I have the next question:
Find Green's function for $$ \frac{\mathrm{d} }{\mathrm{d} x}\left [ x^{2} \cdot \frac{\mathrm{d} G}{\mathrm{d} x}\right ]-n(n+1)\cdot G=\delta (x-x^{'}) $$ It is given that the functions are not entertained at zero and infinity.
After solving a second-order differential equation (by guessing a solution of $x^a$), I came to: $$G(x,x^{'})=A_{1,2}x^{n}+B_{1,2}x^{-(n+1)}.$$
I know I will find one coefficient from $x=x'$, and one more from the integration of both sides of $$ \frac{\mathrm{d} }{\mathrm{d} x}\left [ x^{2} \cdot \frac{\mathrm{d} G}{\mathrm{d} x}\right ]-n(n+1)\cdot G=\delta (x-x^{'}). $$
But how can I find the two remaining coefficients?
Thank you for your help.
To get a finite solution, bounded on $(0,\infty)$, you need $G(x,x')=Ax^n$ for $x<x'$ and $G(x,x')=Bx^{-n-1}$ for $x>x'$. Then continuity and derivative jump of $(x')^{-2}$ at $x=x'$ demand $$ A(x')^n=B(x')^{-(n+1)}\iff B=A(x')^{2n+1},\\ nA(x')^{n-1}+(x')^{-2}=-(n+1)B(x')^{-(n+2)}\iff -(n+1)B=nA(x')^{2n+1}+(x')^{n} $$ so that $B=-\dfrac{(x')^{n}}{2n+1}$ and $A=-\dfrac{(x')^{-(n+1)}}{2n+1}$.
On the connecting conditions
$g(x)=G(x,x')$ satisfies the homogeneous equation outside $x=x'$. One can parametrize this as $g(x)=y_1(x)+u(x-x')(y_2(x)-y_1(x))$ where $u$ is the unit step function. Then the first derivative is $$ g'(x)=y_1'(x)+u(x-x')(y_2'(x)-y_1'(x))+δ(x−x')(y_2(x')-y_1(x')) $$ so that the continuity in $x'$ cancels the singular term. The second derivative is then $$ g''(x)=y_1''(x)+u(x-x')(y_2''(x)-y_1''(x))+δ(x−x')(y_2'(x')-y_1'(x')) $$ Inserting into the defining equation for $g(x)=G(x,x')$, the first terms cancel as solutions of the homogeneous equation, what remains is only the identity of the coefficient of the Dirac-delta, $$ (x')^2(y_2'(x')-y_1'(x'))=1. $$