I have this problem:
Solve the integral equation
$$e^{-t}=y(t) +2 \int_0^t \cos(t-u)y(u) \, du$$
So I'm thinking that Laplace transform will be the way to go here, so I get
$$\frac{1}{s+1}=Y(s)+2\frac{s}{s^2+1}Y(s)= \left(1+\frac{2s}{s^2+1}\right)Y(s)$$
Is this right so far? After this it's just algebraic manipulation and then use inverse transformation back to get $y(t)$. I have no answer or solution to go on here so I just had to ask, is the any way I can check that my answer is correct when I'm done?
Here's my answer if you want to check
$$ \frac{1}{s+1} = \frac{s^2+2s+1}{s^2+1}Y(s) = \frac{(s+1)^2}{s^2+1}Y(s) $$ $$ Y(s) = \frac{s^2+1}{(s+1)^3} = \frac{(s+1-1)^2 + 1}{(s+1)^3} \\ = \frac{(s+1)^2-2(s+1)+2}{(s+1)^3} \\ = \frac{1}{s+1} - \frac{2}{(s+1)^2} + \frac{2}{(s+1)^3} $$
Taking the inverse transform $$ y(t) = e^{-t} - 2te^{-t} + t^2e^{-t} $$