Question

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

\(17.{X^\prime } = \left( {\begin{array}{*{20}{r}}0&2\\{ - 1}&3\end{array}} \right)X + \left( {\begin{array}{*{20}{r}}1\\{ - 1}\end{array}} \right){e^t}\)

Short answer

The general solution of the system for \({X^\prime } = \left( {\begin{array}{*{20}{r}}0&2\\{ - 1}&3\end{array}} \right)X + \left( {\begin{array}{*{20}{r}}1\\{ - 1}\end{array}} \right){e^t}\) is \(X(t) = {c_1}\left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right){e^{2t}} + {c_2}\left( {\begin{array}{*{20}{l}}2\\1\end{array}} \right){e^t} + \left( {\begin{array}{*{20}{l}}3\\3\end{array}} \right){e^t} + \left( {\begin{array}{*{20}{l}}4\\2\end{array}} \right)t{e^t}\)

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

\({X^\prime } = \left( {\begin{array}{*{20}{r}}0&2\\{ - 1}&3\end{array}} \right)X + \left( {\begin{array}{*{20}{r}}1\\{ - 1}\end{array}} \right){e^t}\)

Where

\(A = \left( {\begin{array}{*{20}{r}}0&2\\{ - 1}&3\end{array}} \right)\)

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

\(\det (A - \lambda I) = 0\;\;\; \to \left| {\begin{array}{*{20}{c}}{0 - \lambda }&2\\{ - 1}&{ - 3 - \lambda }\end{array}} \right| = 0\)

\( - \lambda (3 - \lambda ) + 2 = 0\)

\({\lambda ^2} - 3\lambda + 2 = 0\)

\({\lambda ^2} - 2\lambda + \frac{3}{4} = 0\)

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

So our eigenvalues are\({\lambda _1} = 2\), and\({\lambda _2} = 1\).

Determine the Complementary function

For \({\lambda _1} = 2\) :

\((A - 2I\mid 0){\rm{ }} = \left( {\begin{array}{*{20}{c}}{0 - 2}&2&0\\{ - 1}&{3 - 2}&0\end{array}} \right)\)

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

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

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

So here we have a single equation,

\( - 2{k_1} + 2{k_2} = 0\;\;\; \to \;\;\;{k_1} = {k_2}\)

Choosing \({k_2} = 1\)yields\({k_1} = 1\).

This gives an eigenvector and a corresponding solution vector:

\({K_{\bf{1}}} = \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right),\;\;\;{X_{\bf{1}}} = \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right){e^{2t}} = \left( {\begin{array}{*{20}{l}}{{e^{2t}}}\\{{e^{2t}}}\end{array}} \right)\)

For\({\lambda _2} = 1\):

\((A - I\mid 0) = \left( {\begin{array}{*{20}{c}}{0 - 1}&2&0\\{ - 1}&{3 - 1}&0\end{array}} \right)\)

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

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

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

So here we have a single equation,

\( - {k_1} + 2{k_2} = 0\;\;\; \to \;\;\;{k_1} = 2{k_2}\)

Choosing \({k_2} = 1\)yields\({k_1} = 2\).

This gives an eigenvector and a corresponding solution vector:

\({K_2} = \left( {\begin{array}{*{20}{l}}1\\2\end{array}} \right),\;\;\;{X_2} = \left( {\begin{array}{*{20}{l}}2\\1\end{array}} \right){e^t} = \left( {\begin{array}{*{20}{c}}{2{e^t}}\\{{e^t}}\end{array}} \right)\)

Hence, the complementary function is

\({X_{\bf{c}}} = {c_1}\left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right){e^{2t}} + {c_2}\left( {\begin{array}{*{20}{l}}2\\1\end{array}} \right){e^t}\)

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,

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

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

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

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

\( = {e^{3t}} - 2{e^{3t}}\)

\( = - {e^{3t}} \ne 0\)

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

So now,

\({\Phi ^{ - 1}}(t) = \frac{1}{{ - {e^{3t}}}}\left( {\begin{array}{*{20}{c}}{{e^t}}&{ - 2{e^t}}\\{ - {e^{2t}}}&{{e^{2t}}}\end{array}} \right)\)

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

Obtaining the particular solution,

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

\( = \left( {\begin{array}{*{20}{c}}{{e^{2t}}}&{2{e^t}}\\{{e^{2t}}}&{{e^t}}\end{array}} \right)\smallint \left( {\begin{array}{*{20}{c}}{ - {e^{ - 2t}}}&{2{e^{ - 2t}}}\\{{e^{ - t}}}&{ - {e^{ - t}}}\end{array}} \right)\left( {\begin{array}{*{20}{r}}1\\{ - 1}\end{array}} \right){e^t}dt\)

\( = \left( {\begin{array}{*{20}{c}}{{e^{2t}}}&{2{e^t}}\\{{e^{2t}}}&{{e^t}}\end{array}} \right)\smallint \left( {\begin{array}{*{20}{c}}{ - {e^{ - t}} - 2{e^{ - t}}}\\{1 + 1}\end{array}} \right)dt\)

\( = \left( {\begin{array}{*{20}{c}}{{e^{2t}}}&{2{e^t}}\\{{e^{2t}}}&{{e^t}}\end{array}} \right)/\left( {\begin{array}{*{20}{c}}{ - 3{e^{ - t}}}\\2\end{array}} \right)dt\)

\( = \left( {\begin{array}{*{20}{c}}{{e^{2t}}}&{2{e^t}}\\{{e^{2t}}}&{{e^t}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{3{e^{ - t}}}\\{2t}\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{l}}{3{e^t} + 4t{e^t}}\\{3{e^t} + 2t{e^t}}\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{l}}3\\3\end{array}} \right){e^t} + \left( {\begin{array}{*{20}{l}}4\\2\end{array}} \right)t{e^t}\)

Finally, the general solution of the system is

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

\( = {c_1}\left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right){e^{2t}} + {c_2}\left( {\begin{array}{*{20}{l}}2\\1\end{array}} \right){e^t} + \left( {\begin{array}{*{20}{l}}3\\3\end{array}} \right){e^t} + \left( {\begin{array}{*{20}{l}}4\\2\end{array}} \right)t{e^t}\)

Therefore, the general solution of the system for \({X^\prime } = \left( {\begin{array}{*{20}{r}}0&2\\{ - 1}&3\end{array}} \right)X + \left( {\begin{array}{*{20}{r}}1\\{ - 1}\end{array}} \right){e^t}\) is \(X(t) = {c_1}\left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right){e^{2t}} + {c_2}\left( {\begin{array}{*{20}{l}}2\\1\end{array}} \right){e^t} + \left( {\begin{array}{*{20}{l}}3\\3\end{array}} \right){e^t} + \left( {\begin{array}{*{20}{l}}4\\2\end{array}} \right)t{e^t}\)