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O.D.E. Undetermined coefficients solution

  1. Oct 23, 2009 #1
    1. The problem statement, all variables and given/known data
    i have this problem

    [tex] y'' - y' - 2y= -6xe^{-x} [/tex]

    2. Relevant equations

    this is the homogenous solution : [tex] y_{h}= c_{1} e^{-x} + c_{2} e^{2x} [/tex]

    3. The attempt at a solution first i can do the problem but what is perplexing me is why of these two tentative particular solutions the latter is chosen, [tex] y_{p1}= Axe^{-x} [/tex]
    y_{p2}= x(A + Bx)e^{-x} [/tex]
    the former leads to an impossible situation but none the less its not appearing in the homogenous solution , the latter looks like the choice made when the roots of the auxilliary equation of the differential equation are real and the same , but in this case they are not real and the same , but this is what works , why would one choose the latter?
  2. jcsd
  3. Oct 23, 2009 #2


    Staff: Mentor

    This is the short answer, which might not be satisfactory to you. If the diff. equation happened to be y'' - y' -2y = -6e-x, then the first choice for a particular solution would be the one that would work. But because the right side is -6xe-x, you're going to need the next higher power of x to take that into account. IOW, yp = (Ax + Bx2)e-x. Hope that makes sense.
  4. Oct 23, 2009 #3
    thanks for the reply, makes perfect sense for the most part, but why do you have to take the next higher power of x into account?
  5. Oct 23, 2009 #4


    Staff: Mentor

    OK, here's a bit more detail that involves the concept of annihilators. If we look at the homogeneous version of your DE, the characteristic equation is r2 - r -2 = 0, which can be factored into (r - 2)(r + 1) = 0.

    Looking at the homogeneous equation in terms of derivative operators, it's not too hard to see that it can be written as (D2 - D -2)y = 0. This equation can be factored into (D - 2)(D + 1)y = 0. The D + 1 operator annihilates e-x or any constant multiple of it. In case you aren't following, (D + 1)e-x means d/dx(e-x + e-x), which is clearly equal to zero. Similarly, the operator D - 2 annihilates e2x.

    When you have repeated roots in the characteristic equation, you have repeated annihilator factors. For example, if the homogeneous DE were y'' + 2y' + y = 0, the char. equation would be r2 + 2r + 1 = 0, or (r + 1)2 = 0.

    The operator form of this DE would be (D + 1)2y = 0. One solution to this DE is e-x, and as it turns out, xe-x is another linearly independent solution. All solutions of this DE are of the form y = c1e-x + c2xe-x. The D + 1 operator annihilates e-x, and the (D + 1)2 operator annihilates e-x and xe-x.

    If the DE were y''' + 3y'' + 3y' + y = 0, the characteristic equation would be (r + 1)3 = 0. Following the logic as before, three linearly independent solutions are e-x, xe-x, and x2e-x. The (D + 1)3 operator annihilates e-x, xe-x, and x2e-x.

    OK, now back to your original nonhomogeneous problem: y'' - y' - 2y = -6xe-x. This can be rewritten as (D2 - D - 2)y = -6xe-x, or equivalently as (D - 2)(D + 1)y = -6xe-x.

    My previous work shows that the (D + 1)2 operator annihilates xe-x, so we'll tack another couple of factors of (D + 1) to make the nonhomogeneous equation a homogeneous one.

    (D - 2)(D + 1)3)y = (D + 1)2(-6xe-x) = 0.

    For this homogeneous equation, the set of linearly independent solutions is {e2x, e-x, xe-x, x2e-x}.

    The first two are the solutions to the original homogeneous equation; the last two are solutions to the new homogeneous equation, and should therefore be selected as particular solutions to your nonhomogeneous problem.
  6. Oct 24, 2009 #5
    Cool thanks, i had not met this perspective, thanks, il look at it further.
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