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2nd Order nonhomogeneous ODE using Undetermined Coefficients

  1. Aug 4, 2009 #1
    1. The problem statement, all variables and given/known data

    Find General Solution:
    y"+6y'+9y=e-3x-27x2

    2. Relevant equations



    3. The attempt at a solution

    I know you have yh which is the general solution to the left side of the equation set to 0 and then fine the particular solution.

    When i try to find yp1 I get yp1=Ae-3x, y'p1=-3Ae-3x, and y"p1=9Ae-3x

    Substituting that into the left side we get
    9Ae-3x+6(-3Ae-3x)+9(Ae-3x)=Ae-3x
    Canceling out Ae-3x we get
    9A-18A+9A=1
    0A=1

    SO does this mean when i'm finding the general solution y=yh+yp1+yp2 that yp1 is gonna be 0?

    I Know this is a very simple question but my mind is running in circles!
     
  2. jcsd
  3. Aug 4, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    roots of the characteristic polynomial are -3 and -2, so your PI for the e-3x is not Ae-3x but Axe-3x, the PI for -27x2 is Bx2+Cx+D, so combining those two, your PI would be

    [tex]y_{PI} = Axe^{-3x}(Bx^2+Cx+D)[/tex]
     
  4. Aug 4, 2009 #3
    The characteristic polynomial being the left side of the equation?

    Isn't it a double root -3? (y+3)2?
     
  5. Aug 4, 2009 #4

    Mark44

    Staff: Mentor

    No, -2 is not a root of the characteristic equation. The only root is -3, and this is a double root.
     
  6. Aug 4, 2009 #5

    rock.freak667

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    Homework Helper

    oh right it is, I did that wrong in my head...in that case, Axe-3x would be Ax2e-3x
     
  7. Aug 4, 2009 #6

    Mark44

    Staff: Mentor

    Yes.

    Your solution to the homogeneous equation is yh = Ae-3x + Bxe-3x.

    The particular solution to the nonhomogeneous problem is yp = Cx2e-3x + D + Ex + Fx2.

    The general solution is y = yh + yp.
     
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