1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor
    Homework Helper

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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook