Use the Laplace transform to solve the initial value problem. $$
y''-3y'+2y=e^{-t}; \quad\text{where}~ ~ y(2)=1, y'(2)=0
$$
Hint given: consider a translation of $y(x)$.
I am stuck on this problem on our homework. I don't understand what they mean by a "translation". Do they just mean a substitution?
Update - after spending a bit more time on this is where I'm stuck:
$$
L(y)=\frac{\delta(s-1)}{s^2-3s+2}
$$
Where do I go from here?

I cant understand where you get the $\delta$-function, too. According to my work:
$$y''(t+2)-3y'(t+2)+2y(t+2)=e^{-(t+2)}\qquad y(0)=1\,,\,y'(0)=0$$
$$s^2\mathcal{L}[y(t+2)]-sy(0)-y'(0)-3(s\mathcal{L}[y(t+2)]-y(0))+2\mathcal{L}[y(t+2))]=e^{2s}\mathcal{L}[e^{-t}]$$
$$s^2(e^{2s}\mathcal{L}[y])-s-3(se^{2s}\mathcal{L}[y]-1)+2e^{2s}\mathcal{L}[y]=\frac{e^{2s}}{s+1}$$
$$e^{2s}(s^2-3s+2)\mathcal{L}[y]=\frac{e^{2s}}{s+1}+s-3$$
$$\mathcal{L}[y]=e^{-2s}(\frac{1}{(s-1)(s-2)})(\frac{e^{2s}}{s+1}+s-3)$$
$$\mathcal{L}[y]=e^{-2s}(\frac{1}{s-2}-\frac{1}{s-1})(\frac{e^{2s}}{s+1}+s-3)$$
The laplace inverse can be computed with convolution theorem.
$$\mathcal{L}^{-1}[e^{-2s}(\frac{1}{s-2}-\frac{1}{s-1})]=e^{2(t-2)}-e^{(t-2)}$$
$$\mathcal{L}^{-1}[\frac{e^{2s}}{s+1}]=e^{-(t+2)}$$
I just don't know how to find the laplace inverse of $s$ and the calculate $s-3$ with the transform theorem. any help?
