# Seperable and Bernoulli?

• Feb 13th 2011, 04:16 PM
slapmaxwell1
Seperable and Bernoulli?
ok here is the other problem...

dy/dx = (y^2+y)/(x^2+x)

i see that it is separable, but could it also be Bernoulli because of the y squared?
• Feb 13th 2011, 04:37 PM
FernandoRevilla
Yes, it is also a Bernouilli equation.

Fernando Revilla
• Feb 13th 2011, 04:42 PM
slapmaxwell1
my book says that is only separable..??
• Feb 13th 2011, 05:17 PM
FernandoRevilla
Quote:

Originally Posted by slapmaxwell1
my book says that is only separable..??

Don't worry, we quickly make another book. :)

The equation is equivalent to:

$y'+\left(\dfrac{-1}{x^2+1}\right)y=\left(\dfrac{1}{x^2+1}\right)y^2$

That is, it has the form $y'+p(x)y=q(x)y^n$ (Bernouilli equation).

Fernando Revilla
• Feb 13th 2011, 06:34 PM
slapmaxwell1
Quote:

Originally Posted by FernandoRevilla
Don't worry, we quickly make another book. :)

The equation is equivalent to:

$y'+\left(\dfrac{-1}{x^2+1}\right)y=\left(\dfrac{1}{x^2+1}\right)y^2$

That is, it has the form $y'+p(x)y=q(x)y^n$ (Bernouilli equation).

Fernando Revilla

hahaha if we could get it in print before my next exam that would great