Bernoulli's equation: dy/dx + P(x)y = Q(x)$y^n$ where n is not 0 or 1. The trick is to divide by $y^n$: y^(-n)dy/dx + Py^(1-n) = Q. Substitute v = y^(1-n), then dv/dx = (1-n)*y^(-n)*dy/dx. So dv/dx + (1-n)Pv = (1-n)Q, which is a standard linear equation in v. Solve for v, then back-substitute y = v^($\frac{1}{1-n}$). Example: dy/dx + y/x = $x^2$$y^3$. Here n = 3, so v = y^(-2), dv/dx = -2y^(-3)*dy/dx. Dividing original by $y^3$: y^(-3)*dy/dx + y^$\frac{-2}{x}$ = $x^2$. So -$\frac{dv}{2dx}$ + v/x = $x^2$, i.e., dv/dx - 2v/x = -2$x^2$. This is linear in v with IF = 1/$x^2$.