1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Euler's Equation

  1. Nov 11, 2008 #1
    1. The problem statement, all variables and given/known data
    For the functional [tex]J(y(x))=\int^{x1}_{x2}F(x,y,y')dx[/tex], write out the curve [tex]y=y(x)[/tex] for finding the extremas of J where [tex]F(x,y,y')=y'^2+y^2[/tex].

    2. Relevant equations
    Euler's Equations:
    [tex]\frac{\partial f}{\partial y} - \frac{d}{dx}\frac{\partial f}{\partial y'}=0[/tex]
    [tex]\frac{\partial f}{\partial x} - \frac{d}{dx}(f-y' \frac{\partial f}{\partial y'})=0[/tex]

    3. The attempt at a solution
    Using [tex]\frac{\partial f}{\partial y} - \frac{d}{dx}\frac{\partial f}{\partial y'}=0[/tex],
    [tex]\frac{\partial f}{\partial y}=2y[/tex]
    [tex]2y=\frac{d}{dx}\frac{\partial f}{\partial y'}[/tex]
    [tex]y=C*e^x[/tex] Where C is a constant.

    Is this correct? Using the 2nd equation, I get an ugly answer that involves Sinh.
    Last edited: Nov 11, 2008
  2. jcsd
  3. Nov 11, 2008 #2

    The general solution will be y=Ce^(lambda)x.

    So you do y''-y=0

    Which will look like: c(lambda)^2*e^x-c*e^x=0.

    You can find lambda to be + or - 1. So your general solution will be C1*e^x+C2*e^-x=y.

    Also, on the rest of our Homework III and IV, you can assume that it is a constant even if it has an x in it, Sophie said so.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Euler's Equation