"L−1[s(s+3)2+9]" solved and explained

Marlee Yates

Answered question

2022-02-12

L^{- 1} [\frac{s}{{(s + 3)}^{2} + 9}]

Answer & Explanation

Justice Jacobson

Beginner2022-02-13Added 17 answers

The given function is
$L^{- 1} [\frac{s}{{(s + 3)}^{2} + 9}]$
In this problem we use the shifting properties of Inverse Laplace Transformation.
$= L^{- 1} [\frac{s + 3 - 3}{{(s + 3)}^{2} + 9}] {∴ add and subtract 3}$
$= L^{- 1} [\frac{s + 3}{{(s + 3)}^{2} + 9}] - 3 L^{- 1} [\frac{1}{{(s + 3)}^{2}} + 4]$
Property of inverse Laplace Transform
$L^{- 1} {f (s)} = f (t)$
then, $L^{- 1} {\overset{―}{f} (s - a)} = e^{a t} f (t) \to shifting property$ .
$= e^{- 3 t} L^{- 1} [\frac{S}{S^{2} + 3^{2}}] - 3 e^{- 3 t} L^{- 1} [\frac{1}{S^{2} + 3^{2}}]$
$= e^{- 3 t} \cos 3 t - ⧸ 3 e^{- 3 t} \cdot \frac{1}{⧸ 3} \sin 3 t$
$= e^{- 3 t} \cos 3 t - e^{- 3 t} \sin 3 t$
$= e^{- 3 t} (\cos 3 t - \sin 3 t)$
or $e^{- 3 t} (- \sin 3 t + \cos 3 t)$
This is the required answer.

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