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Nonhomogeneous diff equations method of undeterined coeff.

  1. Oct 20, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the general solution to the diff equation using undetermined coefficients
    y''-2y'-3y = 3te^-1


    2. Relevant equations



    3. The attempt at a solution
    r^2 - 2r -3 = 0
    r = -1, 3
    so y = c1 e^-t + c2e^3t + yp
    since e^-t already exists as a solution, i have to multiply my Yp by t to make sure i don't' end up with the same solution.
    So my Yp with the unknown coeffcient should be,
    yp = At^2e^-t
    should it be (At^2+bt+c)e^-t instead or something else? i've done it both ways and came up with wrong answers.
    I know what to do after figuring out yp, its just i can't get it right with ideas i have right now.
    Any help is greatly appreciated.
    Thank you!
     
  2. jcsd
  3. Oct 20, 2009 #2

    Mark44

    Staff: Mentor

    Yes, go with (At2 + Bt + C)e-t for your particular solution. Show us your work, and we'll figure out what's going wrong.
     
  4. Oct 20, 2009 #3
    thanks for responding, Lets see if i have the yp's correct
    yp = (At^2+Bt+c)e^-t
    y'p = (2At + B)e^-t -(At^2+Bt+c)e^-t
    y''p = 2Ae^-t - (2At + B)e^-t - (2At+B)e^-t -(At^2+Bt+c)e^-t
     
  5. Oct 20, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    In general, it you have a power of t on the right side, you will need to try a polynomial up to that power. If your right side were [itex]te^{at}[/itex] and [itex]e^{at}[/itex] were not already a solution, you would try [itex](At+ B)e^{at}[/itex]. Since, here, [itex]e^{at}[/itex] is a solution, multiply that by t: try [itex](At^2+ Bt)e^{at}[/itex].

    You should find that you do NOT need that "c". (Using it will just give C= 0.)

    You can add those two middle terms and the last term should be "+".
    y"/= 2Ae^-t- 2(2At+ B)e^-t+ (At^2+ Bt+ C)e^-t
     
  6. Oct 20, 2009 #5
    Thanks alot HallsofIvy, it seems i just subbed in wrong for y''p and the "c" ended up canceling out anyways.
     
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