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Math Help - Solution of the ODE

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    Solution of the ODE

    Solution of the ODE, dy/dx = 2xy^2, for which y(0)=1?
    And what is x when y reaches infinity?
    Last edited by tykim; August 6th 2012 at 01:01 AM.
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    Re: Solution of the ODE

    Quote Originally Posted by tykim View Post
    Solution of the ODE, dy/dx = 2xy^2, for which y(0)=1?
    And what is x when y reaches infinity?
    \displaystyle \begin{align*} \frac{dy}{dx} &= 2x\,y^2 \\ y^{-2}\,\frac{dy}{dx} &= 2x \\ \int{y^{-2}\,\frac{dy}{dx}\,dx} &= \int{2x\,dx} \\ \int{y^{-2}\,dy} &= x^2 + C_1 \\ -y^{-1} + C_2 &= x^2 + C_1 \\ y^{-1} &= C - x^2 \textrm{ where }C = C_2 - C_1 \\ y &= \frac{1}{C - x^2} \end{align*}

    When \displaystyle \begin{align*} x = 0, y = 1 \end{align*}, so

    \displaystyle \begin{align*} 1 &= \frac{1}{ C - 0^2 } \\ 1 &= \frac{1}{C} \\ 1 &= C \end{align*}

    So the solution is \displaystyle \begin{align*} y = \frac{1}{1 - x^2} = \frac{1}{(1 - x)(1 + x)} \end{align*}, and if \displaystyle \begin{align*} y \to \infty, x \to -1 \textrm{ or }x \to 1 \end{align*}.
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