Question

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

\[29.{X^\prime } = \left( {\begin{array}{*{20}{r}}1&2\\{ - \frac{1}{2}}&1\end{array}} \right)X + \left( {\begin{array}{*{20}{l}}{\csc t}\\{\sec t}\end{array}} \right){e^t}\]

Short answer

The general solution of \[{X^\prime } = \left( {\begin{array}{*{20}{r}}1&2\\{ - \frac{1}{2}}&1\end{array}} \right)X + \left( {\begin{array}{*{20}{l}}{\csc t}\\{\sec t}\end{array}} \right){e^t}\] is \[X(t) = {c_1}\left( {\begin{array}{*{20}{c}}{2\cos t}\\{ - \sin t}\end{array}} \right){e^t} + {c_2}\left( {\begin{array}{*{20}{c}}{2\sin t}\\{\cos t}\end{array}} \right){e^t} + \left( {\begin{array}{*{20}{c}}{\cos t}\\{ - \frac{1}{2}\sin t}\end{array}} \right){e^t}\ln |\sin t| + \left( {\begin{array}{*{20}{c}}{3\sin t}\\{\frac{3}{2}\cos t}\end{array}} \right)t{e^t}\]

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

\[{X^\prime } = \left( {\begin{array}{*{20}{c}}1&2\\{ - 1/2}&1\end{array}} \right)X + \left( {\begin{array}{*{20}{c}}{\csc t}\\{\sec t}\end{array}} \right){e^t}\]

Where

\[A = \left( {\begin{array}{*{20}{c}}1&2\\{ - 1/2}&1\end{array}} \right)\]

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

\[(1 - \lambda )(1 - \lambda ) + 1 = 0\]

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

\[1 - 2\lambda + {\lambda ^2} + 1 = 0\]

\[{\lambda ^2} - 2\lambda + 2 = 0\]

Where,

\[\lambda = \frac{{2 \pm \sqrt {4 - 4(1)(2)} }}{2}\]

\[ = \frac{{2 \pm 2i}}{2}\]

\[ = 1 \pm i\]

Find the eigenvector and a corresponding solution vector:

For\[{\lambda _1} = 1 + i\]:

\[ = \left( {\begin{array}{*{20}{c}}{ - i}&2&0\\{ - 1/2}&{ - i}&0\end{array}} \right)\]

Apply row operation\[(2i){R_2} - {R_1} \to {R_2}\]:

\[ = \left( {\begin{array}{*{20}{c}}{ - i}&2&0\\0&0&0\end{array}} \right)\]

So here we have a single equation,

\[ - i{k_1} + 2{k_2} = 0\;\;\; \to \;\;\;{k_1} = - 2i{k_2}\]

Choosing\[{k_2} = i\]yields\[{k_1} = 2\]. This gives an eigenvector:

\[K = \left( {\begin{array}{*{20}{l}}2\\i\end{array}} \right) = \left( {\begin{array}{*{20}{l}}2\\0\end{array}} \right) + i\left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right)\]

And column vectors:

\[{B_1} = \left( {\begin{array}{*{20}{l}}2\\0\end{array}} \right),\;\;\;{\rm{ and }}\;\;\;{B_{\bf{2}}} = \left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right)\]

Also,

\[\lambda = \alpha + \beta i\;\;\; \to \;\;\;\lambda = 1 + i\]

Where\[\alpha = 1\]and \[\beta = 1\]

Therefore,

\[{X_1} \& = \left[ {{B_1}cos\beta t - {B_2}sin\beta t} \right]{e^{\alpha t}}\]

\[ = \left[ {\left( {\begin{array}{*{20}{l}}2\\0\end{array}} \right)\cos t - \left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right)\sin t} \right]{e^t} = \left( {\begin{array}{*{20}{c}}{2\cos t}\\{ - \sin t}\end{array}} \right){e^t}\]

And

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

\[ = \left[ {\left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right)\cos t + \left( {\begin{array}{*{20}{l}}2\\0\end{array}} \right)\sin t} \right]{e^t} = \left( {\begin{array}{*{20}{c}}{2\sin t}\\{\cos t}\end{array}} \right){e^t}\]

Hence, the complementary function is

\[{X_{\bf{c}}} = {c_1}\left( {\begin{array}{*{20}{c}}{2\cos t}\\{ - \sin t}\end{array}} \right){e^t} + {c_2}\left( {\begin{array}{*{20}{c}}{2\sin t}\\{\cos t}\end{array}} \right){e^t}\]

Find the invertible matrix:

The entries in\[{X_1}\]form the first column of\[\Phi (t)\], and the entries in\[{X_{\bf{2}}}\]form the second column of\[\Phi (t)\$ .\]

Therefore,

\[\Phi (t) = \left( {\begin{array}{*{20}{c}}{2{e^t}\cos t}&{2{e^t}\sin t}\\{ - {e^t}\sin t}&{{e^t}\cos t}\end{array}} \right)\]

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

\[|\Phi (t)| = \left| {\begin{array}{*{20}{c}}{2{e^t}\cos t}&{2{e^t}\sin t}\\{ - {e^t}\sin t}&{{e^t}\cos t}\end{array}} \right|\]

\[ = 2{e^{2t}}{\cos ^2}t + 2{e^{2t}}{\sin ^2}t{\rm{ }}\]

\[ = 2{e^{2t}}\left( {{{\cos }^2}t + {{\sin }^2}t} \right){\rm{ }}\]

\[ = 2{e^{2t}} \ne 0\]

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

So now,

\[{\Phi ^{ - 1}}(t) = \frac{1}{{2{e^{2t}}}}\left( {\begin{array}{*{20}{c}}{{e^t}\cos t}&{ - 2{e^t}\sin t}\\{{e^t}\sin t}&{2{e^t}\cos t}\end{array}} \right) = \left( {\begin{array}{*{20}{l}}{\frac{1}{2}{e^{ - t}}\cos t}&{ - {e^{ - t}}\sin t}\\{\frac{1}{2}{e^{ - t}}\sin t}&{{e^{ - t}}\cos t}\end{array}} \right)\]

Find the general solution of a Nonhomogeneous system:

Obtaining the particular solution,

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

\[ = \left( {\begin{array}{*{20}{c}}{2{e^t}\cos t}&{2{e^t}\sin t}\\{ - {e^t}\sin t}&{{e^t}\cos t}\end{array}} \right)\smallint \left( {\begin{array}{*{20}{c}}{\frac{1}{2}{e^{ - t}}\cos t}&{ - {e^{ - t}}\sin t}\\{\frac{1}{2}{e^{ - t}}\sin t}&{{e^{ - t}}\cos t}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\csc t}\\{\sec t}\end{array}} \right){e^t}\]

\[ = \left( {\begin{array}{*{20}{c}}{2{e^t}\cos t}&{2{e^t}\sin t}\\{ - {e^t}\sin t}&{{e^t}\cos t}\end{array}} \right)\smallint \left( {\begin{array}{*{20}{l}}{\frac{1}{2}\cos t\csc t - \sin t\sec t}\\{\frac{1}{2}\sin t\csc t + \cos t\sec t}\end{array}} \right)\]

\[ = \left( {\begin{array}{*{20}{c}}{2{e^t}\cos t}&{2{e^t}\sin t}\\{ - {e^t}\sin t}&{{e^t}\cos t}\end{array}} \right)\smallint \left( {\begin{array}{*{20}{c}}{\frac{1}{2}\cot t - \tan t}\\{\frac{3}{2}}\end{array}} \right)\]

\[ = \left( {\begin{array}{*{20}{c}}{2{e^t}\cos t}&{2{e^t}\sin t}\\{ - {e^t}\sin t}&{{e^t}\cos t}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{\frac{1}{2}\ln |\sin t| - \ln |\sec t|}\\{\frac{3}{2}t}\end{array}} \right)\]

\[ = \left( {\begin{array}{*{20}{c}}{{e^t}\cos t\ln |\sin t| - 2{e^t}\cos t\ln |\sec t| + 3t{e^t}\sin t}\\{ - \frac{1}{2}{e^t}\sin t\ln |\sin t| + {e^t}\sin t\ln |\sec t| + \frac{3}{2}t{e^t}\cos t}\end{array}} \right)n\]

\[ = \left( {\begin{array}{*{20}{c}}{\cos t}\\{ - \frac{1}{2}\sin t}\end{array}} \right){e^t}\ln |\sin t| + \left( {\begin{array}{*{20}{c}}{ - 2\cos t}\\{\sin t}\end{array}} \right){e^t}\ln |\sec t| + \left( {\begin{array}{*{20}{c}}{3\sin t}\\{\frac{3}{2}\cos t}\end{array}} \right)t{e^t}\]

Finally, we conclude that the general solution is

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

\[ = {c_1}\left( {\begin{array}{*{20}{c}}{2\cos t}\\{ - \sin t}\end{array}} \right){e^t} + {c_2}\left( {\begin{array}{*{20}{c}}{2\sin t}\\{\cos t}\end{array}} \right){e^t} + \left( {\begin{array}{*{20}{c}}{\cos t}\\{ - \frac{1}{2}\sin t}\end{array}} \right){e^t}\ln |\sin t| + \left( {\begin{array}{*{20}{l}}{3\sin t}\\{\frac{3}{2}\cos t}\end{array}} \right)t{e^t}\]