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Math Help - [SOLVED] Solve Separable ODE

  1. #1
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    [SOLVED] Solve Separable ODE

    Im stuck on this question, been trying to figure out on the train, just ended up with crossed off working out.


    dy/dx = y^2 x

    Ans: y = 2/C-x^2

    I end up separating it to

    | dy/y^2 = | xdx

    | 1/y^2 dy = | xdx

    |1/y^2 dy = 1/2 x^2 + C


    but i get stuck with trying to integrate y, im missing something, do i have to use the substitution rule?

    help appreciated
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  2. #2
    Eater of Worlds
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    Let's just step through it. Okey-doke.

    \frac{dy}{dx}=xy^{2}

    Separate:

    \frac{1}{y^{2}}dy=xdx

    Integrate:

    \int\frac{1}{y^{2}}dy=\int xdx

    \frac{-1}{y}=\frac{1}{2}x^{2}+C

    y=\frac{-2}{x^{2}+2C}
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  3. #3
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    Prove It's Avatar
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    Quote Originally Posted by galactus View Post
    Let's just step through it. Okey-doke.

    \frac{dy}{dx}=xy^{2}

    Separate:

    \frac{1}{y^{2}}dy=xdx

    Integrate:

    \int\frac{1}{y^{2}}dy=\int xdx

    \frac{-1}{y}=\frac{1}{2}x^{2}+C

    y=\frac{-2}{x^{2}+2C}
    Using alternative notation...


    \frac{dy}{dx} = xy^2

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

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

    \int{y^{-2}\,dy} = \int{x\,dx}


    I'm sure you can go from here...
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  4. #4
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    Quote Originally Posted by Prove It View Post
    Using alternative notation...


    \frac{dy}{dx} = xy^2

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

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

    \int{y^{-2}\,dy} = \int{x\,dx}


    I'm sure you can go from here...
    Yep sure can guys, thanks alot
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