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Math Help - Initial value problem

  1. #1
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    Initial value problem

    I cant seem to get the right answer:

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

    Thanks
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  2. #2
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    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.
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  3. #3
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    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
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  4. #4
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    Quote Originally Posted by dmbocci View Post
    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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