Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Superposition Approach HELP PLEASE!

  1. Mar 11, 2007 #1
    Solve this equation by superposition approach (undetermined coefficients); pg. 154 in Zill, according to the formula Y = Yc + Yp

    y''' -2y'' -4y' +8y = 6xe^(2x)

    I got:
    Yc = C1 e^(-2x) + C2 e^(2x) + C3 xe^(2x)
    for the complementary function Yc...

    I'm having trouble finding the particular solutions Yp

    Any help is greatly appreciated
  2. jcsd
  3. Mar 11, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    Your book should give you a standard way to find a particular solution when the right sides is of the form P(x)ekx where P(x) is a polynomial and k is some real number. I believe the particular solution should take the form Q(x)ekx where Q and P have the same degree. Since in your problem, P(x) = 6x is a degree 1 polynomial, let Q(x) = ax + b be an arbitrary degree 1 polynomial. Plug in y = (ax+b)e2x into your differential equation, and solve for a and b.
  4. Mar 11, 2007 #3
    since 6xe^(2x) matches the general term C3xe^(2x), the 6x has to be raised in power to 6x^2; in that case the equation is (Ax^2 + Bx +C)e^(2x)... but when I try to take all 3 derivatives of this function and then substituting them back into the y''' - 2y'' - 4y' + 8y equation, it doesn't match up...

    i get 8Ae^(2x) = 6x^2e^(2x), but the A constant needs an X^(2) in order to work, otherwise you can't solve for it

    any way I can resolve that?
  5. Mar 11, 2007 #4


    User Avatar
    Science Advisor
    Homework Helper

    Show all your work, otherwise how can we tell where you're messing up.
  6. Mar 11, 2007 #5
    The solution formula is Y = Yc + Yp

    Yc = (C1 e^-2x) + (C2 e^2x) + (C3 xe^-2x) <--- since this C3 matches 6xe^2x, the rule is that it has to increased in power and it becomes 6x^2e^2x

    the formula to find Yp when the polynomial in front of e^2x is of 2nd power is
    (Ax^2 + Bx + C) e^2x

    y = (Ax^2 + Bx + C) e^2x
    y' = 2Axe^2x + 2Ax^2e^2x + Be^2x +2Bxe^2x + 2Ce^2x
    y'' = 2Ae^2x + 8Axe^2x + 4Ax^2e^2x + 4Be^2x + 4Bxe^2x + 4Ce^2x
    y''' = 12Ae^2x + 24Axe^2x + 8Ax^2e^2x + 12Be^2x + 8Bxe^2x + 8Ce^2x

    So i take that new form y, and it's three derivatives (y', y'', y''') and plug it back into the original y''' - 2y'' - 4y' + 8y and set it equal to 6x^2e^2x

    everything cancels and i'm left with 8Ae^2x = 6x^2e^2x

    I can't solve for A however, because it doesn't have equal powers with 6x^2
    if it were 8Ax^2e^2x then I could cancel the X's and solve for the A constant
  7. Mar 11, 2007 #6


    User Avatar
    Science Advisor
    Homework Helper

    Okay, I did a little reading and I think this is the approach: if the right hand side is P(x)ekx, then your particular solution takes the form xsQ(x)ekx where deg(Q) = deg(P), and s is the multiplicity of k as a root of the characteristic polynomial of the differential equation. In your case, the characteristic polynomial is:

    z3 - 2z2 - 4z + 8
    = (z - 2)2(z + 2)

    k = 2, and P(x) = 6x. So set Q(x) = (ax+b), and set s = 2, since k (which is 2) is a double root of your characteristic polynomial.

    yp = x2(ax+b)e2x = (ax3 + bx2)e2x.

    Plug this into your diff. eq. and solve for a and b.
  8. Mar 12, 2007 #7
    o wow! that works out... thank's a lot for all your help :smile:
  9. Mar 12, 2007 #8


    User Avatar
    Science Advisor
    Homework Helper

    I always go back to this site when I get rusty with basic differential equation theory.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook