Question

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

\(30.{X^\prime } = \left( {\begin{array}{*{20}{l}}1&{ - 2}\\1&{ - 1}\end{array}} \right)X + \left( {\begin{array}{*{20}{c}}{\tan t}\\1\end{array}} \right)\)

Short answer

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

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

\({X^\prime } = \left( {\begin{array}{*{20}{l}}1&{ - 2}\\1&{ - 1}\end{array}} \right)X + \left( {\begin{array}{*{20}{c}}{\tan t}\\1\end{array}} \right)\)

Where

\(A = \left( {\begin{array}{*{20}{l}}1&{ - 2}\\1&{ - 1}\end{array}} \right)\)

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

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

\((1 - \lambda )( - 1 - \lambda ) + 2 = 0\)

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

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

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

\(\lambda = \pm i\)

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

Find the eigenvector and a corresponding solution vector:

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

\((A - (i)I\mid 0) = \left( {\begin{array}{*{20}{c}}{1 - i}&{ - 2}&0\\1&{ - 1 - i}&0\end{array}} \right)\)

Apply row operation\((1 - i){R_2} - {R_1} \to {R_2}\$ :\)

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

So here we have a single equation,

\((1 - i){k_1} - 2{k_2} = 0\;\;\; \to \;\;\;{k_1} = \frac{2}{{1 - i}}{k_2}\)

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

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

And column vectors:

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

Also,

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

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

Therefore,

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

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

And

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

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

Hence, the complementary function is

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

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,

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

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

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

\( = \sin t(\cos t - \sin t) - \cos t(\cos t + \sin t)\)

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

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

\( = - 1 \ne 0\)

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

So now,

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

Find the general solution of a Nonhomogeneous system:

Back to obtaining the particular solution,

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

\(\left( {\begin{array}{*{20}{c}}{\cos t - \sin t}&{\cos t + \sin t}\\{\cos t}&{\sin t}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - \ln |\tan t + \sec t| + 2\sin t - \cos t}\\{ - \sin t - 2\cos t}\end{array}} \right)\)

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

Finally, we conclude that the general solution is

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

\( = {c_1}\left( {\begin{array}{*{20}{c}}{cost - sint}\\{cost}\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}{cost + sint}\\{sint}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ln|tant + sect|(sint - cost) - 3co{s^2}t - 3si{n^2}t}\\{ - ln|tant + sect|cost - co{s^2}t - si{n^2}t}\end{array}} \right)\)