Initial value problem

**dmbocci** · February 21st 2011, 10:07 PM

I cant seem to get the right answer:

dy/dx = 2y^2 + xy^2 where y(0)=1

Thanks

**Prove It** · February 21st 2011, 10:23 PM

This is separable...

$\displaystyle \frac{dy}{dx} = 2y^2 + x\,y^2$

$\displaystyle \frac{dy}{dx} = y^2(2 + x)$

$\displaystyle y^{-2}\,\frac{dy}{dx} = 2 + x$

$\displaystyle \int{y^{-2}\,\frac{dy}{dx}\,dx} = \int{2 + x\,dx}$

$\displaystyle \int{y^{-2}\,dy} = \int{2 + x\,dx}$ .

Go from here.

**dmbocci** · February 21st 2011, 10:35 PM

Ok, I follow that and I got close to something like that originally. Although, I dont quite get the books answer of y= -1/(x^2/2+2x-1) x=-2

**Prove It** · February 21st 2011, 10:52 PM

Originally Posted by dmbocci

Ok, I follow that and I got close to something like that originally. Although, I dont quite get the books answer of y= -1/(x^2/2+2x-1) x=-2

And what did you get? Show your working and I'll check what you have done wrong, if anything...

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