Question

Find the real solution of the system $\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\left[\begin{array}{cc}0& 4\\ -9& 0\end{array}\right]\overrightarrow{\mathbf{x}}$

Short answer

The solution is$\overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}2sin\left(6t\right)& 2cos\left(6t\right)\\ 3cos\left(6t\right)& -3sin\left(6t\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$

Step-by-step solution

Definition of the theorem

Continuous dynamical system with eigenvalues

Consider the linear system $\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{A}\overrightarrow{\mathbf{x}}$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 eigenvectorrole="math" localid="1659877431239" $\overrightarrow{\mathbf{v}}\mathbf{+}\mathit{i}\overrightarrow{\mathbf{w}}$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.

$\frac{d\overrightarrow{x}}{dt}=\left[\begin{array}{cc}0& 4\\ -9& 0\end{array}\right]\overrightarrow{x}$

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

$$\begin{gathered} \left[\begin{array}{cc}0-\lambda & 4\\ -9& 0-\lambda \end{array}\right]=0 \\ \left(0-\lambda \right)\left(0-\lambda \right)-\left(-9\right)\left(4\right)=0 \\ {\lambda }^2+36=0 \\ {\lambda }^2=-36 \end{gathered}$$

Simplify further as follows

$$\begin{gathered} {\lambda }^2=-36 \\ \lambda =\pm 6i \end{gathered}$$

Therefore, the eigenvalues are${\lambda }_1=6i$and${\lambda }_2=-6i$

To find the eigenvectors

To find a formula for trajectory as follows.

$$\begin{gathered} E_{6i}=ker\left[\begin{array}{cc}0-6i& 4\\ -9& 0-6i\end{array}\right] \\ =span\left(\left[\begin{array}{c}2\\ 0\end{array}\right]+i\left[\begin{array}{c}0\\ 3\end{array}\right]\right) \end{gathered}$$

Therefore, the eigenvectors are as follows.

$$\begin{gathered} \overrightarrow{v}=\left[\begin{array}{c}2\\ 0\end{array}\right] \\ \overrightarrow{w}=\left[\begin{array}{c}0\\ 3\end{array}\right] \end{gathered}$$

Then by the theorem, the general solution for the system $\frac{d\overrightarrow{x}}{dt}=\left[\begin{array}{cc}0& 4\\ -9& 0\end{array}\right]\overrightarrow{x}$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.

$$\begin{gathered} p=0 \\ q=6 \\ S=\left[\begin{array}{cc}0& 2\\ 3& 0\end{array}\right] \end{gathered}$$

Substitute the value 0 for p and 6 for q and $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ for S in $\overrightarrow{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}cos\left(qt\right)& -sin\left(qt\right)\\ sin\left(qt\right)& cos\left(qt\right)\end{array}\right]S^{-1}{\overrightarrow{x}}_0$ as follows.

$$\begin{gathered} \overrightarrow{x}\left(t\right)=e^{pt}S\left[\begin{array}{cc}cos\left(qt\right)& -sin\left(qt\right)\\ sin\left(qt\right)& cos\left(qt\right)\end{array}\right]S^{-1}{\overrightarrow{x}}_0 \\ \overrightarrow{x}\left(t\right)=e^{0t}\left[\begin{array}{cc}0& 2\\ 3& 0\end{array}\right]\left[\begin{array}{cc}cos\left(6t\right)& -sin\left(6t\right)\\ sin\left(6t\right)& cos\left(6t\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right] \\ \overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}0\left(cos\left(6t\right)\right)+2\left(sin\left(6t\right)\right)& 0\left(-sin\left(6t\right)\right)+2\left(cos\left(6t\right)\right)\\ 3\left(cos\left(6t\right)\right)+0\left(sin\left(6t\right)\right)& 3\left(-sin\left(6t\right)\right)+0\left(cos\left(6t\right)\right)\end{array}\right] \\ \overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}2sin\left(6t\right)& 2cos\left(6t\right)\\ 3cos\left(6t\right)& -3sin\left(6t\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right] \end{gathered}$$

Hence, the solution for the system$\frac{d\overrightarrow{x}}{dt}=\left[\begin{array}{cc}0& 4\\ -9& 0\end{array}\right]\overrightarrow{x}$is$\overrightarrow{x}\left(t\right)=\left[\begin{array}{cc}2sin\left(6t\right)& 2cos\left(6t\right)\\ 3cos\left(6t\right)& -3sin\left(6t\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$

where x and y are arbitrary constants.