> [!NOTE] Theorem > The [[Solution to Scalar Ordinary Differential Equation|solution]] of the [[Bernoulli Equation]] $\frac{d}{dt} x(t) + p(t) x(t) = q(t) x(t)^{n}$for $n\in\{2,3,\dots\},$ satisfies $\frac{d}{dt}u(t)+(1-n)p(t)u(t) =(1-n)q(t)$where $u(t)=[x(t)]^{1-n}.$ Proof: Let us consider the variable $u = x^{1-n}$. If $x(t)$ now denotes a solution then $\begin{align} \frac{d}{dt} u(t) &= (1-n)x^{-n}(t) \frac{d}{dt} x(t) \\ &= (1-n)x^{-n}(t) (-p(t) x(t) + q(t) x^{n} (t)) \\ &= (1-n) (-p(t) x^{1-n}(t) +(1-n) q(t)) \\ &= (n-1) p(t) u(t) + (1-n) q(t) \end{align}$