Question

In Problems 13-32 use variation of parameters to solve the given nonhomogeneous system.

\(27.{X^\prime } = \left( {\begin{array}{*{20}{r}}0&1\\{ - 1}&0\end{array}} \right)X + \left( {\begin{array}{*{20}{c}}0\\{\sec t\tan t}\end{array}} \right)\)

Short answer

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

Step-by-step solution

Variation of parameters for nonhomogeneous linear systems:

Parameter variation is a generic approach for identifying a specific solution of a differential equation by substituting the constants in the solution of a related (homogeneous) equation with functions and defining these functions so that the original differential equation is fulfilled.

Find the characteristic equation of the coefficient matrix:

We have

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

where

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

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\({\lambda _2} = \overline {{\lambda _1}} = - i\).

Find the eigenvector and a corresponding solution vector:

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,

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

choosing\({k_1} = 1\)yields\({k_2} = i\). This gives an eigenvector:

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

and column vectors:

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

also,

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

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

Therefore,

\({X_{\bf{1}}} = \left[ {{B_{\bf{1}}}\cos \beta t - {B_2}\sin \beta t} \right]{e^{\alpha t}} = \left( {\begin{array}{*{20}{l}}1\\0\end{array}} \right)\cos t - \left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right)\sin t = \left( {\begin{array}{*{20}{r}}{\cos t}\\{ - \sin t}\end{array}} \right)\)

and

\({X_2} = \left[ {{B_2}\cos \beta t + {B_1}\sin \beta t} \right]{e^{\alpha t}} = \left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right)\cos t + \left( {\begin{array}{*{20}{l}}1\\0\end{array}} \right)\sin t = \left( {\begin{array}{*{20}{c}}{\sin t}\\{\cos t}\end{array}} \right)\)

Hence, the complementary function is

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

Find the invertible matrix:

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,

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

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

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

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

\( = 1 \ne 0\)

since the determinant does not equal zero, the matrix is in fact, an invertible matrix.

So now,

\({\Phi ^{ - 1}}(t) = \left( {\begin{array}{*{20}{r}}{\cos t}&{ - \sin t}\\{\sin t}&{\cos t}\end{array}} \right)\)

Find the general solution of a Nonhomogeneous system:

obtaining the particular solution,

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

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

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

\( = \left( {\begin{array}{*{20}{r}}{\cos t}&{\sin t}\\{ - \sin t}&{\cos t}\end{array}} \right)\smallint \left( {\begin{array}{*{20}{r}}{ - {{\tan }^2}t}\\{\tan t}\end{array}} \right)dt\)

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

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

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

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

Finally, the general solution of the system is

\(X(t) = {X_c} + {X_p}\)

\( = {c_1}\left( {\begin{array}{*{20}{r}}{cost}\\{ - sint}\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}{sint}\\{cost}\end{array}} \right) + \left( {\begin{array}{*{20}{r}}{cost}\\{ - sint}\end{array}} \right)t + \left( {\begin{array}{*{20}{r}}{ - sint}\\{sinttant}\end{array}} \right) - \left( {\begin{array}{*{20}{l}}{sint}\\{cost}\end{array}} \right)ln|cost|\)