In Exercises 3-6, solve the initial value problem \(x'\left( t \right) = Ax\left( t \right)\) for \(t \ge 0\), with \(x\left( 0 \right) = \left( {3,2} \right)\). Classify the nature of the origin as an attractor, repeller, or saddle point of the dynamical system described by \(x' = Ax\). Find the directions of greatest attraction and/or repulsion. When the origin is a saddle point, sketch typical trajectories.

6. \(A = \left( {\begin{aligned}{ {20}{c}}1&{ - 2}\\3&{ - 4}\end{aligned}} \right)\)

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:

\(A = \left( {\begin{aligned}{ {20}{c}}1&{ - 2}\\3&{ - 4}\end{aligned}} \right)\)

For eigenvalues:

\(\begin{aligned}{c}\det (A - \lambda I) = 0\\(1 - \lambda )( - 4 - \lambda ) + 6 = 0\\2 + 3\lambda + {\lambda ^2} = 0\\{\lambda _1} = - 2,{\lambda _2} = - 1\end{aligned}\)

The obtained eigenvalues are:\( - 2\)and\( - 1\).

Hence, the origin is an attractor.

Now, for\({\lambda _1} = - 2\),we have:

\(A = \left( {\begin{aligned}{ {20}{c}}3&{ - 2}&0\\3&{ - 2}&0\end{aligned}} \right)\~\left( {\begin{aligned}{ {20}{c}}1&{ - \frac{2}{3}}&0\\0&0&0\end{aligned}} \right)\)

So,\({x_1} = \frac{2}{3}{x_2}\)with\({x_2}\)free.

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

\({v_1} = \left( {\begin{aligned}{ {20}{l}}2\\3\end{aligned}} \right)\)

Similarly,for\({\lambda _2} = - 1\)we have:

\(A = \left( {\begin{aligned}{ {20}{c}}2&{ - 2}&0\\3&{ - 3}&0\end{aligned}} \right)\~\left( {\begin{aligned}{ {20}{l}}1&{ - 1}&0\\0&0&0\end{aligned}} \right)\)

So\({x_1} = {x_2}\)with\({x_2}\)free.

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

\({v_2} = \left( {\begin{aligned}{ {20}{l}}1\\1\end{aligned}} \right)\)

Using both eigenvectors, we have:

\(\left( {\begin{aligned}{ {20}{l}}{{v_1}}&{{v_2}}&{x(0)}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}2&1&3\\3&1&2\end{aligned}} \right)\~\left( {\begin{aligned}{ {20}{c}}1&0&{ - 1}\\0&1&5\end{aligned}} \right)\)

Thus,\({c_1} = - 1\)and\({c_2} = 5\),

Now, the general solution will be:

\(x(t) = - \left( {\begin{aligned}{ {20}{l}}2\\3\end{aligned}} \right){e^{ - 2t}} + 5\left( {\begin{aligned}{ {20}{l}}1\\1\end{aligned}} \right){e^{ - t}}\)

Hence, this is the required solution.

Since both eigenvalues are less than 1, so the origin is an attractor of the dynamical system described by\(x\prime = Ax\).

The direction of greatest repulsion is through the origin, and the eigenvector is \({v_1} = \left( {\begin{aligned}{ {20}{c}}2\\3\end{aligned}} \right)\).