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?

Let $f(t)=y(t+2)$. That way $f(0)=y(2)$ and $f'(0)=y'(2)$. Equivalently, $f(t-2)=y(t)$, $y'(t)=f'(t-2)$ and $y''(t)=f''(t-2)$. Then
$$
        f''(t-2)-3f'(t-2)+2f(t-2)=e^{-t},\;\;\; f(0)=1, f'(0)=0.
$$
Substitute $t$ with $t+2$:
$$
        f''(t)-3f'(t)+2f(t) = e^{-t}e^{-2}, \;\;\; f(0)=1, f'(0)=0.
$$
