Solution of one dimensional wave equation by variable separation method When solving the One dimensional wave equation by variable separable method, we equate left-hand side and right-hand side to a constant which is negative in nature. Why has the constant be only negative in nature?
 A: I am assuming you mean that if you split the assumed solution up like $\psi(z,\,t)=Z(z)\,T(t)$ you end up with:
$$\frac{Z^{''}(z)}{Z(z)}=\frac{1}{c^2}\frac{T^{''}(t)}{T(t)} = -k^2$$
and you're asking why $-k^2$ is assumed to be negative. This is simply because we know the wave must have bounded energy. There is no mathematical reason for this. But if we allow solutions with $+k^2$ on the RHS then you would have solutions of the form 
$$Z(z) = A_{+,z}\,e^{k\,z} + A_{-,z} e^{-k\,z}$$
$$T(t) = A_{+,t}\,e^{k\,c\,t} + A_{-,t} e^{-k\,c\,t}$$
which diverge to infinity as $t\to\pm\infty$, $z\to\pm\infty$, which is almost always unphysical behaviour, i.e. the "waves" concerned would have energies that increase exponentially with time. Sometimes if we have semi-infinite mediums and are interested in solutions only for $t>0$, then we might consider the solution $\psi\propto e^{-k\,(z+c\,t)}$ but this is very seldom. Also, when we consider waves in resonators or waveguides (often the reason for using separation of variables in the first place) to find standing waves, we need the mode functions $\psi(z,\,t)$ to have zeros at two different, constant values of $z$, so that we can match boundary conditions. If $-k^2$ were positive and we had $\cosh,\,\sinh$ behaviour instead of the $\cos,\,\sin$ behaviour that prevails when $-k^2$ is negative, then we could only in general force a zero of $\psi(z,\,t)$ at one constant value of $z$ and thus such functions cannot match waveguide boundary conditions.
A: This solution:
$\varphi(x,t)=(a_1\cosh(\lambda t)+a_2\sinh(\lambda t))(a_3\cosh(c \lambda x)+a_4\sinh(c \lambda x))$
is about as useful as this solution:
$\varphi(x,t)=(a_1+a_2 t)(a_3+a_4 x)$.
They're both valid, they're just not useful in most physical problems. (Except in the exponentially decreasing case $(a_1\cosh(\lambda t)+a_2\sinh(\lambda t))=c_5 e^{-\lambda t}$, you're going to have $\varphi(x,\infty)\to \pm \infty$, which is just not physical behavior in most cases)
(valid solutions of $\varphi_{tt}=c^2 \varphi_{xx}$, with $\varphi_x$ being shorthand for $\frac{\partial \varphi}{\partial x}$)
