Solve $y'' + 2y' + y = 4e^t - 4e^tu(t-1)$, where $u(t-1)$ is the unit step function at $1$, $y(0)=0$ and $y'(0)=0$

Question

I need to solve $$y'' + 2y' + y = 4e^t - 4e^tu(t-1)$$where $u(t-1)$ is the unit step function at $1$, $y(0)=0$ and $y'(0)=0$

I used the Laplace transform and got

$$Y(s) = \frac{4}{(s-1)(s+1)^{2}} - \frac{4e^{1-s}}{(s-1)(s+1)^{2}}$$

then by the Inverse Laplace Transform,

$$y(t) = e^{t} - e^{-t} -2te^{-t} + (-e^{t} + e^{2-t} +2e^{2-t}(t-1))u(t-1)$$

is the answer correct?