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Math Help - general and singular solution

  1. #1
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    general and singular solution

    How to proceed to find the general and singular solution of the equation
    3xy=2px2-2p2, p=dy/dx
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  2. #2
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    Re: general and singular solution

    Hey Suvadip.

    For those unfamiliar with non-linear DE's, what techniques have you covered in your course? (Analytic as opposed to numeric which could be applied here)
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  3. #3
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    Re: general and singular solution

    When you write \displaystyle \begin{align*} p^2 \end{align*}, are you meaning \displaystyle \begin{align*} \left( \frac{dy}{dx} \right) ^2 \end{align*} or \displaystyle \begin{align*} \frac{d^2y}{dx^2} \end{align*}?
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    Re: general and singular solution

    (dy/dx)^2
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    Re: general and singular solution

    Quote Originally Posted by Suvadip View Post
    How to proceed to find the general and singular solution of the equation
    3xy=2px2-2p2, p=dy/dx
    I would probably try something like this:

    \displaystyle \begin{align*} 3x\,y &= 2x^2\,\frac{dy}{dx} - 2\left( \frac{dy}{dx} \right) ^2 \\ 2 \left( \frac{dy}{dx} \right) ^2 - 2x^2\,\frac{dy}{dx} + 3x\,y &= 0 \\ \left( \frac{dy}{dx} \right) ^2 - x^2\,\frac{dy}{dx} + \frac{3}{2}x\,y &= 0 \\ \left( \frac{dy}{dx} \right) ^2 - x^2 \, \frac{dy}{dx} + \left( -\frac{1}{2}x^2 \right) ^2 - \left( -\frac{1}{2}x^2 \right) ^2 + \frac{3}{2}x\,y &= 0 \\ \left( \frac{dy}{dx} - \frac{1}{2}x^2 \right) ^2 - \frac{1}{4}x^2 + \frac{3}{2}x\,y &= 0 \\ \left( \frac{dy}{dx} - \frac{1}{2}x^2 \right) ^2 &= \frac{x^2 - 6x\,y}{4} \\ \frac{dy}{dx} - \frac{1}{2}x^2 &= \frac{\pm \sqrt{x^2 - 6x\,y}}{2} \\ \frac{dy}{dx} &= \frac{x^2 \pm \sqrt{x^2 - 6x\,y}}{2} \end{align*}

    I don't know if this looks any easier to solve...
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    Re: general and singular solution

    how to find y from the last expression?
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    Re: general and singular solution

    That is NOT a 'linear' differential equation and so may NOT have a "general" solution at all.
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  8. #8
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    Re: general and singular solution

    y=px+p^2 is also not a linear diff eq, but its general solution is y=cx+c^2
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