Endpoint problems & eigenvalues

[cue928]

Homework Statement

y'' + (lambda)y = 0, y'(0) = 0, y(1) = 0
We are told that all eigenvalues are nonnegative.

Even with looking at the solution manual, I am unsure how to start setting these up. I've been starting by doing the following:
y(x) = A cos cx + B sin dx
y'(x) = -Ac sin(cx) + Bd cos(dx)
Subbing in the initial values:
y'(0)=0: 0 = B
This leaves y(x) = A cos (cx)
But this is also where I'm breaking down. I understand from y(1) = 0 that I am looking for a value of c such that the result is 0?

I mean, what should I be looking for on setting these up? That problem seems relatively straight forward but the next one I tried, y''+(lambda)y = 0, y(-pi) = 0, y(pi) = 0, has me baffled. Any guidance would be greatly appreciated.

 

[Dustinsfl]

[tex]m^2+\lambda=0\Rightarrow m^2=-\lambda\Rightarrow m=\pm i\sqrt{\lambda}[/tex]

[tex]y=C_1\cos(x\sqrt{\lambda})+C_2\sin(x\sqrt{\lambda})[/tex]

We obtain the basics solutions from this:
[tex]y_1(0)=1, \ y_1'(0)=0, \ y_2(0)=0, \ y_2'(0)=1[/tex]

[tex]y_1(0): \ C_1=1, \ y_1'(0): \ C_2=0, \ y_2(0): \ C_1=0, \ C_2=\frac{1}{\sqrt{\lambda}}[/tex]

[tex]y=y_1+y_2=A\cos(x\sqrt{\lambda})+\frac{B\sin(x\sqrt{\lambda})}{\sqrt{\lambda}}[/tex]

Can you take it from here?

 

[cue928]

To be honest, that doesn't look like anything I saw in the solution manual or notes. I guess what I am trying to understand is what the end game is on this. What am I looking for?

 

[Dustinsfl]

From where I left off, we now apply the conditions to determine the eigenfunction.

[tex]y=A\cos(x\sqrt{\lambda})+\frac{B\sin(x\sqrt{\lambda})}{\sqrt{\lambda}}[/tex]

[tex]y'=B\cos(x\sqrt{x})-A\sqrt{\lambda}\sin(x\sqrt{\lambda})[/tex]

[tex]y'(0): \ B=0[/tex]

[tex]y(1): \ A\cos(\sqrt{\lambda})=0[/tex]

Now, solve

[tex]\cos(\sqrt{\lambda_n})=0[/tex]