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Homework Help: 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]2y=\frac{d}{dx}2y'[/tex]
    [tex]y=\frac{d^2y}{dx^2}[/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
    Yo,

    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.
     
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