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Solve Seperable diff eq with substitution

  1. Jun 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Solve the differential equation:

    [tex] x \frac {dy}{dx} = y + e^{\frac {y}{x}} [/tex]

    with the change of variable:

    [tex] v = \frac {y}{x} [/tex]

    2. Relevant equations

    3. The attempt at a solution

    I would just like to know if I have successfully solved the problem. Thanks!

    First if we divide by x the equation becomes:

    [tex] \frac {dy}{dx} = \frac {y}{x} + \frac {e^{\frac{y}{x}}}{x} [/tex]

    then if:

    [tex] v = \frac {y}{x} [/tex]


    [tex] \frac {dv}{dx} = \frac {x*\frac{dy}{dx} - y}{x^{2}}= \frac {\frac{dy}{dx}}{x}-\frac{y}{x^{2}} [/tex]


    [tex] \frac {dy}{dx}= \frac{dv}{dx}x+\frac {y}{x}=\frac{dv}{dx}x+v [/tex]

    now I can substitute this back into the original differential equation:

    [tex] \frac {dv}{dx}x+v=v+\frac{e^{v}}{x} [/tex]

    the v on both sides will cancel out with subtraction so you get:

    [tex] \frac{dv}{dx}x=\frac{e^{v}}{x} [/tex]


    [tex] \frac{dv}{dx} = \frac {e^{v}}{x^{2}} [/tex]

    this can be separated like so:

    [tex] \frac {dv}{e^{v}}=\frac {dx}{x^{2}} [/tex]

    [tex] \int e^{-v}dv= \int x^{-2}dx [/tex]

    [tex] -e^{-v}=-\frac {1}{x} + c [/tex]

    [tex] e^{-v}=\frac {1}{x} - c [/tex]

    [tex] -v = ln ( \frac {1}{x} -c) [/tex]

    [tex] v = -ln ( \frac {1}{x} -c) [/tex]

    then to get the y of the original equation:

    [tex] \frac {y}{x} = -ln ( \frac {1}{x} -c) [/tex]

    [tex] y = -xln ( \frac {1}{x} -c) [/tex]
  2. jcsd
  3. Jun 27, 2011 #2

    Char. Limit

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    Gold Member

    Looks good to me.
  4. Jun 27, 2011 #3


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    Science Advisor

    You might find it simpler to write that y= xv so that y'= xv'+ v
    [tex]y'= xv'+ v= v+ \frac{e^v}{x}[/tex]
    [tex]xv'= \frac{e^v}{x}[/tex]
    [tex]e^{-v}dv= \frac{dx}{x^2}[/tex]

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