Vipul: Created page with "==Definition== 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..."

2012-07-05T18:43:39Z

Created page with "==Definition== In normalized form, this first-order first-degree differential equation looks like: <math>y' + p(x)y = q(x)y^n</math> where <math>n \ne 0,1</math>. (Note..."

New page

==Definition==

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

<math>y' + p(x)y = q(x)y^n</math>

where <math>n \ne 0,1</math>. (Note that the cases <math>n = 0,1</math> give [[first-order linear differential equation]]s).

===Solution method and formula===

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

We get:

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

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

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

Multiply by <matH>1 - n</math> to get:

<math>w' + (1 - n)p(x)w = (1 - n)q(x)</math>

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

Bernoulli differential equation - Revision history

Vipul: Created page with "==Definition== 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..."

Vipul: Created page with "==Definition== 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..."