Solve Dirac Delta Function IVP: y''-2y'-3y=2\delta (t-1)-\delta (t-3)

[Precursor]

Homework Statement
Solve the given symbolic initial value problem: $$y''-2y'-3y=2\delta (t-1)-\delta (t-3) ;y(0)=2,y'(0)=2$$


The attempt at a solution

Let Y(s):= L{y(t)}(s)

Taking laplace transform of both sides:


$$[s^{2}Y(s)-2s-2]-2[sY(s)-2]-3Y(s)=2e^{-s}-e^{-3s}$$
$$s^{2}Y(s)-2sY(s)-3Y(s)=2e^{-s}-e^{-3s}+2s-2$$
$$Y(s)=\frac{2e^{-s}-e^{-3s}+2s-2}{s^{2}-2s-3}$$
$$Y(s)=\frac{2e^{-s}}{s^{2}-2s-3}- \frac{e^{-3s}}{s^{2}-2s-3}+\frac{2s-2}{s^{2}-2s-3}$$
$$y(t)=e^{-(t-1)}\frac{1}{2}e^{-(t-1)}(e^{4(t-1)}-1)u(t-1)-e^{-(t-3)}e^{-(t-3)}(e^{4(t-3)}-1)u(t-3)+e^{-t}+e^{3t}$$

And my final answer:


$$y(t)=e^{1-t}(e^{4t-4}-1)u(t-1)-2e^{3-t}(e^{4t-12}-1)u(t-3)+e^{-t}+e^{3t}$$

Is this correct?

 

[hunt_mat]

It looks to be right sort of thing to be doing.