Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Nonhomogeneous: Undetermined coefficients

  1. Apr 30, 2005 #1
    (d^2x/dt^2)+(w^2)x=Fsin(wt), x(0)=0,x'(0)=0

    Hope that's readable. First part is second derivative of x with respect to t. w is a constant and F is a constant. I need to find a solution to this using method of undetermined coeffecients and I'm confused with all the different variables. Anyone get me started at least?
  2. jcsd
  3. Apr 30, 2005 #2


    User Avatar
    Homework Helper

    Well, first off start by solving the homogenous equation to find the fundamental solution.

    [tex] \ddot{x} + \omega^{2}x = 0 [/tex]

    After that try a Particular solution of the type

    [tex] y_{p} = A x \sin(\omega t) + B x\cos(\omega t) [/tex]

    Remember that if the fundamental solution has already sin and cos, you will need to try a xsin and xcos, like this case.
    Last edited: Apr 30, 2005
  4. Apr 30, 2005 #3
    I got my homogenous equation x''+(w^2)x=0 but I can't find my roots with that w^2 in there.
  5. Apr 30, 2005 #4


    User Avatar
    Homework Helper

    What seems to be the problem? Show me your work.
  6. Apr 30, 2005 #5


    User Avatar
    Homework Helper

    Here, i will start you off

    [tex] \ddot{x} + \omega^{2}x = 0 [/tex]

    we assume a as a solution

    [tex] x(t) = e^{rt} [/tex]

    So we substitute in our ODE

    [tex] r^{2}e^{rt} + \omega^{2}e^{rt} = 0 [/tex]


    [tex] e^{rt}(r^{2} + \omega^{2}) = 0 [/tex]

    because [itex] e^{rt} [/itex] cannot be equal to 0

    [tex] r^{2} + \omega^{2} = 0 [/tex]

    which ends up as

    [tex] r = \pm \omega i [/tex]
    Last edited: Apr 30, 2005
  7. May 1, 2005 #6
    I figured it out, thanks a lot for your help, I was just being dumb.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook