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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 $$\mathcal{L} \{ y'' \} - 4 \mathcal{L} \{ y \} = 8 \mathcal{L} \{ t \} - 20 \mathcal{L} \{ e^{-2t} \}$$ $$s^2 y(s) - s\ y(0) - y'(0) - 4y(s) = \frac{8}{s^2} - \frac{20}{s+2}$$ $$s^2 y(s) - 19 - 4y(s) = \frac{8}{s^2} - \frac{20}{s+2}$$ $$y(s) (s^2-4) = \frac{8}{s^2} - \frac{20}{s+2} + 19$$ $$y(s) = \frac{8}{s^2(s^2-4)} - \frac{20}{(s+2)(s^2-4)} + \frac{19}{s^2-4}$$

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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 $A = -5/4$.

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