The inverse Laplace transform of (s^(3/2)-a-bs)/(s^(3/2)+a+bs)

Antwan Perez

Antwan Perez

Answered question

2022-10-16

How can I solve the inverse Laplace transform as below:
L 1 ( s 3 / 2 a b s s 3 / 2 + a + b s )
where a and b are constants.

Answer & Explanation

Shyla Maldonado

Shyla Maldonado

Beginner2022-10-17Added 15 answers

By rationalizing the denominator, we have
L 1 { s 3 2 a b s s 3 2 + a + b s }
= L 1 { ( s 3 2 a b s ) 2 ( s 3 2 + a + b s ) ( s 3 2 a b s ) }
= L 1 { s 3 2 s 3 2 ( b s + a ) + b 2 s 2 + 2 a b s + a 2 s 3 b 2 s 2 2 a b s a 2 }
= L 1 { 1 } L 1 { 2 s 3 2 ( b s + a ) s 3 b 2 s 2 2 a b s a 2 } + L 1 { 2 ( b 2 s 2 + 2 a b s + a 2 ) s 3 b 2 s 2 2 a b s a 2 }
The first term is the inverse Laplace transform of 1 and it is equal to δ ( t ) .
The third term is the inverse Laplace transform of rational function.
The second term can be considered by this way:
L 1 { 2 s 3 2 ( b s + a ) s 3 b 2 s 2 2 a b s a 2 }
= L 1 { 1 s × 2 s 2 ( b s + a ) s 3 b 2 s 2 2 a b s a 2 }
= 1 π t L 1 { 2 b s 3 + 2 a s 2 s 3 b 2 s 2 2 a b s a 2 }

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