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.

10. \(A = \left( {\begin{aligned}{ {20}{c}}3&1\\{ - 2}&1\end{aligned}} \right)\)

Short answer

The requiredcomplex solution is:

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

And, the real general solution of is:

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

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,\(A = \left( {\begin{aligned}{ {20}{r}}3&1\\{ - 2}&1\end{aligned}} \right)\).

For eigenvalues, we have:

\(\begin{aligned}{c}\det \left( {x{I_2} - A} \right) = 0\\(3 - x)(1 - x) + 2 = 0\\{x^2} - 4x + 3 + 2 = 0\\{x^2} - 4x + 5 = 0\\x = 2 \pm i\end{aligned}\)

Find eigenvectors for both eigenvalues

Now, forthe eigenvalue\(2 + i\), we have:

\(\left( {\begin{aligned}{ {20}{c}}{1 - i}&1\\{ - 2}&{ - 1 - i}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{l}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{l}}0\\0\end{aligned}} \right)\)

So,\({x_1} = - \frac{{\left( {1 + i} \right)}}{2}{x_2}\)with\({x_2}\)free.

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

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

Similarly,forthe eigenvalue\(2 - i\), we have:

\(\left( {\begin{aligned}{ {20}{c}}{1 + i}&1\\{ - 2}&{ - 1 + i}\end{aligned}} \right)\left( {\begin{aligned}{ {20}{l}}{{y_1}}\\{{y_2}}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{l}}0\\0\end{aligned}} \right)\)

So\({y_1} = - \frac{{\left( {1 - i} \right)}}{2}{y_2}\)with\({y_2}\)free.

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

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

Using both eigenvectors, we have:

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

Hence, this is the required solution.

Find the real general solution

Now, the real general solution will be given as:

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

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

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