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S S 2 S (s+3) (S 2) : Evaluate L-1

The document provides links to videos and examples for taking the inverse Laplace transform of various functions, including s/(s^2+a^2)^2, s/(s^2+1)(s^2+4), and s/s^4+s^2+1. It also gives an example of solving a pair of first order differential equations and evaluating the inverse Laplace transform of s(s+3)(s-2)/(s^2+s-2) using partial fractions. Finally, it links to a video for using the Laplace transform to solve a second order differential equation.

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yasin muhamed
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457 views2 pages

S S 2 S (s+3) (S 2) : Evaluate L-1

The document provides links to videos and examples for taking the inverse Laplace transform of various functions, including s/(s^2+a^2)^2, s/(s^2+1)(s^2+4), and s/s^4+s^2+1. It also gives an example of solving a pair of first order differential equations and evaluating the inverse Laplace transform of s(s+3)(s-2)/(s^2+s-2) using partial fractions. Finally, it links to a video for using the Laplace transform to solve a second order differential equation.

Uploaded by

yasin muhamed
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as DOCX, PDF, TXT or read online on Scribd
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1) laplace inverse of s/(s^2+a^2)^2 --> in notes

2) convolution theorem of the Laplace inverse of s/(s^2+1)(s^2+4)

https://www.quora.com/How-do-I-do-the-convolution-theorem-of-the-Laplace-inverse-of-s-s-2+1-s-2+4

3) Inverse Laplace of s/s^4+s^2+1

https://www.youtube.com/watch?v=45opyCFuCIk&t=11s

4)

https://www.youtube.com/watch?v=7uHkfustHo4

5) Solve dy/dt+x=cos t dx/dt+y=sin t x(0)=2 y(0)=0

s2 +s−2

6) Evaluate L-1
{ s( s+3 )(s−2 ) }
s2 +s−2

Evaluate L-1
{ s( s+3 )(s−2 ) }
s2 +s−2

Solution: To find L-1


{ s( s+3 )(s−2 ) }
s2 +s−2 B

By partial fractions, L-1


{ s( s+3 )(s−2 ) }  L-1
{}A
s
 L-1
{ }
( s+3 )
 L-1

C
{ }
( s−2)

1 4 2
3 ; B = 15 ; C = 5
We get A =

s2 +s−2 4 2

L-1
{ s( s+3 )(s−2 ) }  L-1
1
{ }
3s
 L-1
{
15 (s+3) }  L-1
{ 5( s−2) }
1 1


1
3
L-1
{1s } 
4
15
L-1
{ }
( s+3 )

2
5
L-1
{ }
( s−2)

1 4 2
3 15 5
  e-3t  e2t

s2 +s−2

 L-1
{ s( s+3 )(s−2 ) } 
1
3

4
15
e-3t 
2
5
e2t is required solution.

7) Using Laplace Transform , Solve y"-3y`+2y = e^3x where y(0)=0 y`(0)=0

https://www.youtube.com/watch?v=A5qYTcssOPY

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