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General solution to differential equation

  1. Mar 6, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the general solution to differential equation [itex]y''-6y'+9y=e^x((2x+1)\cos x+(x+3)\sin x)[/itex]

    2. Relevant equations
    -Non homogeneous differential equation
    -Homogeneous differential equation with constant coefficients
    -Method of undetermined coefficients

    3. The attempt at a solution
    First, we find the solution for homogeneous equation [itex]y''-6y'+9y=0\Rightarrow (\lambda -3)^2=0\Rightarrow \lambda_1=\lambda_2=3[/itex].
    Roots are real and multiple [itex]\Rightarrow y_h=c_1e^{3x}+c_2xe^{3x}[/itex].

    We can find a particular solution for the equation by method of undetermined coefficients:

    [tex]y_p=e^x((Ax+B)\cos x+(Cx+D)\sin x)=Axe^x\cos x+Be^x\cos x+Cxe^x\sin x+De^x\sin x[/tex]
    [tex]{y'}_p=xe^x\cos x(A+C)+e^x\cos x(A+B+D)+xe^x\sin x(C-A)+(e^x\sin x(D+C-B))[/tex]
    [tex]y′′p=2Ce^x\cos x+e^x\cos x(2D+2C+2A)−2Axe^x\sin x+e^x\sin x(2C−2B−2A)[/tex]

    [tex]y′′_p−6y′_p+9y_p=xe^x\cos x(4C+3A)+e^x\cos x(−4A+3B+2C−4D)+xe^x\sin x(4A+3C)+e^x\sin x(−2A+4B−4C+3D)=2xe^x\cos x+e^x\cos x+xe^x\sin x+3e^x\sin x[/tex]

    Solving the system:
    [itex]4C+3A=2[/itex]
    [itex]−4A+3B+2C−4D=1[/itex]
    [itex]4A+3C=1[/itex]
    [itex]−2A+4B−4C+3D=3[/itex]

    gives [itex]A=−2/7,B=23/35,C=5/7,D=31/35[/itex]
    that are wrong results.

    Could someone check for possible errors?
    Coefficients should be [itex]A=2/5,B=21/25,C=−1/5,D=−3/25[/itex]
     
    Last edited: Mar 6, 2016
  2. jcsd
  3. Mar 6, 2016 #2

    Ray Vickson

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    Homework Helper

    You can check this for yourself (and you should always do that): just substitute your "solution" into the DE, to see if it works.
     
  4. Mar 6, 2016 #3

    RUber

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    Homework Helper

    I checked your evaluation for y', and it matches mine. I was able to get the answer you put as the solution, so your mistake must be in either the evaluation of y'' or the solution step.
     
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