I was hoping someone could explain something to me that I appear to be missing. It seems to be rather fundamental.
Given $u(x,t)$ can be expressed as $X(x)T(t)$, where $X(x)$ is only a function of $x$ and $T(t)$ is only a function of $t$. After some juggling we get:
$$\frac{X''}{X} = k \qquad\text{and}\qquad \frac{1}{c^2}\frac{T''}{T} = k $$
If $k = 0$ then $X'' = 0$, then $X' = a$ and $X = ax + b$.
The above line is where I am confused. If we integrate $X''$ twice how do we get $ax + b$? This would imply we are integrating wrt to t and hence adding a constant back in each case (a function of $x$). Is this line of thinking correct?
Thanks for your help!
I am hoping this is just a brain freeze! First, since X is function of x, you are certainly not "integrating wrt t"! And the rest is basic Calculus. If $\frac{d^2X}{dx^2}= 0$ then, integrating, $\frac{dX}{dx}= a$ for some constant a since the derivative of any constant is 0. Then integrating again, [tex]X(x)= ax+ b[/tex] where b is also a constant. As I said, it is basic Calculus that the derivative, with respect to x, of ax+ b is a.