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Math Help - undetermined coefficients

  1. #1
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    undetermined coefficients

    I'm supposed to find the solution this initial value problem using the method of undetermined coefficients
    <br />
y'' - 2y' - 3y = 3te^{2t}, y(0)=1, y'(0)=0.

    My guess was to use Y(t) = Ate^{2t} + Be^{2t} for the particular solution, but this led me to the wrong answer.
    I have a hunch that I should use Y(t) = t(Ate^{2t} + Be^{2t}) but I have no idea why, since e^{2t} is not part of the complimentary solution.

    I appreciate any help.
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  2. #2
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    the P.I. should be c1.e^-t+c2.e^3t
    and the A.E.=3.e^2t(t-6)
    are they correct???

    we get the P.I. by solving the quadratic in y' ie. D^2-2D-3=0
    which gives D=-1,3

    and the A.E. is calculated by separating the exponential part then........(it should be given in any text book).hope this was helpful!
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  3. #3
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    Quote Originally Posted by magdaddy101 View Post
    I'm supposed to find the solution this initial value problem using the method of undetermined coefficients
    <br />
y'' - 2y' - 3y = 3te^{2t}, y(0)=1, y'(0)=0.

    My guess was to use Y(t) = Ate^{2t} + Be^{2t} for the particular solution, but this led me to the wrong answer.
    What did you do that gave you a wrong answer?

    If y= Ate^{2t}+ Be^{2t}, then
    y"= Ae^{2t}+ 2Ate^{2t}+ 2Be^{2t}= (A+ 2B)e^{2t}+ 2Ate^{2t}
    y"= 2(A+ 2B)e^{2t}+ 2Ae^{2t}+ 4Ate^{2t}= (6A+ 2B)e^{2t}+ 4Ate^{2t}
    Putting those into the equation gives
    [(6A+ 2B)-2(A+ 2B)- 3B]e^{2t}+ [4A- 4A- 3A]te^{2t} = (4A- 5B)e^{2t}+ (-3A)te^{2t}= 3te^{2t}

    We must have 4A- 5B= 0 and -3A= 3. A= 1 so 4- 5B= 0 and B= 4/5.
    That is a perfectly good solution.

    I have a hunch that I should use Y(t) = t(Ate^{2t} + Be^{2t})<br />
but I have no idea why, since e^{2t} is not part of the complimentary solution.

    I appreciate any help.
    Follow Math Help Forum on Facebook and Google+

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