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Math Help - Exact differential.

  1. #1
    Junior Member ceode's Avatar
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    Exact differential.

    (1+e^x/y)dx + e^x/y ( 1-x/y )dy =0
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    Is this

    \left(1 + \frac{e^x}{y}\right)\,dx + \frac{e^x}{y}\left(1 - \frac{x}{y}\right)\,dy = 0

    or

    \left(1 + e^{\frac{x}{y}}\right)\,dx + e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)\,dy = 0?
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    Junior Member ceode's Avatar
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    2nd one
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    \displaystyle{\left(1 + e^{\frac{x}{y}}\right)\,dx + e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)\,dy = 0}

    If this is an exact equation of the form M\,dx + N\,dy = 0 then \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}.


    \frac{\partial M}{\partial y} = -\frac{x}{y^2}e^{\frac{x}{y}}.


    \frac{\partial N}{\partial x} = -\frac{1}{y}e^{\frac{x}{y}} + \frac{1}{y}e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)

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

     = -\frac{x}{y^2}e^{\frac{x}{y}}

     = \frac{\partial M}{\partial y} as required.


    So it's an exact equation.

    That means that the solution F(x, y) has the property that

    \frac{\partial F}{\partial x} = M and \frac{\partial F}{\partial y} = N.


    Thus \frac{\partial F}{\partial x} = 1 + e^{\frac{x}{y}}

    F = \int{1 + e^{\frac{x}{y}}\,dx}

     = x + y\,e^{\frac{x}{y}} + g(y).


    \frac{\partial F}{\partial y} = e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)

    F = \int{e^{\frac{x}{y}}\left(1 - \frac{x}{y}\right)\,dy}

      = y\,e^{\frac{x}{y}} + h(x) (according to Wolfram)...


    Comparing these two results, it's easy to see that h(x) = x + C and g(y) = C.


    Therefore F =x +  y\,e^{\frac{x}{y}} + C.
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  5. #5
    Junior Member ceode's Avatar
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    Thank you very much
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