Method for taking this inverse laplace transform

Question

i am having trouble taking the answer to this problem that i found a book called "Differential equations with applications and historical notes":

$$ e^{-x} = y(x) + 2 \int_0^x\cos(x-t)y(t) dt $$

First i took the Laplace transform:

$$ L[e^{-x}] = L[y] + 2 L[\cos(x)] L[y] $$

$$ \frac{1}{p+1} = L[y](1+\frac{2p}{p^2+1}) $$

$$ \frac{p^2+1}{(p+1)^3} = L[y] $$

Then i need to find the Inverse Laplace Transform, so i did partial fraction decomposition:

$$ \frac{p^2+1}{(p+1)^3} = \frac{a}{p+1} + \frac{b}{(p+1)^2}+\frac{c}{(p+1)^3} $$

$$ p^2+1 = a(p+1)^2 + b(p+1) + c $$

$$ p^2 + 1 = ap^2 + p(2a+b) + (a+b+c) $$

it is clear that $a=1$, $b=-2$, $c=2$ therefore:

$$ L[y] = \frac{1}{p+1} - \frac{2}{(p+1)^2}+\frac{2}{(p+1)^3} $$

But here i get stuck,

$$ y = e^{-x} + L^{-1}[\frac{2}{(p+1)^3}] - 2 L^{-1}[\frac{1}{(p+1)^2}] $$

Any advice or methods that may help me would be greatly appreciated.

Answer 1

You need the following property of the Laplace transform

$$ \mathcal{L}(x^nf(x)) = (-1)^{n} F^{n}(s). $$

You can see that

$$ \mathcal{L}(x^2 e^{-x}) = \frac{2}{(1+s)^3}. $$

You can find the other one with the same technique.