How to solve this Initial boundary value PDE problem? [SOLVED] Today I came across a question on PDE which makes me really frustrating.
The question is to solve this initial boundary value problem using method of separation variables:
$$u_{tt}=9u_{xx}\text{ for } x>0, t>0$$
$$u(x,0)=x^2 ,\ u_t(x,0)=e^{-x} \text{ for } x>0$$
$$u(0,t)=0 \text{ for }t>0$$
Given answer: $u(x,t)=x^2+9t^2+\dfrac{1}{3}e^{-x}\sinh3t$ for $x-3t>0$
$$u(x,t)=\begin{cases}x^2+9t^2+\dfrac{1}{3}e^{-x}\sinh3t&x-3t>0\\x^2+9t^2+\dfrac{1}{3}e^{-3t}\sinh x&x-3t<0\end{cases}$$
I really dont know how to solve this question because they only provide 1 boundary condition only. 
Could someone please show some workings on this problem so that I can undesrtand clearly. Furthermore, I dont have example of this question in my textbook. And my teacher also not able to do it because he said he is not good in PDE chapter. Im really dead now because no one able to help me and my friends also not sure about this. So please please help me to solve this only question so that I can use this solution as reference to solve my other PDEs problems by my ownself.
Lastly, if you can help me, I will really be very grateful to you.
 A: Note that when without the condition $u(0,t)=0$ this is in fact a just-determining problem and the solution can be expressed by using D’Alembert’s formula $u(x,t)=\dfrac{(x+3t)^2+(x-3t)^2}{2}+\dfrac{1}{6}\int_{x-3t}^{x+3t}e^{-s}~ds=\dfrac{x^2+6xt+9t^2+x^2-6xt+9t^2}{2}-\dfrac{1}{6}[e^{-s}]_{x-3t}^{x+3t}=\dfrac{2x^2+18t^2}{2}-\dfrac{e^{-x-3t}-e^{-x+3t}}{6}=x^2+9t^2+\dfrac{e^{-x}\sinh3t}{3}$
Check for $u(0,t)$ :
$u(0,t)=9t^2+\dfrac{\sinh3t}{3}\neq0$
$\therefore$ We should need to use the result in http://eqworld.ipmnet.ru/en/solutions/lpde/lpde201.pdf:
The solution is $u(x,t)=\begin{cases}\dfrac{(x+3t)^2+(x-3t)^2}{2}+\dfrac{1}{6}\int_{x-3t}^{x+3t}e^{-s}~ds&\text{when}~x-3t>0\\\dfrac{(x+3t)^2-(3t-x)^2}{2}+\dfrac{1}{6}\int_{3t-x}^{x+3t}e^{-s}~ds&\text{when}~x-3t<0\end{cases}=\begin{cases}\dfrac{x^2+6xt+9t^2+x^2-6xt+9t^2}{2}-\dfrac{1}{6}[e^{-s}]_{x-3t}^{x+3t}&\text{when}~x-3t>0\\\dfrac{x^2+6xt+9t^2-(9t^2-6xt+x^2)}{2}-\dfrac{1}{6}[e^{-s}]_{3t-x}^{x+3t}&\text{when}~x-3t<0\end{cases}=\begin{cases}\dfrac{2x^2+18t^2}{2}-\dfrac{e^{-x-3t}-e^{-x+3t}}{6}&\text{when}~x-3t>0\\\dfrac{12xt}{2}-\dfrac{e^{-x-3t}-e^{-3t+x}}{6}&\text{when}~x-3t<0\end{cases}=\begin{cases}x^2+9t^2+\dfrac{e^{-x}\sinh3t}{3}&\text{when}~x-3t>0\\6xt+\dfrac{e^{-3t}\sinh x}{3}&\text{when}~x-3t<0\end{cases}$
