I have the ODE:
$G(x,s)$ satisfies $$xG''-(2x+1)G'+(x+1)G= \delta (x-s)$$ with $G(0,s)=G(1,s)=0$
I have that the general solution to the ODE $$xy''-(2x+1)y'+(x+1)y=0$$ is $y(x)=c_1 e^x + c_2 e^x x^2$.
So my Green's function is: $f\left(x\right) = \left\{ \begin{array}{lr} G_+ = A(s)e^x+B(s)e^x x^2 & : 0 \le s < x \\ G_- = C(s)e^x+D(s)e^x x^2 & : x < s \le 1 \end{array} \right.\\ $
Using the initial boundary conditions, I get $C=0$ so $G_-=Dx^2e^x$ and $-A=B$ so $G_+=Ae^x-Ae^xx^2=Ae^x(1-x^2$) (I am assuming these are correct).
Using the jump conditions at $x=s$, where $[G]=G_+(s,s)-G_-(s,s)=0$ and $[\delta G/ \delta x]=1$
With this, I get $D=\frac {A(1-s^2)}{s^2}$ so now, $G_-=Ae^x(1-x^2)$
However, this means that $G_-=G_+$ and this doesn't make sense since $[\delta G/ \delta x]=1$ so I will get $0=1$.
I understand this may be a long question, but I am really struggling to see where I am going wrong. I have checked my values for $A,B,C,D$ over and I get them the same every time.
Any helo would be appreciated, thanks.