Question

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.

4. \(A = \left( {\begin{aligned}{ {20}{c}}{ - 2}&{ - 5}\\1&4\end{aligned}} \right)\)

Short answer

The required solution is:

\({\rm{x}}\left( t \right) = \frac{{13}}{4}\left( {\begin{aligned}{ {20}{c}}{ - 1}\\1\end{aligned}} \right){e^{3t}} - \frac{5}{4}\left( {\begin{aligned}{ {20}{c}}{ - 5}\\1\end{aligned}} \right){e^{ - t}}\)

The origin is a saddle point.

The direction of greatest repulsion is through the origin, and the eigenvector is\({v_1} = \left( {\begin{aligned}{ {20}{c}}{ - 1}\\1\end{aligned}} \right)\).

The direction of greatest attraction is through the origin and eigenvector is \({v_2} = \left( {\begin{aligned}{ {20}{c}}{ - 5}\\1\end{aligned}} \right)\).

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.

Find the eigenvalues.

According to the question, we have:

\(A = \left( {\begin{aligned}{ {20}{c}}{ - 2}&{ - 5}\\1&4\end{aligned}} \right)\)

For eigenvalues:

\(\begin{aligned}{c}\det (A - \lambda I) = 0\\( - 2 - \lambda )(4 - \lambda ) + 5 = 0\\ - 3 - 2\lambda + {\lambda ^2} = 0\\{\lambda _1} = 3,{\lambda _2} = - 1\end{aligned}\)

The obtained eigenvalues are 3 and\( - 1\).

Hence, the origin is a saddle point.

Find eigenvectors for both eigenvalues

Now, for\({\lambda _1} = 3\)we have:

\(A = \left( {\begin{aligned}{ {20}{c}}{ - 5}&{ - 5}&0\\1&1&0\end{aligned}} \right)\~\left( {\begin{aligned}{ {20}{c}}1&1&0\\0&0&0\end{aligned}} \right)\)

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

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

\({v_1} = \left( {\begin{aligned}{ {20}{l}}{ - 1}\\1\end{aligned}} \right)\)

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

\(A = \left( {\begin{aligned}{ {20}{c}}{ - 1}&{ - 5}&0\\1&5&0\end{aligned}} \right)\~\left( {\begin{aligned}{ {20}{l}}1&5&0\\0&0&0\end{aligned}} \right)\)

So\({x_1} = - 5{x_2}\)with\({x_2}\)free.

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

\({v_2} = \left( {\begin{aligned}{ {20}{l}}{ - 5}\\1\end{aligned}} \right)\)

Using both eigenvectors, we have:

\(\left( {\begin{aligned}{ {20}{l}}{{v_1}}&{{v_2}}&{x(0)}\end{aligned}} \right) = \left( {\begin{aligned}{ {20}{c}}{ - 1}&{ - 5}&3\\1&1&2\end{aligned}} \right)\~\left( {\begin{aligned}{ {20}{c}}1&0&{\frac{{13}}{4}}\\0&1&{ - \frac{5}{4}}\end{aligned}} \right)\)

Thus,\({c_1} = \frac{{13}}{4}\)and\({c_2} = - \frac{5}{4}\).

Now, the general solution will be:

\(x(t) = \frac{{13}}{4}\left( {\begin{aligned}{ {20}{c}}{ - 1}\\1\end{aligned}} \right){e^{3t}} - \frac{5}{4}\left( {\begin{aligned}{ {20}{c}}{ - 5}\\1\end{aligned}} \right){e^{ - t}}\)

Hence, this is the required solution.

Find the directions of greatest attraction and/or repulsion

Since one eigenvalue is less than 1 and the other is greater, so the origin is a saddle point of the dynamical system described by\(x\prime = Ax\).

The direction of greatest repulsion is through the origin and eigenvector:\({v_1} = \left( {\begin{aligned}{ {20}{c}}{ - 1}\\1\end{aligned}} \right)\).

The direction of greatest attraction is through the origin and eigenvector: \({v_2} = \left( {\begin{aligned}{ {20}{c}}{ - 5}\\1\end{aligned}} \righ