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Math Help - Innocent looking 1st order DE

  1. #1
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    Innocent looking 1st order DE

    2y e^{y/x} \frac{dy}{{dx}} - 2x - y = 0

    I'm not sure how difficult you guys at this forum might find this. It took me a long long long time to figure out (I am not a math major). Should need at least two substitutions.

    (The solution, in implicit form, is y^2 - x^2 e^{y/x} = C).
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  2. #2
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    Quote Originally Posted by milodino View Post
    2y e^{y/x} \frac{dy}{{dx}} - 2x - y = 0

    I'm not sure how difficult you guys at this forum might find this. It took me a long long long time to figure out (I am not a math major). Should need at least two substitutions.

    (The solution, in implicit form, is y^2 - x^2 e^{y/x} = C).
    If you divide your ODE by x then

    2\frac{y}{x} \,e^{\frac{y}{x}} \,\frac{dy}{{dx}} - 2 - \frac{y}{x} = 0

    which suggests the substitution u = \frac{y}{x} and the ODE separates. Not really sure how you got your answer. An ODE that looks a lot like your and gives rise to your solution is

     <br />
\left( 2y \,e^{-\frac{y}{x}} -x\right) y' - 2x + y = 0<br />
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  3. #3
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    u = y/x is one of the substitutions, but it doesn't actually neatly separate after that.

    x \frac{du}{dx} + u = \frac{2+u}{2u e^u + 1}

    Admittedly it does reach a pretty nice form eventually but it looked pretty nasty here.
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