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Differential equations

  1. Apr 11, 2014 #1

    wel

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    Show that the nonlinear oscillator $$y" + f(y) =0$$ is equivalent to the system
    $$y'= -z $$,
    $$z'= f(y)$$
    and that the solutions of the system lie on the family of curves
    $$2F(y)+ z^2 = constant $$
    where $$F_y= f(y)$$. verify that if $$f(y)=y$$ the curves are circle.

    =>
    nonlinear oscillator $$y" + f(y) =0$$

    where
    $$y'= -z $$,
    $$z'= f(y)$$

    so that means
    $$z''+z =0$$


    for the solution of the system lie on the family of curves, i was thinking

    $$\frac{d}{dt}[2F(y(t))+z^2(t)]= 2F \frac{dy}{dt} + 2z \frac{dz}{dt}$$

    $$=-2Fz +2zf(y)$$

    $$=-2f(y)z+2zf(y)$$

    $$\frac{d}{dt}[2F(y)+z^2]=0$$

    $$2F(y)+ z^2 = constant $$

    for last part,
    if $$f(y)=y$$ , then the differential equation is $y'' + y =0$, meaning that
    $$y=A cosx +B Sinx$$ and $$z=-y'= - A sinx +B cosx$$.
    And I also know $$z'' + z =0$$
    im trying to connect the above equations.
    So, I can get $$cos^2x + sin^2x =1$$ which show the curves on the circles.

    can someone please check my first,second and last part of answer.
     
    Last edited: Apr 11, 2014
  2. jcsd
  3. Apr 11, 2014 #2

    HallsofIvy

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    This shows that the single equation ##y''+ f(y)= 0## can be written as the two equations ##y'= -z## and ##z'= f(y)##. To show they are equivalent you must also show the other way: that if ##y'= -z## and ##z'= f(y)## then ##y''+ f(y)= 0##.

    where did you get that ##y''= -y##? That would be true ony iin the very simple case that ##f(y)= -y##. If this is a "nonlinear oscillator then ##f## must be nonlinear.


    This isn't true. You are missing a factor of ##2##. Further you have "lost" F on the right. ##\frac{dF}{dy}## is not necessarily equal to ##y##.

    Where did the ##-yz+ zy## come from? What you had before was a single expression that was not equal to anything. Why do

    No, if ##y= A \cos(x)+ B \sin(x)## then ##z= -y'= A sin(x)- B cos(x)##.

     
    Last edited by a moderator: Apr 11, 2014
  4. Apr 11, 2014 #3

    micromass

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  5. Apr 11, 2014 #4

    wel

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    i have edited my answers for the first and second part. Would you someone don't mind checking it please.
     
  6. Apr 11, 2014 #5

    wel

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    correction for last part of answer
    if [itex]f(y)=y[/itex], then the differential equation is [itex]y'' + y =0[/itex], meaning that
    [itex]y=A cosx +B Sinx[/itex] and [itex]z=-y'= - A sinx +B cosx[/itex] are the rotate axes.
    [itex]pA^2+qAB+rB^2=1[/itex]
    [itex]p,q,r[/itex] depends on [itex]x[/itex]
    choose [itex]x[/itex] such that [itex]q=0[/itex]
    [itex]pA^2+rB^2=1[/itex]

    what can i do after that?
     
    Last edited: Apr 11, 2014
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