2nd order Eigenvalue Problem Consider the problem $$ 2t^2f(t)+\lambda(tf''(t)-f'(t))=0 $$ where $f(0)=f'(0)=0$. What can you say about the solutions? Preferrably a closed form solution of course.
The problem arise when trying to solve the eigenvalue problem $$ \lambda f(t)=\int_0^1f(x)\min\{x^2,t^2\}dx $$.
 A: The equation is, in selfadjoint form,
$$
                  -tf''(t)+f'(t)=\frac{1}{\lambda}2t^2 f(t) \\
           -t^2\frac{d}{dt}\left(\frac{1}{t}\frac{df}{dt}\right)=
\frac{1}{\lambda}2t^2f(t) \\
            -\frac{d}{dt}\left(\frac{1}{t}\frac{df}{dt}\right)=\frac{2}{\lambda}f
$$
The usual eigenvalue problem is where $\lambda$ is replaced by $1/\lambda$. The solutions of
$$
                 -\frac{d}{dt}\left(\frac{1}{t}\frac{df}{dt}\right)=0
$$
are
$$
              f= Ct^2+D.
$$
All of the solutions will have the same basic behavior $f \sim Ct^2+D$ near the singular endpoint at $t=0$, which means that the restriction $f'(0)=0$ is basically meaningless while the restriction $f(0)=0$ is not superfluous.
Start with $f_0=t^2$. For the general inductive setep, solve the following with $0$ constants of integration
$$
                     -\frac{d}{dt}\left(\frac{1}{t}\frac{df_{n+1}}{dt}\right)=f_n \\
         -\frac{1}{t}\frac{df_{n+1}}{dt} = \int_{0} f_n dt \\
              -\frac{df_{n+1}}{dt} = t\int_{0}f_n dt \\
                f_{n+1}(t) = -\int_{0}t\int_{0}f_n dt
$$
For example,
$$
                 f_1 = -\int_{0}t\int_{0}t^2 dt = -\frac{1}{(3)(5)}t^5 \\
          f_2 = \frac{(-1)^2}{(3)(5)(6)(8)}t^8 \\
          f_3 = \frac{(-1)^3}{(3)(5)(6)(8)(9)(11)}t^{11}
$$
The solution you want is any constant multiple of
$$
                       f_{\lambda}(x)=\sum_{n=0}^{\infty}\left(\frac{2}{\lambda}\right)^{n}f_{n+1}(x)
$$
From the construction above, you can show that
$$
      -\frac{d}{dt}\left(\frac{1}{t}\frac{d}{dt}\right)f_{\lambda}=\frac{2}{\lambda}f_{\lambda}, \\
            f_{\lambda}(0)=0,\;\; f_{\lambda}'(0)=0.
$$
