Can I consider these equations as Sturm-Liouville problem?

Question

I have two ODE's and I want to solve it but I am not confident that can I consider it to Sturm-Liouville type? if I can then what will be the general solution for these equations. \begin{aligned} F'(X) + F(X)\, \lambda &= 0 \\ G''(Y) + \frac{1}{2}\left(Y-Y^2\right) G(Y)\, \lambda &= 0 \end{aligned} These equations come from the PDE $$\partial_X\theta - \frac{2}{Y-Y^2} \partial_{YY}\theta = 0$$ where the separation of variables $\theta(X,Y) = F(X)\, G(Y)$ has been assumed. The boundary conditions are $\theta(X,0) = 0$ and $\partial_Y\theta(X,1) = 0$ for $X\geq 0$. The initial condition is $\theta(0,Y)=1$ for $0<Y<1$.

Answer 1

The first equation is first-order, so it is not a Sturm-Liouville problem. The solution is $F(X) = a_1 e^{-\lambda X}$.

The second one is a Sturm-Liouville problem over the domain where $Y(1−Y)$ is positive, i.e. $Y\in [0,1]$. Computer algebra gives $$ G(Y)=c_1\, D_{(\sqrt{2\lambda}−8)/16}\left(\frac{(2Y−1)\lambda^{1/4}}{2^{3/4}}\right) + c_2\, D_{−(\sqrt{2\lambda}+8)/16}\left(\text{i}\frac{(2Y−1)\lambda^{1/4}}{2^{3/4}}\right) $$ where $D_n(z)$ is the parabolic cylinder function.

The initial condition gives $a_1G(Y) = 1$. The BCs give $a_1e^{-\lambda X} G(0) = 0$ and $a_1e^{-\lambda X} G'(1) = 0$. It seems hard to solve it, but this is another problem.