= Solution
For $n\ne1$ and a nonzero solution, set $v=y^{1-n}$. Since $v'=(1-n)y^{-n}y'$, multiplying the <Bernoulli differential equation> by $(1-n)y^{-n}$ gives the <linear ordinary differential equation>
$$
\boxed{v'+(1-n)p(x)v=(1-n)q(x).}
$$
An <integrating factor> $\mu=\exp((1-n)\int p(x)\,dx)$ gives $v=\mu^{-1}(C+(1-n)\int\mu q\,dx)$. Recover $y$ from $y^{1-n}=v$ on an appropriate real or complex branch. This substitution can omit the zero solution, which must be checked in the original equation. For $n=1$, no substitution is needed: $y'+(p-q)y=0$ is already a <linear ordinary differential equation>.
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