Question

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

\(28.{X^\prime } = \left( {\begin{array}{*{20}{r}}0&1\\{ - 1}&0\end{array}} \right)X + \left( {\begin{array}{*{20}{c}}1\\{\cot t}\end{array}} \right)\)

Short answer

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

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

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

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:

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

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

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

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:

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

Also,

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

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

Therefore,

\({X_1} = \left[ {{B_1}cos\beta t - {B_2}sin\beta t} \right]{e^{\alpha t}} = \left( {\begin{array}{*{20}{l}}1\\0\end{array}} \right)cost - \left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right)sint = \left( {\begin{array}{*{20}{r}}{cost}\\{ - sint}\end{array}} \right)\)

And

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

Hence, the complementary function is

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

Find the invertible matrix:

The entries in\({X_{\bf{1}}}\)form the first column of\(\Phi (t)\), and the entries in\({X_{\bf{2}}}\)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

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

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

\( = 1 \ne 0\)

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,

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

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

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

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

\( = \left( {\begin{array}{*{20}{r}}{\cos t}&{\sin t}\\{ - \sin t}&{\cos t}\end{array}} \right)\left( {\begin{array}{*{20}{r}}0\\{\ln |\csc t - \cot t|}\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{l}}{\sin t\ln |\csc t - \cot t|}\\{\cos t\ln |\csc t - \cot t|}\end{array}} \right)\)

Finally, the general solution of the system is

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

\( = {c_1}\left( {\begin{array}{*{20}{r}}{cost}\\{ - sint}\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}{sint}\\{cost}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{sintln|csct - cott|}\\{costln|csct - cott|}\end{array}} \right)\)