Question

Solve the system$\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{\left[}\begin{array}{cc}\mathbf{-}\mathbf{1}& \mathbf{1}\\ \mathbf{-}\mathbf{2}& \mathbf{1}\end{array}\mathbf{\right]}\overrightarrow{\mathbf{x}}$with$\overrightarrow{\mathbf{x}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{\left[}\begin{array}{c}\mathbf{0}\\ \mathbf{1}\end{array}\mathbf{\right]}$. Give the solution in real form. Sketch the solution.

Short answer

The solution of the system is$\overrightarrow{x}\left(t\right)=\left[\begin{array}{c}\sin \left(t\right)\\ \sin \left(t\right)+\cos \left(t\right)\end{array}\right]$and the graph is

Step-by-step solution

Find the Eigen values of the matrix. 

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

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

$A=\left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right]$

Assume $\lambda$is an Eigen value of the matrix$\left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right]$ implies $|A-\lambda I|=0$.

Substitute the values$\left[\begin{array}{cc}-1& 1\\ -2& 1\end{array}\right]$ 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.

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

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

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

Further, simplify the equation as follows.

$\begin{array}{c}(-1-\lambda )(1-\lambda )+2=0\\ -1+{\lambda }^2+2=0\\ {\lambda }^2+1=0\\ {\lambda }^2=-1\end{array}$

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

Determine the Eigen vector corresponding to the Eigen value λ=i.

Substitute the values$i$ for$\lambda$ in the equation $\left|\left[\begin{array}{cc}-1-\lambda & 1\\ -2& 1-\lambda \end{array}\right]\right|=0$as follows.

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

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

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

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

Determine the solution for dx→dt=[-11-21]x→.

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

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

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$\left[\begin{array}{cc}0& 1\\ 1& 1\end{array}\right]$for $S$, $\left[\begin{array}{cc}-1& 1\\ 1& 0\end{array}\right]$for ,$S^{-1}$$\left[\begin{array}{c}0\\ 1\end{array}\right]$for ,${\overrightarrow{x}}_0$

$0$for$p$ and$1$ 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.

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

Further, simplify the equation as follows.

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

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

Sketch the solution. 

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

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