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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 489
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

In Problems \(5 - 14\) solve the given linear system.

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

The solution for the linear system \({\bf{X'}} = \left( {\begin{array}{*{20}{c}}{{\rm{ 0 2 1}}}\\{1{\rm{ }}1{\rm{ }} - 2}\\{2{\rm{ }}2{\rm{ }} - 1}\end{array}} \right){\bf{X}}\) is\({{\bf{X}}_{\bf{c}}} = {c_1}\left( {\begin{array}{*{20}{r}}{ - 7}\\5\\4\end{array}} \right){e^{ - 2t}} + {c_2}\left( {\begin{array}{*{20}{c}}{\cos \sqrt 2 t}\\{ - \frac{1}{2}\sin \sqrt 2 t}\\{\cos \sqrt 2 t}\end{array}} \right){e^t} + {c_3}\left( {\begin{array}{*{20}{c}}{\sin \sqrt 2 t}\\{\frac{1}{2}\cos \sqrt 2 t}\\{\sin \sqrt 2 t}\end{array}} \right){e^t}\).

See the step by step solution

Step by Step Solution

Step 1: 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,

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

Step 2: Find the eigenvalue of the given linear system.

We are given that,

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

where,

\(A = \left( {\begin{array}{*{20}{c}}{{\rm{ 0 2 1}}}\\{1{\rm{ }}1{\rm{ }} - 2}\\{2{\rm{ }}2{\rm{ }} - 1}\end{array}} \right)\)

Now we find the characteristic equation of the coefficient matrix,

\(\begin{array}{*{20}{c}}{\det ({\bf{A}} - \lambda {\bf{I}}) = 0 \to \left| {\begin{array}{*{20}{c}}{0 - \lambda }&2&1\\1&{1 - \lambda }&{ - 2}\\2&2&{ - 1 - \lambda }\end{array}} \right| = 0}\\{ - \lambda \left| {\begin{array}{*{20}{c}}{1 - \lambda }&{ - 2}\\2&{ - 1 - \lambda }\end{array}} \right| - 2\left| {\begin{array}{*{20}{c}}1&{ - 2}\\2&{ - 1 - \lambda }\end{array}} \right| + \left| {\begin{array}{*{20}{c}}1&{1 - \lambda }\\2&2\end{array}} \right| = 0}\end{array}\)

\(\begin{array}{*{20}{r}}{ - \lambda [(1 - \lambda )( - 1 - \lambda ) + 4] - 2[( - 1 - \lambda ) + 4] + [2 - 2(1 - \lambda )] = 0}\\{ - \lambda \left[ { - 1 - \lambda + \lambda + {\lambda ^2} + 4} \right] - 2(3 - \lambda ) + 2\lambda = 0}\\{ - \lambda \left[ {{\lambda ^2} + 3} \right] - 6 + 2\lambda + 2\lambda = 0}\end{array}\)

\(\begin{array}{*{20}{r}}{ - {\lambda ^3} - 3\lambda - 6 + 4\lambda = 0}\\{ - {\lambda ^3} + \lambda - 6 = 0}\\{ - (\lambda + 2)\left( {{\lambda ^2} - 2\lambda + 3} \right) = 0}\end{array}\)

So we have eigenvalues of \({\lambda _1} = - 2,{\lambda _2} = 1 + \sqrt 2 i,{\rm{\;}}{\lambda _3} = \overline {{\lambda _2}} = 1 - \sqrt 2 i\).

Step 3: Find the eigenvector and solution vector of the given linear system.

For \({\lambda _1} = - 2\)

\(\begin{array}{*{20}{c}}{({\bf{A}} + 2{\bf{I}}\mid 0) = \left( {\begin{array}{*{20}{c}}{0 - ( - 2)}&2&1&0&{}\\{}&1&{1 - ( - 2)}&{ - 2}&0\\{}&2&2&{ - 1 - ( - 2)}&0\end{array}} \right)}\\{ = \left( {\begin{array}{*{20}{c}}2&2&1&0\\1&3&{ - 2}&0\\2&2&1&0\end{array}} \right)}\end{array}\)

From first row,

\(2{k_1} + 2{k_2} + {k_3} = 0 \to {k_3} = - 2{k_2} - 2{k_1}\)

From second row,

\(\begin{array}{*{20}{c}}{{k_1} + 3{k_2} - 2{k_3} = 0 \to {k_1} + 3{k_2} - 2\left( { - 2{k_2} - 2{k_1}} \right) = 0}\\{{k_1} + 3{k_2} + 4{k_2} + 4{k_1} = 0}\\{5{k_1} + 7{k_2} = 0}\\{{k_1} = - \frac{7}{5}{k_2}}\end{array}\)

Choosing \({k_2} = 5\)we have\({k_1} = - 7\). For \({k_3}\),

\({k_3} = - 2(5) - 2( - 7) = 4\)

This gives an eigenvector and a corresponding solution vector \({{\bf{K}}_1} = \left( {\begin{array}{*{20}{r}}{ - 7}\\5\\4\end{array}} \right),\quad {{\bf{X}}_1} = \left( {\begin{array}{*{20}{r}}{ - 7}\\5\\4\end{array}} \right){e^{ - 2t}}\) respectively.

Step 4: Find the eigenvector and solution vector of the given linear system.

For \({\lambda _2} = 1 + \sqrt 2 i\),

\(\begin{array}{*{20}{c}}{({\bf{A}} - (1 + \sqrt 2 i){\bf{I}}\mid 0) = \left( {\begin{array}{*{20}{c}}{0 - (1 + \sqrt 2 i)}&2&1&0\\1&{1 - (1 + \sqrt 2 i)}&{ - 2}&0\\2&2&{ - 1 - (1 + \sqrt 2 i)}&0\end{array}} \right)}\\{ = \left( {\begin{array}{*{20}{c}}{ - 1 - \sqrt 2 i}&2&1&0\\1&{ - \sqrt 2 i}&{ - 2}&0\\2&2&{ - 2 - \sqrt 2 i}&0\end{array}} \right)}\end{array}\)

From first row,

\( - (1 + \sqrt 2 i){k_1} + 2{k_2} + {k_3} = 0 \to {k_3} = (1 + \sqrt 2 i){k_1} - 2{k_2}\)

From second row,

\(\begin{array}{*{20}{c}}{{k_1} - \sqrt 2 i{k_2} - 2{k_3} = 0 \to {k_1} - \sqrt 2 i{k_2} - 2\left( {(1 + \sqrt 2 i){k_1} - 2{k_2}} \right) = 0}\\{{k_1} - \sqrt 2 i{k_2} - (2 + 2\sqrt 2 i){k_1} + 4{k_2} = 0}\\{ - (1 + 2\sqrt 2 i){k_1} + (4 - \sqrt 2 i){k_2} = 0}\\{{k_1} = - \sqrt 2 i{k_2}}\end{array}\)

Choosing \({k_2} = \frac{i}{{\sqrt 2 }}\)we have\({k_1} = 1\). For \({k_3}\),

\({k_3} = (1 + \sqrt 2 i) - 2\left( {\frac{i}{2}} \right) = 1\)

This gives an eigenvector \({{\bf{K}}_2} = \left( {\begin{array}{*{20}{c}}1\\{\frac{i}{2}}\\1\end{array}} \right) = \left( {\begin{array}{*{20}{l}}1\\0\\1\end{array}} \right) + i\left( {\begin{array}{*{20}{c}}0\\{\frac{1}{2}}\\0\end{array}} \right)\).

Step 5: Find the general solution of the given linear system.

From the eigenvector obtained from the previous step, we can calculate the column vectors, \({B_1} = \left( {\begin{array}{*{20}{l}}1\\0\\1\end{array}} \right),{B_2} = \left( {\begin{array}{*{20}{c}}0\\{\frac{1}{2}}\\0\end{array}} \right)\).

Also,

\(\lambda = \alpha + \beta i \to \lambda = 1 + \sqrt 2 i\)

where,

\(\alpha = 1{\rm{,\;}}\beta = \sqrt 2 \)

The corresponding solution vectors,

\(\begin{array}{*{20}{c}}{{{\bf{X}}_2} = \left[ {{{\bf{B}}_1}\cos \beta t - {{\bf{B}}_2}\sin \beta t} \right]{e^{\alpha t}}}\\{ = \left[ {\left( {\begin{array}{*{20}{c}}1\\0\\1\end{array}} \right)\cos \sqrt 2 t - \left( {\begin{array}{*{20}{c}}0\\{\frac{1}{2}}\\0\end{array}} \right)\sin \sqrt 2 t} \right]{e^t} = \left( {\begin{array}{*{20}{c}}{\cos \sqrt 2 t}\\{ - \frac{1}{2}\sin \sqrt 2 t}\\{\cos \sqrt 2 t}\end{array}} \right){e^t}}\end{array}\)

and

\(\begin{array}{*{20}{c}}{{{\bf{X}}_3} = \left[ {{{\bf{B}}_2}\cos \beta t + {{\bf{B}}_1}\sin \beta t} \right]{e^{\alpha t}}}\\{ = \left[ {\left( {\begin{array}{*{20}{c}}0\\{\frac{1}{2}}\\0\end{array}} \right)\cos \sqrt 2 t + \left( {\begin{array}{*{20}{c}}1\\0\end{array}} \right)\sin \sqrt 2 t} \right]{e^t} = \left( {\begin{array}{*{20}{c}}{\sin \sqrt 2 t}\\{\frac{1}{2}\cos \sqrt 2 t}\\{\sin \sqrt 2 t}\end{array}} \right){e^t}}\end{array}\)

Therefore the general solution of the system is\({{\bf{X}}_{\bf{c}}} = {c_1}\left( {\begin{array}{*{20}{r}}{ - 7}\\5\\4\end{array}} \right){e^{ - 2t}} + {c_2}\left( {\begin{array}{*{20}{c}}{\cos \sqrt 2 t}\\{ - \frac{1}{2}\sin \sqrt 2 t}\\{\cos \sqrt 2 t}\end{array}} \right){e^t} + {c_3}\left( {\begin{array}{*{20}{c}}{\sin \sqrt 2 t}\\{\frac{1}{2}\cos \sqrt 2 t}\\{\sin \sqrt 2 t}\end{array}} \right){e^t}\).

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