# Inverse Laplace Transform of Discontinuous Function

I'm currently studying transform of discontinuous and periodic functions (Differential Equations.) I was presented with the following question.

$$\dfrac{se^{-3s}}{s^2+4s+5}$$

(Sorry, I couldn't get the formatting to work properly. Feel free to fix it. )

I've identified $F(s)$ as:

$$\dfrac{s}{s^2+4s+5}$$

but I'm a little stuck on how to find the inverse Laplace transform of this. Do I complete the square? I'd appreciate some advice here. Thanks.

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Thanks @Nasusama – codedude Jul 18 '13 at 23:51

## 2 Answers

Completing the square is a great way to go. We could say \begin{align} \frac{s}{s^2+4s+5} &= \frac{s}{(s+2)^2+1} \\ &= \frac{(s+2)-2}{(s+2)^2+1} \\ &= \frac{(s+2)}{(s+2)^2+1} - \frac{2}{(s+2)^2+1} \end{align} Where could you go from there?

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Wow. How did I miss that? Well it's certainly been a long day. Thanks! – codedude Jul 18 '13 at 23:54
Happens to the best of us :) glad to help – Omnomnomnom Jul 19 '13 at 0:35

Ha dude I happened to learn this week in my ODE class! You complete the square first $\frac{s}{s^2+4s+5}=\frac{s}{(s+2)^2+1}$, then split this thing into cosine and sine's Laplace transforms: $\frac{s}{(s+2)^2+1} = \frac{s+2}{(s+2)^2+1} - \frac{2}{(s+2)^2+1}$.

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haha nice! Appreciate your time. :) – codedude Jul 18 '13 at 23:56