Laplace transform of the wave equation I started of with the wave equation $$\frac{\partial^2 u}{\partial x^2}=\frac{\partial^2 u}{\partial t^2}$$
with boundary conditions $u=0$ at $x=0$ and $x=1$ and initial condition $u=sin(\pi x)$ and $\frac{\partial u}{\partial t}=0$ when $t=0$
I have gotten to the stage $$\frac{d^2 v}{dx^2}-s^2v=-s* sin(\pi x)$$
Where do I go from here?
My notes say find the complimentary function using the axillary function $m^2-s^2$
to get $$v_c=Ae^{sx}+Be^{-sx}$$ Then it says find the particular integral for the right hand side and use the trial function $$v_p=a*sin(\pi x)+b*cos(\pi x)$$
But I do not quite understand this. 
Why do we need the complimentary function? 
What is the trial function and why do we need it?

The language seems to be geared toward the annihilator method.
$$
                (D^{2}-s^{2})v = -s \sin(\pi x) \\
                (D^{2}+\pi^{2})(D-s)(D+s)v = 0.
$$
The general solution of the second equation is, for $s > 0$,
$$
                   v(x) = Ae^{-sx}+Be^{sx}+E\sin(\pi x)+F\cos(\pi x).
$$
Applying $(D-s^{2})$ to the right annihilates $Ae^{-sx}+Be^{sx}$, which leaves
\begin{align}
       -s\sin(\pi x) & = (D^{2}-s^{2})(E\sin(\pi x)+F\cos(\pi x)) \\
           & = (-\pi^{2}-s^{2})\{E\sin(\pi x)+F\cos(\pi x)\}
\end{align}
Hence, $F=0$ and $E=s/(\pi^{2}+s^{2})$. Therefore,
$$
                 v(x) = Ae^{-sx}+Be^{sx}+\frac{s}{\pi^{2}+s^{2}}\sin(\pi x).
$$
And $v(0)=0=v(1)$ implies
$$
                      A+B = 0\\
                      Ae^{-s}+Be^{s} = 0
$$
Unless the equations are dependent, $A=B=0$. They are dependent iff $s=0$. For $s > 0$,
$$
                    v(x) = \frac{s}{\pi^{2}+s^{2}}\sin(\pi x).
$$
