= Solution
{c}
For $y=\alpha x+\beta x^2$, the derivatives are $y'=\alpha+2\beta x$ and $y''=2\beta$. 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> $x^2\neq0$ 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 solution
$$
z_p(x)=-x\int_1^x\frac{f(t)}{t^2}dt
+x^2\int_1^x\frac{f(t)}{t^3}dt.
$$
Thus $z=Ax+Bx^2+z_p$. The first <boundary condition> gives $A=-B$. The integral endpoint terms cancel when differentiating $z_p$, so the second gives
$$
0=3B-\int_1^2\frac{f(t)}{t^2}dt
+4\int_1^2\frac{f(t)}{t^3}dt.
$$
Consequently
$$
z(x)=\frac{x^2-x}{3}\int_1^2
\left(\frac1{t^2}-\frac4{t^3}\right)f(t)dt
-x\int_1^x\frac{f(t)}{t^2}dt
+x^2\int_1^x\frac{f(t)}{t^3}dt.
$$
Collect the coefficient of $f(t)$ on either side of $t=x$. A suitable <Green function> is
$$
\boxed{G(x,t)=\frac1{3t^3}
\begin{cases}
(x^2-x)(t-4),&x\leq t,\\
(t-1)(x^2-4x),&t\leq x.
\end{cases}}
$$
The 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 is
$$
G_x(t+,t)-G_x(t-,t)=\frac1{t^2}.
$$
Thus 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 $Ax+Bx^2$. Its <boundary conditions> give $A+B=0$ and $A+4B=0$, forcing $A=B=0$. Hence \b[the solution is unique]. This is the <mixed Green function for an Euler operator with exponents one and two>.
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