Question

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.

13. \(A = \left( {\begin{aligned}{ {20}{c}}4&{ - 3}\\6&{ - 2}\end{aligned}} \right)\)

Short answer

The requiredcomplex solution is:

\(x(t) = {c_1}\left( {\begin{aligned}{ {20}{c}}{1 + i}\\2\end{aligned}} \right){e^{(1 + 3i)t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{1 - i}\\2\end{aligned}} \right){e^{(1 - 3i)t}}\)

And, the real general solution of is:

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

Step-by-step solution

System of Differential Equations

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.

Calculation for eigenvalues

It is given that,\(A = \left( {\begin{aligned}{ {20}{l}}4&{ - 3}\\6&{ - 2}\end{aligned}} \right)\).

For eigenvalues, we have:

\(\begin{aligned}{c}\det \left( {x{I_2} - A} \right) = 0\\(4 - x)( - 2 - x) + 18 = 0\\{x^2} - 2x + 12 = 0\\{x_1} = 1 + 3i,{\rm{ }}{x_2} = 1 - 3i\end{aligned}\)

Find eigenvectors for both eigenvalues

Now, forthe eigenvalue\(1 + 3i\), we have:

\(\left( {\begin{aligned}{ {20}{c}}{3 - 3i}&{ - 3}\\6&{ - 3 - 3i}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{l}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{l}}0\\0\end{aligned}} \right)\)

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

\({v_1} = \left( {\begin{aligned}{ {20}{c}}{1 + i}\\2\end{aligned}} \right)\)

Similarly,forthe eigenvalue\(1 - 3i\), we have:

\({v_2} = \left( {\begin{aligned}{ {20}{c}}{1 - i}\\2\end{aligned}} \right)\)

Using both eigenvectors, we have:

\(\begin{aligned}{c}x(t) = {c_1}{v_1}{e^{{\lambda _1}t}} + {c_2}{v_2}{e^{{\lambda _2}t}}\\ = {c_1}\left( {\begin{aligned}{ {20}{c}}{1 + i}\\2\end{aligned}} \right){e^{(1 + 3i)t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{1 - i}\\2\end{aligned}} \right){e^{(1 - 3i)t}}\end{aligned}\)

Hence, this is the required solution.

Find the real general solution

Now, the real general solution will be given as:

\(\begin{aligned}{c}x(t) = {c_1}\left( {\begin{aligned}{ {20}{c}}{1 + i}\\2\end{aligned}} \right){e^{(1 + 3i)t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}{1 - i}\\2\end{aligned}} \right){e^{(1 - 3i)t}}\\ = {c_1}\left( {\begin{aligned}{ {20}{c}}{1 + i}\\2\end{aligned}} \right)(\cos 3t + i\sin 3t){e^t} + {c_2}\left( {\begin{aligned}{ {20}{c}}{1 - i}\\2\end{aligned}} \right)(\cos 3t - i\sin 3t){e^t}\\ = {c_1}\left( {\begin{aligned}{ {20}{c}}{\cos 3t - \sin 3t}\\{2\cos 3t}\end{aligned}} \right){e^t} + {c_2}\left( {\begin{aligned}{ {20}{c}}{\cos 3t + \sin 3t}\\{2\sin 3t}\end{aligned}} \right){e^t}\end{aligned}\)

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

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

Hence, this is the required general real solution. And, the trajectories are spiral as the eigenvalues are complex numbers.