Find the Green's function for the BVP $$y''-\frac1xy'=0 \ \ ; \ \ y(0)=y(1)=0$$
Clearly the operator is not self-adjoint, so the equivalent self-adjoint equation is $$\left(\frac{y'}{x}\right)'=0$$ Therefore the Green's function could be taken as $$G(x,t)=\begin{cases}A+Bx^2 & \text{if} \ \ 0\leq x<t \\ C+Dx^2 & \text{if} \ \ t<x\leq1\end{cases}$$ Now boundary conditions give $A=C+D=0$. Giving $$G(x,t)=\begin{cases}Bx^2 & \text{if} \ \ 0\leq x<t \\ C(1-x^2) & \text{if} \ \ t<x\leq1\end{cases}$$ Continuity of $G$ at $x=t$ gives $Bt^2=C(1-t^2)$. Also the jump discontinuity of $\displaystyle\frac{\partial G}{\partial x}$ at $x=t$ gives $$-2Ct-2Bt=t\implies B+C=-\frac12$$ Therefore $C=-\frac{t^2}2$ and $B=\frac{t^2-1}2$. Hence $$G(x,t)=-\begin{cases}\dfrac{(1-t^2)x^2}2 & \text{if} \ \ 0\leq x<t \\ \dfrac{(1-x^2)t^2}2 & \text{if} \ \ t<x\leq1\end{cases}$$ which is clearly wrong since the correct answer is $$G(x,t)=-\begin{cases}\dfrac{(1-t^2)x^2}{2t} & \text{if} \ \ 0\leq x<t \\ \dfrac{(1-x^2)t}2 & \text{if} \ \ t<x\leq1\end{cases}$$ I checked that taking the jump $1$ instead of $t$ would give the correct answer, but here in self-adjoint form the coefficient is $\displaystyle\frac1x$, whose reciprocal is what is taken as jump. What is possibly wrong here? Any help is appreciated.
In order that the solution to the non-homogeneous ODE $$ y''-\frac{1}{x}y'=f(x), \qquad y(0)=y(1)=0 \tag{1} $$ be given by $$ y(x)=\int_0^1G(x,t)f(t)\,dt, \tag{2} $$ the Green's function $G(x,t)$ must be the solution to $$ \left(\frac{\partial^2}{\partial x^2}-\frac{1}{x}\frac{\partial}{\partial x}\right)G(x,t)=\delta(x-t), \qquad G(0,t)=G(1,t)=0 \qquad(0<t<1). \tag{3} $$ Eq. $(3)$ can be rewritten as $$ \frac{\partial}{\partial x}\left(\frac{1}{x}\frac{\partial}{\partial x}G(x,t)\right)=\frac{1}{x}\delta(x-t), \tag{4} $$ which implies $$ \lim_{\epsilon\to 0^{+}}\int_{t-\epsilon}^{t+\epsilon}\frac{\partial}{\partial x}\left(\frac{1}{x}\frac{\partial}{\partial x}G(x,t)\right)dx =\lim_{\epsilon\to 0^{+}}\int_{t-\epsilon}^{t+\epsilon}\frac{1}{x}\delta(x-t)dx =\frac{1}{t} $$ $$ \implies \lim_{\epsilon\to 0^{+}}\left.\frac{1}{x}\frac{\partial}{\partial x}G(x,t)\right|_{t-\epsilon}^{t+\epsilon}=\frac{1}{t} \implies \lim_{\epsilon\to 0^{+}}\left.\frac{\partial}{\partial x}G(x,t)\right|_{t-\epsilon}^{t+\epsilon}=1. \tag{5} $$ Therefore, the jump discontinuity of $\frac{\partial G}{\partial x}$ at $x=t$ is $1$, not $t$.