Question

Solve the system$\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{\left[}\begin{array}{cc}\mathbf{0}& \mathbf{1}\\ \mathbf{-}\mathbf{4}& \mathbf{0}\end{array}\mathbf{\right]}\overrightarrow{\mathbf{x}}$with$\overrightarrow{\mathbf{x}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{1}\\ \mathbf{0}\end{array}\mathbf{\right]}$. Give the solution in real form. Sketch the solution.

Short answer

The solution of the system is and the graph is

Step-by-step solution

Find the Eigen values of the matrix.

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

Compare the equations $\frac{d\overrightarrow{x}}{dt}=\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]\overrightarrow{x}$and$\frac{d\overrightarrow{x}}{dt}=A\overrightarrow{x}$ as follows.

$A=\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]$

Assume$\lambda$ is an Eigen value of the matrix $\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]$implies$|A-\lambda I|=0$ .

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

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

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

$\begin{array}{c}|\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]-\lambda \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]|=0\\ |\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]-\left[\begin{array}{cc}\lambda & 0\\ 0& \lambda \end{array}\right]|=0\\ \left|\left[\begin{array}{cc}-\lambda & 1\\ -4& -\lambda \end{array}\right]\right|=0\\ {\lambda }^2+4=0\end{array}$

Therefore, the Eigen values of $A$are $\lambda =\pm i2$.

Determine the Eigen vector corresponding to the Eigen valueλ=i2 .

Substitute the values$i2$ for$\lambda$ in the equation$\left|\left[\begin{array}{cc}-\lambda & 1\\ -4& -\lambda \end{array}\right]\right|=0$ as follows.

$\begin{array}{c}\left|\left[\begin{array}{cc}-\lambda & 1\\ -4& -\lambda \end{array}\right]\right|=0\\ \left|\left[\begin{array}{cc}-2i& 1\\ -4& -2i\end{array}\right]\right|=0\end{array}$

As $E_{-1+2i}=\ker \left[\begin{array}{cc}-2i& 1\\ -4& -2i\end{array}\right]=span\left[\begin{array}{c}-2i\\ 1\end{array}\right]$, the values $\overrightarrow{v}+i\overrightarrow{w}$is defined as follows.

$\overrightarrow{v}+i\overrightarrow{w}=\left[\begin{array}{c}0\\ 1\end{array}\right]+i\left[\begin{array}{c}-2\\ 0\end{array}\right]$

Therefore, the value of $S$is$S=\left[\begin{array}{cc}-2& 0\\ 0& 1\end{array}\right]$ .

Determine the solution for dx→dt=[01-40]x→

The inverse of the matrix$S=\left[\begin{array}{cc}-2& 0\\ 0& 1\end{array}\right]$ is defined as follows.

$\begin{array}{c}{S}^{-1}=\frac{1}{-2}\left[\begin{array}{cc}1& 0\\ 0& -2\end{array}\right]\\ S^{-1}=\left[\begin{array}{cc}\frac{1}{-2}& 0\\ 0& 1\end{array}\right]\end{array}$

As $\overrightarrow{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}\cos \left(qt\right)& -\sin \left(qt\right)\\ \sin \left(qt\right)& \cos \left(qt\right)\end{array}\right]S^{-1}{\overrightarrow{x}}_0$, Substitute the value$\left[\begin{array}{cc}-2& 0\\ 0& 1\end{array}\right]$for$S$, $\left[\begin{array}{cc}\frac{1}{-2}& 0\\ 0& 1\end{array}\right]$for $S^{-1}$, $\left[\begin{array}{c}1\\ 0\end{array}\right]$for ${\overrightarrow{x}}_0$,$0$for$p$and$2$for$q$in the equation

$\overrightarrow{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}\cos \left(qt\right)& -\sin \left(qt\right)\\ \sin \left(qt\right)& \cos \left(qt\right)\end{array}\right]S^{-1}{\overrightarrow{x}}_0$as follows.

$\begin{array}{c}\overrightarrow{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}\cos \left(qt\right)& -\sin \left(qt\right)\\ \sin \left(qt\right)& \cos \left(qt\right)\end{array}\right]S^{-1}{\overrightarrow{x}}_0\\ \overrightarrow{x}\left(t\right)=e^{\left(0\right)t}\left[\begin{array}{cc}-2& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}\cos \left(2t\right)& -\sin \left(2t\right)\\ \sin \left(2t\right)& \cos \left(2t\right)\end{array}\right]\left[\begin{array}{cc}\frac{1}{-2}& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}-2\cos \left(2t\right)& 2\sin \left(2t\right)\\ \sin \left(2t\right)& \cos \left(2t\right)\end{array}\right]\left[\begin{array}{cc}\frac{1}{-2}& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}\cos \left(2t\right)& 2\sin \left(2t\right)\\ -\frac{\sin \left(2t\right)}{2}& \cos \left(2t\right)\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\end{array}$

Further, simplify the equation as follows.

$\begin{array}{l}\overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}\cos \left(2t\right)& 2\sin \left(2t\right)\\ -\frac{\sin \left(2t\right)}{2}& \cos \left(2t\right)\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\\ \overrightarrow{x}\left(t\right)=\left[\begin{array}{c}\cos \left(2t\right)\\ -\frac{\sin \left(2t\right)}{2}\end{array}\right]\end{array}$

Therefore, the solution of the system is$\overrightarrow{x}\left(t\right)=\left[\begin{array}{c}\cos \left(2t\right)\\ -\frac{\sin \left(2t\right)}{2}\end{array}\right]$ .

 Step 4: Sketch the solution.

As${\lambda }_{1,2}=\pm 2i$, draw the graph of the solution$\overrightarrow{x}\left(t\right)=\left[\begin{array}{c}\cos \left(2t\right)\\ -\frac{\sin \left(2t\right)}{2}\end{array}\right]$as follows.

Hence, the solution of the system$\frac{d\overrightarrow{x}}{dt}=\left[\begin{array}{cc}0& 1\\ -4& 0\end{array}\right]\overrightarrow{x}$with the initial value$\overrightarrow{x}\left(0\right)=\left[\begin{array}{c}1\\ 0\end{array}\right]$is$\overrightarrow{x}\left(t\right)=\left[\begin{array}{c}\cos \left(2t\right)\\ -\frac{\sin \left(2t\right)}{2}\end{array}\right]$and the graph of the solution is an ellipse in counterclockwise direction.