The constant $c$ when deriving the general series solution for the heat and wave equations When deriving the general solution for the heat and wave equations, we have a constant $c$ that is carried in our calculations. Take the wave equations, for example:
$$\frac{\partial^2{u}}{\partial^2{t}} = c^2 \frac{\partial^2{u}}{\partial^2{x}}$$
Let's say we have boundary conditions $u(0, t) = 0$ and $u(1, t) = 0$.
Now assume a solution of the form 
$$u(x, t) = X(x)T(t)$$
The boundary conditions imply that 
$$X(0) = X(1) = 0$$
We substitute the assumed form of the solution into the PDE and use separation of variables:
$$\frac{1}{c^2} \frac{1}{T(t)} \frac{d^2T}{dt^2} = \frac{1}{X} \frac{d^2 X}{dx^2}$$
This must be equal to a constant. Make the constant $- \lambda$:
$$\frac{1}{c^2} \frac{1}{T(t)} \frac{d^2T}{dt^2} = \frac{1}{X} \frac{d^2 X}{dx^2} = -\lambda$$
The eigenvalue problems become either
$$c^2 \frac{1}{X} \frac{d^2 X}{dx^2} = - \lambda$$
OR
$$\frac{1}{c^2} \frac{1}{T(t)} \frac{d^2T}{dt^2} = - \lambda$$
Does it matter which equation takes the constant $c$? Will we get the same solutions either way?
We are trying to derive the general series solution, so why would it matter where we put the $c$? Shouldn't it produce the same general solution either way? Otherwise, I don't think it would be 'general'?
 A: In general you should get the following
$$\frac{1}{c^{2}}\frac{1}{h}\frac{d^{2}h}{dt^{2}} = \frac{1}{\phi} \frac{d^{2}\phi}{dx^{2}} = -\lambda \tag{1} $$
which gives you two eigenvalue problems 
$$ \frac{d^{2}h}{dt^{2}} = -\lambda c^{2} h  \tag{2} $$
and
$$ \frac{d^{2}\phi}{dx^{2}} = -\lambda \phi  \tag{3} $$
then you end up with your boundary conditions
$$ BC1: \phi(0) = 0 \\ BC2 : \phi(1) =1 \tag{4} $$
Now if $\lambda >0$
$$ h(t) = c_{1} \cos(c\sqrt{\lambda}t) + c_{2}\sin(c\sqrt{\lambda}t) \tag{5}$$
when $\lambda = 0$ we have instead
$$ h(t) = c_{1} +c_{2}t \tag{6}$$
If $\lambda< 0$ then we get hyperbolics, addressing the question. If we change this does it matter? What is the actual solution? Consider when the boundary conditons below
$$ BC1: u(0,t) = 0 \\  BC2 : u(L,t) = 0 \tag{7} $$
Ok if we solve for them.
$$ \frac{d^{2}\phi}{dx^{2}} = -\lambda \phi  \tag{8} $$
becomes 
$$ \frac{d^{2}\phi}{dx^{2}} + \lambda \phi = 0  \tag{9} $$
The solutions are simply of the forms
$$ \phi(x) = c_{1}\cos(\sqrt{\lambda}x) + c_{2}\sin(\sqrt{\lambda}x) \tag{10} $$
when we use the boundary conditions we get sines with these eigenvalues
$$ \lambda = \big(\frac{n\pi }{L}\big)^{2} \tag{11} $$
so we have
$$ \phi(x) = c_{1}\sin(\frac{n\pi x}{L}) \tag{12} $$
similarly, when we find $h(t)$ comes out as above.
$$ h(t) = c_{1} \cos(c\sqrt{\lambda}t) + c_{2}\sin(c\sqrt{\lambda}t) \tag{13}$$
Note that boundary conditions made on $x$ not on $t$. So the ansatz is a product of these 
$$  u(x,t) = \phi(x)h(t) \tag{14}$$
we denoted that it was any linear combinations of them. What that really means is take the product of these eigenfunctions as a basis and sum them infinitely. 
$$ u(x,t) = \sum_{n=1}^{\infty} \bigg( A_{n}\sin(\frac{n\pi x}{L}) \cos(\frac{n\pi ct}{L}) + B_{n} \sin(\frac{n\pi x}{L}) \cos(\frac{n\pi ct}{L})\bigg) \tag{15} $$
Ok, so if you change where the $c^{2}$ is we have 
$$ u(x,t) = \sum_{n=1}^{\infty} \bigg( A_{n}\sin(\frac{n\pi c x}{L}) \cos(\frac{n\pi t}{L}) + B_{n} \sin(\frac{n\pi cx}{L}) \cos(\frac{n\pi t}{L})\bigg) \tag{16} $$
Which I will take a moment to note that this is not the same equation. 
Typically we have a initial condition right. It goes like this
$$ IC1: u(x,0) = f(x) \\ IC2 :\frac{\partial u}{\partial t}u(x,0) =g(x) \tag{17} $$
Now, not what we haven't solved for above. The coefficients. When we solve for them the major differnce appears. If I use the equation one line $15$
$$ A_{n} = \frac{2}{L} \int_{0}^{L} f(x) \sin(\frac{n \pi x}{L})dx \tag{18} $$
$$ B_{n}\frac{n \pi c}{L} = \frac{2}{L} \int_{0}^{L} g(x) \sin(\frac{n \pi x}{L})dx \tag{19} $$
This changes when you switch that clearly.
