Seperable and Bernoulli?

**slapmaxwell1** · February 13th 2011, 03:16 PM

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?

**FernandoRevilla** · February 13th 2011, 03:37 PM

Yes, it is also a Bernouilli equation.

Fernando Revilla

**slapmaxwell1** · February 13th 2011, 03:42 PM

my book says that is only separable..??

**FernandoRevilla** · February 13th 2011, 04:17 PM

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

**slapmaxwell1** · February 13th 2011, 05:34 PM

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

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