# Solve the initial value problem below using the method of Laplace transform

I am able to get up to a certian point this problem but get stuck when doing the Laplace transforms. The problem is

Solve the initial value problem below using the method of Laplace transform : $$y''-4y=8t-20e^{-2t}$$ with $$y(0)=0$$ and $$y'(0)=19$$.

My try :

Taking Laplace to both sides we get $$\mathcal{L} \{ y'' \} - 4 \mathcal{L} \{ y \} = 8 \mathcal{L} \{ t \} - 20 \mathcal{L} \{ e^{-2t} \}$$ $$s^2 y(s) - s\ y(0) - y'(0) - 4y(s) = \frac{8}{s^2} - \frac{20}{s+2}$$ $$s^2 y(s) - 19 - 4y(s) = \frac{8}{s^2} - \frac{20}{s+2}$$ $$y(s) (s^2-4) = \frac{8}{s^2} - \frac{20}{s+2} + 19$$ $$y(s) = \frac{8}{s^2(s^2-4)} - \frac{20}{(s+2)(s^2-4)} + \frac{19}{s^2-4}$$

Since you already have the expression for $$y(s)$$, you can further simplify each term using partial fractions.
First, we have $$\frac{8}{s^2 (s^2 - 4)} = \frac{A}{s^2} + \frac{B}{s^2 - 4}$$, which gives us $$8 = A (s^2 - 4) + Bs^2$$. Thus, $$A = -2$$ and $$B = 2$$.
Then we have $$\frac{20}{(s + 2)(s^2 - 4)} = \frac{A}{s + 2} + \frac{B}{(s + 2)^2} + \frac{C}{s - 2}$$, which gives us $$20 = A (s - 2)(s + 2) + B (s - 2) + C (s + 2)^2$$. Set $$s = -2$$, and we get $$B(-4) = 20$$, or $$B = -5$$. Then set $$s = 2$$ to get $$20 = 16C$$, which gives us $$C = 5/4$$. Arbitrarily set $$s = 1$$, and we find $$20 = -3A + 5 + 45/4$$, which gives us $$A = -5/4$$.