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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
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2 Answers

up vote 1 down vote accepted

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
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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
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