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Solve x'' = cox(x)

  1. Feb 10, 2007 #1
    im new to 2nd order differential eqns and im getting myself all confused.
    ive tried so hard but i just cant solve it.
    can someone please help me with tips to solve/solutions?

    x'' = cos(x)
  2. jcsd
  3. Feb 10, 2007 #2
    Try using [tex]\ddot{x} = \dot{x}\frac{d\dot{x}}{dx}[/tex]

    That will allow you to turn it into a first order equation at the very least.
  4. Feb 10, 2007 #3
    thankyou!! i didnt think of that!
  5. Feb 12, 2007 #4


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    I don't think that's correct.
    You should have written

    [tex] \ddot{x}=\frac{1}{\dot{x}}\frac{d}{dt}\left(\frac{\dot{x}^{2}}{2}\right) [/tex]
  6. Feb 12, 2007 #5


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    No, Dextercioby, the first response was correct. If you let u= dx/dt, then du/dt= d2x/dt2. But we can apply the chain rule to the left side: du/dt= (du/dx)(dx/dt)= u du/dx since u= dx/dt. d2u/dt2= u du/dx.

    This method is called "quadrature" (because the integral involves u2) and can be used to reduce the order of a differential equation when the independent variable, t, does not appear explicitely in the equation.
  7. Mar 30, 2007 #6
    x" = cosx

    its x-missing

    v = x'

    v' = x" = v dv/dx

    v dv/dx = cosx

    vdv = cosx dx

    v^2 = 2 sinx

    (x')^2 = 2 sinx

    x' = ( 2 sinx )^1/2

    dx / (sinx )^1/2 = 2^1/2 dy

    Q = 2^1/2 y + C

    such that Q = http://integrals.wolfram.com/index.jsp
    Last edited: Mar 30, 2007
  8. Mar 30, 2007 #7
    You cannot neglect the constant of integration.


    [tex]y^{\prime \prime} = \cos{y}[/tex]

    multiply by [tex]y^{\prime}[/tex]

    gives you

    [tex]y^{\prime} y^{\prime \prime} = y^{\prime} \cos{y}[/tex]

    integrate, and you get

    [tex] \frac{1}{2}y^{\prime 2} = \sin{y} + C[/tex]

    Note the constant of integration which must not be ignored.

    Therefore, you have

    [tex]y^{\prime} = \sqrt{2 \sin{y} + C} [/tex]

    which isn't easily soluble -- the only way to get an answer is to use elliptic integrals -- have a look at this Wolfram page.
  9. Apr 4, 2007 #8


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    It can't be, since [tex] \frac{d\dot{x}}{dx}=0 [/tex].
  10. Apr 4, 2007 #9


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    Where in the world did you get that idea?

    If x(t)= et, then x'= et= x. In other words,
    [tex]\frac{dx'}{dx}= 1[/tex]
    not 0.
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