# Initial value problem

• Feb 21st 2011, 10:07 PM
dmbocci
Initial value problem
I cant seem to get the right answer:

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

Thanks
• Feb 21st 2011, 10:23 PM
Prove It
This is separable...

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

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

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

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

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

Go from here.
• Feb 21st 2011, 10:35 PM
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
• Feb 21st 2011, 10:52 PM
Prove It
Quote:

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...