= Solution
{c}
With $y=xv$, direct differentiation gives
$$
x^2y''-2xy'+(x^2+2)y=x^3(v''+v).
$$
The homogeneous <ordinary differential equation> therefore has general solution $y=x(A\cos x+B\sin x)$. Choose
$$
y_1(x)=x\cos x,\qquad y_2(x)=x\sin x,
$$
so $y_1(\pi/2)=0$ and $y_2(\pi)=0$. Their <Wronskian> in the convention $W=y_1y_2'-y_1'y_2$ is $W(x)=x^2$. There is no nonzero homogeneous solution satisfying both <boundary conditions>, since the first forces $B=0$ and the second forces $A=0$.
For the operator as printed, whose leading coefficient is $x^2$, the <Green's function> is
$$
\boxed{G(x,s)=\begin{cases}
\displaystyle\frac{x\cos x\sin s}{s^3},&\pi/2\leq x\leq s\leq\pi,\\
\displaystyle\frac{x\sin x\cos s}{s^3},&\pi/2\leq s\leq x\leq\pi.
\end{cases}}
$$
It satisfies the endpoint <boundary conditions>, is continuous across $x=s$, and its first derivative has jump $1/s^2$. Multiplying that jump by the leading coefficient $s^2$ produces the unit source required by a <Green's function>. This factor matters because the operator was not normalized to leading coefficient one; it is an instance of a <Green function under dependent-variable rescaling>.
Thus $y(x)=\int_{\pi/2}^\pi G(x,s)f(s)\,ds$. For $f(s)=s^3$, split the integral at $s=x$ to obtain
$$
\begin{aligned}
y(x)&=x\sin x\int_{\pi/2}^x\cos s\,ds+x\cos x\int_x^\pi\sin s\,ds\\
&=x\sin x(\sin x-1)+x\cos x(1+\cos x).
\end{aligned}
$$
Therefore
$$
\boxed{y(x)=x(1+\cos x-\sin x).}
$$
The substitution above gives $v''+v=1$, so the forcing is indeed $x^3$, and both endpoint values vanish.
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