# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-6 => Topic started by: yangqi40 on November 15, 2019, 02:40:49 PM

Title: Quiz 6 Lec5101
Post by: yangqi40 on November 15, 2019, 02:40:49 PM
.a) Consider the system of equations:
\begin{equation}
\textbf{x'}=\left(\begin{array}{cc}3&6\\-1&-2\end{array}\right)\textbf{x}
\end{equation}

Assume that $\textbf{x}= \xi e^n$.and substitute for $\textbf{x}$ in the above equation, then obtain the following algebraic system.
\begin{equation}
\left(\begin{array}{cc}3-r&6\\-1&-2-r\end{array}\right)
\left(\begin{array}{c}\xi_1\\ \xi_2\end{array}\right)=
\left(\begin{array}{c}0\\ 0\end{array}\right)
\end{equation}

To find the Eigen values, set $\left|\begin{array}{cc}3-r&6\\-1&-2-r\end{array}\right|=0$

That is

\begin{aligned}(3-r)(-2-r)+6&=0\\ r^2-r-6+6&=0\end{aligned}

\begin{aligned}r^2-r&=0\\ (r-1)r&=0\\ r&=0,1\end{aligned}

So, the Eigen values are $r_1 = 0$ and $r_2 = 1$

For $r=r_1=0$ , equation (2) gives,

$\left(\begin{array}{cc}3&6\\-1&-2\end{array}\right) \left(\begin{array}{c}\xi_1\\ \xi_2\end{array}\right)= \left(\begin{array}{c}0\\ 0\end{array}\right)$

$\left(\begin{array}{cc}1&2\\1&2\end{array}\right) \left(\begin{array}{c}\xi_1\\ \xi_2\end{array}\right)= \left(\begin{array}{c}0\\ 0\end{array}\right)$

Each row in this vector equation generates the condition, $\xi_1=-2\xi_2$

So, the corresponding Eigen vector is $\xi^{(1)}=\left(\begin{array}{c}-2\\ 1\end{array}\right)$.

Now for $r=r_2=1$, then equation (2) gives

$\left(\begin{array}{cc}2&6\\-1&-3\end{array}\right) \left(\begin{array}{c}\xi_1\\ \xi_2\end{array}\right)= \left(\begin{array}{c}0\\ 0\end{array}\right)$

Each row in this vector equation gives condition, $\xi =-3\xi_2$

Then the corresponding Eigen vector is $\xi^{(2)} =\left(\begin{array}{c}-3\\ 1\end{array}\right)$

Thus the fundamental set of solutions of the system (1) is

$\textbf{x}^{(1)}(t) =\left(\begin{array}{c}-2\\ 1\end{array}\right)$,
$\textbf{x}^{(2)}(t) =\left(\begin{array}{c}-3\\ 1\end{array}\right)e^t$

And hence the general solution is $\textbf{x}=c_1\textbf{x}^{(1)}(t)+c_2\textbf{x}^{(2)}(t)$.

$\textbf{x} =c_1\left(\begin{array}{c}-2\\ 1\end{array}\right)+ c_2\left(\begin{array}{c}-3\\ 1\end{array}\right)e^t$

here $c_1$ and $c_2$, are arbitrary constants.

b) The sketch of the direction field and trajectories is shown below