Question

Solve the system$\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{\left[}\begin{array}{cc}\mathbf{4}& \mathbf{3}\\ \mathbf{4}& \mathbf{8}\end{array}\mathbf{\right]}\overrightarrow{\mathbf{x}}$with the given initial value$\overrightarrow{\mathbf{x}}\left(0\right)\mathbf{=}\left[\begin{array}{c}1\\ 1\end{array}\right]$.

Short answer

The solution of the system is $\overrightarrow{x}\left(t\right)=0.25e^{-2t}\left[\begin{array}{c}1\\ -2\end{array}\right]+0.75e^{10t}\left[\begin{array}{c}1\\ 2\end{array}\right]$.

Step-by-step solution

Determine the Eigen values

Consider the equation of dynamical system $\frac{\mathrm{d}\overrightarrow{\mathrm{x}}}{\mathrm{dt}}=\left[\begin{array}{cc}4& 3\\ 4& 8\end{array}\right]\overrightarrow{\mathrm{x}}$with the initial value $\overrightarrow{x}\left(0\right)=\left[\begin{array}{c}1\\ 1\end{array}\right]$.

Compare the equations$\frac{\mathrm{d}\overrightarrow{\mathrm{x}}}{\mathrm{dt}}=\left[\begin{array}{cc}4& 3\\ 4& 8\end{array}\right]\overrightarrow{\mathrm{x}}$and$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$as follows.

$A=\left[\begin{array}{cc}4& 3\\ 4& 8\end{array}\right]$

Assume$\lambda$is an Eigen value of the matrix role="math" localid="1668579048604" $\left[\begin{array}{cc}4& 3\\ 4& 8\end{array}\right]$implies $\left|A-\lambda I\right|=0$.

Substitute the values$\left[\begin{array}{cc}4& 3\\ 4& 8\end{array}\right]$for A and$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$for I in the equation$\left|A-\lambda I\right|=0$as follows.

$$\begin{gathered} \left|A-\lambda I\right|=0 \\ \left|\left[\begin{array}{cc}4& 3\\ 4& 8\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|=0 \end{gathered}$$

Simplify the equation $\left|\left[\begin{array}{cc}4& 3\\ 4& 8\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|=0$as follows.

$\begin{array}{rcl}\left|\left[\begin{array}{cc}4& 3\\ 4& 8\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|& =& 0\\ \left|\left[\begin{array}{cc}4& 3\\ 4& 8\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]\right|& =& 0\\ \left|\left[\begin{array}{cc}4-\lambda & 3\\ 4& 8-\lambda \end{array}\right]\right|& =& 0\\ \left(4-\lambda \right)\left(8-\lambda \right)-12& =& 0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{rcl}\left(4-\lambda \right)\left(8-\lambda \right)-12& =& 0\\ 32-4\lambda -8\lambda +{\lambda }^2-12& =& 0\\ {\lambda }^2-12\lambda +20& =& 0\\ {\lambda }^2-10\lambda +2\lambda +20& =& 0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{rcl}{\lambda }^2-10\lambda +2\lambda +20& =& 0\\ \lambda \left(\lambda -10\right)+2\left(\lambda -10\right)& =& 0\\ \left(\lambda +2\right)\left(\lambda -10\right)& =& 0\end{array}$

Therefore, the Eigen values of A are $\lambda =-2,10$.

Determine the Eigen vector corresponding to the Eigen values

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 $\lambda =-2,10$implies $\left|A-{\lambda }_{1}I\right|v_1=0$and $\left|A-{\lambda }_{2}I\right|v_2=0$.

Substitute the values role="math" localid="1668580359171" $\left[\begin{array}{cc}4& 3\\ 4& 8\end{array}\right]$for A, -2 for role="math" localid="1668580413744" ${\lambda }_1,\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]$ for v₁ and $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for I in the equation $\left|A-{\lambda }_{1}I\right|v_1=0$as follows.

role="math" localid="1668580106593" $\begin{array}{rcl}\left|A-{\lambda }_{1}I\right|v_1& =& 0\\ \left|\left[\begin{array}{cc}4& 3\\ 4& 8\end{array}\right]+2\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\\ \left|\left[\begin{array}{cc}4& 3\\ 4& 8\end{array}\right]+\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\\ \left|\left[\begin{array}{cc}6& 3\\ 4& 10\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{rcl}\left|\left[\begin{array}{cc}6& 3\\ 4& 10\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]& =& 0\\ 6x_1+3y_1& =& 0\\ 4x_1+10y_1& =& 0\end{array}$

Simplify the equations $6x_1+3y_1=0$and $4x_1+10y_1=0$as follows.

$\begin{array}{rcl}6x_1+3y_1& =& 0\\ 6x_1& =& -3y_1\end{array}$

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

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

Substitute the values $\left[\begin{array}{cc}4& 3\\ 4& 8\end{array}\right]$for A, 10 for ${\lambda }_2,\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]$for v₂ and$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$for I in the equation$\left|A-{\lambda }_{2}I\right|v_2=0$as follows.

role="math" localid="1668580713360" $\begin{array}{rcl}\left|A-{\lambda }_{2}I\right|v_2& =& 0\\ \left|\left[\begin{array}{cc}4& 3\\ 4& 8\end{array}\right]-10\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\\ \left|\left[\begin{array}{cc}4& 3\\ 4& 8\end{array}\right]-\left[\begin{array}{cc}10& 0\\ 0& 10\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\\ \left|\left[\begin{array}{cc}-6& 3\\ 4& -2\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\end{array}$

Further, simplify the equation as follows.

$\begin{array}{rcl}\left|\left[\begin{array}{cc}-6& 3\\ 4& -2\end{array}\right]\right|\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]& =& 0\\ -6x_2+3y_2& =& 0\\ 4x_2-2y_2& =& 0\end{array}$

Simplify the equations$-6x_2+3y_2=0$and$4x_2-2y_2=0$as follows.

$$\begin{gathered} -6x_2+3y_2=0 \\ 6x_2=3y_2 \\ 2x_2=y_2 \end{gathered}$$

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

Therefore, the Eigen vector corresponding to${\lambda }_2=10$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 $v_1=\left[\begin{array}{c}1\\ -2\end{array}\right]$and $v_2=\left[\begin{array}{c}1\\ 2\end{array}\right]$corresponding to the Eigen values ${\lambda }_1=-2$and${\lambda }_2=10$respectively implies$S^{-1}AS=B$where$S=\left[1\begin{array}{cc}1& 1\\ -2& 2\end{array}\right]$and $B=\left[\begin{array}{cc}-2& 0\\ 0& 10\end{array}\right]$.

Determine he solution for dx→dt=[4348]x→

Substitute the value SBS^-1 for A in the equation$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$as follows.

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

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

The solution of the equation$\frac{d}{dt}\overrightarrow{c}=B\overrightarrow{c}$is role="math" localid="1668581649314" $\overrightarrow{c}\left(t\right)=\left[\begin{array}{cc}{e}^{-2t}& c_1\\ e^{10t}& c_2\end{array}\right]$.

Substitute the valuerole="math" localid="1668581869674" width="77" height="63">e-2tc1e10tc2for $\overrightarrow{c}\left(t\right)$androle="math" localid="1668581850330" width="61" height="47">11-22for S in the equation $\overrightarrow{x}=S\overrightarrow{c}$as follows.

role="math" localid="1668582393345" $$\begin{gathered} \overrightarrow{x}\left(t\right)=S\overrightarrow{c} \\ \overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}1& 1\\ -2& 2\end{array}\right]\left[\begin{array}{cc}{e}^{-2t}& c_1\\ e^{10t}& c_2\end{array}\right] \\ \overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}{c}_1{e}^{-2t}& +e^{10t}{c}_2\\ -2e^{-2t}{c}_1& +2e^{10t}{c}_2\end{array}\right] \end{gathered}$$

Substitute the value 0 for t in the equation$\overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}{c}_1{e}^{-2t}& +e^{10t}{c}_2\\ -2e^{-2t}{c}_1& +2e^{10t}{c}_2\end{array}\right]$as follows.

role="math" localid="1668582630380" $$\begin{gathered} \overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}{c}_1{e}^{-2t}& +e^{10t}{c}_2\\ -2e^{-2t}{c}_1& +2e^{10t}{c}_2\end{array}\right] \\ \overrightarrow{x}\left(0\right)=\left[\begin{array}{cc}{c}_1{e}^{-2\left(0\right)}& +e^{10\left(0\right)}{c}_2\\ -2e^{-2\left(0\right)}{c}_1& +2e^{10\left(0\right)}{c}_2\end{array}\right] \\ \overrightarrow{x}\left(0\right)=\left[\begin{array}{c}{c}_1+c_2\\ -2c_1+2c_2\end{array}\right] \end{gathered}$$

As $\overrightarrow{x}\left(0\right)=\left[\begin{array}{c}1\\ 1\end{array}\right]$, substitute the value $\left[\begin{array}{c}1\\ 1\end{array}\right]$for $\overrightarrow{x}\left(0\right)$in the equation $\overrightarrow{x}\left(0\right)=\left[\begin{array}{c}{c}_1+c_2\\ -2c_1+2c_2\end{array}\right]$as follows.

role="math" localid="1668582748528" width="149" height="120">x→0=c1+c2-2c1+2c211=c1+c2-2c1+2c2

Find the values of c1,c2

Compare the equation both side in the equation$\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}{c}_1+c_2\\ -2c_1+2c_2\end{array}\right]$as follows.

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

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

$\begin{array}{rcl}2-1& =& 2\left\{c_1+c_2\right\}-\left\{-2c_1+2c_2\right\}\\ 1& =& 4c_1\\ c_1& =& 0.25\end{array}$

Substitute the value 0.25 for c₁ in the equation$1=c_1+c_2$as follows.

$\begin{array}{rcl}1& =& c_1+c_2\\ 1& =& 0.25+c_2\\ c_2& =& 0.75\end{array}$

Therefore, the value of$c_1,c_2$are$c_1=0.25$and $c_2=0.75$.

Determine the general solution for x→(t)

The general solution of$\overrightarrow{x}\left(t\right)$is $\overrightarrow{x}\left(t\right)=c_1{e}^{-2t}{v}_1+c_2{e}^{10t}{v}_2$.

Substitute the value$\left[\begin{array}{c}1\\ -2\end{array}\right]$for v₁ and$\left[\begin{array}{c}1\\ 2\end{array}\right]$for v₂ in the equation$\overrightarrow{x}\left(t\right)=c_1{e}^{-2t}{v}_1+c_2{e}^{4t}{v}_2$as follows.

role="math" localid="1668583249388" $$\begin{gathered} \overrightarrow{x}\left(t\right)=c_1{e}^{-2t}{v}_1+c_2{e}^{10t}{v}_2 \\ \overrightarrow{x}\left(t\right)=c_1{e}^{-2t}\left[\begin{array}{c}1\\ -2\end{array}\right]+c_2{e}^{10t}\left[\begin{array}{c}1\\ 2\end{array}\right] \end{gathered}$$

Substitute the values 0.25 for c₁ and 0.75 for c₂ in the equation$\overrightarrow{x}\left(t\right)=c_1{e}^{-2t}\left[\begin{array}{c}1\\ -2\end{array}\right]+c_2{e}^{10t}\left[\begin{array}{c}1\\ 2\end{array}\right]$as follows.

$$\begin{gathered} \overrightarrow{x}\left(t\right)=c_1{e}^{-2t}\left[\begin{array}{c}1\\ -2\end{array}\right]+c_2{e}^{10t}\left[\begin{array}{c}1\\ 2\end{array}\right] \\ \overrightarrow{x}\left(t\right)=0.25e^{-2t}\left[\begin{array}{c}1\\ -2\end{array}\right]+0.75e^{10t}\left[\begin{array}{c}1\\ 2\end{array}\right] \end{gathered}$$

Hence the solution of the system$\frac{d\overrightarrow{x}}{dt}=\left[\begin{array}{cc}4& 3\\ 4& 8\end{array}\right]\overrightarrow{x}$with the initial value$\overrightarrow{x}\left(0\right)=\left[\begin{array}{c}1\\ 1\end{array}\right]$is $\overrightarrow{x}\left(t\right)=c_1{e}^{-2t}\left[\begin{array}{c}1\\ -2\end{array}\right]+c_2{e}^{10t}\left[\begin{array}{c}1\\ 2\end{array}\right]$.