91Ó°ÊÓ

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

Consider the equation of dynamical system $\frac{\mathrm{d}\stackrel{→}{\mathrm{x}}}{\mathrm{dt}}=\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]\stackrel{→}{\mathrm{x}}$.

Compare the equations $\frac{\mathrm{d}\stackrel{→}{\mathrm{x}}}{\mathrm{dt}}=\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]\stackrel{→}{\mathrm{x}}$and $\frac{d\stackrel{→}{x}}{dt}=A\stackrel{→}{x}$as follows.

role="math" localid="1668663561959" $A=\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]$

Assume $\mathrm{λ}$is an Eigen value of the matrix role="math" localid="1668663652708" $\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]$implies $\left|A-\mathrm{λ}I\right|=0$.

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

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

Simplify the equation $\left|\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]-\mathrm{λ}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|=0$as follows.

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

Further, simplify the equation as follows.

$$\begin{gathered} \left(1-\mathrm{λ}\right)\left(4-\mathrm{λ}\right)-4=0 \\ 4-\mathrm{λ}-4\mathrm{λ}+{\mathrm{λ}}^2-4=0 \\ {\mathrm{λ}}^2-5\mathrm{λ}=0 \\ \mathrm{λ}\left(\mathrm{λ}-5\right)=0 \end{gathered}$$

Therefore, the Eigen values of A are $\mathrm{λ}=0,5$.

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 $\mathrm{λ}=0,5$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 $\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]$for A, 0 for ${\mathrm{λ}}_1,\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]$ for v₁ and $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for I in the equation $\left|A-{\mathrm{λ}}_{1}I\right|v_1=0$as follows.

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

Further, simplify the equation as follows.

$\begin{array}{rcl}\left|\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\\ x_1+2y_1& =& 0\\ 2x_1+4y_1& =& 0\end{array}$

Simplify the equations $x_1+2y_1=0$and $2x_1+4y_1=0$as follows.

$\begin{array}{rcl}{x}_1+2y_1& =& 0\\ x_1& =& -2y_1\end{array}$

For $x_1=2$implies $y_1=-1$.

Therefore, the Eigen vector corresponding to ${\mathrm{λ}}_1=0$is $\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]=\left[\begin{array}{c}2\\ -1\end{array}\right]$.

Substitute the values $\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]$for A, 5 for ${\mathrm{λ}}_2,\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]$ for v₂ and $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for I in the equation $\left|A-{\mathrm{λ}}_{2}I\right|v_2=0$as follows.

$\begin{array}{rcl}\left|A-{\mathrm{λ}}_{2}I\right|v_2& =& 0\\ \left|\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]-5\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\\ \left|\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]+\left[\begin{array}{cc}-5& 0\\ 0& -5\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\\ \left|\left[\begin{array}{cc}-4& 2\\ 2& -1\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{rcl}\left|\left[\begin{array}{cc}-4& 2\\ 2& -1\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\\ -4x_2+2y_2& =& 0\\ 2x_2-1y_2& =& 0\end{array}$

Simplify the equations $-4x_2+2y_2=0$and $2x_2-1y_2=0$as follows.

$\begin{array}{rcl}-4x_2+2y_2& =& 0\\ 4x_2& =& 2y_2\\ 2x_2& =& y_2\end{array}$

For $x_2=1$implies $y_2=2$.

Therefore, the Eigen vector corresponding to ${\mathrm{λ}}_2=5$is $\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]=\left[\begin{array}{c}1\\ 2\end{array}\right]$.

The Eigen vectors are $v_1=\left[\begin{array}{c}2\\ -1\end{array}\right]$and $v_2=\left[\begin{array}{c}1\\ 2\end{array}\right]$corresponding to the Eigen values ${\mathrm{λ}}_1=0$and ${\mathrm{λ}}_2=5$respectively implies $S^{-1}AS=B$where $S=\left[\begin{array}{cc}2& 1\\ -1& 2\end{array}\right]$and $B=\left[\begin{array}{cc}0& 0\\ 0& 5\end{array}\right]$.

Substitute the value SBS^-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 $\stackrel{→}{c}\left(t\right)=\left[\begin{array}{c}{c}_1\\ e^{5t}{c}_2\end{array}\right]$.

Substitute the value $\left[\begin{array}{c}{c}_1\\ e^{5t}{c}_2\end{array}\right]$for $\stackrel{→}{c}\left(t\right)$and $\left[\begin{array}{cc}2& 1\\ -1& 2\end{array}\right]$ for S in the equation $\stackrel{→}{x}=S\stackrel{→}{c}$as follows.

$\begin{array}{rcl}\stackrel{→}{x}\left(t\right)& =& S\stackrel{→}{c}\\ \stackrel{→}{x}\left(t\right)& =& \left[\begin{array}{cc}2& 1\\ -1& 2\end{array}\right]\left[\begin{array}{c}{c}_1\\ e^{5t}{c}_2\end{array}\right]\\ \stackrel{→}{x}\left(t\right)& =& \left[\begin{array}{c}2c_1+e^{5t}{c}_2\\ -c_1+2e^{5t}{c}_2\end{array}\right]\end{array}$

Substitute the value 0 for t in the equation $\stackrel{→}{x}\left(t\right)=\left[\begin{array}{c}2c_1+e^{5t}{c}_2\\ -c_1+2e^{5t}{c}_2\end{array}\right]$as follows.

$$\begin{gathered} \stackrel{→}{x}\left(t\right)=\left[\begin{array}{c}2c_1+e^{5t}{c}_2\\ -c_1+2e^{5t}{c}_2\end{array}\right] \\ \stackrel{→}{x}\left(0\right)=\left[\begin{array}{c}2c_1+e^{5\left(0\right)}{c}_2\\ -c_1+2e^{5\left(0\right)}{c}_2\end{array}\right] \\ \stackrel{→}{x}\left(0\right)=\left[\begin{array}{c}2c_1+c_2\\ -c_1+2c_2\end{array}\right] \end{gathered}$$

As role="math" localid="1668667020454" $\stackrel{→}{x}\left(0\right)=\left[\begin{array}{c}2\\ -1\end{array}\right]$, substitute the value $\left[\begin{array}{c}2\\ -1\end{array}\right]$for $\stackrel{→}{x}\left(0\right)$in the equation $\stackrel{→}{x}\left(0\right)=\left[\begin{array}{c}2c_1+c_2\\ -c_1+2c_2\end{array}\right]$as follows.

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

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

$\begin{array}{rcl}2& =& 2c_1+c_2\\ -1& =& -c_1+2c_2\end{array}$

Subtract the equation $4=2\left(2c_1+c_2\right)$from $-1=-c_1+2c_2$as follows.

$$\begin{gathered} 4-\left(-1\right)=2\left\{2c_1+c_2\right\}-\left\{-c_1+2c_2\right\} \\ 5=4c_1+c_1 \\ {c}_1=1 \end{gathered}$$

Substitute the value 1 for c₁ in the equation $2=2c_1+c_2$as follows.

$\begin{array}{rcl}1& =& 2c_1+c_2\\ 1& =& 2\left(1\right)+c_2\\ c_2& =& -1\end{array}$

Therefore, the value of $c_1,c_2$are $c_1=1$and $c_2=-1$.

The general solution of $\stackrel{→}{x}\left(t\right)$is $\stackrel{→}{x}\left(t\right)=c_1{v}_1+c_2{e}^{5t}{v}_2$.

Substitute the value $\left[\begin{array}{c}2\\ -1\end{array}\right]$ for v₁ and $\left[\begin{array}{c}1\\ 2\end{array}\right]$ for v₂ in the equation $\stackrel{→}{x}\left(t\right)=c_1{v}_1+c_2{e}^{5t}{v}_2$as follows.

$$\begin{gathered} \stackrel{→}{x}\left(t\right)=c_1{v}_1+c_2{e}^{5t}{v}_2 \\ \stackrel{→}{x}\left(t\right)=c_1\left[\begin{array}{c}2\\ -1\end{array}\right]+c_2{e}^{5t}\left[\begin{array}{c}1\\ 2\end{array}\right] \end{gathered}$$

Substitute the values 1 for c₁ and -1 for c₂ in the equation $\stackrel{→}{x}\left(t\right)=c_1\left[\begin{array}{c}2\\ -1\end{array}\right]+c_2{e}^{5t}\left[\begin{array}{c}1\\ 2\end{array}\right]$as follows.

$$\begin{gathered} \stackrel{→}{x}\left(t\right)=c_1\left[\begin{array}{c}2\\ -1\end{array}\right]+c_2{e}^{5t}\left[\begin{array}{c}1\\ 2\end{array}\right] \\ \stackrel{→}{x}\left(t\right)=\left[\begin{array}{c}2\\ -1\end{array}\right]+e^{5t}\left[\begin{array}{c}1\\ 2\end{array}\right] \end{gathered}$$

Hence the solution of the system $\frac{d\stackrel{→}{x}}{dt}=\left[\begin{array}{cc}1& 2\\ 2& 4\end{array}\right]\stackrel{→}{x}$with the initial value $\stackrel{→}{x}\left(0\right)=\left[\begin{array}{c}2\\ -1\end{array}\right]$is $\stackrel{→}{x}\left(t\right)=\left[\begin{array}{c}2\\ -1\end{array}\right]+e^{5t}\left[\begin{array}{c}1\\ 2\end{array}\right]$.