Question

Find the real solution of the system$\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\left[\begin{array}{cc}2& 4\\ -4& 2\end{array}\right]\overrightarrow{\mathbf{x}}$

Short answer

• The solution is $\overrightarrow{x}\left(t\right)=e^{2t}\left[\begin{array}{cc}sin\left(4t\right)& cos\left(4t\right)\\ cos\left(4t\right)& -sin\left(4t\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$
  

Step-by-step solution

Definition of the theorem

Continuous dynamical system with eigenvalues

Consider the linear system $\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}\mathbf{}$where A is a real $\mathbf{2}\mathbf{\times }\mathbf{2}$matrix with complex eigenvalues $\mathit{p}\mathbf{\pm }\mathit{i}\mathit{q}$(and $\mathit{q}\mathbf{\ne }\mathbf{0}$).

Consider an eigenvector $\overrightarrow{\mathit{v}}\mathit{+}\mathit{i}\overrightarrow{\mathit{w}}$with eigenvalue $\mathit{p}\mathbf{\pm }\mathit{i}\mathit{q}$.

Then $\overrightarrow{\mathbf{x}}\left(t\right)\mathbf{=}{\mathit{e}}^{\mathbf{p}\mathbf{t}}\mathit{S}\left[\begin{array}{cc}cos\left(qt\right)& -sin\left(qt\right)\\ sin\left(qt\right)& cos\left(qt\right)\end{array}\right]{\mathit{S}}^{\mathbf{-}\mathbf{1}}{\mathit{x}}_{\mathbf{0}}$where $\mathit{S}\mathbf{=}\left[\overrightarrow{w} \overrightarrow{v}\right]$. Recall that ${\mathit{S}}^{\mathbf{-}\mathbf{1}}{\mathit{x}}_{\mathbf{0}}$is the coordinate vector of ${\overrightarrow{\mathbf{x}}}_{\mathbf{0}}$with respect to basis $\overrightarrow{\mathbf{w}}\mathbf{,}\overrightarrow{\mathbf{v}}$.

To find the eigenvalues

Consider the given system as follows.

$\frac{d\overrightarrow{x}}{dt}=\left[\begin{array}{cc}2& 4\\ -4& 2\end{array}\right]\overrightarrow{x}$

No, to find the eigenvalues of the coefficient matrix as follows.

$$\begin{gathered} \left[\begin{array}{cc}2-\lambda & 4\\ -4& 2-\lambda \end{array}\right]=0 \\ \left(2-\lambda \right)\left(2-\lambda \right)-\left(-4\right)\left(4\right)=0 \\ \left({\lambda }^2-4\lambda +4\right)+16=0 \\ {\lambda }^2-4\lambda +20=0 \end{gathered}$$

Simplify further as follows

$$\begin{gathered} \lambda =\frac{4\pm \sqrt{{\left(4\right)}^2-4\left(1\right)\left(20\right)}}{2\left(1\right)} \left\{Qa=1,b=4,c=20\right\} \\ \lambda =\frac{4\pm \sqrt{16-80}}{2} \\ \lambda =\frac{4\pm \sqrt{-64}}{2} \\ \lambda =\frac{4\pm i8}{2} \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \lambda =\frac{4\pm i8}{2} \\ \lambda =2\pm i4 \\ {\lambda }_1=2+i4 \\ {\lambda }_2=2-4i \end{gathered}$$

Therefore, the eigenvalues are ${\lambda }_1=2+4iand{\lambda }_2=2-4i$

To find the eigenvectors

To find a formula for trajectory as follows.

$$\begin{gathered} E_{2+4i}=ker\left[\begin{array}{cc}2-\left(2+4i\right)& 4\\ -4& 2-\left(2-4i\right)\end{array}\right] \\ =ker\left[\begin{array}{cc}-4i& 4\\ -4& -4i\end{array}\right] \\ =span\left(\left[\begin{array}{c}1\\ 0\end{array}\right]+i\left[\begin{array}{c}0\\ 1\end{array}\right]\right) \end{gathered}$$

Therefore, the eigenvectors are as follows.

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

Then by the theorem, the general solution for the system $\frac{d\overrightarrow{x}}{dt}=\left[\begin{array}{cc}2& 4\\ -4& 2\end{array}\right]\overrightarrow{x}$is as follows.

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

Similarly, the values of p,q and S are as follows.

$$\begin{gathered} p=2 \\ q=4 \\ S=\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right] \end{gathered}$$

Substitute the value 2 for p and 4 for q and $\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]forSin$

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

$$\begin{gathered} \overrightarrow{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}cos\left(qt\right)& -sin\left(qt\right)\\ sin\left(qt\right)& cos\left(qt\right)\end{array}\right]S^{-1}{\overrightarrow{x}}_0 \\ \overrightarrow{x}\left(t\right)=e^{2t}\left[\begin{array}{cc}0& 1\\ 1& 0\end{array}\right]\left[\begin{array}{cc}cos\left(4t\right)& -sin\left(4t\right)\\ sin\left(4t\right)& cos\left(4t\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right] \\ \overrightarrow{x}\left(t\right)=e^{2t}\left[\begin{array}{cc}0\left(cos\left(4t\right)\right)+1\left(sin\left(4t\right)\right)& 0\left(-sin\left(4t\right)\right)+1\left(cos\left(4t\right)\right)\\ 1\left(cos\left(4t\right)\right)+0\left(sin\left(4t\right)\right)& 1\left(-sin\left(4t\right)\right)+0\left(cos\left(4t\right)\right)\end{array}\right] \\ \overrightarrow{x}\left(t\right)=e^{2t}\left[\begin{array}{cc}sin\left(4t\right)& cos\left(4t\right)\\ cos\left(4t\right)& -sin\left(4t\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right] \end{gathered}$$

Hence, the solution for the system$\frac{\mathrm{d}\overrightarrow{\mathrm{x}}}{\mathrm{dt}}=\left[\begin{array}{cc}2& 4\\ -4& 2\end{array}\right]\overrightarrow{\mathrm{x}}$ is $\overrightarrow{x}\left(t\right)=e^{2t}\left[\begin{array}{cc}sin\left(4t\right)& cos\left(4t\right)\\ cos\left(4t\right)& -sin\left(4t\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$, where x and y are arbitrary constants.