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: Differential Equations - Method of Undetermined Coefficients

  1. Sep 25, 2011 #1
    1. The problem statement, all variables and given/known data
    I've been having problems with a number of these things, here's the first one:
    y'' -2y' -3y = -3te-t

    2. Relevant equations
    I know that the general solution will be
    y = yh + yp
    where yh is the general solution to the homogeneous equation, and yp
    is the particular solution of the non-homogeneous equation.

    3. The attempt at a solution
    I got the homogeneous solutions very easily, but I'm tricked by how to solve for yp. I understand the principle of the solution if the right side of the equation was simply -3e-t, but when the extra t is thrown in there, my understanding breaks down.

    I tried yp=t2e-t, differentiated twice, and then input these expressions into the equation. I just realized that I was supposed to use an unknown coefficient (A we'll call it) with my guess at yp. Since the terms on the right side are a product, will I just use A, or will I need some B as well? Thanks for any help.
  2. jcsd
  3. Sep 25, 2011 #2


    User Avatar
    Homework Helper

    You need to use unknown coefficients, in plural :wink:. Try the solution in the form yp=(at2+bt+c)e-t.

  4. Sep 25, 2011 #3


    User Avatar
    Science Advisor

    IF [itex]e^{-t}[/itex] were not already a solution to the associated homogeneous equation, since the right side is [itex]te^{-t}[/itex], you would try [itex]y_p= (At+ B)e^{-t}[/itex]. Because [itex]e^{-t}[/itex] IS a solution to the associated homogeneous equation, you should try [itex]y_p= (At^2+ Bt)e^{-t}[/itex].

    (You don't really need the "c" in ehild's suggestion. It wouldn't hurt, but you would find that c= 0.)
  5. Sep 25, 2011 #4
    Thanks y'all! I'll be trying a little later this afternoon.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook