How to find the inverse laplace. ccL^(-1){(1)/(−s^2−2s+37)}=?

Sluisu4 2022-10-10 Answered
Finding the inverse laplace L 1 { 1 s 2 2 s + 37 } = ?
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Answers (1)

Emmalee Reilly
Answered 2022-10-11 Author has 6 answers
Well, consider the Laplace transform of the following function:
(1) F ( s ) := L t [ 2 b 2 4 a c e b t 2 a sinh ( t b 2 4 a c 2 a ) ] ( s )
We can set the 'constant' in front:
(2) F ( s ) = 2 b 2 4 a c L t [ e b t 2 a sinh ( t b 2 4 a c 2 a ) ] ( s )
Using the 'frequency shifting' property of the Laplace transform, we get:
(3) F ( s ) = 2 b 2 4 a c L t [ sinh ( t b 2 4 a c 2 a ) ] ( s + b 2 a )
Now, using the Laplace transform of the hyperbolic sine function, we get:
(4) F ( s ) = 2 b 2 4 a c b 2 4 a c 2 a ( s + b 2 a ) 2 ( b 2 4 a c 2 a ) 2
When ( s + b 2 a ) > | b 2 4 a c 2 a |
And (4), simplifies to:
(5) F ( s ) = 1 a s 2 + b s + c
So, when a = 1 , b = 2 , c = 37
(6) f ( t ) = 1 38 exp ( t ) sinh ( 38 t )
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