Ok, so I decided to solve via a separable solution. From D\frac {\partial^2 \phi}{\partial x^2} = \frac {\partial \phi}{\partial t}, assume that \phi(x,t) = X(x)T(t). I can write D\frac{X''(x)}{X(x)}=\frac{T'(t)}{T(t)}=-\lambda . A solution for X(x) is X(x) = A\cos(\sqrt{\frac{\lambda}{D}}x) +...