10 Points

10 5! (1)( 3)(5 ) π 10 840 2 π 45 π π + 7 5+ 1 − 4 3 + 3 = + − 3 + 7 +1 s s s s 5+ 1 2s 2 23 s 3 2 8s 2 s2

2

(s + 4 )

2( s2 − 14 )

− 3

(s

2

+ 41 )

2

5. (2 pts) There’s no need for partial fractions with this one.

G ( s) =

7 ( s − 32 + 32 ) + 3 7 ( s − 32 ) + 272 7s + 3 7s + 3 7s + 3 = = = = 2 2 s − 6 s + 10 2( s 2 − 3s + 5) 2( s 2 − 3s + 94 − 94 + 5) 2 ( s − 23 )2 + 114 2 ( s − 23 ) + 114

(

2

27 11  1  7 ( s − 32 ) 2 11 =  +  2 2 2  ( s − 32 ) + 114 ( s − 23 ) + 114   

g (t ) =

⇒

1 2

(7e

3t 2

) (

cos

( t) + 11 2

27 11

3t

e 2 sin

( t) ) 11 2

6. (2 pts) Here’s the partial fraction work for this one.

F ( s) =

A B C + + s + 2 s −4 2s −3 s 2 − 1 = A ( s − 4 )( 2s − 3) + B ( s + 2 )( 2s − 3) + C ( s + 2 )( s − 4 )

Picking value of s and solving gives,

s = −2 3 = A ( − 6 )(− 7 ) s = 4 15 = B (6 )(5 ) 5 s = 32 = C ( 72 )( − 25 ) 4

⇒

A = 141 B = 12 C = − 71

The inverse transform is then,

F (s ) =

1 14

s +2

+

1 2

s −4

−

1 71 2 s − 32

⇒

f (t ) = 141 e−2 t + 21 e 4 t − 141 e 2

3t

9. (2 pts) Here’s the partial fraction work for this one.

F (s ) =

A B Cs + D + + s s 2 s 2 + 16

⇒

9 − 2 s = As ( s 2 + 16 ) + B ( s 2 + 16 ) + s 2 ( Cs + D ) = ( A + C ) s 3 + ( B + D ) s 2 + 16 As + 16 B

Setting the coefficients equal and solving gives, 3 s : A+C = 0 2 s : B+D =0 1 16 A = − 2 s : 0 16 B = 9 s :

The inverse transform is then,

⇒

A = − 81 B = 169 C = 18 D = − 169

)

Math 3301

Homework Set 7 – Solutions

F (s ) =

10 Points

1  − 2 9 2 s− 9  1  − 2 9 2s 9 44  + + = + + − 16  s s 2 s 2 + 16  16  s s 2 s 2 + 16 s 2 + 16 

f (t ) =

1 16

(−2 + 9t + 2cos (4t ) −

9 4

sin (4t ))

Not Graded 2. H ( s ) =

3. G ( s ) =

s −6 14 8 + 2 − s − 6 s + 4 ( s − 6) 2 − 4

6( s cos ( −3 ) + sin ( −3 )) 8 2s − 2 + s − 64 s + 16 s2 + 1 2

7. Here’s the partial fraction work for this one.

G ( s) =

A B C + + 3 1 s+ s − ( s −1 )2

⇒

10 s2 − 9 = A ( s − 1) + B ( s − 1)( s + 3 ) + C ( s + 3 ) 2

Picking value of s and solving gives,

s = −3 81 = A (16) 1 = C ( 4) s =1 s = 0 −9 = A (1) + B ( −1)( 3) + C ( 3)

⇒

A= B= C=

81 16 79 16 1 4

The inverse transform is then,

G ( s) =

81 16

s +3

+

79 16

s −1

+

1 4

(s − 1 )

2

−3 t t t 1 g ( t ) = 81 + 79 16 e 16 e + 4 te

⇒

8. Here’s the partial fraction work for this one.

H (s ) =

As + B Cs + D + 2 2 s + 3 s + 6s + 7 8 − 4s = ( s 2 + 6s + 7 ) ( As + B ) + ( s2 + 3) ( Cs + D ) = ( A + C ) s3 + ( 6 A + B + D ) s2 + ( 7 A + 6 B + 3C ) s + 7 B + 3 D

Setting the coefficients equal and solving gives,

s3 : A+C =0 2 6 A+ B + D = 0 s : s1 : 7 A + 6 B + 3C = −4 s0 : The inverse transform is then,

7 B+3D = 8

⇒

16 A = − 31 10 B = − 31 C = 1631 D = 106 31

Math 3301

Homework Set 7 – Solutions

10 Points

1  − 16s − 10 16 ( s + 3 − 3 )+ 106  +   2  31  s2 + 3 3 2 + − s ( )   3  58 55 16 ( s + 3) 1  − 16 s 10 3 =  2 − 2 + +  2 2 31  s + 3 s + 3 ( s + 3 ) − 2 ( s + 3 ) − 2   

H ( s) =

h (t ) =

1 31

(− 16 cos ( 3 t ) −

10 3

sin

( 3t ) + 16e

−3 t

cosh

(

)

2t +

58 2

e −3 t sinh

(

2t

))...