 Solution

For $y=\alpha x+\beta x^2$, the derivatives are $y^{\prime }=\alpha +2\beta x$ and $y^{\prime \prime }=2\beta$. Substituting gives $2\beta x^2-2x(\alpha +2\beta x)+2(\alpha x+\beta x^2)=0$. The independent solutions $x,x^2$ have Wronskian $x^2\ne 0$ on $[1,2]$, so they form a fundamental system of solutions.

Divide the inhomogeneous linear differential equation by $x^2$ and use variation of parameters. Choosing integrals from one gives the particular solutionThus $z=Ax+B{x}^2+z_p$. The first boundary condition gives $A=-B$. The integral endpoint terms cancel when differentiating $z_p$, so the second givesConsequentlyCollect the coefficient of $f(t)$ on either side of $t=x$. A suitable Green function isThe two pieces agree at $x=t$, $G(1,t)=0$, and $G_x(2,t)=0$ for $t<2$. For an interior source point its derivative jump isThus multiplication by the leading coefficient $t^2$ gives unit in a ring source strength. Away from $x=t$, both pieces solve the homogeneous linear differential equation. With continuous $f$, the integral representation has the required second derivative and solves the equation pointwise in $(1,2)$.

For uniqueness, the difference of two solutions is $Ax+B{x}^2$. Its boundary conditions give $A+B=0$ and $A+4B=0$, forcing $A=B=0$. Hence the solution is unique. This is the mixed Green function for an Euler operator with exponents one and two.