Solving Sturmian Equation How to solve this linear pde for $y(x)$ (other functions are known and $\lambda$ is a constant):$$\frac{d}{d x}(g(x)y'(x))=\lambda ^2g(x)y(x)$$ Everything I know about this equation is that it is called the Sturmian equation. I did some research, but the theory, which is for a more general form of the equation, is too hard for me to understand.
 A: Consider the operator
$$
                Lf = \frac{1}{g}\frac{d}{dx}\left(g\frac{df}{dx}\right).
$$
You want to solve $Lf=\lambda^2 f$. Start by solving for $f_0$ such that
$$
     Lf_0=0,\;\; f_0(0)=A,\; f_0'(0)=B.
$$
This is done by integrating the following in $x$:
$$
          \frac{d}{dx}\left(g\frac{df_0}{dx}\right)=0 \\
         g\frac{df_0}{dx}-g(0)B=0 \\
        \frac{df_0}{dx} = \frac{Bg(0)}{g(x)} \\
          f_0(x)=A+B\int_{0}^{x}\frac{g(0)}{g(x)}dx.
$$
Then recursively solve
$$
                  Lf_n = f_{n-1},\;\; f_n(0)=0,\;\; f_n'(0)=0,\;\;\; n \ge 1.
$$
This is done by integrating:
$$
            \frac{1}{g}\frac{d}{dx}\left(g\frac{df_{n}}{dx}\right)=f_{n-1},\;\; f_n(0)=0,\;f_n'(0)=0 \\
            g\frac{df_n}{dx}=\int_{0}^{x}g(y)f_{n-1}(y)dy \\
        f_n(x) = \int_{0}^{x}\frac{1}{g(z)}\int_{0}^{z}g(y)f_{n-1}(y)dy dz
$$
Now form the sum
$$
              f(x,\lambda)=\sum_{n=0}^{\infty}\lambda^{2n}f_n(x).
$$
You can verify that $f(0)=f_0(0)=A$ and $f'(0)=f_0'(0)=B$. And,
$$
       Lf = \sum_{n=1}^{\infty}\lambda^{2n}f_{n-1}=\lambda^2\sum_{n=0}^{\infty}\lambda^{2n}f_n=\lambda^2 f.
$$
