Question

Use \((1)\) to find the general solution of

\({\bf{X'}} = \left( {\begin{array}{*{20}{c}}{ - 4{\rm{ }}0{\rm{ }}6{\rm{ }}0}\\{{\rm{ 0 }} - 5{\rm{ }}0{\rm{ }} - 4}\\{ - 1{\rm{ }}0{\rm{ }}1{\rm{ }}0}\\{{\rm{ 0 }}3{\rm{ }}0{\rm{ }}2}\end{array}} \right){\bf{X}}\)

Use MATLAB or a CAS to find \({e^{{\bf{A}}t}}\).

Short answer

The general solution of \({e^{{\bf{A}}t}}\)found using MATLAB is \(X = \left( {\begin{array}{*{20}{c}}{{c_1}\left( {3{e^{ - 2t}} - 2{e^{ - t}}} \right) + {c_3}\left( {6{e^{ - t}} - 6{e^{ - 2t}}} \right)}\\{{c_2}\left( {4{e^{ - 2t}} - 3{e^{ - t}}} \right) + {c_4}\left( {4{e^{ - 2t}} - 4{e^{ - t}}} \right)}\\{{c_1}\left( {{e^{ - 2t}} - {e^{ - t}}} \right) + {c_3}\left( {3{e^{ - t}} - 2{e^{ - 2t}}} \right)}\\{{c_2}\left( {3{e^{ - t}} - 3{e^{ - 2t}}} \right) + {c_4}\left( {4{e^{ - t}} - 3{e^{ - 2t}}} \right)}\end{array}} \right)\).

Step-by-step solution

Define matrix exponential.

Consider a square matrix \(A\)of size \(n \times n\). This matrix can contain either complex numbers or real numbers. The matrix can be calculated as

\({A^0} = I,\;\;\;{A^1} = A,\;\;\;{A^2} = A \cdot A,\;\;\;{A^3} = {A^2} \cdot A, \ldots ,{A^k} = \mathop {\mathop {A \cdot A \cdots }\limits_{} }\limits_{{\rm{k\;times\;}}} \)

where \(I\)is the unit matrix of order\(n\).

Therefore the infinite matrix power series is

\(I + \frac{t}{{1!}}A + \frac{{{t^2}}}{{2!}}{A^2} + \frac{{{t^3}}}{{3!}}{A^3} + \cdots + \frac{{{t^k}}}{{k!}}{A^k} + \cdots \)

Now, the matrix exponential is defined as the sum of the infinite matrix power series. It is denoted by the expression\({e^{{\bf{A}}t}}\). It is given by the formula,

\({e^{tA}} = \mathop \sum \limits_{k = 0}^\infty \frac{{{t^k}}}{{k!}}{A^k}\)

Find the value of \({e^{{\bf{A}}t}}\).

It is given that,

\({\bf{X'}} = \left( {\begin{array}{*{20}{c}}{ - 4{\rm{ }}0{\rm{ }}6{\rm{ }}0}\\{{\rm{ 0 }} - 5{\rm{ }}0{\rm{ }} - 4}\\{ - 1{\rm{ }}0{\rm{ }}1{\rm{ }}0}\\{{\rm{ 0 }}3{\rm{ }}0{\rm{ }}2}\end{array}} \right){\bf{X}}\)

Using MATLAB we obtained,

\({e^{{\bf{A}}t}} = \left( {\begin{array}{*{20}{c}}{3{e^{ - 2t}} - 2{e^{ - t}}}&0&{6{e^{ - t}} - 6{e^{ - 2t}}}&0\\0&{4{e^{ - 2t}} - 3{e^{ - t}}}&0&{4{e^{ - 2t}} - 4{e^{ - t}}}\\{{e^{ - 2t}} - {e^{ - t}}}&0&{3{e^{ - t}} - 2{e^{ - 2t}}}&0\\0&{3{e^{ - t}} - 3{e^{ - 2t}}}&0&{4{e^{ - t}} - 3{e^{ - 2t}}}\end{array}} \right)\)

The general solution of the system is

\(\begin{array}{*{20}{c}}{{\bf{X}} = {e^{{\bf{A}}t}}{\bf{C}}}\\{ = \left( {\begin{array}{*{20}{c}}{3{e^{ - 2t}} - 2{e^{ - t}}}&0&{6{e^{ - t}} - 6{e^{ - 2t}}}&0\\0&{4{e^{ - 2t}} - 3{e^{ - t}}}&0&{4{e^{ - 2t}} - 4{e^{ - t}}}\\{{e^{ - 2t}} - {e^{ - t}}}&0&{3{e^{ - t}} - 2{e^{ - 2t}}}&0\\0&{3{e^{ - t}} - 3{e^{ - 2t}}}&0&{4{e^{ - t}} - 3{e^{ - 2t}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{c_1}}\\{{c_2}}\\{{c_3}}\\{{c_4}}\end{array}} \right)}\end{array}\)

\( = \left( {\begin{array}{*{20}{c}}{{c_1}\left( {3{e^{ - 2t}} - 2{e^{ - t}}} \right) + {c_3}\left( {6{e^{ - t}} - 6{e^{ - 2t}}} \right)}\\{{c_2}\left( {4{e^{ - 2t}} - 3{e^{ - t}}} \right) + {c_4}\left( {4{e^{ - 2t}} - 4{e^{ - t}}} \right)}\\{{c_1}\left( {{e^{ - 2t}} - {e^{ - t}}} \right) + {c_3}\left( {3{e^{ - t}} - 2{e^{ - 2t}}} \right)}\\{{c_2}\left( {3{e^{ - t}} - 3{e^{ - 2t}}} \right) + {c_4}\left( {4{e^{ - t}} - 3{e^{ - 2t}}} \right)}\end{array}} \right)\)

Hence the value of \({e^{{\bf{A}}t}}\)is \(X = \left( {\begin{array}{*{20}{c}}{{c_1}\left( {3{e^{ - 2t}} - 2{e^{ - t}}} \right) + {c_3}\left( {6{e^{ - t}} - 6{e^{ - 2t}}} \right)}\\{{c_2}\left( {4{e^{ - 2t}} - 3{e^{ - t}}} \right) + {c_4}\left( {4{e^{ - 2t}} - 4{e^{ - t}}} \right)}\\{{c_1}\left( {{e^{ - 2t}} - {e^{ - t}}} \right) + {c_3}\left( {3{e^{ - t}} - 2{e^{ - 2t}}} \right)}\\{{c_2}\left( {3{e^{ - t}} - 3{e^{ - 2t}}} \right) + {c_4}\left( {4{e^{ - t}} - 3{e^{ - 2t}}} \right)}\end{array}} \right)\).

The MATLAB code to find value of \({e^{{\bf{A}}t}}\).

The MATLAB code is

\(\begin{array}{*{20}{c}}{{\rm{\; > \; > }}A = syn\left( {\left[ {\begin{array}{*{20}{c}}{ - 4}&0&6&{0;0}&{ - 5}&0&{ - 4; - 1}&0&1&{0;0}&3&0&2\end{array}} \right]} \right)}\\{A = }\\{[ - 4,0,6,0]}\\{[0, - 5,0, - 4]}\\{[ - 1,0,1,0]}\\{[0,3,0,2]}\\{{\rm{\; > > \;s}}ymst}\\{ > > expn\left( {{A^ * }t} \right)}\end{array}\)

\(ans = \)

\(\begin{array}{*{20}{c}}{\left[ {\begin{array}{*{20}{c}}{\left[ {{3^ * }\exp \left( { - {2^ * }t} \right) - {2^ * }\exp ( - t),} \right.}&{0,{6^ * }\exp ( - t) - {6^ * }\exp \left( { - {2^ * }t} \right),}&\theta \end{array}} \right]}\\{\left[ {\begin{array}{*{20}{c}}{0,{4^ * }\exp \left( { - {2^ * }t} \right) - {3^ * }\exp ( - t),}&{\left. {\theta ,{4^ * }\exp \left( { - {2^ * }t} \right) - {4^ * }\exp ( - t)} \right]}\\{\left[ {\quad \exp \left( { - {2^ * }t} \right) - \exp ( - t),} \right.}&\theta \end{array}} \right]}\\{\left[ {\quad 0,{3^ * }\exp ( - t) - {3^ * }\exp \left( { - {2^ * }t} \right),\quad 0,{4^ * }\exp ( - t) - {3^ * }\exp \left( { - {2^ * }t} \right)} \right]}\end{array}\)\(\).