Try to calculate L − 1 ( 3 s 3 − 3 s 2 + 3 s − 5 s 2 ( s 2 + 2 s + 5 ) )

Name: Try to calculate L − 1 ( 3 s 3 − 3 s 2 + 3 s − 5 s 2 ( s 2 + 2 s + 5 ) )
Uploaded: 2022-02-23T17:22:57+00:00
Description: Try to calculate L − 1 ( 3 s 3 − 3 s 2 + 3 s − 5 s 2 ( s 2 + 2 s + 5 ) )

Question

Try to calculate            L              −      1            (                  3                  s          3                −        3                  s          2                +        3        s        −        5                              s          2                (                  s          2                +        2        s        +        5        )              )

Phiplyrhypelw0 · Accepted Answer

We can work directly with the inverse Laplace transform and poles of

s = 0, - 1 \pm 2 i

L^{- 1} {F (s)} (t) = \frac{1}{2 π i} \int_{γ - i \infty}^{γ + i \infty} \frac{3 s^{3} - 3 s^{2} + 3 s - 5}{s^{2} (s^{2} + 2 s + 5)} e^{s t} d s = \sum Res {F (s); z_{j}}

Then the sum of residues are

\begin{matrix} lim_{s \to 0} \frac{d}{d s} s^{2} \frac{3 s^{3} - 3 s^{2} + 3 s - 5}{s^{2} (s^{2} + 2 s + 5)} e^{s t} + lim_{s \to - 1 + 2 i} (s + 1 - 2 i) \frac{3 s^{3} - 3 s^{2} + 3 s - 5}{s^{2} (s^{2} + 2 s + 5)} e^{s t} + lim_{s \to - 1 + 2 i} (s + 1 - 2 i) \frac{3 s^{3} - 3 s^{2} + 3 s - 5}{s^{2} (s^{2} + 2 s + 5)} e^{s t} d s \\ = 1 - t - \frac{1}{2} e^{(- 1 + 2 i) t} - \frac{1}{2} e^{- (1 + 2 i) t} \end{matrix}

which can be reduced however you see fit.

Try to calculate L^(-1)((3s^3−3s^2+3s−5)/(s^2(s^2+2s+5)))

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