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: 2nd order non-homgenous DE

  1. Oct 31, 2012 #1


    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data

    4y''+4y'+y = cos(2t), y(0)=0, y'(0)=0

    2. Relevant equations


    3. The attempt at a solution

    characteristic roots are repeated, m=-1/2, so

    [itex] y_h = A_1 e^{\frac{-t}{2}}+A_2 te^{\frac{-t}{2}}[/itex]

    undetermined coefficients:

    yp = Acos(2t)+Bsin(2t)

    plugging into original equation and solving, I got A= -15/289 and B= 8/289

    using initial conditions, I got A_1 = 15/289 and A_2 = -47/578


    [itex] y(t) = \frac{15}{289}e^{\frac{-t}{2}} - \frac{47}{578}te^{\frac{-t}{2}} - \frac{15}{289}sin(2t) + \frac{8}{289}cos(2t) [/itex]

    should be the answer, but apparently it's wrong. I have checked over my work and can't find the fault.
    Last edited: Oct 31, 2012
  2. jcsd
  3. Oct 31, 2012 #2


    User Avatar
    Homework Helper

    Your process is correct. You solved L[y] = 0 and your assumption for the solution of L[y] = cos(2t) is also good. It's more than likely an arithmetic error.
  4. Oct 31, 2012 #3


    User Avatar
    Gold Member

    Checked it a second time. I must've typed something into the calculator wrong at the very end of the question, just my luck.
    [itex]-\frac{47}{578}[/itex] should be [itex]-\frac{1}{34}[/itex] instead.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook