Solve differential equation using Laplace transform with weird condition

Question

$$y'' + 2y' + y = 0$$

Where $y(0) = 0$ and $y(1)=2$.

I first try to apply Laplace transform in both sides which I get:

$$s^2 Y(s) - sy(0) - y'(0) + 2(sY(s)-y(0)) + Y(s) = 0$$

After this I substitute the $y(0)=0$ so I get:

$$Y(s) = y'(0)/(s^2+2s+1)$$

But now I am stuck. What should I do now?

Thank you in advance. And sorry I do not know how to write in LaTex.

Answer 1

Note that by linearity, $$y(t)=\mathcal{L}^{-1}\left(\frac{y'(0)}{s^2+2s+1}\right)=y'(0)\,\mathcal{L}^{-1}\left(\frac{1}{(s+1)^2}\right)=y'(0)te^{-t}.$$ Now find the constant $y'(0)$ by imposing $y(1)=2$.