I'm currently trying to solve an rSLP which looks like this (Euler equation): $$x^2f''(x)+xf'(x)+cf(x)=0,\ f(1)=f'(2)=0,\ x\in [1,2] \text{ and }c\neq0.$$ It's simple that $f(x)=x^{\pm\sqrt{-c}}$ solves the equation for $c\neq0$. By theorem, $\{x^{\sqrt{-c}},x^{-\sqrt{-c}}\}$ will be a basis for solutions to the equation. This means that all solutions can be written as a linear comb. of of the basis: $$f(x)=Ax^{\sqrt{-c}}+Bx^{\sqrt{-c}}=Ae^{ln(x)\sqrt{-c}}+Be^{-ln(x)\sqrt{-c}}.$$ First boundary condition: $f(1)=0\Leftrightarrow A=-B\ \Rightarrow \ f(x)=e^{ln(x)\sqrt{-c}}-e^{-ln(x)\sqrt{-c}} \text{ up to some constant.}$
The second BC: $f'(2)=0$. Derivative is $$f'(x)=\frac{\sqrt{-c}}{r}e^{ln(x)\sqrt{-c}}+\frac{\sqrt{-c}}{r}e^{-ln(x)\sqrt{-c}}=\left\{\begin{matrix} \frac{\sqrt{c}}{r}\text{cos}(ln(x)\sqrt{c}),& c>0\\ \frac{\sqrt{-c}}{r}\text{cosh}(ln(x)\sqrt{-c}),& c<0 \end{matrix}\right. \text{ (again up to constants)}.$$ Since $\text{cosh}(x)\neq0 \text{ and }c\neq0$ we can tell that $c>0 $ and that $f'(x)=\frac{\sqrt{c}}{r}\text{cos}(ln(x)\sqrt{c})$ which also implies that $f(x)=\text{sin}(ln(x)\sqrt{c})$ up to some constant. So $f'(2)=0$: $$f'(2)=\frac{\sqrt{c}}{2}\text{cos}(ln(2)\sqrt{c})=0\ \Leftrightarrow \ \text{cos}(ln(2)\sqrt{c})=0\ \Leftrightarrow \ c=\frac{\pi^2(2n+1)^2}{4(ln(2))^2}.$$ So we have the eigenfunctions $\{\text{sin}(\frac{\pi(2n+1)}{2ln(2)}ln(x))\}_{n\geq 1}$ with eigenvalues $\frac{\pi^2(2n+1)^2}{4(ln(2))^2}$ to this regular SLP.
HOWEVER, what I don't understand is that if you put in $f_n(x)=\text{sin}(\frac{\pi(2n+1)}{2ln(2)}ln(x))$ in the Euler equation which we are solving, the $f_n(x)$ term and the $f_n''(x)$ term will cancel each other out. Then we are left with the equation $$\frac{\pi(2n+1)}{2ln(2)}\text{cos}(\frac{\pi(2n+1)}{2ln(2)}ln(x))=0.$$ This is obviously not true for all $x\in [1,2]$. What am I getting wrong?
The homogeneous equation $x^2f''(x) + xf'(x) + cf(x) = 0$ on $1 < x < 2$ has solution \begin{equation*} f(x) = \begin{cases} A\cos(\sqrt{c}\log x) + B\sin(\sqrt{c}\log x) & \text{ $c > 0$}, \\ A\cosh(\sqrt{-c}\log x) + B\sinh(\sqrt{-c}\log x) & \text{ $c < 0$}, \\ A\log x + B & \text{ $c = 0$}. \end{cases} \end{equation*} The boundary conditions $f(1) = f'(2) = 0$ constrain nontrivial solutions. Simple substitution shows $A$ and $B$ must vanish in the $c = 0$ case, so we'll ignore it. In the other two cases the condition $f(1) = 0$ requires $A = 0$, so \begin{equation*} f(x) = \begin{cases} B\sin(\sqrt{c}\log x) & \text{ $c > 0$}, \\ B\sinh(\sqrt{-c}\log x) & \text{ $c < 0$}. \\ \end{cases} \end{equation*} The condition $f'(2) = 0$ cannot be satisfied by the second, so we're left with the restriction that $c > 0$ and $\sqrt{c}\log 2 = (2n+1)\pi/2$ for $n=0, 1, \dots$.