(Laplace Method) $y'' - 4y' = 6e^{3t} - 3e^{-t}$

Question

For this problem $y(0) = 1$ and $y'(0) = -1$

I need to solve this problem using this:

\begin{align*} y(t) &\longrightarrow Y(s)\\ y'(t) &\longrightarrow sY(s) - y(0)\\ y''(t) &\longrightarrow s^2Y(s) - sy(0) - y'(0) \end{align*}

And partial fractions.

I do all the procedure and I get $$Y(s) = \frac{s^3 + s^2 - 6s - 6 }{s(s-4)(s -3)(s+1)}$$ which converts to $$\frac As + \frac B{s-4} + \frac C{s-3} + \frac D{s+1}$$

Then I get $$s^3 + s^2 - 6s - 6 = A + B + C + D (s^3) - 6A -2B -3C -7D (s^2) + 5A - 3B - 4C + 12D (s) + 12A$$

However the resulting system of equations will NOT give me the correct answer.

The correct answers is $$y(t) = \frac{11}{10} e^{4t} - 2e^{3t} -\frac 35e^{-t} + \frac 52$$

From $Y(s)$ to $y(t)$ is the Inverse Laplace, so looking at what I have seems like I would get right the inverse but not the coefficients of $A$, $B$, $C$ and $D$.

Help.

Answer 1

Taking the Laplace Transform to both sides of the ODE we get \begin{align*} s^2Y(s)-s(1)-(-1)-4[sY(s)-1]&=\frac 6{s-3}-\frac 3{s+1}\\[4pt] (s^2-4s)Y(s)&=\frac 6{s-3}-\frac 3{s+1}+s-5\\[4pt] Y(s)&=\frac{6(s+1)-3(s-3)+(s-5)(s-3)(s+1)}{s(s-4)(s-3)(s+1)}\\[4pt] Y(s)&=\frac{s^3-7s^2+10s+30}{s(s+1)(s-3)(s-4)} \end{align*}