Sturm Liouville problem for $x^2φ''+ xφ' + λφ = 0$ Consider the eigenvalue problem 
$$x^2φ''+ xφ' + λφ = 0, \quad 1<x<2, \quad φ(1) = 0, φ(2) = 0$$
(a) Write the problem in Sturm-Liouvile form, identifying $p, q,$ and $σ$.
(b) Is the problem regular? Explain.
(c) Find all eigenvalues and eigenfunctions.
(d) Find the orthogonal expansion of $f(x) = x^5(2 − x)^3, 1<x<2$, in terms of these eigenfunctions.
(e) Find the smallest $N$ such that the average (over $1<x<2$) of the point wise error between the $N$th  partial sum in (e)  and $f(x)$ is less than $0.1$. Plot $f$ and $F_N(x)$ on the same coordinate plane for this value of $N$.
My work:
(a) $(xφ')'+ \frac{1}{x} λφ = 0$ where $p(x)=x, q(x)=0, σ(x)=\frac{1}{x}$
(b) yes the problem is regular
(c) $φ(x)=\sum_{n=1}^∞ A\sin(\frac{2nπ}{x})+B\cos(\frac{2nπ}{x})$ and $λ=4n^2π^2$
for (d) and (e) I'm not sure how to do them or where to start.   
 A: The eigenfunction equation is
$$
               x^2\varphi''+x\varphi'+\lambda\varphi=0 \\
               x\varphi''+\varphi'+\lambda\frac{1}{x}\varphi =0 \\
                 -\frac{d}{dx}\left(x\frac{d}{dx}\varphi\right)=\lambda\frac{1}{x}\varphi.
$$
So the Sturm-Liouville problem is studied on $L^2_{\sigma}[0,1]$ where $\sigma=1/x$. To be specific, the inner product on this space is
$$
                   (f,g)_{\sigma} = \int_{1}^{2}f(x)\overline{g(x)}\frac{1}{x}dx.
$$
Here $p(x)=x$ and $q(x)=0$.
The problem is a regular problem because the interval is $[1,2]$ where $p$ does not vanish and the coefficients are non-singular.
By the way, there is a nice trick for getting rid of $p$ from such problems: make the substitution
$$
                   \varphi(x) = \psi\left(\int\frac{1}{p}dx\right)
$$
Then $p\frac{d}{dx}\varphi(x)= \psi'\left(\int\frac{1}{p}dx\right)$. In this case the substitution works out really well because the weight $\sigma$ is the reciprocal of $p$:
$$
                -\frac{1}{x}\psi''\left(\int\frac{1}{p}dx\right)=\lambda\frac{1}{x}\psi\left(\int\frac{1}{p}dx\right) \\
                   -\psi''(y) = \lambda \psi(y).
$$
Here, $y=\int\frac{1}{x}dx=\ln(x)$ ranges over $[0,\ln(2)]$ as $x$ ranges over $[1,2]$. And, when you make the substitution, you're now working on $L^2[0,\ln(2)]$ where the weight is now unity.
Without changing variables, the eigenfunction equation is Euler's equation
$$
                      x^2\varphi''+x\varphi'+\lambda\varphi=0,
$$
which always has a solution $x^{\alpha}$ where
$$
                      \alpha(\alpha-1)+\alpha+\lambda=0 \\
                        \alpha^2+\lambda=0.
$$
This leads to two linearly independent solutions if $\lambda \ne 0$:
$$
      \varphi(x) = Ae^{i\sqrt{\lambda}x}+Be^{-i\sqrt{\lambda}x}.
$$
If $\lambda=0$, then $\varphi_{\lambda}=1$ is a solution, and the second independent solution is $\ln x$, which is easily obtained by solving $x^2\varphi'+x\varphi=0$ or $\varphi'/\varphi= -1/x$. $\lambda=0$ is not an eigenvalue because $A+B\ln x$ cannot vanish at $x=1$ and $x=2$ unless $A=B=0$. For non-zero $\lambda$, the solution of
$$
                          x^2\varphi''+x\varphi'+\lambda\varphi=0 \\
                        \varphi(1)=0,\;\; \varphi'(1)=1
$$
is $\varphi(x) = Ae^{i\sqrt{\lambda}\ln x}+Be^{-i\sqrt{\lambda}\ln x}$ where
$$
                    A+B=0, \;\;\; A-B=\frac{1}{i\sqrt{\lambda}}\\
                        A = \frac{1}{2i\sqrt{\lambda}},\;\;\; B=-\frac{1}{2i\sqrt{\lambda}},\\
              \varphi_{\lambda}(x) = \frac{1}{2i\sqrt{\lambda}}(e^{i\sqrt{\lambda}\ln x}-e^{-i\sqrt{\lambda}\ln x}).
$$
The eigenvalues $\lambda$ are the values of $\lambda$ for which $\varphi_{\lambda}(2)=0$, which happens iff
$$
                  \sin(\sqrt{\lambda}\ln 2)=0
                   \sqrt{\lambda}\ln 2 = \pi,2\pi,3\pi,\cdots \\
                \sqrt{\lambda} = \frac{n\pi}{\ln 2} \\
                \varphi_{n}(x) = C_n\sin\left(n\pi\frac{\ln x}{\ln 2}\right).
$$
The Fourier series for a function $f$ on $[1,2]$ is
$$
               f(x) = \sum_{n=1}^{\infty}\frac{(f,\varphi_n)_w}{(\varphi_n,\varphi_n)_w}\varphi_n(x)
            = \sum_{n=1}^{\infty}\frac{\int_{1}^{2}f(t)\varphi_n(t)\frac{1}{t}dt}{\int_{1}^{2}\varphi_n(t)^2\frac{1}{t}dt}\varphi_{n}(x).
$$
The integrals are resolved by the substitution $x=\ln t$ so that $\frac{1}{t}dt=dx$ (as suggested by the substitution mentioned above.)
