0
$\begingroup$

I am having a hard time finding the inverse Laplace transform of $$\frac{1}{(s^2+1)^2} - \frac{1}{s^2(s^2+1)^2}$$ and would appreciate some guidance. I have tried breaking it down to partial fractions but got stuck.

$\endgroup$
0
$\begingroup$

First get rid of the ${ 1\over s^2}$ part as in: $\frac{1}{(s^2+1)^2} - \frac{1}{s^2(s^2+1)^2} = \frac{1}{{s}^{2}+1}+\frac{2}{{( {s}^{2}+1) }^{2}}-\frac{1}{{s}^{2}}$.

Then using $s^2+1 = (s-i)(s+i)$ we write $\frac{1}{{s}^{2}+1}+\frac{2}{{( {s}^{2}+1) }^{2}} = \frac{a}{s+i}+\frac{b}{s-i}+\frac{c}{{( s+i) }^{2}}+\frac{d}{{( s-i) }^{2}}$, and solving gives $a=i, b=-i, c=d = -{1 \over 2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.