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Solving differential equations using substitution

  1. Feb 3, 2006 #1
    Solve the differential equation by making an appropriate substitution

    [tex]x^2dy=(xy+x^2e^\frac{y}{x})[/tex]

    Here was my first attempt:

    Let y=ux, dy=udx+xdu

    [tex]x^2(udx+xdu)=(ux^2+x^2e^\frac{ux}{x})dx[/tex]
    [tex]x^2(udx+xdu)=(ux^2+x^2e^u)dx[/tex]

    [tex]x^2udx+x^3du=x^2udx+x^2dx+e^udx[/tex]

    [tex]x^3du=x^2dx+e^udx[/tex]

    And after that I got stuck.
    My teacher said to pick the differential that would be the easiest to integrate and find the substitutions for that. So naturally that would be the dy differential.


    So then I did it the other way:

    Let x=vy, dx=vdy+ydu

    [tex]v^2y^2dy=(y^2v+y^2v^2e^2\frac{y}{vy})(vdy+ydu)[/tex]

    [tex]vy^2(vdy)=vy^2(1+ve^\frac{1}{v})(vdy+ydu)[/tex]

    [tex]vdy=vdy+v^2e^\frac{1}{v}dy+ydv+yve^\frac{1}{v}dv[/tex]

    [tex]-v^2e^\frac{1}{v}dy=ydv+yve^\frac{1}{v}dv[/tex]

    [tex]-v^2e^\frac{1}{v}dy=y(1+ve^\frac{1}{v})dv[/tex]

    [tex]\frac{1}{y}dy=(\frac{1+ve^\frac{1}{v}}{-v^2e^\frac{1}{v}})dv[/tex]

    Integrating this gives me the following result:

    [tex]ln|y|=-ln|v|-\frac{1}{e^\frac{1}{v}}[/tex]

    substituting [tex]\frac{x}{y}[/tex] back in for v we get

    [tex]ln|y|=-ln|\frac{x}{y}|-\frac{1}{e^\frac{1}{v}}[/tex]
     
    Last edited: Feb 3, 2006
  2. jcsd
  3. Feb 5, 2006 #2
    is this your diff eq
    [tex]x^2dy=(xy+x^2e^\frac{y}{x})[/tex]
    or do you mean this
    [tex]x^2 \frac{dy}{dx} =(xy+x^2 e^\frac{y}{x})[/tex]
    if the first one was your question and u didnt make a typing eerror then in taht case i cant get a closed form solution the best i can come up with is
    [tex] y = \frac{y}{x} + e^{\frac{y}{x}} [/tex]
    but i could be wrong!
     
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