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Homework Help: Differential Annihilator

  1. Mar 13, 2010 #1
    1. The problem statement, all variables and given/known data
    1. y''' + y'' = 8x^2
    2. y'' - 2y' - 3y = 4e^x - 9

    2. Relevant equations
    Annihilator for any polinomial D^n+1

    annihilator for e functions = e^ax = (D-a)^n


    3. The attempt at a solution

    1. getting the complementary eq.
    values of m = 0, 0, -1
    yc= c1e^-x +c2 + xc3

    then we multiply both sides by the annihilator
    D^3<D^2(D+1)> = 8x^2<D^3>

    we get y= c1e^-x +c2 + xc3 + x^2c4 + x^3c5

    and from the original eq. the x^2 term has a coefficient of 8
    therefore c4 = 8
    sice there is no x^3 term then c5 should be zero ?
    and what about c4?

    the answer on the back says the answer is:
    c1+c2x+c3e^-x + 2/3 x^4 + 8/3x^3 + 8x^2

    ===========

    2.values of m=3, 2

    yc=c1e^3x + c2e^-x
    if we multiply by the annihilator alpha = 1 in this case therefore annihilator would be (D-1)
    <D-1><D^2-2D-3>=4e^x-9 <D-1>
    so we have
    yc=c1e^3x + c2e^-x + c3e^x

    how do I go from there?

    thank you!
     
  2. jcsd
  3. Mar 13, 2010 #2
    Use the wronskian and cramer's rule.
     
  4. Mar 13, 2010 #3

    vela

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    You should have one more power of x.
    You're thinking y=8x^2, but it's actually y'''+y''=8x^2. You need to plug your trial solution back into the differential equation and then solve for the coefficients.
     
  5. Mar 13, 2010 #4

    vela

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    m=2?

    You need a slightly different annihilator since

    (D-1)(4 ex-9) = 9

    Again, once you get your trial solution, plug it back into the differential equation and then solve for the coefficients.
     
  6. Mar 17, 2010 #5
    Thank you all. I have one more

    y'' + y' + y = xsinx

    The annihilator for xsinx is (D2+1)2 which would yield 4 answers that are imaginary (+i,-i,+i,-i)
    when you do the particular solution, since the answers are repeating do you put an x before the constant?
    so far i have that yc=e-x/2(c_1cos<(sqr3)/2> + c_2sin<(sqr3)/2>)
    then the annihilator values for D
    yp= c_3cosx + c_4sinx + c_5cosx + c_6sinx
    or is it
    yp= c_3cosx + c_4sinx + xc_5cosx + xc_6sinx

    and if its the latter, to get the coefficients you have to take up to the third derivative. . .using product rules? (...im hoping not) or is there another "easier" way?


    thank you
     
    Last edited: Mar 18, 2010
  7. Mar 17, 2010 #6

    vela

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    Your complementary solution is wrong. You have a third-order differential equation, so you should have three independent solutions. Yours only has two.

    You need the second particular solution you wrote down because you need four linearly independent solutions. You multiply by powers of x to make the repeated solutions independent.
     
  8. Mar 18, 2010 #7
    my bad. . .it is double prime not triple.
    *edited*
    thanks for pointing that out
     
  9. Mar 18, 2010 #8

    vela

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    Well, at least you only have to differentiate yp twice.

    The arguments of the cosine and sine in the complementary solution should have an x in it.
     
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