I been trying to find the Green's function for a particular problem. For the equation $$ q(x) \frac{\mathrm{d}^2u(x)}{\mathrm{d}x^2} + p(x) u(x) =0 \tag{1} $$ where $q(x)$ and $p(x)$ are some functions of $x$. I have solved these and have two linearly independent results. $$ u(x) = c_1 u_1(x) + c_2 u_2(x) $$ with $c_1, c_2$ being arbitrary constants and both $u_1, u_2$ solve equation (1). Now for the boundary conditions. At $x\to +\infty$ we require that the solution goes to zero, only $u_2$ satisfies, another condition is at $x\to1_+$. Only $u_1$ satisfies this. Thus our Green's function can be written as $$ G(x,y) = c_1 \mathcal{H}(y-x) u_1(x) + c_2 \mathcal{H}(x-y) u_2(x). $$ Where $\mathcal{H}$ is the Heaviside step function and $y>1$.
Now the Green's function must be continous at $x = y$, thus we have condition for (for example) $c_1$. But what is the other condition? Does the condition $$ \lim_{x\to y_-} \frac{\partial G(x,y) }{\partial x} -\lim_{x\to y_+} \frac{\partial G(x,y) }{\partial x} = 1 $$ hold even for our special ODE (1)?