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Found in: Page 441

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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

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

See the step by step solution

## Step 1: Definition of the theorem

Continuous dynamical system with eigenvalues

Consider the linear system $\frac{\mathbf{d}\stackrel{\mathbf{\to }}{\mathbf{x}}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}{\mathbit{A}}\stackrel{\mathbf{\to }}{\mathbf{x}}$where A is a real ${\mathbf{2}}{\mathbf{×}}{\mathbf{2}}$ matrix with complex eigenvalues ${\mathbit{p}}{\mathbf{±}}{\mathbit{i}}{\mathbit{q}}$ (and ${\mathbit{q}}{\mathbf{\ne }}{\mathbf{0}}$ ).

Consider an eigenvector role="math" localid="1659877431239" $\stackrel{\mathbf{\to }}{\mathbf{v}}{\mathbf{+}}{\mathbit{i}}\stackrel{\mathbf{\to }}{\mathbf{w}}$with eigenvalue ${\mathbit{p}}{\mathbf{±}}{\mathbit{i}}{\mathbit{q}}$.

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

## Step 2: To find the eigenvalues

Consider the given system as follows.

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

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

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

Simplify further as follows

${\lambda }^{2}=-36\phantom{\rule{0ex}{0ex}}\lambda =±6i$

Therefore, the eigenvalues are ${\lambda }_{1}=6i$ and ${\lambda }_{2}=-6i$

## Step3: To find the eigenvectors

To find a formula for trajectory as follows.

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

Therefore, the eigenvectors are as follows.

$\stackrel{\to }{v}=\left[\begin{array}{c}2\\ 0\end{array}\right]\phantom{\rule{0ex}{0ex}}\stackrel{\to }{w}=\left[\begin{array}{c}0\\ 3\end{array}\right]\phantom{\rule{0ex}{0ex}}$

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

$\stackrel{\to }{x}\left(t\right)={e}^{pt}S\left[\begin{array}{cc}cos\left(qt\right)& -sin\left(qt\right)\\ sin\left(qt\right)& cos\left(qt\right)\end{array}\right]{S}^{-1}{\stackrel{\to }{x}}_{0}$ where $S=\left[\stackrel{\to }{w} \stackrel{\to }{v}\right]$

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

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

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 $\stackrel{\to }{x}\left(t\right)={e}^{pt}S\left[\begin{array}{cc}cos\left(qt\right)& -sin\left(qt\right)\\ sin\left(qt\right)& cos\left(qt\right)\end{array}\right]{S}^{-1}{\stackrel{\to }{x}}_{0}\phantom{\rule{0ex}{0ex}}$ as follows.

$\stackrel{\to }{x}\left(t\right)={e}^{pt}S\left[\begin{array}{cc}cos\left(qt\right)& -sin\left(qt\right)\\ sin\left(qt\right)& cos\left(qt\right)\end{array}\right]{S}^{-1}{\stackrel{\to }{x}}_{0}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{x}\left(t\right)={e}^{0t}\left[\begin{array}{cc}0& 2\\ 3& 0\end{array}\right]\left[\begin{array}{cc}cos\left(6t\right)& -sin\left(6t\right)\\ sin\left(6t\right)& cos\left(6t\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{x}\left(t\right)=\left[\begin{array}{cc}0\left(cos\left(6t\right)\right)+2\left(sin\left(6t\right)\right)& 0\left(-sin\left(6t\right)\right)+2\left(cos\left(6t\right)\right)\\ 3\left(cos\left(6t\right)\right)+0\left(sin\left(6t\right)\right)& 3\left(-sin\left(6t\right)\right)+0\left(cos\left(6t\right)\right)\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{x}\left(t\right)=\left[\begin{array}{cc}2sin\left(6t\right)& 2cos\left(6t\right)\\ 3cos\left(6t\right)& -3sin\left(6t\right)\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]$

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

where x and y are arbitrary constants.