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What kind of ODE is this?by StewartHolmes
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#1
Oct310, 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(ts)} \left( (1)^n j(s)  h(s) \right) \, ds [/tex] 


#2
Oct310, 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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