# Toronto Math Forum

## MAT244-2013S => MAT244 Math--Tests => Quiz 3 => Topic started by: Victor Ivrii on February 27, 2013, 07:46:41 PM

Title: Day Section Problem 2
Post by: Victor Ivrii on February 27, 2013, 07:46:41 PM
Post problem and solution
Title: Re: Day Section Problem 2
Post by: Changyu Li on February 27, 2013, 09:17:36 PM
I think the question was
$$y'''-y= 2 \sin t$$
solution:
$$r^3 - 1 = 0 \\ (r+1)(r^2+r+1) = 0\\ r = -1, \frac{-1 \pm \sqrt{3} i}{2}\\ y_h = c_1 e^{-t} + c_2 e^{ \frac{-1 +\sqrt{3} i}{2}t} + c_3 e^{ \frac{-1 - \sqrt{3} i}{2}t} \\ y_p = A \sin t + B \cos t \\ y_p' = A \cos t - B \sin t \\ y_p'' = -A \sin t - B \cos t \\ y_p''' = - A \cos t + B \sin t \\ A = -1, B = 1\\ y = c_1 e^{-t} + c_2 e^{ \frac{-1 +\sqrt{3} i}{2}t} + c_3 e^{ \frac{-1 - \sqrt{3} i}{2}t} + \cos t - \sin t$$
Title: Re: Day Section Problem 2
Post by: Jason Hamilton on February 27, 2013, 10:09:57 PM
NVM   8)
Title: Re: Day Section Problem 2
Post by: Victor Ivrii on February 28, 2013, 02:08:59 AM
Just got confirmation that this is a correct problem.