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Math Help - First order differential equation

  1. #1
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    First order differential equation

    Hi,

    I'm new to this forum, so please let me know if I'm posting in the wrong place and I'll move this thread.

    I have the equation:
    (x-y)dx + (x+y)dy = 0

    So far I've rearranged it to get:
    y' = (-x+y) / (x+y)

    But I'm not sure where to go from here.

    Help would be appreciated,

    Thanks
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: First order differential equation

    Quote Originally Posted by dwally89 View Post
    Hi,

    I'm new to this forum, so please let me know if I'm posting in the wrong place and I'll move this thread.

    I have the equation:
    (x-y)dx + (x+y)dy = 0

    So far I've rearranged it to get:
    y' = (-x+y) / (x+y)

    But I'm not sure where to go from here.

    Help would be appreciated,

    Thanks
    The DE equation...

    y^{'} = \frac{y-x}{y+x} (1)

    ... is of the 'Homogeneous type'. Setting \frac{y}{x}= t \implies \frac{dy}{dx}= t + x \frac{dt}{dx} You arrive at
    the DE...

    x\ \frac{dt}{dx} = -\frac{1+t^{2}}{1+t} (2)

    ... where the variables are separated...

    Kind regards

    \chi \sigma
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  3. #3
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    Re: First order differential equation

    Quote Originally Posted by chisigma View Post
    Setting \frac{y}{x}= t \implies \frac{dy}{dx}= t + x \frac{dt}{dx} You arrive at
    the DE...

    x\ \frac{dt}{dx} = -\frac{1+t^{2}}{1+t} (2)
    After you set t=y/x, I don't understand how you get to the next part,

    Thanks

    EDIT: Understand now, thanks
    Last edited by dwally89; July 4th 2011 at 06:44 AM.
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  4. #4
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    Re: First order differential equation

    Quote Originally Posted by dwally89 View Post
    After you set t=y/x, I don't understand how you get to the next part,

    Thanks

    EDIT: Understand now, thanks
    \displaystyle t = \frac{y}{x} \implies y = tx. This is a product of functions, so to get \displaystyle \frac{dy}{dx} you need to use the product rule on \displaystyle tx.
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