MHB How can I solve this LaPlace Transform using Laplace Transforms?

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To solve the Laplace Transform problem $$\mathcal{L}^{-1} \left( \frac{\frac{5s}{4} + \frac{13}{4}}{s^2+5s+8} \right),$$ the expression can be rewritten to facilitate the application of inverse transforms. It simplifies to $$\frac{1}{4} \left[ \frac{5s + 13}{\left( s + \frac{5}{2} \right)^2 + \frac{7}{4}} \right],$$ allowing for the use of the shift theorem. The numerator can be adjusted to match the standard forms of sine and cosine transforms. This approach leads to the final solution through the application of inverse Laplace Transform techniques.
shamieh
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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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nvm, i was doing something wrong
 
shamieh said:
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

$\displaystyle \begin{align*} \frac{\frac{5s}{4} + \frac{13}{4}}{s^2 + 5s + 8} &= \frac{1}{4} \left( \frac{5s + 13}{s^2 + 5s + 8} \right) \\ &= \frac{1}{4} \left[ \frac{5s + 13}{s^2 + 5s + \left( \frac{5}{2} \right) ^2 - \left( \frac{5}{2} \right) ^2 + 8} \right] \\ &= \frac{1}{4} \left[ \frac{5s + 13}{ \left( s + \frac{5}{2} \right) ^2 + \frac{7}{4} } \right] \\ &= \frac{1}{4} \left[ \frac{5 \left( s + \frac{5}{2} \right) + \frac{1}{2}}{\left( s + \frac{5}{2} \right) ^2 + \left( \frac{\sqrt{7}}{2} \right) ^2} \right] \end{align*}$

This is now in a form where you can apply a shift, and then take the transform.
 

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