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Solve by Laplace Transforms.

So I'm stuck on how to find this $\mathcal{L}^{-1}$ $( \frac{\frac{5s}{4} + \frac{13}{4}}{s^2+5s+8} ) $

I'm not sure what t odo. I was thinking I need to use the $\cos(at)$ and $\sin(at)$ formulas but i'm not sure... Any help would be great

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  • $\begingroup$ Maybe completing the square on the denominator will help? $\endgroup$
    – GFauxPas
    Apr 6, 2015 at 21:05
  • $\begingroup$ Nope, didn't help at all... Any other advice? $\endgroup$
    – Yusha
    Apr 6, 2015 at 21:12
  • $\begingroup$ See that's what I had initially thought to do. $\endgroup$
    – Yusha
    Apr 6, 2015 at 21:12

1 Answer 1

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$$\mathcal{L}^{-1} \left\{ \frac{\frac{5s}{4} + \frac{13}{4}}{s^2+5s+8} \right\}$$

$$ = \mathcal{L}^{-1} \left\{{\frac {\frac{5s}4 + \frac 5 4\frac 5 2 - \frac 5 4\frac 5 2 + \frac {13}4}{\left({s + \frac 5 2}\right)^2 + \frac 7 4}}\right\}$$

You'll need:

$$\mathcal{L}^{-1}\left\{{F(s + \alpha)}\right\} = e^{-\alpha t}\mathcal L^{-1} \left\{ {F(s)}\right\}$$

where $\alpha$ is constant.

Let me know if you need more help.

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