Question

Question: Consider the system $\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\left[\begin{array}{cc}0& 1\\ -1& 0\end{array}\right]\mathit{A}\overrightarrow{\mathbf{x}}$ where$\mathit{A}\mathbf{=}\mathbf{\left[}\begin{array}{cc}\mathbf{0}& \mathbf{1}\\ 0& \mathbf{0}\end{array}\mathbf{\right]}$. Sketch a direction field for Base on your sketch, describe the trajectories geometrically. Can you find the solution analytically?

Short answer

Answer

The solution of the system is$\overrightarrow{x}\left(t\right)=\left[\begin{array}{c}{c}_1+c_2+c_{2}t\\ c_2\end{array}\right]$ .

Step-by-step solution

Determine the Eigen values of the matrix.

Consider the equation $\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$with $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$.

Assume is an Eigen value of the matrix $\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$ implies .

Substitute the values $\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$for A androle="math" localid="1660643543411" $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for in the equation as follows$\left|A-\lambda I\right|=0$.

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

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

$$\begin{gathered} \left|\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|=0 \\ \left|\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]\right|=0 \\ \left|\left[\begin{array}{cc}-\lambda & 0\\ 0& -\lambda \end{array}\right]\right|=0 \\ {\lambda }^2=0 \end{gathered}$$

Therefore, the Eigen values of A are $\lambda =0$.

Determine the Eigen vector corresponding to the Eigen value λ=0

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

Substitute the values $\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$for A,0 for role="math" localid="1660644995182" ${\lambda }_1\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]$, for v₁ and for in the equation$\left|A-{\lambda }_{1}I\right|v_1=0$as follows.

$$\begin{gathered} \left|A-{\lambda }_{1}I\right|v_1=0 \\ \left|\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]-0\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]=0 \\ \left|\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\right|\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]=0 \\ \left[\begin{array}{c}{y}_1\\ 0\end{array}\right]=\left[\begin{array}{c}0\\ 0\end{array}\right] \end{gathered}$$

As is chosen to be arbitrary, assume x₁ = 1 implies$\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]=\left[\begin{array}{c}1\\ 0\end{array}\right]$

Therefore, the Eigen vector corresponding to role="math" localid="1660645072341" ${\lambda }_1=0is\left[\begin{array}{c}{x}_1\\ y_1\end{array}\right]=\left[\begin{array}{c}1\\ 0\end{array}\right]$.

Substitute the values$\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$ for A, 0 for , for and$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for in the equation as follows.

As is chosen to be arbitrary, assume implies $\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$

Therefore, the Eigen vector corresponding to ${\lambda }_2=0is\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]=\left[\begin{array}{c}1\\ 1\end{array}\right]$.

The Eigen vectors are and corresponding to the Eigen valu${\lambda }_2\left[\begin{array}{c}{x}_2\\ y_2\end{array}\right]$es $v_1=\left[\begin{array}{c}1\\ 0\end{array}\right]and{v}_2\left[\begin{array}{c}1\\ 1\end{array}\right]$respectively.

 Step 3: Find the general solution for x→(t)

The general solution of$\overrightarrow{x}\left(t\right)$ is$\overrightarrow{x}\left(t\right)=c_1{v}_1+c_2\left\{t{v}_2+v_1\right\}$ .

Substitute the value $\left[\begin{array}{c}1\\ 0\end{array}\right]$for v1 and $\left[\begin{array}{c}1\\ 1\end{array}\right]$for v₂ in the equation $\overrightarrow{x}\left(t\right)=c_1{v}_1+c_2\left\{t{v}_2+v_1\right\}$as follows.

$$\begin{gathered} \overrightarrow{x}\left(t\right)=c_1{v}_1+c_2\left\{t{v}_2+v_1\right\} \\ \overrightarrow{x}\left(t\right)=c_1\left[\begin{array}{c}1\\ 0\end{array}\right]+c_2\left\{t\left[\begin{array}{c}1\\ 1\end{array}\right]+\left[\begin{array}{c}1\\ 0\end{array}\right]\right\} \\ \overrightarrow{x}\left(t\right)=\left[\begin{array}{c}{c}_1\\ 0\end{array}\right]+c_2\left\{t\left[\begin{array}{c}{c}_{2}t\\ c_2\end{array}\right]+\left[\begin{array}{c}{c}_2\\ 0\end{array}\right]\right\} \\ \overrightarrow{x}\left(t\right)=\left[\begin{array}{c}{c}_1+c_2+c_{2}t\\ c_2\end{array}\right] \end{gathered}$$

Sketch the direction field graph ofAx→ .

As${\lambda }_{1,2}=0$ , draw the direction field graph of the function as follows.

Hence, the formula for the solution of the system$\frac{d\overrightarrow{x}}{dt}=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\overrightarrow{x}$ is and the direction field graph is$\overrightarrow{x}\left(t\right)=\left[\begin{array}{c}{c}_1+c_2+c_{2}t\\ c_2\end{array}\right]$ sketched.