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Math Help - Particular Solution....

  1. #1
    Junior Member
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    May 2009
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    Particular solution 2

    d^2x/[dt]^2 -4 dx/dt=7-3e^4t


    Could someone also help me with this? Find the particular solution. I find trouble knowing what trial to use because of the 7-3e^4t. My Tutor said to use a constant with the particular trial >>xp(t)=Ate^4t. Can you show me how to get the A,B and constant value please. Thanx
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  2. #2
    Super Member Deadstar's Avatar
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    Let x_p(t) = Ate^{4t} then

    \frac{dx}{dt} = 4Ate^{4t} + Ae^{4t} and

    \frac{d^2x}{dt^2} = 4Ae^{4t} + 16Ate^{4t} + 4Ae^{4t}

    then you have

    \frac{d^2x}{dt^2} - 4\frac{dx}{dt} = 4Ae^{4t} + 16Ate^{4t} + 4Ae^{4t} -4(4Ate^{4t} + Ae^{4t}) = 4Ae^{4t} = -3e^{4t}, so A = -\frac{3}{4}. Hence x_p(t) = -\frac{3}{4}te^{4t}.

    For the complementary solution try x_c(t) = Bt.

    Then -4\frac{dx}{dt} = -4B = 7 so B = -\frac{7}{4}.

    Hence x(t) = x_p(t) + x_c(t) = -\frac{3}{4}te^{4t} - \frac{7}{4}t.
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