1. ## 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?

2. Yes, it is also a Bernouilli equation.

Fernando Revilla

3. my book says that is only separable..??

4. 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

5. 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