Question

In Problems \({\bf{13}} - {\bf{32}}\)use variation of parameters to solve the given nonhomogeneous system.

\(23.{{\bf{X}}^\prime } = \left( {\begin{array}{*{20}{r}}0&{ - 1}\\1&0\end{array}} \right){\bf{X}} + \left( {\begin{array}{*{20}{c}}{\sec t}\\0\end{array}} \right)\)

Short answer

The general solution of the system for \({X^\prime } = \left( {\begin{array}{*{20}{r}}0&{ - 1}\\1&0\end{array}} \right)X + \left( {\begin{array}{*{20}{c}}{\sec t}\\0\end{array}} \right)\) is \({\bf{X}}(t) = {c_1}\left( {\begin{array}{*{20}{c}}{ - \sin t}\\{\cos t}\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}{\cos t}\\{\sin t}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{\cos t}\\{\sin t}\end{array}} \right)t + \left( {\begin{array}{*{20}{r}}{ - \sin t}\\{\cos t}\end{array}} \right)\ln |\cos t|\)

Step-by-step solution

Determine the complex Eigen values

The polynomial of degree \(n\) of the variable \(\lambda \) is the characteristic polynomial of the \(n \times n\) matrix\(A\).

\(p(\lambda ) = \det (\lambda I - A)\)

Its root is the eigenvalue of\(A\).

Determine the Eigen values

Given

\({X^\prime } = \left( {\begin{array}{*{20}{r}}0&{ - 1}\\1&0\end{array}} \right)X + \left( {\begin{array}{*{20}{c}}{\sec t}\\0\end{array}} \right)\)

\(A = \left( {\begin{array}{*{20}{r}}0&{ - 1}\\1&0\end{array}} \right)\)

Now, we find the characteristic equation of the coefficient matrix,

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

\({\lambda ^2} + 1 = 0\)

\({\lambda ^2} = - 1\)

\(\lambda = \pm i\)

So our eigenvalues are\({\lambda _1} = i\), and\({\lambda _2} = \overline {{\lambda _1}} = - i\).

Determine the Complementary function

For \({\lambda _1} = i:\)

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

\( = \left( {\begin{array}{*{20}{c}}{ - i}&{ - 1}&0\\1&{ - i}&0\end{array}} \right)\)

Apply row operation\(( - i){R_2} - {R_1} \to {R_2}\):

\( = \left( {\begin{array}{*{20}{c}}{ - i}&{ - 1}&0\\0&0&0\end{array}} \right)\)

So here we have a single equation,

\( - i{k_1} - {k_2} = 0\;\;\; \to \;\;\;{k_1} = i{k_2}\)

Choosing \({k_2} = 1\)yields\({k_1} = i\).

This gives an eigenvector:

\(K = \left( {\begin{array}{*{20}{l}}i\\1\end{array}} \right) = \left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right) + i\left( {\begin{array}{*{20}{l}}1\\0\end{array}} \right)\)

And column vectors:

\({B_1} = \left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right),\)and \({B_2} = \left( {\begin{array}{*{20}{l}}1\\0\end{array}} \right)\) also,

\(\lambda = \alpha + \beta i\;\;\; \to \;\;\;\lambda = 0 + i\)

Where \(\alpha = 0\)and \(\beta = 1\)

Therefore,

\({X_{\bf{1}}} = \left[ {{B_1}\cos \beta t - {B_{\bf{2}}}\sin \beta t} \right]{e^{\alpha t}}\)

\( = \left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right)\cos t - \left( {\begin{array}{*{20}{l}}1\\0\end{array}} \right)\sin t\)

And

\( = \left( {\begin{array}{*{20}{r}}{ - \sin t}\\{\cos t}\end{array}} \right)\)

\({X_2} = \left[ {{B_2}\cos \beta t + {B_1}\sin \beta t} \right]{e^{\alpha t}}\)

\( = \left( {\begin{array}{*{20}{l}}1\\0\end{array}} \right)\cos t + \left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right)\sin t\)

\( = \left( {\begin{array}{*{20}{c}}{\cos t}\\{\sin t}\end{array}} \right)\)

Hence, the complementary function is

\({X_{\bf{c}}} = {c_1}\left( {\begin{array}{*{20}{r}}{ - \sin t}\\{\cos t}\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}{\cos t}\\{\sin t}\end{array}} \right)\)

Determine the general solution of the system

The entries in \({X_1}\)form the first column of\(\Phi (t)\), and the entries in \({X_2}\)form the second column of\(\Phi (t)\).

Therefore,

\(\Phi (t) = \left( {\begin{array}{*{20}{r}}{ - \sin t}&{\cos t}\\{\cos t}&{\sin t}\end{array}} \right)\)

We want to make sure that \(\Phi (t)\)is an invertible matrix by checking the determinant, where

\(|\Phi (t)|{\rm{ }} = \left| {\begin{array}{*{20}{r}}{ - \sin t}&{\cos t}\\{\cos t}&{\sin t}\end{array}} \right|\)

\( = - {\sin ^2}t - {\cos ^2}t\)

\( = - \left( {{{\sin }^2}t + {{\cos }^2}t} \right)\)

\( = - 1 \ne 0\)

Since the determinant does not equal zero, the matrix is in fact, an invertible matrix. \({\Phi ^{ - 1}}(t) = - 1\left( {\begin{array}{*{20}{c}}{\sin t}&{ - \cos t}\\{ - \cos t}&{ - \sin t}\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}{ - \sin t}&{\cos t}\\{\cos t}&{\sin t}\end{array}} \right)\)

Obtaining the particular solution,

\({X_{\bf{p}}} = \Phi (t)\int {{\Phi ^{ - 1}}} (t)F(t)dt\)

\( = \left( {\begin{array}{*{20}{r}}{ - \sin t}&{\cos t}\\{\cos t}&{\sin t}\end{array}} \right)\smallint \left( {\begin{array}{*{20}{r}}{ - \sin t}&{\cos t}\\{\cos t}&{\sin t}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\sec t}\\0\end{array}} \right)dt\)

\( = \left( {\begin{array}{*{20}{r}}{ - \sin t}&{\cos t}\\{\cos t}&{\sin t}\end{array}} \right)d\left( {\begin{array}{*{20}{r}}{ - \sin t\sec t}\\{\cos t\sec t}\end{array}} \right)dt\)

\( = \left( {\begin{array}{*{20}{c}}{ - \sin t}&{\cos t}\\{\cos t}&{\sin t}\end{array}} \right)\smallint \left( {\begin{array}{*{20}{c}}{ - \tan t}\\1\end{array}} \right)dt\)

\( = \left( {\begin{array}{*{20}{c}}{ - \sin t}&{\cos t}\\{\cos t}&{\sin t}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\ln |\cos t|}\\t\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{r}}{ - \ln |\cos t|\sin t + t\cos t}\\{\ln |\cos t|\cos t + t\sin t}\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{r}}{\cos t}\\{\sin t}\end{array}} \right)t + \left( {\begin{array}{*{20}{r}}{ - \sin t}\\{\cos t}\end{array}} \right)\ln |\cos t|\)

Finally, the general solution of the system is

\(X(t) = {X_{\bf{c}}} + {X_{\bf{p}}}\)

\( = {c_1}\left( {\begin{array}{*{20}{r}}{ - \sin t}\\{\cos t}\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}{\cos t}\\{\sin t}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{\cos t}\\{\sin t}\end{array}} \right)t + \left( {\begin{array}{*{20}{r}}{ - \sin t}\\{\cos t}\end{array}} \right)\ln |\cos t|\)

Therefore, the general solution of the system for \({X^\prime } = \left( {\begin{array}{*{20}{r}}0&{ - 1}\\1&0\end{array}} \right)X + \left( {\begin{array}{*{20}{c}}{\sec t}\\0\end{array}} \right)\) is \({\bf{X}}(t) = {c_1}\left( {\begin{array}{*{20}{c}}{ - \sin t}\\{\cos t}\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}{\cos t}\\{\sin t}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{\cos t}\\{\sin t}\end{array}} \right)t + \left( {\begin{array}{*{20}{r}}{ - \sin t}\\{\cos t}\end{array}} \right)\ln |\cos t|\)