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Differential equations

  1. Apr 11, 2014 #1

    wel

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    Gold Member

    Simple harmonic motion: ##y'= -z,~z'= f(y)##


    the modified explicit equation are


    $$y'=-z+\frac {1}{2} hf(y)$$


    $$y'=f(y)+\frac {1}{2} hf_y z$$
    and deduce that the coresponding approximate solution lie on the family of curves
    $$2F(y)-hf(y)y+z^2=\textrm{constant}$$


    where ##f_y= f(y)##.
    What are the curves when verify ##f(y)=y##


    =>


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


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


    but i am not sure


    if ##f(y)=y##, then the differential equation is ##y'' + y =0##, meaning that ##y=A \cos x +B \sin x## and ##z=-y'= - A \sin x +B \cos x##.
     
    Last edited by a moderator: Apr 11, 2014
  2. jcsd
  3. Apr 11, 2014 #2

    micromass

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    2016 Award

    Dear wel,

    This thread involves derivatives and diff eq. Thus it belongs in the Calculus & Beyond homework and not precalculus. Please post there in the future.
     
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