I'm having some trouble with this problem:
Solve the integral equation with Laplace transform
$$e^{-t}=y(t)+\int_0^t(t-u)y(u)du$$
I know how to use the Laplace transform for more "normal" equations but I don't understand this step here below in my solution manual. So they use the transform and get this
$$\frac{1}{s+1}=Y(s)+\frac{1}{s^2}Y(s)$$
can anyone derive or explain that last term for me? I have no idea how to get from that integral to that.
We have: $$L\left[\int_0^t(t-u)y(u)du\right]=Y(s)\cdot L[t]=Y(s)\cdot\frac{1}{s^2},$$ where in the first equality we used the convolution identity:
$$L\left[\int_0^t f(u)g(t-u)du\right]=F(s)\cdot G(s)$$
To clarify how we get the $\frac{1}{s^2}$: For $n\in\mathbb{N}$, we have:
$$L[t^n]=\frac{n!}{s^{n+1}}$$