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Math Help - 4th Order - Using Method of Undetermined Coeffecients

  1. #1
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    4th Order - Using Method of Undetermined Coeffecients

    I'm having trouble with one of my homework problems. The instructions are to use the method of undetermined coefficients to find the particular solution, then form the general solution.
    The equation in question is: y''''-3y'''+2y''=3e^(-t) + 6e^(2t) - 6t

    I factored the auxiliary equation to get roots of 0 (multiplicity 2),2, and 1 so that:
    y1=1
    y2=t
    y3=e^(2t)
    y4=e^t

    In forming the undetermined coefficients part of the problem, I have Ae^(-t) + Bte ^(2t) + Ct^2
    (Using the modification/multiplication rule)

    The problem is that when I find the derivatives of this and plug them back into the equation, most of it cancels out and I get 6Ae^(-t) + 4C = 3e^(-t) + 6e^(2t) - 6t, which does not help me solve for the coefficients. Where am I going wrong here? Thanks!
    Last edited by vesperka; March 3rd 2010 at 02:52 PM.
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  2. #2
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    Quote Originally Posted by vesperka View Post
    I'm having trouble with one of my homework problems. The instructions are to use the method of undetermined coefficients to find the particular solution, then form the general solution.
    The equation in question is: y''''-3y'''+2y''=3e^(-t) + 6e^(2t) - 6t

    I factored the auxiliary equation to get roots of 0 (multiplicity 2),2, and 1 so that:
    y1=1
    y2=t
    y3=e^(2t)
    y4=e^t

    In forming the undetermined coefficients part of the problem, I have Ae^(-t) + Bte ^(2t) + Ct^2
    (Using the modification/multiplication rule)

    The problem is that when I find the derivatives of this and plug them back into the equation, most of it cancels out and I get 6Ae^(-t) + 4C = 3e^(-t) + 6e^(2t) - 6t, which does not help me solve for the coefficients. Where am I going wrong here? Thanks!
    The polynomial part of your particlular solution should be of the form
    y_p=Ct^2+Dt^3

    So if there were not reapted part of the complimentry and particular solution you would get y_p=C+Dt but since zero is a root of multiplicity 2 you must muliply by t^2 to get
    y_p=Ct^2+Dt^3
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  3. #3
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    Lol this is Andy, who posted on my account? Please tell me we went to HS together, otherwise i got keylogged
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  4. #4
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    Quote Originally Posted by vesperka View Post
    Lol this is Andy, who posted on my account? Please tell me we went to HS together, otherwise i got keylogged
    It's Chris. LOLZ. Your password was still saved on my computer from like a year ago and I couldn't figure out this problem.

    Realized that I was missing a step in taking the derivatives. Simple stuff. Thanks for the help.
    Last edited by vesperka; March 3rd 2010 at 08:13 PM. Reason: Figured it out
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