Normalizing a Sturm-Liouville theory problem I have the following D.E.:
$$y''(x) + \lambda y(x) = 0$$ on $[0,L]$
I've solved for
1) $\lambda =0$
2) $\lambda > 0$ 
3) $\lambda <0$
And found that $\lambda > 0$  gives me the only non-trivial solution, where 
$$ \lambda = \left(\frac{\pi}{2L}(2k+1) \right)^2 $$
and:
$$f_n(x) = B_n\cos\left( \frac{\pi}{2L}(2k+1)x \right)  $$ 
I now need to normalize my function for $n=0$ and $n>0$.
But am I normalizing $f_n(x)$ with itself?
i.e. for $n=0$: 
$$ \left< B_0\cos\left(\frac{\pi}{2L}x\right),B_0\cos\left(\frac{\pi}{2L}x\right)\right>$$
I prefer hints rather than solutions.
 A: You've solved correctly for $\sqrt{\lambda} L = \frac{\pi}{2}+n\pi$ or $\sqrt{\lambda}=(n+\frac{1}{2})\pi/L$. So the eigenfunctions are
$$
         Y_n(x)=\cos((n+1/2)\pi x/L)
$$
The normalization comes from trying to expand a function $f$ in a Fourier series of this eigenfunctions
$$
                         f \sim c_1 Y_1 + c_2 Y_2 + \cdots
$$
These functions are orthogonal with respect to the integral. So, multiplying by $Y_n$ and integrating both sides gives only one term on the right
$$
               \int_{0}^{L}f(x)Y_n(x)dx = c_n \int_{0}^{L}Y_n(x)^2dx \\
              c_n = \frac{\int_{0}^{L}f(x)Y_n(x)dx}{\int_{0}^{L}Y_n(x)^2dx}
$$
Everything simplies if you normalize $Y_n$ so that $\int_{0}^{1}Y_n(x)^2dx=1$, which means dividing the original $Y_n$ by $\sqrt{\int_{0}^{1}Y_n(x)^2dx}$, which you can see from the final series
$$
              f \sim \sum_{n} \frac{\int_{0}^{L}f(x')Y_n(x')dx'}{\int_{0}^{L}Y_n(x')^2dx'}Y_n(x)
$$
Assuming $Y_n$ has been so normalized, the series expansion is
$$
          f \sim \sum_{n} \left(\int_{0}^{L}f(x')Y_n(x')dx'\right) Y_n(x).
$$
