Past exam of the mathematical methods I course of the University of Oxford / 2018 / 4 / a / Solution

For $n\ne 1$ and a nonzero solution, set $v=y^{1-n}$. Since $v^{\prime }=(1-n)y^{-n}{y}^{\prime }$, multiplying the Bernoulli differential equation by $(1-n)y^{-n}$ gives the linear ordinary differential equationAn integrating factor $\mu =\exp ((1-n)\int p(x)dx)$ gives $v={\mu }^{-1}(C+(1-n)\int \mu qdx)$. 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: $y^{\prime }+(p-q)y=0$ is already a linear ordinary differential equation.