Bernoulli Substitution

Differential Equations · Axiom Academy

One change of variable turns a whole family of nonlinear equations into linear ones you already know how to solve. A Bernoulli equation is one that can be written in this form: Bernoulli form — genuinely nonlinear whenever and It is almost the standard linear form y' + P(x)y = Q(x) from the integrating-factor lesson. The only difference is that exponent n on the right. To see why that single exponent matters so much, freeze x and graph the right-hand side purely as a function of y . "Linear in y " means that graph is a straight line . Watch what the exponent does to it. Only two values of n leave the graph straight, and both of them collapse the problem back to something already covered: y' + P(x)y = Q(x) — already linear. Use the integrating factor directly. — linear and separable. No substitution needed. 2. The Substitution v = y^ 1-n The fix is to stop tracking y and start tracking a stretched version of it. Define a new unknown: That exponent is not a guess. Differentiating gives v' = (1-n)y^ -n y' , so the rate at which v changes relative to y carries a factor of y^ -n — precisely the factor needed to cancel the troublesome y^ n . The animation makes that stretch visible: evenly spaced values of y do not map to evenly spaced values of v , and the live ratio is the conversion factor. Now run the algebra. Multiply the whole Bernoulli equation by (1-n)y^ -n :

This is the written version of the interactive lesson above. See the full Differential Equations course.