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
$$\mathcal{L} \{ y'' \} – 4 \mathcal{L} \{ y \} = 8 \mathcal{L} \{ t \} – 20 \mathcal{L} \{ e^{-2t} \}$$
$$s^2 y(s) – s\ y(0) – y'(0) – 4y(s) = \frac{8}{s^2} – \frac{20}{s+2}$$
$$s^2 y(s) – 19 – 4y(s) = \frac{8}{s^2} – \frac{20}{s+2}$$
$$y(s) (s^2-4) = \frac{8}{s^2} – \frac{20}{s+2} + 19$$
$$y(s) = \frac{8}{s^2(s^2-4)} – \frac{20}{(s+2)(s^2-4)} + \frac{19}{s^2-4}$$
Best Answer
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 $A = -5/4$.