Solving PDE with Separation of Variables I have done a few of these now but I'm stuck at this one
$$\begin{array}{}
u_{xx}=u_{tt}+2u_t \\
u(0,t)= u(\pi,t)= 0 \\
u(x,0)=0, \, u_t(x,0)=\sin ^3x \\
\end{array}
$$
for $0<x< \pi$ and $t>0$.
So I rewrite $u(x,t)=X(x)T(t)$, plug it in and separate the variables. That gives me
$$
\frac{X''(x)}{X}=\frac{T''(t)}{T(t)}+2\frac{T'(t)}{T(t)}=- \lambda
$$ 
which leads me to these two ODEs
$$
T''(t) +2 T'(t)+ \lambda T(t) = 0 \\
X''(x)+\lambda X(x) = 0
$$
These I can solve but I am a little bit confused how to combine the Boundary- and Initial conditions for it. Should I just solve these and put them together and then check the conditons or can i do it simultaneously? What about $\lambda$, do I get different solutions depending on the sign? Can anyone show me how I should begin with this?
EDIT:
Ok so I took some advice from below and got to this:
For $\lambda≤0$ there is only the trivial solution (got that from the boundary conditions), so consider $\lambda>0$. We then write $\lambda=\omega^2$ and that would give me the solution
$$
X(x)=A\cos(\omega x)+ B\sin(\omega x)
$$
And from the boundry conditions I would end up with
$$
X(x)=B \sin(nx), \, for\, n=1,2,3,...
$$
Now what? What is $B$?
 A: As a general rule, first pay attention to the variable whose boundary conditions are homogeneous. Here, the variable that you may consider first is $x$. Since
$$\begin{array}{}
u(0,t)=X(0)T(t)=0 & \to X(0)=0 \\
u(\pi,t)=X(\pi)T(t)=0 & \to X(\pi)=0 \\
\end{array}
$$
and hence you should start by first solving the following eigen-value problem
$$\begin{array}{}
X''+ \lambda X =0  \\
X(0)=0 \\
X(\pi)=0 \\
\end{array}
$$
The solution to this eigen-value problem will be
$$\begin{array}{}
X(x)=\sin(nx), & n=1,2,3,... \\
\lambda=n^2 \\
\end{array}
$$
and hence the differential equation for $T(t)$ will be
$$T''(t) + 2 T'(t) + n^2 T(t) = 0$$
and the solution to this ODE is
$$T(t)=A_n e^{-t} \cos(\sqrt{n^2-1}t) + B_n e^{-t} \sin(\sqrt{n^2-1}t)$$
and hence
$$u(x,t)=\sum_{n=0}^{\infty} \left( A_n e^{-t} \cos(\sqrt{n^2-1}t) + B_n e^{-t} \sin(\sqrt{n^2-1}t) \right) \sin(nx)$$
and the derivative of $u(x,t)$ with respective to time will be
$$
u_t(x,t) 
= \sum_{n=0}^{\infty} \left( A_n e^{-t} \left( -\cos(\sqrt{n^2-1}t)+ \sqrt{n^2-1} \sin(\sqrt{n^2-1}t) \right) \\ 
\qquad \qquad \qquad \qquad \quad
 + B_n e^{-t} \left( -\sin(\sqrt{n^2-1}t)+ \sqrt{n^2-1} \cos(\sqrt{n^2-1}t) \right) \right) \sin(nx)
$$
and finally applying the intial conditions will lead you to
$$\begin{array}{}
u(x,0)=\sum_{n=0}^{\infty} A_n \sin(nx) = 0 \to &  A_n=0
\end{array}$$
$$\begin{array}{}
u_t(x,0) = \sum_{n=0}^{\infty} B_n \sqrt{n^2-1} \sin(nx)
\end{array} = \sin^3(x)$$
and I just leave finding the coefficients $B_n$ for you. :)
