= Solution
For two solutions $y_1,y_2$, define their <Wronskian> by
$$
W=y_1y_2'-y_1'y_2.
$$
Differentiation and substitution of the <second-order linear differential equation> give
$$
W'=y_1y_2''-y_1''y_2
=y_1(-py_2'-qy_2)-(-py_1'-qy_1)y_2=-pW.
$$
This proves the <Abel identity> and hence
$$
\boxed{W(x)=C\exp\left(-\int_{x_0}^x p(r)\,dr\right).}
$$
On an interval where the known solution $y_1$ does not vanish,
$$
\left(\frac{y_2}{y_1}\right)'=\frac{W}{y_1^2}.
$$
Integrating gives the <reduction of order> formula
$$
\boxed{y_2(x)=y_1(x)\int_{x_0}^x
\frac{\exp(-\int_{x_0}^s p(r)\,dr)}{y_1(s)^2}\,ds.}
$$
Choosing the normalization $C=1$ makes the <Wronskian> nonzero, so this solution is linearly independent of $y_1$. To verify it really solves the equation, write $y_2=y_1h$. Its equation reduces to
$$
y_1h''+(2y_1'+py_1)h'=0;
$$
the integrand gives $h'=\exp(-\int p)/y_1^2$, which satisfies this equation on differentiation. Adding an integration constant only adds a multiple of $y_1$. The formula is local where $y_1\ne0$; at a zero of $y_1$ with regular coefficients, the actual solution can be continued by initial-value existence and uniqueness, even if its integral representation has an apparent singularity.
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