# Reducing the Bernoulli Equation

1. May 29, 2012

### phion

I'm practicing first-order linear differential equations, and have come across something I find interesting - being able to reduce nonlinear equations to linear equation with appropriate substitutions. I'll start with the well-known Bernoulli equation, and if there are other ways to do this technique please share!

$y' + P(x)y = Q(x)y^n$

This equation is linear when $n=0$ and has seperable variables if $n=1$. So, in the following development, and assuming that $n≠0$ and $n≠1$, we can multiply by $y^{-n}$ and $(1-n)$ to obtain

$y^{-n}y' + P(x)y^{1-n}=Q(x)$

$(1-n)y^{-n}y'+(1-n)P(x)y^{1-n}=(1-n)Q(x)$

$\frac{d}{dx}$$[y^{1-n}]+(1-n)P(x)y^{1-n}=(1-n)Q(x)$

which is a linear equation with the variable $y^{1-n}$, and if we let $z=y^{1-n}$ we then get

$\frac{dz}{dx}+(1-n)P(x)z=(1-n)Q(x)$.

Now, by multiplying by the integrating factor $e^{∫P(x)dx}$ we can convert the left side of the equation into the derivative of the product $ye^{∫P(x)dx}$, and we get the general solution of the Bernoulli equation!

$y^{(1-n)P(x)dx}=∫(1-n)Q(x)e^{∫(1-n)P(x)dx}dx+C$

That's freakin' qute, right?

2. May 29, 2012

### JJacquelin

3. May 29, 2012

### phion

That it is! Wolfram demonstrates it a bit more eloquently, naturally.