= Solution
Substitute $y_2=vy_1$ and use $y_1''+py_1'+qy_1=0$ to cancel the terms proportional to $v$. The remaining <ordinary differential equation> is
$$
\boxed{y_1v''+(2y_1'+py_1)v'=0.}
$$
On an interval where $y_1\ne0$, divide through and put $w=v'$:
$$
w'+\left(2\frac{y_1'}{y_1}+p\right)w=0,\qquad w=C\frac{e^{-\int p(x)\,dx}}{y_1(x)^2}.
$$
Thus <reduction of order> gives
$$
\boxed{y_2(x)=y_1(x)\int^x\frac{e^{-\int^s p(r)\,dr}}{y_1(s)^2}\,ds.}
$$
A nonzero multiplicative constant gives an independent solution; adding a constant inside the integral adds a multiple of $y_1$. The <Wronskian> is proportional to $e^{-\int p}$ and is nonzero on the interval. The undivided equation remains meaningful at a zero of $y_1$, but this integral representation then needs continuation rather than direct division.
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