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MAT244--2018F => MAT244--Lectures & Home Assignments => Topic started by: Meiyi Lu on November 27, 2018, 05:06:07 PM

Title: Final review question
Post by: Meiyi Lu on November 27, 2018, 05:06:07 PM
How to solve the question
Find the general solution by method of the underterminded coefficients:
$y^{'''} - 2y^{''} + 4y' - 8y = 16e^{2t} + 30cos(t)$
Title: Re: Final review question
Post by: Tianfangtong Zhang on November 27, 2018, 05:18:44 PM
$r^3 - 2r^2 + 4r - 8 = 0$

$(r-2)r^2 + 4(r-2) = 0$

$(r-2)(r^2+4) = 0$

r =2, r =$\pm$2i

Then

$y_c(t) = c_1e^{2t} + c_2\cos2t + c_3\sin2t$

$y^{'''} - 2y^{''} + 4y' - 8y = 16e^{2t}$

$y_p(t) = Ae^{2t} = Ate^{2t}$

$(Ate^{2t})^{'''} - 2(Ate^{2t})^{''} + 4(Ate^{2t})' - 8(Ate^{2t}) = 16e^{2t}$

Then we get A = 2

Then $y_p(t) = 2e^{2t}$

$y^{'''} - 2y^{''} + 4y' - 8y = 30\cos t$

$y_p(t) = B\cos t + C\sin t$

$(B\cos t + C\sin t)^{'''} - 2(B\cos t + C\sin t)^{''} + 4(B\cos t + C\sin t)' - 8(B\cos t + C\sin t) = 30\cos t$

Then we get B=-4 and C=2

$y_p(t) = 2 \sin t - 4 \cos t$

Thus, $y(t) = c_1e^{2t} +c_2\cos(2t)+c_3\sin(2t)+2te^t+2\sin t - 4\cos t$