# Toronto Math Forum

## MAT244-2013S => MAT244 Math--Tests => Quiz 3 => Topic started by: Victor Ivrii on February 27, 2013, 07:45:30 PM

Title: Night Sections Problem 2
Post by: Victor Ivrii on February 27, 2013, 07:45:30 PM
4.3 p 239, # 4

Find a particular solution and then the general solution of the following ODE
\begin{equation*}
y'''-y'= 2 \sin t .
\end{equation*}
Title: Re: Night Sections Problem 2
Post by: Victor Lam on February 27, 2013, 08:31:03 PM
General solution is the summation of the homogeneous and particular solutions. See attachment.
Title: Re: Night Sections Problem 2
Post by: Rudolf-Harri Oberg on February 27, 2013, 10:17:22 PM
We start by solving $r^3-r=0$ which gives that $r_1=0, r_2=1, r_3=-1$.
Variation of parameters is not a good method to guess a particular solution here. You can try guessing that the particular solution is $Y_p=A\sin(t)+B\cos(t)$ or just look at the equation and deduce that $Y_p=\cos(t)$

So, general solution to the equation is
$Y_G=\cos(t)+c_1+c_2e^t+c_3e^{-t}$.
Title: Re: Night Sections Problem 2
Post by: Victor Ivrii on February 28, 2013, 02:36:14 AM
Observing that the r.h.e. is an odd function and equation contains only odd derivatives we look for even solution: $y_p= A\cos(t)$ which makes easy problem even easier.