Bernoulli differential equation

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In normalized form, this first-order first-degree differential equation looks like:

y' + p(x)y = q(x)y^n

where n \ne 0,1. (Note that the cases n = 0,1 give first-order linear differential equations).

Solution method and formula

Divide both sides by y^n. If n > 0, this means that we may be potentially discarding the stationary solution y = 0, and must remember to add that back to the solution family at the end.

We get:

\frac{y'}{y^n} + \frac{p(x)}{y^{n-1}} = q(x)

Now put w = 1/y^{n-1} to get:

\frac{w'}{1 - n} + p(x)w = q(x)

Multiply by 1 - n to get:

w' + (1 - n)p(x)w = (1 - n)q(x)

This is now a first-order linear differential equation in w, and can be solved to get a family of functional solutions for w in terms of x. Plugging back w = 1/y^{n-1} gives a family of functional solutions for y in terms of x. We can now add back y = 0.