Author Topic: Day Section Problem 2  (Read 2753 times)

Victor Ivrii

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Day Section Problem 2
« on: February 27, 2013, 07:46:41 PM »
Post problem and solution

Changyu Li

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Re: Day Section Problem 2
« Reply #1 on: February 27, 2013, 09:17:36 PM »
I think the question was
$$
y'''-y= 2 \sin t
$$
solution:
$$
r^3 - 1 = 0 \\
(r+1)(r^2+r+1) = 0\\
r = -1, \frac{-1 \pm \sqrt{3} i}{2}\\
y_h = c_1 e^{-t} + c_2 e^{ \frac{-1 +\sqrt{3} i}{2}t} + c_3 e^{ \frac{-1 - \sqrt{3} i}{2}t} \\
y_p = A \sin t + B \cos t \\
y_p' = A \cos t - B \sin t \\
y_p'' = -A \sin t - B \cos t \\
y_p''' = - A \cos t + B \sin t \\
A = -1, B = 1\\
y  = c_1 e^{-t} + c_2 e^{ \frac{-1 +\sqrt{3} i}{2}t} + c_3 e^{ \frac{-1 - \sqrt{3} i}{2}t} + \cos t - \sin t
$$
« Last Edit: February 28, 2013, 02:03:43 AM by Victor Ivrii »

Jason Hamilton

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Re: Day Section Problem 2
« Reply #2 on: February 27, 2013, 10:09:57 PM »
NVM   8)
« Last Edit: February 28, 2013, 01:02:09 PM by Jason Hamilton »

Victor Ivrii

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Re: Day Section Problem 2
« Reply #3 on: February 28, 2013, 02:08:59 AM »
Just got confirmation that this is a correct problem.
« Last Edit: February 28, 2013, 11:03:23 AM by Victor Ivrii »