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How to go about solving this first-order nonlinear differential equation?

  1. Sep 23, 2012 #1
    I saw this post at stackexchange:
    I ran across this post when trying to solve a homework problem. But I have no idea how he got that solution for that. When I use the Euler-Lagrange, I get this diff eq below.

    Here is the simplest form I have managed to get it in:
    [itex]\frac{\alpha}{2\pi}=\frac{\eta \quad \dot{\eta}^2}{\sqrt{1+\dot{\eta}^2}}-\sqrt{1+\dot{\eta}^2}[/itex]

    Where [itex]\alpha[/itex] is a constant. For the simplification, [itex]\eta = (y-y_0)[/itex] & [itex]λ \equiv -2\pi y_0[/itex] is used from the quote above.

    I don't know where to even begin to solve this. Any help would be good.

    Thanks.

    -edit
    Some details that I thought I should add.

    Since [itex]h=f+\lambda g≠h(x)[/itex], I used the special case of the E-L equation:
    [itex]\dot{y}\frac{\partial h}{\partial \dot{y}}-h=const.[/itex]

    But after the substituions to trim it up the special E-L turns into this:
    [itex]\dot{\eta}\frac{\partial h}{\partial \dot{\eta}}-h=const.=\alpha[/itex]
    Where [itex]h[/itex] now is: [itex]h=2\pi\eta\sqrt{1+\dot{\eta}^2}[/itex]
     
    Last edited: Sep 23, 2012
  2. jcsd
  3. Sep 24, 2012 #2
    That should be [tex]
    \frac{\alpha}{2\pi}=\frac{\eta \dot{\eta}^2}{\sqrt{1+\dot{\eta}^2}}-\eta\sqrt{1+\dot{\eta}^2} = \eta [\frac{\quad \dot{\eta}^2}{\sqrt{1+\dot{\eta}^2}}-\sqrt{1+\dot{\eta}^2}]
    [/tex]

    Then you square the equation: [tex]
    (\frac{\alpha}{2\pi})^2= \eta^2 [\frac{\quad \dot{\eta}^4}{1+\dot{\eta}^2} -2 \dot{\eta}^2 + 1 + \dot{\eta}^2] = \eta^2 [\frac{\quad \dot{\eta}^4}{1+\dot{\eta}^2} -\dot{\eta}^2 + 1]

    \\ (\frac{\alpha}{2\pi})^2= \eta^2 \frac{\quad \dot{\eta}^4 -\dot{\eta}^2(1+\dot{\eta}^2) + (1+\dot{\eta}^2)} {1+\dot{\eta}^2}
    = \eta^2 \frac{\quad \dot{\eta}^4 -\dot{\eta}^2 - \dot{\eta}^4 + 1+\dot{\eta}^2} {1+\dot{\eta}^2}
    = \eta^2 \frac{1} {1+\dot{\eta}^2}[/tex]

    I assume you can continue from here.
     
  4. Sep 24, 2012 #3
    Thanks a lot! That made the problem a lot simpler.

    I can't believe I didn't carry that [itex]\eta[/itex] over.
     
    Last edited: Sep 24, 2012
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