### Author Topic: Final review question  (Read 1203 times)

#### Meiyi Lu

• Jr. Member
• Posts: 5
• Karma: 1
##### Final review question
« on: November 27, 2018, 05:06:07 PM »
How to solve the question
Find the general solution by method of the underterminded coefficients:
$y^{'''} - 2y^{''} + 4y' - 8y = 16e^{2t} + 30cos(t)$

#### Tianfangtong Zhang

• Full Member
• Posts: 16
• Karma: 15
##### Re: Final review question
« Reply #1 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, $y(t) = c_1e^{2t} +c_2\cos(2t)+c_3\sin(2t)+2te^t+2\sin t - 4\cos t$
« Last Edit: November 27, 2018, 05:21:51 PM by Tianfangtong Zhang »