1. ## Eigenvalues and eigenfunctions

Find the eigenvalues and eigenfunctions for

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

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

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

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

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

2. Originally Posted by chella182
Find the eigenvalues and eigenfunctions for

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

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

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

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

and now I'm stuck and not even sure if this can be done. Obviously $\displaystyle \omega=0$ would work, but wouldn't that just give a trivial solution?
I assume that you have put $\displaystyle \omega = \sqrt{-\lambda}$, so that the general solution of the equation is $\displaystyle y = A\cos\omega x + B\sin\omega x$. The boundary condition y(0)=0 tells you that A=0, so that $\displaystyle y = B\sin\omega x$. Then you differentiate to get the other boundary condition in the form $\displaystyle \omega\cos\omega = 0$. You can disregard the solution $\displaystyle \omega=0$, because that leads to the trivial solution y=0. So the eigenvalues are given by $\displaystyle \cos\omega=0$, which has an infinite family of solutions.

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