# Toronto Math Forum

## MAT244-2013F => MAT244 Math--Tests => Quiz 3 => Topic started by: Victor Ivrii on November 06, 2013, 08:11:53 PM

Title: Problem 1 (night sections)
Post by: Victor Ivrii on November 06, 2013, 08:11:53 PM
Find the general solution of the given differential equation.
\begin{equation*}
y'''-y''-y'+ y = 0.
\end{equation*}
Title: Re: Problem 1 (night sections)
Post by: Ka Hou Cheok on November 06, 2013, 08:32:38 PM
\begin{equation*}
y'''-y''-y'+ y = 0
\end{equation*}
The responding characteristic equation is $$r^3-r^2-r+1=0$$
$$(r^3-r)-(r^2-1)=0$$
$$r(r^2-1)-(r^2-1)=0$$
$$(r-1)(r^2-1)=0$$
$$r_1=1, r_2=1, r_3=-1$$
So the general solution is $$y=c_1e^t+c_2te^t+c_3e^{-t}$$
Title: Re: Problem 1 (night sections)
Post by: Mark Kazakevich on November 06, 2013, 08:33:04 PM
For the differential equation:
\begin{equation} y'''-y''-y'+y=0 \end{equation}

We assume that $y = e^{rt}$.
Therefore, we must solve the characteristic equation:

\begin{equation} r^3 - r^2 - r + 1 = 0 \end{equation}

We find:
$r^3 - r^2 - r + 1 = 0 \implies (r^2-1)(r-1) = 0 \implies (r+1)(r-1)^2 = 0$

This means the roots of this equation are:

$r_1 = 1, r_2=1, r_3=-1$
(We have a repeated root at r = 1)

So the general solution to (1) is:
\begin{equation} y(t) = c_1e^{t} + c_2e^{t}t + c_3e^{-t} \end{equation}