1
$\begingroup$

How would you take the inverse laplace transform of $\frac{3p}{(p^2 + 4)^2} $

I attempted to split it as $\frac{3p}{p^2 + 4} $ x $\frac{1}{p^2 + 4} $

Then i know $\frac{3p}{p^2 + 4} $ has laplace tranform $ 3cos(2t) $ and $\frac{1}{p^2 + 4} $ has laplace transform $\frac{1}{2}sin(2t)$

So via convolution theorem the function i want is:

$$\frac{3}{2} \int_{0}^{t} sin(2s)cos(2(t-s)) \, ds$$

Which im not too sure how to integrate?

Or is there a simpler way?

$\endgroup$
2
  • 1
    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$
    – zoli
    Mar 5, 2017 at 18:25
  • $\begingroup$ Integrals like this can be made simpler by use of the cosine and sine addition/subtraction formulas $\endgroup$
    – Triatticus
    Mar 5, 2017 at 18:34

1 Answer 1

2
$\begingroup$

"Or is there a simpler way?"

HINT:

$$\frac{d}{dp}\left( \frac{-3/2}{p^2+4}\right)=\frac{3p}{(p^2+4)^2}$$

$\endgroup$
1
  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$
    – Mark Viola
    Mar 6, 2017 at 5:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .