Question

In Problems \({\bf{33}}\) and \({\bf{34}}\) use (14) to solve the given initial-value problem.

\({\bf{33}}\). \({{\bf{X}}^\prime } = \left( {\begin{array}{*{20}{r}}3&{ - 1}\\{ - 1}&3\end{array}} \right){\bf{X}} + \left( {\begin{array}{*{20}{l}}{4{e^{2t}}}\\{4{e^{4t}}}\end{array}} \right),\;\;\;{\bf{X}}(0) = \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right)\)

Short answer

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

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

\({X^\prime } = \left( {\begin{array}{*{20}{r}}3&{ - 1}\\{ - 1}&3\end{array}} \right)X + \left( {\begin{array}{*{20}{l}}{4{e^{2t}}}\\{4{e^{4t}}}\end{array}} \right)\)

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

And

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

Obtaining the eigenvalues of the coefficient matrix,

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

\({(3 - \lambda )^2} - 1 = 0\)

\(9 - 6\lambda + {\lambda ^2} - 1 = 0\)

\({\lambda ^2} - 6\lambda + 8 = 0\)

\((\lambda - 4)(\lambda - 2) = 0\)

So our eigenvalues are \({\lambda _1} = 4\)and\({\lambda _2} = 2\).

Determine the eigenvector and a corresponding solution vector

For\({\lambda _1} = 4\):

\((A - 4I\mid 0) = \left( {\begin{array}{*{20}{c}}{3 - 4}&{ - 1}&0\\{ - 1}&{3 - 4}&0\end{array}} \right)\)

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

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

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

Here we have

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

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

This gives an eigenvector and a corresponding solution vector

\({K_{\bf{1}}} = \left( {\begin{array}{*{20}{r}}{ - 1}\\1\end{array}} \right),\;\;\;{X_{\bf{1}}} = \left( {\begin{array}{*{20}{r}}{ - 1}\\1\end{array}} \right){e^{4t}}\)

For \({\lambda _2} = 2:\)

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

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

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

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

Here we have

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

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

This gives an eigenvector and a corresponding solution vector

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

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,

\(\Phi (t) = \left( {\begin{array}{*{20}{r}}{ - {e^{4t}}}&{{e^{2t}}}\\{{e^{4t}}}&{{e^{2t}}}\end{array}} \right)\)

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

\(|\Phi (t)| = \left| {\begin{array}{*{20}{r}}{ - {e^{4t}}}&{{e^{2t}}}\\{{e^{4t}}}&{{e^{2t}}}\end{array}} \right|\)

\( = - {e^{4t}}{e^{2t}} - {e^{2t}}{e^{4t}}\)

\( = - {e^{6t}} - {e^{6t}}\)

\( = - 2{e^{6t}} \ne 0\)

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

So now,

\({\Phi ^{ - 1}}(t) = - \frac{1}{{2{e^{6t}}}}\left( {\begin{array}{*{20}{c}}{{e^{2t}}}&{ - {e^{2t}}}\\{ - {e^{4t}}}&{ - {e^{4t}}}\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}{ - \frac{1}{2}{e^{ - 4t}}}&{\frac{1}{2}{e^{ - 4t}}}\\{\frac{1}{2}{e^{ - 2t}}}&{\frac{1}{2}{e^{ - 2t}}}\end{array}} \right)\)

Using

\(X = \Phi (t){\Phi ^{ - 1}}(0)X(0) + \Phi (t)\int_0^t {{\Phi ^{ - 1}}} (s)F(s)ds\)

Where

\(X(0) = \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right)\)

First, we have

\(\Phi (t){\Phi ^{ - 1}}(0)X(0) = \left( {\begin{array}{*{20}{r}}{ - {e^{4t}}}&{{e^{2t}}}\\{{e^{4t}}}&{{e^{2t}}}\end{array}} \right)\left( {\begin{array}{*{20}{r}}{ - \frac{1}{2}}&{\frac{1}{2}}\\{\frac{1}{2}}&{\frac{1}{2}}\end{array}} \right)\left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}{ - {e^{4t}}}&{{e^{2t}}}\\{{e^{4t}}}&{{e^{2t}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - \frac{1}{2} + \frac{1}{2}}\\{\frac{1}{2} + \frac{1}{2}}\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{r}}{ - {e^{4t}}}&{{e^{2t}}}\\{{e^{4t}}}&{{e^{2t}}}\end{array}} \right)\left( {\begin{array}{*{20}{l}}0\\1\end{array}} \right)\)

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

Substituting into (1),

\(X = \left( {\begin{array}{*{20}{l}}{{e^{2t}}}\\{{e^{2t}}}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - {e^{4t}}}&{{e^{2t}}}\\{{e^{4t}}}&{{e^{2t}}}\end{array}} \right)\int_0^t {\left( {\begin{array}{*{20}{c}}{ - \frac{1}{2}{e^{ - 4s}}}&{\frac{1}{2}{e^{ - 4s}}}\\{\frac{1}{2}{e^{ - 2s}}}&{\frac{1}{2}{e^{ - 2s}}}\end{array}} \right)} \left( {\begin{array}{*{20}{l}}{4{e^{2s}}}\\{4{e^{4s}}}\end{array}} \right)ds\)

\( = \left( {\begin{array}{*{20}{c}}{{e^{2t}}}\\{{e^{2t}}}\end{array}} \right) + \left( {\begin{array}{*{20}{r}}{ - {e^{4t}}}&{{e^{2t}}}\\{{e^{4t}}}&{{e^{2t}}}\end{array}} \right)\int_0^t {\left( {\begin{array}{*{20}{c}}{ - 2{e^{ - 2s}} + 2}\\{2 + 2{e^{2s}}}\end{array}} \right)} dt\)

\(\left. { = \left( {\begin{array}{*{20}{l}}{{e^{2t}}}\\{{e^{2t}}}\end{array}} \right) + \left( {\begin{array}{*{20}{r}}{ - {e^{4t}}}&{{e^{2t}}}\\{{e^{4t}}}&{{e^{2t}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{e^{ - 2s}} + 2s}\\{2s + {e^{2s}}}\end{array}} \right)} \right]_0^t\)

\( = \left( {\begin{array}{*{20}{l}}{{e^{2t}}}\\{{e^{2t}}}\end{array}} \right) + \left( {\begin{array}{*{20}{r}}{ - {e^{4t}}}&{{e^{2t}}}\\{{e^{4t}}}&{{e^{2t}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{e^{ - 2t}} + 2t - 1}\\{{e^{2t}} + 2t - 1}\end{array}} \right)\)

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

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

\( = \left( {\begin{array}{*{20}{r}}{ - 1}\\1\end{array}} \right){e^{2t}} + \left( {\begin{array}{*{20}{r}}{ - 2}\\2\end{array}} \right)t{e^{4t}} + \left( {\begin{array}{*{20}{l}}2\\0\end{array}} \right){e^{4t}} + \left( {\begin{array}{*{20}{l}}2\\2\end{array}} \right)t{e^{2t}}\)

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