# 2nd Order De Solution

1. Nov 6, 2006

### Peregrine

I am familiar with how to solve a second order, non-homogenous DE with constants, i.e.

$$\frac {\partial^2X(t)}{\partial t^2} + \frac{\partial X(t)}{\partial t} = C$$

by first solving the homogenous eqn, then setting the equation equal to a constant, yielding a sol'n of

$$X(t)= Ae^{0}+ Be^{-t}+ C$$

But how does one solve a 2nd order equation that also has another t variable in it, such as:

$$\frac {\partial^2X(t)}{\partial t^2} + \frac{1}{t} \frac{\partial X(t)}{\partial t} = C$$?

2. Nov 6, 2006

### Matthew Rodman

First of all, you only seem to have one independent variable, so it may suitable to express your equation as

$$\frac{d^2 X(t)}{dt^2} + \frac{1}{t} \frac{d X(t)}{dt} = C$$

(note total derivative, not partial). Also, since no X(t) appears outside a derivative, you effectively have a first order equation, namely

$$\frac{dp}{dt} + \frac{p}{t} = C$$

where

$$p(t) = \frac{dX(t)}{dt}$$

Now, any first order equation of the form

$$\frac{dy}{dx} + s(x) y + r(x) = 0$$

has the solution

$$y(x) = -e^{-\int s(x) dx} \int r(x) e^{\int s(x) dx} dx$$

(just differentiate this and you'll see it works) Hence you can solve for p(t), and then for X(t).

Last edited: Nov 6, 2006
3. Nov 6, 2006

### Peregrine

Ah, that's a very nice way of framing the equation, I hadn't thought of that. Thanks!