Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Need help on a nonlinear first order DE question

  1. Sep 22, 2012 #1
    Hi Everyone

    I tried all ways I can to solve (y^2+x^2+x)y'-y=0, but still cann't find a way to solve it.

    Could anybody help on it?

    Thank you very much!
     
  2. jcsd
  3. Sep 23, 2012 #2
    Try and get it exact. You can write it as:

    [tex](y^2+x^2)dy+xdy-ydx=0[/tex]

    Ok, divide by x^2 to get:

    [tex]\left(1+\frac{y^2}{x^2}\right)dy+\frac{xdy-ydx}{x^2}=0[/tex]

    bingo-bango right?
     
  4. Sep 30, 2012 #3
    you wouldn't go one further?
     
  5. Sep 30, 2012 #4
    Hi !
    Another method consists in converting to polar coordinates, which leads to a linear ODE very easy to solve.
    The result can be expressed on a parametric form : x=(c-t)/tan(t) , y= c-t
    or as x fonction on y :
    x = y / tan(c-y)
     
  6. Sep 30, 2012 #5
    Well, one is a differential of just y and the other is a differential of the quantity y/x. So then if we have:

    [tex]\left(1+\frac{y^2}{x^2}\right)dy+\frac{xdy-ydx}{x^2}=0[/tex]

    we could re-write that as:

    [tex]dy=\frac{d(y/x)}{1+\left(\frac{y}{x}\right)^2}[/tex]

    Now, that right side is equivalent to the expression:

    [tex]\frac{du}{f(u)}[/tex]

    right? So we just integrate it:

    [tex]\int dy=\int\frac{d(y/x)}{1+\left(\frac{y}{x}\right)^2}[/tex]

    and get:

    [tex]y=\arctan(y/x)+c[/tex]
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook