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Solution strange, differential equation

  1. May 28, 2008 #1
    I have big problem with solve this equation:
    I can't go ahead, because I don't know how solve this
  2. jcsd
  3. May 28, 2008 #2


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    Since the independent variable, t, does not appear explicitely in the equation, that is a candidate for "quadrature".

    Let v= dx/dt. Then d2x/dt2= dv/dt. But by the chain rule, dv/dt= (dv/dx)(dx/dt). And dx/dt= dv/dt, of course. That is d2x/dt2= (dx/dt)(dv/dt)= vdv/dx.

    Your differential equation can be reduced to vdv/dx= -ksin(x) which is "separable":
    mvdv= - k sin(x)dx. Integrating both sides, (m/2)v2= k cos(x)+ C. (That square is the reason for the name "quadrature".) Then v2= (2k/m) cos(x)+ C' or
    v= dx/dt= sqrt((2k/m) cos(x)+ C').

    I assume that, in order to get rid of that constant of integration, C', you must have some initial condition on dx/dt.
  4. May 28, 2008 #3
    Great. Thanks Hallsoflvy.
  5. May 28, 2008 #4
    But this equation [tex]\frac{dx}{dt}=\sqrt{\frac{2k}{m}cosx+C^{'}}[/tex] can I solve? Can I obtain x(t)? For t=0 x=0. It's a equation of motion.
    Last edited: May 28, 2008
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