Solving telegrapher's equation, using seperation of variables. This is question 4.2.9 from introduction to partial differential equations by oliver, et al. 
Let $a, c > 0$ be positive constants. 
The telegrapher’s equation $u_{tt} + au_{t} = c^2 u_{xx}$ represents a damped version of the wave equation. 
Consider the Dirichlet boundary value problem
$$u(t, 0) = u(t, 1) = 0,$$ on the interval $$0 ≤ x ≤ 1 ,$$ 
with initial conditions 
$$u(0, x) = f(x)$, $u_t(0, x) = 0 .$$ 
(a) Find all separable solutions to the telegrapher’s equation that satisfy the boundary conditions. 
(b) Write down a series solution for the initial boundary value problem. (c) Discuss the long term behavior of your solution. 
(d) State a criterion that distinguishes over-damped from under-damped versions of the equation.

This question is assigned as part of a test review and not part of routine homework. The test review will not be turned in for credit.
I am having trouble understanding how to use the boundary conditions in this problem. I don't understand separation of variables because it seems like everyone that explains it skips so many steps.
 A: The whole point of separation of variables is to assume that  your function $u(x,t)$ can be written as the product of two functions which each contain only one of the variables of $u$, i.e
$$u(x,t) = X(x)T(t)$$
Substituting this ansatz into our PDE, we find
$$X T'' + a X T' = c^{2} X'' T$$
Notice that if we divide both sides by the product of functions, $c^{2} X(x) T(t)$, we get
$$\frac{1}{c^{2}} \bigg( \frac{T''}{T} + a \frac{T'}{T} \bigg) = \frac{X''}{X}$$
The key here is that the only way that a function of $t$ and a function of $x$ can be equal is if they both equal a constant, known as the separation constant, which I will denote by $- \lambda$. Hence, we have
\begin{align}
\frac{1}{c^{2}} \bigg( \frac{T''}{T} + a \frac{T'}{T} \bigg) &= \frac{X''}{X} \\
&= - \lambda \\
\end{align}
which gives us two ODEs
$$\frac{1}{c^{2}} \bigg( \frac{T''}{T} + a \frac{T'}{T} \bigg) = - \lambda \implies T'' + aT' + \lambda c^{2} T = 0$$
and
$$\frac{X''}{X} = - \lambda \implies X'' + \lambda X = 0$$
So, in other words, we have separated our PDE into two ODEs. It then remains to solve both ODEs, using your boundary conditions. For instance, the BC
$$u(0, t) = 0 \implies X(0) T(t) = 0$$
which implies $X(0) = 0$ (why?) is a condition for your ODE in $x$. Try this for your other BC and see how you go, if you need more help just comment below.
