What kind of ODE is this?


by StewartHolmes
Tags: kind
StewartHolmes
StewartHolmes is offline
#1
Oct3-10, 04:15 AM
P: 2
I'm trying to follow a proof for the solution of the diffusion equation in 0 < x < l with inhomogeneous boundary conditions.

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

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
[tex] 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 [/tex]
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jackmell
jackmell is offline
#2
Oct3-10, 03:30 PM
P: 1,666
Try and learn to encapsulate everything. You have:

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

Now, isn't the term:

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

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

[tex]\frac{dy}{dt}+k\lambda y=v(t)[/tex]

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.


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