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Find the general solution

  1. Apr 26, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the general solution

    y'' + 4y' +4y = 5xe^(-2x)



    3. The attempt at a solution

    I got (5/2)x^3*e^(-2x) as a particular solution. But I checked online at wolfram alpha and it says the particular solution is (5/6)x^3*e^(-2x). Using method of undetermined coefficients.
     
  2. jcsd
  3. Apr 26, 2010 #2

    rock.freak667

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    Solve the homogeneous part of y'' + 4y' +4y = 0. What are the roots of the characteristic equation?
     
  4. Apr 26, 2010 #3
    The homogeneous part gave me

    y=c1e-2x+c2xe-2x

    from the root (D+2)2.

    Multiplying by another (D+2) annihilated the 5xe-2x, resulting in a new solution

    y=c1e-2x+c2xe-2x+c3x2e-2x

    is that right?
     
  5. Apr 26, 2010 #4

    rock.freak667

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    Your right side is xe-2x, since -2 is seen twice in the homogeneous part, you must multiply it by x2.


    Can you show your work on how you got the constant to be 5/2?
     
  6. Apr 26, 2010 #5
    (D2+4D+4)(c3x2e-2x) = 5xe-2x

    c3(4x2e-2x-8xe-2x+2e-2x+4(2xe-2x-2x2e-2x)+4x2e-2x) = 5xe-2x


    Every thing but c3(2e-2x) reduces to zero

    c3(2e-2x)=5xe-2x
    c3=(5/2)x
     
  7. Apr 26, 2010 #6

    rock.freak667

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    I thought you got yp=c3x3e-2x
     
  8. Apr 26, 2010 #7
    yp=c3x2e-2x
    yp=(5/2)x*x2e-2x
    yp=(5/2)x3e-2x
     
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