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?

Your equation $$\dfrac{d^2 v}{dx^2} - s^2 v = - s \sin(\pi x)$$
is a second-order linear differential equation.
Like any linear differential equation, its general solution is of the form (particular solution) + (general solution of the homogeneous equation). Your "complementary function" is the general solution of the homogeneous equation 
$$ \dfrac{d^2 v}{dx^2} - s^2 v = 0$$
One way to find a particular solution (the method of "undetermined coefficients") is to make an educated guess at the form of a solution and then find what coefficients are needed to make this work.  In this case
the right side involves $\sin(\pi x)$, so we try a linear combination of $\sin(\pi x)$ and $\cos(\pi x)$ (the latter turns out not to be needed).
Once you find your general solution, the next step will be to look at it at $x=0$ and $x=1$ and see what $A$ and $B$ you need to make the boundary conditions true.
