# What kind of ODE is this?

1. ### StewartHolmes

2
I'm trying to follow a proof for the solution of the diffusion equation in 0 < x < l with inhomogeneous boundary conditions.

$$\frac{d u_n(t)}{dt} = k( -\lambda_n u_n(t) - \frac{2n\pi}{l}[ (-1)^n j(t) - h(t) ] )$$
$$u_n(0) = 0$$

Now I just plain don't understand what kind of an ODE I have here. If the term in j(t) and h(t) wasn't there, it'd be a simple ODE, but I'm confused as to what can be done now. I know ODEs of the form

y' + p(x)y + q(x) = 0

But I have something like, y' + p(x)y + q(t) where I have a term in the dependent variable.

The book I have gives the solution as
$$u_n(t) = Ce^{-\lambda_n kt} - \frac{2n\pi k}{l}\int\limits_0^t e^{-\lambda_n k(t-s)} \left( (-1)^n j(s) - h(s) \right) \, ds$$

2. ### jackmell

Try and learn to encapsulate everything. You have:

$$\frac{d u_n(t)}{dt} = k( -\lambda_n u_n(t) - \frac{2n\pi}{l}[ (-1)^n j(t) - h(t) ] )$$

Now, isn't the term:

$$-\frac{2n\pi}{l}k[(-1)^n j(t)-h(t)]$$

just some function of t? Say it's v(t). So you have essentially the equation:

$$\frac{dy}{dt}+k\lambda y=v(t)$$

And you know how to integrate that right by finding the integration factor. Change it to u_n if you want, but it's the same equation essentially.

Last edited: Oct 3, 2010