In Exercises 9 and 18,construct the general solution of\(x' = Ax\)involving complex Eigen functions and then obtain the general real solution. Describe the shapes of typical trajectories.

11. \(A = \left( {\begin{aligned}{ {20}{c}}{ - 3}&{ - 9}\\2&3\end{aligned}} \right)\)

The general solutionfor any system of differential equations withthe eigenvalues\({\lambda _1}\)and\({\lambda _2}\)with the respective eigenvectors\({v_1}\)and\({v_2}\)is given by:

\(x(t) = {c_1}{v_1}{e^{{\lambda _1}t}} + {c_2}{v_2}{e^{{\lambda _2}t}}\)

Here,\({c_1}\)and\({c_2}\)are the constants from the initial condition.

As per the question, we have:

\(A = \left( {\begin{aligned}{ {20}{r}}{ - 3}&{ - 9}\\2&3\end{aligned}} \right)\)

For eigenvalues, we have:

\(\begin{aligned}{c}\det \left( {x{I_2} - A} \right) = 0\\( - 3 - x)(3 - x) + 18 = 0\\{x^2} - 9 + 18 = 0\\{x^2} + 9 = 0\\x = \pm 3i\end{aligned}\)

Now, forthe eigenvalue\(3i\), we have:

\(\left( {\begin{aligned}{ {20}{c}}{ - 3 - 3i}&{ - 9}\\2&{3 - 3i}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{l}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{l}}0\\0\end{aligned}} \right)\)

So,\({x_1} = \frac{{\left( { - 3 + 3i} \right)}}{2}{x_2}\)with\({x_2}\)free.

Taking\({x_2} = 1\), we get, eigenvector:

\({v_1} = \left( {\begin{aligned}{ {20}{c}}{ - 3 + 3i}\\2\end{aligned}} \right)\)

Similarly,forthe eigenvalue\( - 3i\), we have:

\(\left( {\begin{aligned}{ {20}{c}}{ - 3 + 3i}&{ - 9}\\2&{3 + 3i}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{l}}{{y_1}}\\{{y_2}}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{l}}0\\0\end{aligned}} \right)\)

So\({y_1} = \frac{{\left( { - 3 - 3i} \right)}}{2}{y_2}\)with\({y_2}\)free.

Taking\({y_2} = 1\), we get:

\({v_2} = \left( {\begin{aligned}{ {20}{c}}{ - 3 - 3i}\\2\end{aligned}} \right)\)

Using both eigenvectors, we have:

\(\begin{aligned}{c}x(t) = {c_1}{v_1}{e^{{\lambda _1}t}} + {c_2}{v_2}{e^{{\lambda _2}t}}\\ = {c_1}\left( {\begin{aligned}{ {20}{c}}{( - 3 + 3i)}\\2\end{aligned}} \right){e^{(3i)t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{( - 3 - 3i)}\\2\end{aligned}} \right){e^{( - 3i)t}}\end{aligned}\)

Hence, this is the required solution.

Now, the real general solution will be given as:

\(\begin{aligned}{c}x(t) = {c_1}\left( {\begin{aligned}{ {20}{c}}{( - 3 + 3i)}\\2\end{aligned}} \right){e^{(3i)t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{( - 3 - 3i)}\\2\end{aligned}} \right){e^{( - 3i)t}}\\ = {c_1}\left( {\begin{aligned}{ {20}{c}}{( - 3 + 3i)}\\2\end{aligned}} \right)(\cos 3t + i\sin 3t) + {c_2}\left( {\begin{aligned}{ {20}{c}}{( - 3 - 3i)}\\2\end{aligned}} \right)(\cos 3t - i\sin 3t)\\ = {c_1}\left( {\begin{aligned}{ {20}{c}}{ - 3\cos 3t - 3\sin 3t}\\{2\cos 3t}\end{aligned}} \right) + {c_2}\left( {\begin{aligned}{ {20}{c}}{3\cos 3t - 3\sin 3t}\\{2\sin 3t}\end{aligned}} \right)\end{aligned}\)

Hence, the real general solution of\(x' = Ax\)is given by;

\(y(t) = {c_1}\left( {\begin{aligned}{ {20}{c}}{ - 3\cos 3t - 3\sin 3t}\\{2\cos 3t}\end{aligned}} \right) + {c_2}\left( {\begin{aligned}{ {20}{c}}{3\cos 3t - 3\sin 3t}\\{2\sin 3t}\end{aligned}} \right)\), where\({c_1},{c_2} \in \mathbb{R}\)

Hence, this is the required general real solution. And, the trajectories are elliptical as the eigenvalues are complex numbers with real part zero.