Question

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

16. (M) \(A = \left( {\begin{aligned}{ {20}{c}}{ - 6}&{11}&{16}\\2&5&{ - 4}\\{ - 4}&{ - 5}&{10}\end{aligned}} \right)\)

Short answer

The general complex solution of \(x' = Ax\) is \(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}7\\{ - 2}\\3\end{aligned}} \right){e^{4t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}3\\{ - 1}\\1\end{aligned}} \right){e^{3t}} + {c_3}\left( {\begin{aligned}{ {20}{c}}2\\0\\1\end{aligned}} \right){e^{2t}}\). It is observed that the origin is a repellor since every eigenvalue is positive. All trajectories point away from the origin.

Step-by-step solution

Determine the eigenvalues and eigenvector of the matrix

The given matrix is \(A = \left( {\begin{aligned}{ {20}{c}}{ - 6}&{ - 11}&{16}\\2&5&{ - 4}\\{ - 4}&{ - 5}&{10}\end{aligned}} \right)\).

Use the MATLAB code to compute the eigenvalues of the matrix \(A\) as shown below:

\(\begin{aligned}{l} > > {\mathop{\rm A}\nolimits} = \left( { - 6\,\,\, - 11\,\,\,\,16;\,2\,\,\,\,5\,\,\, - 4;\, - 4\,\,\, - 5\,\,\,10} \right)\\ > > {\mathop{\rm ev}\nolimits} = {\mathop{\rm eig}\nolimits} \left( {\mathop{\rm A}\nolimits} \right)\end{aligned}\)

\({\mathop{\rm ev}\nolimits} = \left( {\begin{aligned}{ {20}{c}}{4.0000}\\{3.0000}\\{2.0000}\end{aligned}} \right)\)

Use the MATLAB code to compute the eigenvector of the matrix A as shown below:

\( > > {\mathop{\rm nulbasis}\nolimits} \left( {{\mathop{\rm A}\nolimits} - {\mathop{\rm ev}\nolimits} \left( 1 \right) {\mathop{\rm eye}\nolimits} \left( 3 \right)} \right)\)

\(\left( {\begin{aligned}{ {20}{c}}{2.3333}\\{ - 0.6667}\\{1.0000}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _1} = \left( {\begin{aligned}{ {20}{c}}7\\{ - 2}\\3\end{aligned}} \right)\).

\( > > {\mathop{\rm nulbasis}\nolimits} \left( {{\mathop{\rm A}\nolimits} - {\mathop{\rm ev}\nolimits} \left( 2 \right) {\mathop{\rm eye}\nolimits} \left( 3 \right)} \right)\)

\(\left( {\begin{aligned}{ {20}{c}}{3.0000}\\{ - 1.0000}\\{1.0000}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _2} = \left( {\begin{aligned}{ {20}{c}}3\\{ - 1}\\1\end{aligned}} \right)\).

\( > > {\mathop{\rm nulbasis}\nolimits} \left( {{\mathop{\rm A}\nolimits} - {\mathop{\rm ev}\nolimits} \left( 3 \right) {\mathop{\rm eye}\nolimits} \left( 3 \right)} \right)\)

\(\left( {\begin{aligned}{ {20}{c}}{2.0000}\\{0.0000}\\{1.0000}\end{aligned}} \right)\). Therefore, \({{\mathop{\rm v}\nolimits} _3} = \left( {\begin{aligned}{ {20}{c}}2\\0\\1\end{aligned}} \right)\).

Construct the general solution of \(x' = Ax\)

Therefore, the general complex solution of \(x' = Ax\) is \(x\left( t \right) = {c_1}\left( {\begin{aligned}{ {20}{c}}7\\{ - 2}\\3\end{aligned}} \right){e^{4t}} + {c_2}\left( {\begin{aligned}{ {20}{c}}3\\{ - 1}\\1\end{aligned}} \right){e^{3t}} + {c_3}\left( {\begin{aligned}{ {20}{c}}2\\0\\1\end{aligned}} \right){e^{2t}}\), with \({c_1}\) , and \({c_2}\) are arbitrary complex numbers.

Describe the shape of the trajectories

It is observed that the origin is a repellor since every eigenvalue is positive. Every trajectory point away from the origin.