Reducing the Bernoulli Equation

[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!

[itex]y' + P(x)y = Q(x)y^n[/itex]

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

[itex]y^{-n}y' + P(x)y^{1-n}=Q(x)[/itex]

[itex](1-n)y^{-n}y'+(1-n)P(x)y^{1-n}=(1-n)Q(x)[/itex]

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

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

[itex]\frac{dz}{dx}+(1-n)P(x)z=(1-n)Q(x)[/itex].

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

[itex]y^{(1-n)P(x)dx}=∫(1-n)Q(x)e^{∫(1-n)P(x)dx}dx+C[/itex]

That's freakin' qute, right?

 

[JJacquelin]

Hi !
it's nothing more than the classical method to solve Bernoulli ODE, isn't it ? :
http://mathworld.wolfram.com/BernoulliDifferentialEquation.html

 

[phion]

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