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Non Homogenous Differential Equation

  1. Dec 2, 2009 #1
    1. The problem statement, all variables and given/known data
    y"' - 9y" +18y' = 30ex
    y(0) = 16
    y'(0) = 14
    y"(0) = 11

    2. Relevant equations

    n/a

    3. The attempt at a solution
    Factor Out
    r(r2 - 9r +18)
    r = 0; r = 6; r =3
    General Equation
    y(x) = c0 + c1e3x + c2e6x
    y'(x) = 3c1e3x + 6c2e6x
    y"(x) = 9c1e3x + 36c2e6x
    c1 = (11 - 36c2)/9
    y'(x) = 14 = 3c1e3x + 6c2e6x
    y'(x) = 14 = ((11 - 36c2)/9)e3x + 6c2e6x
    c2 = -31/30
    c1 = (11 - 36*(-31/30))/9 = 5.35555556
    y(0) = 16 = c0 + c1e3x + c2e6x
    16 +31/30 - 5.35555556 = c0 = 11.6777778
    Solve For A in Aex
    yp = Aex
    y'p = Aex
    y"p = Aex
    y"'p = Aex
    Inputting into the original equation we get.
    Aex - 9Aex +18Aex = 30ex
    Simplifying we get.
    10Aex = 30ex
    Which gives A = 3 and since 3 is a root of the original equation we add an x to differentiate between the two.
    So the final equation SHOULD BE 3xex + 11.6777778 + 5.35555556e3x -31/30e6x but of course its not.
    So my question is, what am I doing wrong?
     
    Last edited: Dec 2, 2009
  2. jcsd
  3. Dec 2, 2009 #2

    Mark44

    Staff: Mentor

    Most of your work is fine, but I believe you have made an error in your calculations for c_0, c_1, and c_2. I used matrix methods to solve for these constants and got c_2 = 25/18. I'm fairly confident of this value, but didn't check it.

    Also, in your last paragraph you say something that isn't true. You got A = 3, which means that your particular solution is y_p = 3e^x. The fact that you got a value of 3 when you solved for A is irrelevant to anything else in this problem. Your general solution will include 3e^x, not 3xe^x.
     
  4. Dec 2, 2009 #3
    When you say matrix method do you just make a matrix like this?

    1 1 1 16
    0 3 6 14
    0 9 36 11

    And then you row reduce?
     
  5. Dec 2, 2009 #4

    Mark44

    Staff: Mentor

    Yes, and yes.
     
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