Let $G(x;\lambda)$ be the Green's function.
We require that $x^2G''(x;\lambda)+xG'(x;\lambda)-G(x)=\begin{cases} \mbox{$0$} & \mbox{if } x\in(0,\lambda) \\ \mbox{0} & \mbox{if $x\in (\lambda,1)$} \end{cases} $
Trying a solution of the form $G(x;\lambda)=x^r$ gives two linearly indendent solutions $G(x)=x$ and $G(x)=x^{-1}$
Hence $G(x;\lambda)=\begin{cases} \mbox{$\alpha x+\beta x^{-1}$} & \mbox{if } x\in(0,\lambda) \\ \mbox{$\gamma x+\delta x^{-1}$} & \mbox{if $x\in (\lambda,1)$} \end{cases} $
We have $y(0)=0$ and $y(1)=1$ so this means we have $G(0;\lambda)=G(1;\lambda)=0$
This must mean that $\beta=0$ (since $0^{-1}$ is undefined) and also $\gamma=-\delta$.
So $G(x;\lambda)=\begin{cases} \mbox{$\alpha x$} & \mbox{if } x\in(0,\lambda) \\ \mbox{$\gamma x-\gamma x^{-1}$} & \mbox{if $x\in (\lambda,1)$} \end{cases} $
We require two matching conditions:
(i) $\alpha\lambda=\gamma\lambda-\gamma\lambda^{-1} \implies \alpha=\gamma -\gamma\lambda^{-2}$
(ii) Call $G_1$ the first part of $G$ and $G_2$ the second part of $G$. We need $G_2'(\lambda)-G_1'(\lambda)=\frac{1}{\lambda^2}$:
Now $G_1'(x)=\alpha=\gamma -\gamma\lambda^{-2}$ and $G_2'(x)=\gamma+\gamma x^{-2}$
Hence $G_2'(\lambda)-G_1'(\lambda)= \gamma+\gamma\lambda^{-2}-(\gamma -\gamma\lambda^{-2})=2\gamma\lambda^{-2}=\lambda^{-2}$, giving $\gamma=\frac{1}{2}$
In turn this gives $\alpha=\frac{1}{2}(1-\lambda^{-2}) $
Hence I get $$G(x;\lambda)=\begin{cases} \mbox{$\frac{1}{2}(1-\lambda^{-2}) x$} & \mbox{if } x\in(0,\lambda) \\ \mbox{$\frac{1}{2}( x-x^{-1})$} & \mbox{if $x\in (\lambda,1)$} \end{cases} $$
I don't believe that this is the correct answer, however? I cannot see where I would have gone wrong (if I have gone wrong)? Both matching conditions seem fine to me, I can't see where I would have made a mistake?