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Thread: A separable first-order differential equation

  1. #1
    Junior Member
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    A separable first-order differential equation



    Please, help me to solve this problem. I don't know how to deal with (x) near dy and dx...
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  2. #2
    Senior Member Peritus's Avatar
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    $\displaystyle
    \begin{gathered}
    ye^y dy = \frac{1}
    {x}dx \hfill \\
    \int {ye^y dy = ye^y - \int {e^y dy} = e^y \left( {y - 1} \right) = \ln x} + C \hfill \\
    \end{gathered}
    $
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  3. #3
    Super Member
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    One more step just for fun:

    $\displaystyle \frac{1}{e}\left(e^y(y-1)\right)=(ln(x)+c)\frac{1}{e}$

    $\displaystyle (y-1)e^{y-1}=\frac{1}{e}(ln(x)+c)$

    Taking the Lambert-W function of both sides:

    $\displaystyle y-1=\textbf{W}\left[\frac{1}{e}(ln(x)+c)\right]$

    or:

    $\displaystyle y(x)=1+\textbf{W}\left[\frac{1}{e}(ln(x)+c)\right]$
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