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Found in: Page 489

### 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

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