I've not come across this type of problem before, nor does the chapter I'm working through contain similar problems. Can someone help me understand how I can pull out eigenvalues from the equation above, please?
My instinct says to find a characteristic equation, therefore, I have:
$$r^2+μ=0$$ solving I get: $$r=\pm i\sqrt{\mu}$$
giving me a characteristic equation of:
$$y_c(x)=A\cos(\sqrt{\mu}x)+B\sin(\sqrt{\mu}x)$$
$$y'_c(x)=-\sqrt{\mu}A\sin(\sqrt{\mu}x)+\sqrt{\mu}B\cos(\sqrt{\mu}x)$$
If I plug in a value thats given, such as $y(0)=0$, then that gives me $A=0$.
Now I don't know what to do or understand what I'm looking at. Do I use that other value given, $y'(L)=0$?
If I do I get $$\sqrt{\mu}B\cos(\sqrt{\mu}L)=0$$ and that's all I got.
Any help would be greatly appreciated.
Thank you
There are two ways that the expression $B \cos \sqrt{\mu} L = 0$. One way is to take $B=0$ but this gives the trivial solution $y(x) \equiv 0$ (you've already found $A=0$). If you put $\cos \sqrt{\mu} L = 0$ this gives restrictions on the allowed values of $\mu$.