Rall cable theory equation I'm currently studying Rall cable theory which is a concept from neuroscience and I've come across this differential equation that governs change of axial current depending of time and distance:
$$\partial^2 V/\partial x^2 = \partial V/\partial t + V(x, t)$$
I have some basic knowledge of single variable calculus and can solve basic first order differential equations but this is a second order equation and V is a multi-variable function so... how do I go about solving this? What initial conditions should I choose? Would not it be too much to ask someone to solve this with comments?
 A: The first solution technique that occurred to me is separation of variables, a topic one covers in undergraduate differential equations classes, applicable to many constant coefficient parabolic DEs as here.
There's a lengthy chapter (59 pages) by Rall in the Handbook of Physiology -- The Nervous System (Supp. 1), and the "dimensionless" form of the cable equation of neurophysiology appears about one-third of the way through:
$$ \partial^2 V/ \partial X^2 = \partial V/ \partial T + V $$
The solution by separation of variables appears a couple of pages later:
$$ V(X,T) = [A \sin(\alpha X) + B \cos(\alpha X)] \exp(-(1+\alpha^2) T) $$
which involves three "arbitrary constants" $A,B,\alpha$.
Because the cable equation is linear and homogeneous, it obeys the superposition principle.  Thus more general solutions can be obtained by summing ones shown above, suitably chosen to fit initial and boundary conditions.
Here the coefficients of a Fourier expansion of the initial condition ($T=0$) may be expected to appear.  Notice that the individual components are all decaying exponentially with time $T \gt 0$.
A more sophisticated solution approach involves Green's functions, which model a response to an instantaneous impulse.  The usefulness of such an approach is illustrated by a recent (2010) PubMed paper.

Connecting with solutions of the heat equation can help to visualize solutions of the cable equation.  The heat equation says:
$$ \partial^2 U/ \partial X^2 = \partial U/ \partial T $$
and solutions of the cable equation above are related by an integrating factor:
$$ V(X,T) = e^{-T} U(X,T) $$
The Reader is invited to verify this by directly substituting into the cable equation and simplifying, cancelling the nonzero factor of $e^{-T}$ on both sides.  Thus a solution $V$ of the cable equation is exactly a solution of the heat equation times the extra $e^{-T}$, an exponentially decaying-in-time factor.
The solutions of the heat equation are known for smoothing of uneven "temperature" distribution in the initial condition, as "heat" flows from warmer areas to cooler ones.  Given constant Dirichlet boundary conditions, say $U(0,T) = U_0, U(1,T) = U_1$, the solution evolves toward the "steady state" (flat) linear interpolation between these endpoints. 
Solutions of the cable equation will likewise exhibit this "smoothing out" of peaks in the initial condition, combined with the extra factor of simple exponential decay.
