Question

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

\(25.{{\bf{X}}^\prime } = \left( {\begin{array}{*{20}{r}}1&{ - 1}\\1&1\end{array}} \right){\bf{X}} + \left( {\begin{array}{*{20}{c}}{\cos t}\\{\sin t}\end{array}} \right){e^t}\)

Short answer

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

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

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

Where

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

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

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

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

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

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

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

Where,

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

\( = \frac{{2 \pm 2i}}{2}\)

\( = 1 \pm i\)

So our eigenvalues are \({\lambda _1} = 1 + {i_r}\)and \({\lambda _2} = \overline {{\lambda _1}} = 1 - i\)

Determine the Complementary function

For \({\lambda _1} = 1 + i:\)

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

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

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,

\(\lambda = \alpha + \beta i\;\;\; \to \;\;\;\lambda = 1 + i\)

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

Therefore,

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

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

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

And

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

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

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

Hence, the complementary function is

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

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,

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

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

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

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

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

\( = - {e^{2t}} \ne 0\)

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

So now,

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

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

Obtaining the particular solution,

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

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

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

\( = \left( {\begin{array}{*{20}{r}}{ - {e^t}\sin t}&{{e^t}\cos t}\\{{e^t}\cos t}&{{e^t}\sin t}\end{array}} \right)\smallint \left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right)dt\)

\( = \left( {\begin{array}{*{20}{r}}{ - {e^t}\sin t}&{{e^t}\cos t}\\{{e^t}\cos t}&{{e^t}\sin t}\end{array}} \right)\left( {\begin{array}{*{20}{l}}0\\t\end{array}} \right)\)

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

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

Finally, the general solution of the system is

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

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