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: 1st order linear diff eq. problem

  1. Jul 11, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the general solution of

    [tex]\frac{dy}{dt}[/tex] = 2y +sin(2t)

    2. Relevant equations

    The general solution of a nonhomogeneous ode is the particular solution of the nonhomo plus the solution of the homogeneous ode.: y(t)= y[tex]_{p}[/tex](t)+y[tex]_{h}[/tex](t)

    3. The attempt at a solution

    [tex]\frac{dy}{dt}[/tex] - 2y = sin(2t)

    Initial guess, is that y[tex]_{p}[/tex] = a*Sin(2t)+b*Cos(2t), where a and b are unknown constants later to be found

    Then, the d.e. becomes:

    [tex]\frac{d}{dt}[/tex] ( a*Sin(2t)+b*Cos(2t) ) - 2( a*Sin(2t)+b*Cos(2t) ) = Sin2t

    =2aCos(2t)-2bSin(2t) - 2aSin(2t)+2bCos(2t) = Sin2t

    =(2a+2b)*Cos(2t) + (-2b-2a)*Sin(2t) = Sin2t

    Using the fact that polynomial coefficients must be equal:


    And here is my issue. These simultaneous equations are false, and thus I cannot possibly solve the d.e.

    Can someone tell me where I messed up? Thanks

    PS. I know this can also be found using integrating factors, but i need help on the section with "guessing" the solution
  2. jcsd
  3. Jul 11, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You got a sign wrong when you went from this line

    [tex]\frac{d}{dt} (a \sin 2t + b \cos 2t ) - 2(a \sin 2t + b \cos 2t ) = \sin 2t[/tex]

    to your next one. You should have gotten

    [tex]2a\cos 2t - 2b \sin 2t - 2a \sin 2t - 2b \cos 2t = \sin 2t[/tex]
  4. Jul 11, 2010 #3
    Hey thanks. You hw helpers save my life.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook