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^(1/(1-n)). Example: dy/dx + y/x = x^2y^3. Here n = 3, so v = y^(-2), dv/dx = -2y^(-3)*dy/dx. Dividing original by y^3: y^(-3)*dy/dx + y^(-2)/x = x^2. So -dv/(2dx) + v/x = x^2, i.e., dv/dx - 2v/x = -2x^2. This is linear in v with IF = 1/x^2.