Solution to a Given Differential Equation using Laplace Transform, Assignments of Differential Equations

The solution to a given differential equation using the laplace transform method. The application of the laplace transform to the equation, collection of like terms, and the determination of the constants a, b, c, and d. The final solution is presented in the form of a rational function.

Typology: Assignments

2019/2020

Uploaded on 05/19/2020

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1.
2y
''
+3y
'
2y=t e
2t
y
(
0
)
=0, y
'
(
0
)
=−2
Solution
Apply Laplace Transform to the equation given to all terms.
2
[
s2γ
(
s
)
sy
(
0
)
y'
(
0
)
]
+3
[
(
s
)
y
(
0
)
]
2γ
(
s
)
=1
¿¿
¿tneat , n=1,2,3 1
(sa)n+1
t e2t=1!
(s+2)1+11
(s+2)2
So collect like terms of γ (s)
γ
(
s
)
[
2s
2
+3s2
]
2sy
(
0
)
2y
'
(
0
)
3y
(
0
)
=1
¿¿
substitute the IVP
γ
(
s
)
[
2s
2
+3s2
]
+4=1
¿¿
γ
(
s
)
[
2s
2
+3s2
]
=1
¿¿
γ
(
s
)
=¿
γ
(
s
)
=1
(
2s1
)
(s+2)(s+2)24
(
2s1
)
(s+2)
1
(
2s1
)
(s+2)34
(
2s1
)
(s+2)
γ
(
s
)
=14¿¿
γ
(
s
)
=14(s
2
+4s+4)
(
2s1
)
(s+2)
3
pf3
pf4

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2 y

' '

  • 3 y

'

− 2 y = t e

− 2 t

y

= 0 , y

'

Solution

Apply Laplace Transform to the equation given to all terms.

[

s

2

γ ( s )− sy ( 0 )− y

'

]

[

( s )− y ( 0 )

]

− 2 γ ( s ) =

¿ t

n

e

at

, n =1,2,3

( sa )

n + 1

∴ t e

− 2 t

( s + 2 )

1 + 1

( s + 2 )

2

→ So collect like termsof γ ( s )

∴ γ ( s ) [ 2 s

2

+ 3 s − 2 ]− 2 sy ( 0 )− 2 y

'

( 0 )− 3 y ( 0 )=

→ substitute the IVP

γ ( s ) [ 2 s

2

+ 3 s − 2 ]− 2 s ( 0 )− 2 (− 2 )− 3 ( 0 ) =

γ ( s ) [ 2 s

2

+ 3 s − 2 ]+ 4 =

γ ( s ) [ 2 s

2

+ 3 s − 2 ]=

γ ( s )=¿

γ ( s )=

2 s − 1

( s + 2 )( s + 2 )

2

( 2 s − 1 ) ( s + 2 )

2 s − 1

( s + 2 )

3

( 2 s − 1 ) ( s + 2 )

∴ γ ( s )= 1 − 4 ¿ ¿

γ ( s )=

1 − 4 ( s

2

  • 4 s + 4 )

( 2 s − 1 ) ( s + 2 )

3

∴ γ

s

1 − 4 s

2

− 16 s − 16

( 2 s − 1 ) ( s + 2 )

3

− 4 s

2

− 16 s − 15

( 2 s − 1 ) ( s + 2 )

3

∴ γ

s

− 4 s

2

− 16 s − 15

( 2 s − 1 ) ( s + 2 )

3

A

( 2 s − 1 )

B

( s + 2 )

C

( s + 2 )

2

D

( s + 2 )

3

− 4 s

2

− 16 s − 15 ≡ A ( s + 2 )

3

  • B ( 2 s − 1 ) ( s + 2 )

2

  • C ( 2 s − 1 ) ( s + 2 ) + D ( 2 s − 1 )

∴ A [ s

3

  • 6 s

2

  • 12 s + 8 ]+ B [ 2 s

3

  • 7 s

2

  • 4 s − 4 ]+ C

[

2 s

2

  • 3 s − 2

]

  • 2 DsD

( A + 2 B ) s

3

+( 6 A + 7 B + 2 C ) s

2

+( 12 A + 4 B + 3 C + 2 D ) s + 8 A − 4 B − 2 CD

s

3

: A + 2 B = 0 …

i

s

2

: 6 A + 7 B + 2 C =− 4 …

ii

s

1

: 12 A + 4 B + 3 C + 2 D =− 16 …

iii

s

0

: 8 A − 4 B − 2 CD =− 15 ( iv )

[

|

|

|

|

|

| ]

[

]

Δ D

1

∴ For Δ D 1