Eigenvalues and eigenfunctions

**chella182** · March 1st 2010, 07:57 AM

Find the eigenvalues and eigenfunctions for

$y''+\lambda y=0$ , $0\leq x\leq1$

with the homogenous boundary conditions $y(0)=y'(1)=0$ .

So I subbed in $y=e^{\alpha x}$ et cetera, and somewhere down the line ended up needing to solve

$\sin{\omega}+\omega\cos{\omega}=0$

and now I'm stuck and not even sure if this can be done. Obviously $\omega=0$ would work, but wouldn't that just give a trivial solution?

**Opalg** · March 1st 2010, 09:25 AM

Originally Posted by chella182

Find the eigenvalues and eigenfunctions for

$y''+\lambda y=0$ , $0\leq x\leq1$

with the homogenous boundary conditions $y(0)=y'(1)=0$ .

So I subbed in $y=e^{\alpha x}$ et cetera, and somewhere down the line ended up needing to solve

$\sin{\omega}+\omega\cos{\omega}=0$

and now I'm stuck and not even sure if this can be done. Obviously $\omega=0$ would work, but wouldn't that just give a trivial solution?

I assume that you have put $\omega = \sqrt{-\lambda}$ , so that the general solution of the equation is $y = A\cos\omega x + B\sin\omega x$ . The boundary condition y(0)=0 tells you that A=0, so that $y = B\sin\omega x$ . Then you differentiate to get the other boundary condition in the form $\omega\cos\omega = 0$ . You can disregard the solution $\omega=0$ , because that leads to the trivial solution y=0. So the eigenvalues are given by $\cos\omega=0$ , which has an infinite family of solutions.

**chella182** · March 4th 2010, 10:51 AM

Ignore this, I've figured out why you don't have $e^{x}$ in your solution. Damn shoddy notes :@

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