91Ó°ÊÓ

Consider the equation of dynamical system $\frac{\mathrm{d}\stackrel{→}{\mathrm{x}}}{\mathrm{dt}}=\left[\begin{array}{cc}-4& 3\\ 2& -3\end{array}\right]\stackrel{→}{\mathrm{x}}$with the initial value $\stackrel{→}{\mathrm{x}}\left(0\right)=\left[\begin{array}{c}1\\ 0\end{array}\right]$

Compare the equations $\frac{\mathrm{d}\stackrel{→}{\mathrm{x}}}{\mathrm{dt}}=\left[\begin{array}{cc}-4& 3\\ 2& -3\end{array}\right]\stackrel{→}{\mathrm{x}}$and$\frac{d\stackrel{→}{x}}{dt}=A\stackrel{→}{x}$as follows.

$A=\left[\begin{array}{cc}-4& 3\\ 2& -3\end{array}\right]\stackrel{→}{\mathrm{x}}$

Assume is an Eigen value of the matrix localid="1659608839068" $\left[\begin{array}{cc}-4& 3\\ 2& -3\end{array}\right]$ implies $\left|A-\mathrm{λ}I\right|=0$.

Substitute the values $\left[\begin{array}{cc}-4& 3\\ 2& -3\end{array}\right]$for A and $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$for in the equation $\left|A-\mathrm{λ}I\right|=0$as follows.

localid="1659609197618" $\begin{array}{rcl}\left|A-\mathrm{λ}I\right|& =& 0\\ \left[\begin{array}{cc}-4& 3\\ 2& -3\end{array}\right]-\mathrm{λ}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]& =& 0\end{array}$

width="171">-432-3-λ1001=0

Simplify the equation as follows.

Further, simplify the equation $\left[\begin{array}{cc}-4& 3\\ 2& -3\end{array}\right]-\mathrm{λ}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=0$as

$$\begin{gathered} \left[\begin{array}{cc}-4& 3\\ 2& -3\end{array}\right]-\mathrm{λ}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=0 \\ \left[\begin{array}{cc}-4& 3\\ 2& -3\end{array}\right]-\left[\begin{array}{cc}\mathrm{λ}& 0\\ 0& \mathrm{λ}\end{array}\right]=0 \\ \left[\begin{array}{cc}-4-\mathrm{λ}& 3\\ 2& -3-\mathrm{λ}\end{array}\right]=0 \\ \left(-4-\mathrm{λ}\right)\left(-3-\mathrm{λ}\right)-6=0 \end{gathered}$$

follows.

$\begin{array}{rcl}\left(-4-\mathrm{λ}\right)\left(-3-\mathrm{λ}\right)-6& =& 0\\ 12+4\mathrm{λ}+3\mathrm{λ}+{\mathrm{λ}}^2-6& =& 0\\ {\mathrm{λ}}^2+7\mathrm{λ}+6& =& 0\\ {\mathrm{λ}}^2+6\mathrm{λ}+\mathrm{λ}+6& =& 0\\ & & \end{array}$

Further, simplify the equation as follows.

$$\begin{gathered} {\mathrm{λ}}^2+6\mathrm{λ}+\mathrm{λ}+6=0 \\ \mathrm{λ}\left(\mathrm{λ}+6\right)+\left(\mathrm{λ}+6\right)=0 \\ \left(\mathrm{λ}+1\right)\left(\mathrm{λ}+6\right)=0 \end{gathered}$$

Therefore, the Eigen values of A are$\mathrm{λ}=-1,-6$

Assume $v_1=\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]$and $v_2=\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]$are Eigen vector corresponding to$\mathrm{λ}=-1,-6$implies $\left|A-{\mathrm{λ}}_{1}I\right|v_1=0$and $\left|A-{\mathrm{λ}}_{2}I\right|v_2=0$.

Substitute the values $\left[\begin{array}{cc}-4& 3\\ 2& -3\end{array}\right]$forA , -1 for${\mathrm{λ}}_2$, $\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]$for and for l in the equation $\left|A-{\mathrm{λ}}_{2}I\right|v_2=0$as follows.

$\begin{array}{rcl}\left|A-{\mathrm{λ}}_{2}I\right|v_2& =& 0\\ \left[\begin{array}{cc}-4& 3\\ 2& -3\end{array}\right]+1\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\\ \left[\begin{array}{cc}-4& 3\\ 2& -3\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\\ \left[\begin{array}{cc}-3& 3\\ 2& -2\end{array}\right]\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\end{array}$

Further, simplify the equation as follows

$\begin{array}{rcl}\left[\begin{array}{cc}-3& 3\\ 2& -2\end{array}\right]\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\\ -3x_2+3y_2& =& 0\\ 2x_2-2y_2& =& 0\\ & & \end{array}$

Simplify the equations$-3x_2+3y_2=0$and $2x_2-2y_2=0$as follows.

$\begin{array}{rcl}-3x_2+3y_2& =& 0\\ 3x_2& =& 3y_2\\ x_2& =& y_2\\ & & \end{array}$

For$x_2=1$implies $y_2=1$.

Therefore, the Eigen vector corresponding to${\mathrm{λ}}_2=-1$is $\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]=\left[\begin{array}{c}1\\ 1\end{array}\right]$.

Substitute the values $\left[\begin{array}{cc}-4& 3\\ 2& -3\end{array}\right]$forA ,-6 for${\mathrm{λ}}_1$,$\left[\begin{array}{c}{X}_1\\ y_1\end{array}\right]$for$v_1$and $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$for l in the equation$\left|A-{\mathrm{λ}}_{1}I\right|v_1=0$as follows.

localid="1659683666456" $\begin{array}{rcl}\left|A-{\mathrm{λ}}_{1}I\right|v_1& =& 0\\ \left[\begin{array}{cc}-4& 3\\ 2& -3\end{array}\right]+6\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\\ \left[\begin{array}{cc}-4& 3\\ 2& -3\end{array}\right]+\left[\begin{array}{cc}6& 0\\ 0& 6\end{array}\right]\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\\ \left[\begin{array}{cc}2& 3\\ 2& 3\end{array}\right]\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{rcl}\left[\begin{array}{cc}2& 3\\ 2& 3\end{array}\right]\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\\ 2x_1+3y_1& =& 0\\ 2x_1+3y_1& =& 0\\ & & \end{array}$

Simplify the equations $2x_1+3y_1=0$and $2x_1+3y_1=0$as follows.

$$\begin{gathered} 2x_1+3y_1=0 \\ 2x_1=-3y_1 \end{gathered}$$

For$x_1=3$implies$y_1=-2$

Therefore, the Eigen vector corresponding to ${\mathrm{λ}}_1=-6$is $\left[\left.\begin{array}{c}{X}_1\\ y_1\end{array}\right]=\left[\begin{array}{c}3\\ -2\end{array}\right.\right]$.

The Eigen vectors are $V_1=\left[\begin{array}{c}3\\ -2\end{array}\right]$and$V_2=\left[\begin{array}{c}1\\ 1\end{array}\right]$corresponding to the Eigen values ${\mathrm{λ}}_1=-1$ and ${\mathrm{λ}}_2=-6$respectively implies $S^{-1}AS=B$where $S=\left[\begin{array}{cc}3& 1\\ -2& 1\end{array}\right]$and $B=\left[\begin{array}{cc}-6& 0\\ 0& -1\end{array}\right]$.

Substitute the value$SB{S}^{-1}$for A in the equation $\frac{d\stackrel{→}{x}}{dt}=A\stackrel{→}{x}$as follows.

$\begin{array}{rcl}\frac{d\stackrel{→}{x}}{dt}& =& A\stackrel{→}{x}\\ \frac{d\stackrel{→}{x}}{dt}& =& SB{S}^{-1}\stackrel{→}{x}\\ S^{-1}\frac{d}{dt}& =& B{S}^{-1}\stackrel{→}{x}\\ \frac{d}{dt}\left(S^{-1}\stackrel{→}{x}\right)& =& B\left(S^{-1}\stackrel{→}{x}\right)\\ & & \end{array}$

Assume$\stackrel{→}{c}=S^{-1}\stackrel{→}{x}$implies $\frac{d}{dt}\stackrel{→}{c}=B\stackrel{→}{c}$.

The solution of the equation $\frac{d}{dt}\stackrel{→}{c}=B\stackrel{→}{c}$is$\stackrel{→}{c}\left(t\right)=\left[\begin{array}{c}{e}^{-6t}{c}_1\\ e^{-t}{c}_2\end{array}\right]$

Substitute the value $\left[\begin{array}{c}{e}^{-6t}{c}_1\\ e^{-t}{c}_2\end{array}\right]$for $\stackrel{→}{c}\left(t\right)$and$\left[\begin{array}{cc}3& 1\\ -2& 1\end{array}\right]$ for S in the equation $\stackrel{→}{x}=S\stackrel{→}{c}$as follows.

role="math" localid="1659681154121" $$\begin{gathered} \stackrel{→}{x}\left(t\right)=S\stackrel{→}{c} \\ \stackrel{→}{x}\left(t\right)=\left[\begin{array}{cc}3& 1\\ -2& 1\end{array}\right]\left[\begin{array}{c}{e}^{-6t}{c}_1\\ e^{-t}{c}_2\end{array}\right] \\ \stackrel{→}{x}\left(t\right)=\left[\begin{array}{c}3e^{-6t}{c}_1+e^{-t}{c}_2\\ -2e^{-6t}{c}_1+e^{-t}{c}_2\end{array}\right] \end{gathered}$$

Substitute the value for in the equation $\stackrel{→}{x}\left(t\right)=\left[\begin{array}{c}3e^{-6t}{c}_1+e^{-t}{c}_2\\ -2e^{-6t}{c}_1+e^{-t}{c}_2\end{array}\right]$as follows.

role="math" localid="1659681433311" $$\begin{gathered} \stackrel{→}{x}\left(t\right)=\left[\begin{array}{c}3e^{-6t}{c}_1+e^{-t}{c}_2\\ -2e^{-6t}{c}_1+e^{-t}{c}_2\end{array}\right] \\ \stackrel{→}{x}\left(0\right)=\left[\begin{array}{c}3c_1{e}^{-6\left(0\right)}+e^{-\left(0\right)}{c}_2\\ -2e^{-6\left(0\right)}{c}_1+e^{-\left(0\right)}{c}_2\end{array}\right] \\ \stackrel{→}{x}\left(0\right)=\left[\begin{array}{c}3c_1+c_2\\ -2c_1+c_2\end{array}\right] \end{gathered}$$

As ,$\stackrel{→}{x}\left(0\right)=\left[\begin{array}{c}1\\ 0\end{array}\right]$ substitute the value$\left[\begin{array}{c}1\\ 0\end{array}\right]$ for$\stackrel{→}{x}\left(0\right)$ in the equation $\stackrel{→}{x}\left(0\right)=\left[\begin{array}{c}3c_1+c_2\\ -2c_1+c_2\end{array}\right]$as follows.

$$\begin{gathered} \stackrel{→}{x}\left(0\right)=\left[\begin{array}{c}3c_1+c_2\\ -2c_1+c_2\end{array}\right] \\ \left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}3c_1+c_2\\ -2c_1+c_2\end{array}\right] \end{gathered}$$

Compare the equation both side in the equation$\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}3c_1+c_2\\ -2c_1+c_2\end{array}\right]$ as follows.

$$\begin{gathered} 1=3c_1+c_2 \\ 0=-2c_1+c_2 \end{gathered}$$

Subtract the equation$1=3c_1+c_2$ from$0=-2c_1+c_2$ as follows.

$\begin{array}{rcl}1-0& =& \left\{3c_1+c_2\right\}-\left\{-2c_1+c_2\right\}\\ 1& =& 3c_1+2c_1\\ 1& =& 5c_1\\ c_1& =& 0.2\\ & & \end{array}$

Substitute the value$0.2$ for $c_1$in the equation $1=3c_1+c_2$as follows.

$$\begin{gathered} 1=3c_1+c_2 \\ 1=3\left(0.2\right)+c_2 \\ 1=0.6+c_2 \\ {c}_2=0.4 \end{gathered}$$

Therefore, the value of $c_1,c_2$are $c_1=0.2$and $c_2=0.4$

The general solution of$\stackrel{→}{x}\left(t\right)$is$\stackrel{→}{x}\left(t\right)=c_1{e}^{-6t}{v}_1+c_2{e}^{-t}{v}_2$

Substitute the value$\left[\begin{array}{c}3\\ -2\end{array}\right]$ for$v_1$and $\left[\begin{array}{c}1\\ 1\end{array}\right]$for$v_2$in the equation$\stackrel{→}{x}\left(t\right)=c_1{e}^{-6t}{v}_1+c_2{e}^{-t}{v}_2$as follows.

$$\begin{gathered} \stackrel{→}{x}\left(t\right)=c_1{e}^{-6t}{v}_1+c_2{e}^{-t}{v}_2 \\ \stackrel{→}{x}\left(t\right)=c_1{e}^{-6t}\left[\begin{array}{c}3\\ -2\end{array}\right]+c_2{e}^{-t}\left[\begin{array}{c}1\\ 1\end{array}\right] \end{gathered}$$

Substitute the values $0.2$for$c_1$and $0.4$for $c_2$in the equation $\stackrel{→}{x}\left(t\right)=c_1{e}^{-6t}\left[\begin{array}{c}3\\ -2\end{array}\right]+c_2{e}^{-t}\left[\begin{array}{c}1\\ 1\end{array}\right]$as follows.

$$\begin{gathered} \stackrel{→}{x}\left(t\right)=c_1{e}^{-6t}\left[\begin{array}{c}3\\ -2\end{array}\right]+c_2{e}^{-t}\left[\begin{array}{c}1\\ 1\end{array}\right] \\ \stackrel{→}{x}\left(t\right)=0.2e^{-6t\left[\begin{array}{c}3\\ -2\end{array}\right]+0.4e^{-t}}\left[\begin{array}{c}1\\ 1\end{array}\right] \end{gathered}$$

Hence the solution of the system $\frac{d\stackrel{→}{x}}{dt}=\left[\begin{array}{cc}-4& 3\\ 2& -3\end{array}\right]\stackrel{→}{x}$with the initial value$\stackrel{→}{x}\left(0\right)=\left[\begin{array}{c}1\\ 0\end{array}\right]$ is .

$\stackrel{→}{x}\left(t\right)=0.2e^{-6t}\left[\begin{array}{c}3\\ -2\end{array}\right]+0.4e^{-t}\left[\begin{array}{c}1\\ 1\end{array}\right]$