# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-6 => Topic started by: suyichen on November 15, 2019, 02:02:17 PM

Title: TUT0801
Post by: suyichen on November 15, 2019, 02:02:17 PM
a. Find the general solution of given system of equations and describe the behavior as $t \rightarrow+\infty$

b. Draw a direction field and plot a few trajectories of the system

$$x^{\prime}=\left(\begin{array}{cc}{2} & {-1} \\ {3} & {-2}\end{array}\right) x$$
\begin{aligned} \operatorname{det}(A-\lambda I)=&\left|\begin{array}{cc}{2-\lambda} & {-1} \\ {3} & {-2-\lambda}\end{array}\right| =0 \\(2-\lambda)(-2-&\lambda)-(-3) =0 \\-4+\lambda^{2}&+3 =0 \\ \lambda=1 &\quad \lambda=-1 \end{aligned}

$\lambda=1$
$$\left[\begin{array}{rr|r}{1} & {-1} & {0} \\ {3} & {-3} & {0}\end{array}\right] \sim\left[\begin{array}{rr|r}{1} & {-1} & {0} \\ {0} & {0} & {0}\end{array}\right]$$
$$x_{2}=t \quad x_{1}-t=0 \quad \Rightarrow \quad x_{1}=t$$
$$\left[\begin{array}{l}{x_{1}} \\ {x_{2}}\end{array}\right]=\left[\begin{array}{l}{1} \\ {1}\end{array}\right]t$$

$\therefore$ eigenvector $\left(\begin{array}{l}{1} \\ {1}\end{array}\right)$ corresponding to eigenvalue $\lambda=1$

$\lambda=-1$
$$\left[\begin{array}{rr|r}{3} & {-1} & {0} \\ {3} & {-1} & {0}\end{array}\right] \sim\left[\begin{array}{rr|r}{3} & {-1} & {0} \\ {0} & {0} & {0}\end{array}\right]$$
$$x_{2}=t \quad 3 x_{1}-t=0 \quad \Rightarrow \quad x_{1}=\frac{t}{3}$$
$$\left[\begin{array}{l}{x_{1}} \\ {x_{2}}\end{array}\right]=\left[\begin{array}{l}{1} \\ {3}\end{array}\right] t$$

$\therefore$ eigeuvector $\left(\begin{array}{l}{1} \\ {3}\end{array}\right)$ correspouding to eigenvalue $\lambda=-1$

$\therefore$ general solution is

\begin{aligned} x&=c_{1} e^{t}\left(\begin{array}{l}{1} \\ {1}\end{array}\right)+c_{2} e^{-t}\left(\begin{array}{l}{1} \\ {3}\end{array}\right) \\ \lambda &=-1 \end{aligned}