Author Topic: Night Sections Problem 2  (Read 2574 times)

Victor Ivrii

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Night Sections Problem 2
« 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*}

Victor Lam

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Re: Night Sections Problem 2
« Reply #1 on: February 27, 2013, 08:31:03 PM »
General solution is the summation of the homogeneous and particular solutions. See attachment.

Rudolf-Harri Oberg

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Re: Night Sections Problem 2
« Reply #2 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
$Y_G=\cos(t)+c_1+c_2e^t+c_3e^{-t}$.
« Last Edit: February 27, 2013, 10:53:09 PM by Rudolf-Harri Oberg »

Victor Ivrii

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Re: Night Sections Problem 2
« Reply #3 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.