> [!theorem] Strategy > > Let $(L)$ be an [[Ordinary Differential Equation|ODE]]. $(L)$ is a **Bernoulli** equation if it has the form > $ > \frac{dy}{dx} + f(x)y = g(x)y^n > $ > If $n = 0$ or $n = 1$, then this is a [[Linear Ordinary Differential Equation|linear ODE]]. Otherwise, make a substitution $v = y^q$. Then > $ > \frac{dv}{dx} = qy^{q - 1}\frac{dy}{dx} > $ > And we can substitute > $ > qy^{q - 1}\frac{dy}{dx} = qy^{q - 1}\paren{g(x)y^n - f(x)y} > $ > and > $ > \frac{dv}{dx} = qg(x)y^{n + q - 1} - qf(x)y^{q} > $ > with > $ > \frac{dv}{dx} = qg(x)y^{n + q - 1} - qf(x)v > $ > Choose $q = 1 - n$, then we can vanish a term > $ > \frac{dv}{dx} = qg(x) - qf(x)v > $ > This should create a [[First Order Linear ODE]]. > $ > \frac{dv}{dx} + (1 - n)f(x)v = (1 - n)g(x) > $