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Homework Help: 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

    AKG

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    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

    AKG

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    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

    AKG

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    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

    AKG

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    I always go back to this site when I get rusty with basic differential equation theory.
     
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