How to calculate inverse Laplace transform of $\frac{7{s}^{2}+3s+5}{({s}^{2}-4s+29)({s}^{2}+25)}$?

demitereur
2022-09-26
Answered

How to calculate inverse Laplace transform of $\frac{7{s}^{2}+3s+5}{({s}^{2}-4s+29)({s}^{2}+25)}$?

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Marley Stone

Answered 2022-09-27
Author has **13** answers

Note that

$$\frac{As+B}{{s}^{2}+{5}^{2}}=\frac{A/2-iB/10}{s-5i}+\frac{A/2+iB/10}{s+5i}$$

has inverse Laplace transform

$$(A/2-iB/10){e}^{5it}+(A/2+iB/10){e}^{-5it}=A\mathrm{cos}5t+\frac{B}{5}\mathrm{sin}5t.$$

Since ${s}^{2}-4s+29=(s-2{)}^{2}+{5}^{2}$, the other partial fraction admits a similar analysis.

$$\frac{As+B}{{s}^{2}+{5}^{2}}=\frac{A/2-iB/10}{s-5i}+\frac{A/2+iB/10}{s+5i}$$

has inverse Laplace transform

$$(A/2-iB/10){e}^{5it}+(A/2+iB/10){e}^{-5it}=A\mathrm{cos}5t+\frac{B}{5}\mathrm{sin}5t.$$

Since ${s}^{2}-4s+29=(s-2{)}^{2}+{5}^{2}$, the other partial fraction admits a similar analysis.

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$$f(t)=6t{e}^{-9t}\mathrm{sin}(6t)$$

I'm not sure what the protocol is for multiplying t into it.

The Laplace Transform for $f(t)=6{e}^{-9t}\mathrm{sin}(6t)$ is $\frac{6}{(s+9{)}^{2}-36}$

Can't figure out how to add in the t.

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