Author Topic: Problem 1 (night sections)  (Read 834 times)

Victor Ivrii

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Problem 1 (night sections)
« 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*}

Ka Hou Cheok

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Re: Problem 1 (night sections)
« Reply #1 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 $$y=c_1e^t+c_2te^t+c_3e^{-t}$$
« Last Edit: November 06, 2013, 08:34:46 PM by Ka Hou Cheok »

Mark Kazakevich

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Re: Problem 1 (night sections)
« Reply #2 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}