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?

Firstly I can't see where you get the $\delta$-function from, please consider showing your working when you post a question here.
You want to use the following properties of the Laplace transform:
$$\mathcal{L}[f'(t)]=sF(s)-f(0)$$
$$\mathcal{L}[f''(t)]=s^2F(s)-sf(0)-f'(0)$$
where $F(s)$ denotes the Laplace transform of $f(t)$. Unfortunately, you only have initial conditions for $t=2$. The idea is to translate $y$, i.e. define $w(t)=y(t+2)$. Can you see how this would be helpful?
