= Mixed Green function for an Euler operator with exponents one and two
{title2=$Lz=x^2z''-2xz'+2z,\quad z(1)=z'(2)=0$}
<variation of parameters> with the fundamental solutions $x,x^2$ gives
$$
G(x,t)=\frac1{3t^3}
\begin{cases}(x^2-x)(t-4),&x\leq t,\\
(t-1)(x^2-4x),&t\leq x.\end{cases}
$$
The kernel is continuous across $x=t$ and its first derivative jumps by $t^{-2}$, which gives <unit in a ring> source after multiplication by the leading coefficient. It satisfies both endpoint conditions. The homogeneous solution $Ax+Bx^2$ satisfies them only when $A=B=0$, proving uniqueness.
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