How to compute the inverse laplace transform of this term? ${-{{3}\over{10}}s-{{1}\over{5}}\over{(s+1)^2+1}}$

Question

So, I have been asked to solve $y'-2y=e^{-t} *cos(t)$ where $y(0)=-2.$

I applied the Laplace transform, getting $$\mathcal{L}(y)={{-2s^2-3s-3}\over{(s-2)((s+1)^2+1)}}$$

I set up a partial fraction decomposition of the form $${{-2s^2-3s-3}\over{(s-2)((s+1)^2+1)}}={{A}\over{(s-2)}}+{{Bs+C}\over{(s+1)^2+1}}$$

I solved for values of the constants (and checked them using a solver to avoid mistakes) and I ended up with

$$\mathcal{L}(y)={-{{17}\over{10}}\over{(s-2)}}+{-{{3}\over{10}}s-{{1}\over{5}}\over{(s+1)^2+1}}$$

Obviously, the inverse laplace transform of the first term is $-{{17}\over{10}}e^{2t}$, however, I get stuck trying to find the inverse laplace transform of the second term.

It resembles the form ${{s-a}\over{(s-a)^2+b^2}}=e^{at}cos(bt)$, but I cannot figure out how to wrestle this function fully into this form.

What can I do to get this term into a form that I can handle?

Answer 1

$\mathcal{L}(y)={-{{3}\over{10}}s-{{1}\over{5}}\over{(s+1)^2+1}}$

you want to get this into the form.${a(s+1) + b}\over{(s+1)^2+1}$

At which point:

$y = ae^{-t}\cos t + b e^{-t} \sin t$

${-\frac 3{10}(s+1) + \frac 1{10}}\over{(s+1)^2+1}$

Answer 2

Perhaps write $(s+1)^2+1 = s^2+2s+2 =(s-a)(s-b) $ and doing partial fraction decomposition.

$a$ and $b$ are complex, of course.

Answer 3

Hint. One may write $$ {-{{3}\over{10}}s-{{1}\over{5}}\over{(s+1)^2+1}}=-\frac3{10}{(s+1)\over{(s+1)^2+1}}+\frac1{10}{1\over{(s+1)^2+1}} $$ then one may apply the properties of the inverse Laplace transform to each term (have a look here).