Question

(a) Use (1) to find the general solution of \({{\bf{X}}^\prime } = \left( {\begin{array}{*{20}{l}}4&2\\3&3\end{array}} \right){\bf{X}}\)Use a CAS to find \({e^{At}}.\)Then use the computer to find eigenvalues and eigenvectors of the coefficient matrix \({\bf{A}} = \left( {\begin{array}{*{20}{l}}4&2\\3&3\end{array}} \right)\)and form the general solution in the manner of Section 8.2. Finally, reconcile the two forms of the general solution of the system.

(b) Use (1) to find the general solution of \({{\bf{X}}^\prime } = \left( {\begin{array}{*{20}{r}}{ - 3}&{ - 1}\\2&{ - 1}\end{array}} \right){\bf{X}}\) Use a CAS to find \({e^{A{\rm{. }}}}.\) In the case of complex output, utilize the software to do the simplification; for example, in Mathematica, if \({\bf{m}} = \) Matrix Exp[A t] has complex entries, then try the command Simplify[Complex Expand[m]].

Short answer

(a) The general solution of \(X' = \left( {\begin{array}{*{20}{l}}4&2\\3&3\end{array}} \right)X\)

(b)The general solution of \({{\bf{X}}^\prime } = \left( {\begin{array}{*{20}{r}}{ - 3}&{ - 1}\\2&{ - 1}\end{array}} \right){\bf{X}}.\)

Step-by-step solution

Finding the general equation

Matrix manipulation, function and data plotting, algorithm implementation, user interface creation, and interfacing with programmes written in other languages are all possible with MATLAB.

Using MATLAB to find the general solution

Using MATLAB as shown below, we obtained\({e^{{\rm{At }}}},\)where

\({e^{{\rm{At }}}} = \left( {\begin{array}{*{20}{l}}{\frac{3}{5}{e^{6t}} + \frac{2}{5}{e^t}}&{\frac{2}{5}{e^{6t}} - \frac{2}{5}{e^t}}\\{\frac{3}{5}{e^{6t}} - \frac{3}{5}{e^t}}&{\frac{2}{5}{e^{6t}} + \frac{3}{5}{e^t}}\end{array}} \right)\)

And to find the general solution, we have

\(X = {e^{{\bf{At}}t}}C\)

=\(\left( {\begin{array}{*{20}{l}}{\frac{3}{5}{e^{6t}} + \frac{2}{5}{e^t}}&{\frac{2}{5}{e^{6t}} - \frac{2}{5}{e^t}}\\{\frac{3}{5}{e^{6t}} - \frac{3}{5}{e^t}}&{\frac{2}{5}{e^{6t}} + \frac{3}{5}{e^t}}\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{{\rm{c}}_1}}\\{{{\rm{c}}_2}}\end{array}} \right)\)

=\(\left( {\begin{array}{*{20}{l}}{{c_1}\left( {\frac{3}{5}{e^{6t}} + \frac{2}{5}{e^t}} \right) + {c_2}\left( {\frac{2}{5}{e^{6t}} - \frac{2}{5}{e^t}} \right)}\\{{c_1}\left( {\frac{3}{5}{e^{6t}} - \frac{3}{5}{e^t}} \right) + {c_2}\left( {\frac{2}{5}{e^{6t}} + \frac{3}{5}{e^t}} \right)}\end{array}} \right)\)

Where, \(x(t) = {c_1}\left( {\frac{3}{5}{e^{6t}} + \frac{2}{5}{e^t}} \right) + {c_2}\left( {\frac{2}{5}{e^{6t}} - \frac{2}{5}{e^t}} \right)\)

\(y(t) = {c_1}\left( {\frac{3}{5}{e^{6t}} - \frac{3}{5}{e^t}} \right) + {c_2}\left( {\frac{2}{5}{e^{6t}} + \frac{3}{5}{e^t}} \right)\)

Using MATLAB to find the eigenvalues and eigenvectors of the coefficient matrix

We also used MATLAB to find the eigenvalues and eigenvectors of\({\bf{A}},\)we can see that

\({\lambda _1} = 6,\;\;and\;\;{\lambda _2} = 1\)

And the eigenvectors are

\({{\bf{K}}_1} = \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right),\;\;\;and\;\;\;{{\bf{K}}_2} = \left( {\begin{array}{*{20}{r}}{ - 2}\\3\end{array}} \right)\)

In MATLAB we obtained the second eigenvector to be

\({{\bf{K}}_2} = \left( {\begin{array}{*{20}{c}}{ - \frac{2}{3}}\\1\end{array}} \right)\)

However here I multiplied it by 3 to make it look simpler.

So, the general solution of the system is

\({\bf{X}} = {a_1}\left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right){e^{6t}} + {a_2}\left( {\begin{array}{*{20}{r}}{ - 2}\\3\end{array}} \right){e^t}\)

Where,

\(\begin{array}{l}x(t) = {a_1}{e^{6t}} - 2{a_2}{e^t}\\y(t) = {a_1}{e^{6t}} + 3{a_2}{e^t}\end{array}\)

if we let\({a_1} = \left( {\frac{3}{5}{{\bf{c}}_1} + \frac{2}{5}{{\bf{c}}_2}} \right)\)and\({a_2} = \left( { - \frac{1}{5}{{\bf{c}}_1} + \frac{1}{5}{{\bf{c}}_2}} \right)\$ \)we will obtain the same answer we obtained above when we computed\({e^{At}}.\)

Therefore the general solution of \({{\bf{X}}^\prime } = \left( {\begin{array}{*{20}{l}}4&2\\3&3\end{array}} \right){\bf{X}}\) is

\(\begin{array}{l}x(t) = {{\bf{c}}_1}\left( {\frac{3}{5}{e^{6t}} + \frac{2}{5}{e^t}} \right) + {{\bf{c}}_2}\left( {\frac{2}{5}{e^{6t}} - \frac{2}{5}{e^t}} \right)\\y(t) = {{\bf{c}}_1}\left( {\frac{3}{5}{e^{6t}} - \frac{3}{5}{e^t}} \right) + {{\bf{c}}_2}\left( {\frac{2}{5}{e^{6it}} + \frac{3}{5}{e^t}} \right)\end{array}\)

Using MATLAB to find the general solution

As shown below I used MATLAB to compute\({e^{{\rm{At, }}}}\)I obtained a complex output which is why I used the simplify command. So, we have

\({e^{{\bf{A}}t}} = \left( {\begin{array}{*{20}{c}}{{e^{ - 2t}}(\cos t - \sin t)}&{ - {e^{ - 2t}}\sin t}\\{2{e^{ - 2t}}\sin t}&{{e^{ - 2t}}(\cos t + \sin t)}\end{array}} \right)\)

Obtaining the general solution of the system

\({\bf{X}} = {e^{{\bf{A}}t}}{\bf{C}}\)

\( = \left( {\begin{array}{*{20}{c}}{{e^{ - 2t}}(\cos t - \sin t)}&{ - {e^{ - 2t}}\sin t}\\{2{e^{ - 2t}}\sin t}&{{e^{ - 2t}}(\cos t + \sin t)}\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{{\bf{c}}_1}}\\{{{\bf{c}}_2}}\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}{{{\bf{c}}_1}{e^{ - 2t}}\cos t - {{\bf{c}}_1}{e^{ - 2t}}\sin t - {{\bf{c}}_2}{e^{ - 2t}}\sin t}\\{2{{\bf{c}}_1}{e^{ - 2t}}\sin t + {{\bf{c}}_2}{e^{ - 2t}}\cos t + {{\bf{c}}_2}{e^{ - 2t}}\sin t}\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}{{{\bf{c}}_1}{e^{ - 2t}}\cos t - \left( {{{\bf{c}}_1} + {{\bf{c}}_2}} \right){e^{ - 2t}}\sin t}\\{{{\bf{c}}_2}{e^{ - 2t}}\cos t + \left( {2{{\bf{c}}_1} + {{\bf{c}}_2}} \right){e^{ - 2t}}\sin t}\end{array}} \right)\)

Where

\(\begin{array}{l}x(t) = {{\bf{c}}_1}{e^{ - 2t}}\cos t - \left( {{{\bf{c}}_1} + {{\bf{c}}_2}} \right){e^{ - 2t}}\sin t\\y(t) = {{\bf{c}}_2}{e^{ - 2t}}\cos t + \left( {2{{\bf{c}}_1} + {{\bf{c}}_2}} \right){e^{ - 2t}}\sin t\end{array}\)

Therefore, the general solution of \({{\bf{X}}^\prime } = \left( {\begin{array}{*{20}{r}}{ - 3}&{ - 1}\\2&{ - 1}\end{array}} \right){\bf{X}}\) is

\(x(t) = {{\bf{c}}_1}\left( {\frac{3}{5}{e^{6t}} + \frac{2}{5}{e^t}} \right) + {{\bf{c}}_2}\left( {\frac{2}{5}{e^{6t}} - \frac{2}{5}{e^t}} \right)\)

\(y(t) = {{\bf{c}}_2}{e^{ - 2t}}\cos t + \left( {2{{\bf{c}}_1} + {{\bf{c}}_2}} \right){e^{ - 2t}}\sin t\)