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JasonZ
May3-04, 07:11 PM
My problem is:

\frac{d^2\phi}{dx^2} + 6\frac{d\phi}{dx} + \lambda\phi = 0; \\ \frac{d\phi}{dx}(0) = 0, \frac{d\phi}{dx}(L) = 0.

I am told to begin by finding every eigenvalue and corresponding eigenfunction. I missed the last class where we went over this and the book is not giving me much advice on how to begin. Can someone just provide some help on how I begin solving this? I would show what I have but I have nothing yet ;).

Thanks,
-Jason

cookiemonster
May3-04, 07:15 PM
Try a solution of the form

\phi = e^{rx}

then consider cases for lambda, i.e. positive, zero, negative, and match boundary conditions.

cookiemonster

JasonZ
May3-04, 07:32 PM
Try a solution of the form

\phi = e^{rx}

then consider cases for lambda, i.e. positive, zero, negative, and match boundary conditions.

cookiemonster

Sorry for seeming so dense, but I understand how to apply that to normal O.D.E's, but the eigenvalue is throwing me off. What I am doing is:


Let \ \phi = e^{rx}



\frac{d\phi}{dx} = re^{rx}



\frac{d^2\phi}{dx^2} = r^2e^{rx}


After substituting this back in for \phi(x) I get:


r^2 + 6r + \lambda = 0


This is easily solved for \lambda = 0 , but how do I go about solving it for non-zero eigenvalues? I am pretty embarrased for having to ask questions on such a simple problem, but thank you for helping!

cookiemonster
May3-04, 07:42 PM
How about the quadratic formula? =]

cookiemonster

JasonZ
May3-04, 08:50 PM
Yeah I did that originally, and I think i was looking for some sort of fourier series like lambda to come about, and getting confused by that. I got it now though, thanks man.

-Jason