Solving $y'' - 2y' + y = \delta(t-2)$ for y(0) = 0, y'(0) = 0 by using Laplace Transforms. Need help finishing the problem.

Question

So I'm working on this question and I've taken the Laplcace transforms for everything and separated it as $Y(s) = F(s)G(s)$, and then I've put it into the convolution equation to get $y(t)$. My problem is, I don't understand how to fully/properly integrate this when I have the $\delta(T-2)$ inside the integral. Can anyone explain what to do? I'm adding a picture of what I have until now that I am sure is correct so that there's somewhat of an explanation of my train of thought? (the u and v' at the bottom were if I was going to solve with integration by parts, which I also am not sure of how to use here). But yeah, please help and tell me where I'm going wrong...

Answer 1

$$y'' - 2y' + y = \delta(t-2)$$ $$\implies Y(s)=\dfrac {e^{-2s}}{(s-1)^2}$$ The inverse Laplace transform of $\dfrac 1 {(s-1)^2}$ is simply $e^tt$ then apply the formula 27: $$L^{-1} (e^{-cs}F(s))=u(t-c) f(t-c)$$ You should get : $$y(t)=u(t-2)e^{t-2}(t-2)$$

Answer 2

There is nothing wrong with what you did. \begin{align*} y(t) & = \int_0^t \delta(T-2) (t-T)e^{t-T} \ dT \\ & = \begin{cases} 0 & \text{if } t < 2 \\ (t-2)e^{t-2} & \text{if } t \ge 2, \end{cases} \end{align*} where we made use of the sifting property of the delta function $$ \int_a^b \delta(x-x_0) f(x) \ dx = \begin{cases} f(x_0) & \text{if } a < x_0 < b \\ 0 & \text{otherwise}. \end{cases}$$