= Solution
Differentiating the <Wronskian> gives
$$
\begin{aligned}
W'
&=(y_1y_2'-y_2y_1')'\\
&=y_1y_2''-y_2y_1''\\
&=y_1(-py_2'-qy_2)-y_2(-py_1'-qy_1)\\
&=-pW.
\end{aligned}
$$
Thus $W'+pW=0$, the <Abel identity>.
Also,
$$
\left(\frac{y_2}{y_1}\right)'
=\frac{y_1y_2'-y_2y_1'}{y_1^2}
=\frac W{y_1^2}.
$$
If $y_2(x_0)=0$, integration gives the <reduction of order> formula
$$
\boxed{
y_2(x)=y_1(x)\int_{x_0}^x\frac{W(t)}{y_1(t)^2}\,dt}.
$$
For the specified equation, division by $x^2$ gives $p(x)=-1/x$, so $W=Cx$. With $y_1=x^3$, an independent solution is
$$
x^3\int x^{-5}\,dx,
$$
which is a nonzero multiple of $x^{-1}$. Hence
$$
y=Ax^3+\frac Bx.
$$
The conditions at $x=1$ give $A+B=0$ and $3A-B=1$, so
$$
\boxed{y(x)=\frac14\left(x^3-\frac1x\right)}.
$$
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