Determining whether eigenvalues are positive, negative or 0. 
Consider $ u_{tt} = c^2u_{xx}$ for $0 <  x < l,$ 
with the boundary conditions $$u(0, t)
= 0, u_x(l, t) = 0$$
and the initial conditions $$u(x, 0) = x, u_t (x, 0) = 0.$$
  Find the solution explicitly in series form.

In solution manual, writer has computed $$\frac {X''} X=-\mu^2 (*)$$
$$ \frac{T'}T=-c^2\beta ^2(**)$$
which means he/she assumed that eigenvalues are negative. Does that mean when eigenvalues are 0 or positive he/she got trivial solutions that's why only considered negative eigenvalues? Also in $(*)$ why $c^2 $ is eliminated whereas it appears in $(**)$?
My other doubt is that one of the initial conditions is not homogeneous so how do we apply the method of separation of variables ?
 A: You can easily check what happens when the eigenvalues are assumed $0$ or positive. For $0$ eigenvalues:
$$X''(x) = 0 \ \text{ and } \ T''(t) = 0 \\ X(x) = Ax + B \ \text{ and } \ T(t) = Ct + D$$
For some constants $A, B, C, D$ determiend by the boundary conditions.
$X(0) = 0 \implies B = 0 \\ X'(l) = 0 \implies A = 0$
So the only solution is trivial.
Now for positive eigenvalues:
$$X''(x) = \frac{\lambda^2}{c^2} X(x) \\ T''(t) = \lambda^2 T(t)$$
Which has solutions:
$$X(x) = A\cosh(\frac{\lambda}{c}x) + B\sinh(\frac{\lambda}{c}x) \\ T(t) = C\cosh(\lambda t) + D\sinh(\lambda t)$$
$x(0) = 0 \implies A = 0 \\ x'(l) = 0 \implies \frac{\lambda}{c}A \sinh(\frac{\lambda}{c}L) = 0$
$\sinh(x)$ only has $1$ zero at $x = 0$, so either $\lambda = 0$, or $A = 0$. 
Either way, we get the trivial solution, so that we can only have negative eigenvalues. 
Notice that I attached the $c^2$ term with the $X$ function. It is most common to attach it with the $T$ function, as your book did, but this is always choice. It will not be attached to both. This happens directly from separation of variables. You get the equation:
$$c^2 \frac{X''}{X} = \frac{T''}{T} = \text{constant} \ \ \ \ \text{ OR } \ \ \ 
 \frac{X''}{X} = \frac{T''}{c^2 T} = \text{constant}$$
Where the constant will end up being your eigenvalues. The only difference is where which equation the $c^2$ constant appears in, but they are equivalent representations
It should also be noted that Sturm-Liouville theory will give you the answer you want immediately, but it is always good practice to run through things anyways.
A: After separation of variables, you get $\frac{X''}{X}=C$, a constant. The eigenvalues $\mu^2$ are defined as $-C$, because usually (in this case, it is) solutions only exist for $\mu^2\geq 0$ (maybe even only $\mu^2> 0$). Working with $C\leq 0$ is less nice, because you constantly need to keep the minus sign in mind, so we work with $-C$.
After separation of variables, $u_{tt} = c^2u_{xx}$ turns into $XT''=c^2X''T$, which gives us $\frac{X''}{X}=\frac{T''}{c^2T}=C=-\mu^2$, so $\frac{X''}{X}=-\mu^2$ and $\frac{T''}{T}=-c^2\mu^2$.
The method of separation of variables has nothing to do with the initial conditions, so it even works without any initial conditions.
I hope I answered all of your questions.
