What is my error in applying this Laplace Transform? So, our book has the seemingly innocuous problem: $y''-y'-6y=0$.  I was able to solve by hand, and come up with $${\scr L}(y)=\frac{s-2}{s^{2}-s+6}$$.That completed, I factored the bottom to $(s-3)(s+2)$.  Our professor gave us a sheet that said ${\scr L}\mathrm{{}f(t)\mathrm{}}=F(s)$ which corresponds along the rows to a column $f(t)$.  I broke the equation down to $$\frac{s}{(s-3)(s+2)}-\frac{2}{(s-3)(s+2)}$$ and the corresponding $f(t)$ were $$\frac{ae^{at}-be^{bt}}{a-b}$$ and $$\frac{e^{at}-e^{bt}}{a-b}$$, respectively.  The summation of these leads to:
$$\frac{1}{5}(-e^{2t}+5e^{-3t})$$.  However, the answer in the book is: $$\frac{1}{5}(e^{3t}+4e^{-2t})$$.
I have literally no idea why my method is working.  Would someone be so kind as to enlighten me?
 A: If the problem given was to solve $y''(t)-y'(t)-6y(t)=0$ subject to initial conditions $y(0)=1$ and $y'(0)=-1$, using the Laplace transform, then we write $Y(s) =\mathscr{L}\{y(t)\}(s)$ and
$$\begin{align}
\mathscr{L}\{y''(t)-y'(t)-6y(t)\}(s)&=\left(s^2Y(s)-sy(0)-y'(0)\right)\\
&-\left(sY(s)-y(0)\right)\\
&-6Y(s)\\\\
&=(s^2-s-6)Y(s)-(s-2)\\
&=0
\end{align}$$
implies that 
$$Y(s)=\frac{s-2}{s^2-s-6}=\frac15\left(\frac{1}{s-3}+\frac{4}{s+2}\right)$$
whereupon inverting reveals that $y$ is 
$$y(t)=\frac15\left(e^{3t}+4e^{-2t}\right)$$
A: For the differential equation $y'' - y' -6y = 0$ the Laplace transform is
\begin{align}
( s^2 y(s) - y'(0) - s y(0) ) - (s y(s) - y(0)) - 6 y(s) = 0
\end{align}
which leads to $(s^2 - s - 6) y(s) = y'(0) + y(0) (s-1)$ and
\begin{align}
y(s) &= \frac{y'(0) - y(0)}{s^2 - s -6} + \frac{y(0) \, s}{s^2 - s - 6}  \\
&= \frac{y'(0) - y(0)}{5} \left( \frac{1}{s-3} - \frac{1}{s+2} \right) + \frac{y(0)}{5} \left( \frac{2}{s+2} + \frac{3}{s-3} \right) \\
&= \frac{y'(0) + 2 y(0)}{5} \frac{1}{s-3} + \frac{3 y(0) - y'(0)}{5} \frac{1}{s+2}.
\end{align}
Inversion leads to
\begin{align}
y(t) = \frac{y'(0) + 2 y(0)}{5} \, e^{3t} + \frac{3y(0) - y'(0)}{5} \, e^{-2t}.
\end{align}
