Question

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

\({\bf{13}}.\)\(\frac{{dx}}{{dt}} = 3x - 3y + 4\)

\(\frac{{dy}}{{dt}} = 2x - 2y - 1\)

Short answer

The general solution of the system for \({X^\prime } = \left( {\begin{array}{*{20}{l}}3&{ - 3}\\2&{ - 2}\end{array}} \right)X + \left( {\begin{array}{*{20}{r}}4\\{ - 1}\end{array}} \right)\) is \(X = {c_1}\left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{l}}3\\2\end{array}} \right){e^t} + \left( {\begin{array}{*{20}{l}}{ - 11}\\{ - 11}\end{array}} \right)t + \left( {\begin{array}{*{20}{l}}{ - 15}\\{ - 10}\end{array}} \right)\)

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

\(\left\{ {\begin{array}{*{20}{l}}{\frac{{dx}}{{dt}} = 3x - 3y + 4}\\{\frac{{dy}}{{dt}} = 2x - 2y - 1}\end{array}} \right.\)

Which can be written in the matrix form

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

Where

\(A = \left( {\begin{array}{*{20}{l}}3&{ - 3}\\2&{ - 2}\end{array}} \right)\), and \(F(t) = \left( {\begin{array}{*{20}{r}}4\\{ - 1}\end{array}} \right)\)

Obtaining the eigenvalues of the coefficient matrix,

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

\((3 - \lambda )( - 2 - \lambda ) + 6 = 0\)

\( - 6 - 3\lambda + 2\lambda + {\lambda ^2} + 6 = 0\)

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

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

So our eigenvalues are \({\lambda _1} = 0\)and\({\lambda _2} = 1\).

Determine the Eigen vector and the corresponding solution vector

For\({\lambda _1} = 0\):

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

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

Apply row operation\(\frac{3}{2}{R_2} - {R_1} \to {R_1}\):

\( = \left( {\begin{array}{*{20}{c}}3&{ - 3}&0\\0&0&0\end{array}} \right)\)

So here we have

\(3{k_1} - 3{k_2} = 0\;\;\; \to \;\;\;{k_1} = {k_2}\)

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

This gives an eigenvector and a corresponding solution vector

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

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

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

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

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

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

So here we have

\(2{k_1} - 3{k_2} = 0\;\;\; \to \;\;\;{k_1} = \frac{3}{2}{k_2}\)

Choosing \({k_2} = 2\)yields\({k_1} = 3\).

This gives an eigenvector and a corresponding solution vector

\({K_2} = \left( {\begin{array}{*{20}{l}}3\\2\end{array}} \right),\;\;\;{X_2} = \left( {\begin{array}{*{20}{l}}3\\2\end{array}} \right){e^t}\)

Hence, the complementary solution is

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

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,

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

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

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

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

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

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

So now,

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

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

Obtaining the particular solution,

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

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

\( = \left( {\begin{array}{*{20}{l}}1&{3{e^t}}\\1&{2{e^t}}\end{array}} \right)/\left( {\begin{array}{*{20}{c}}{ - 11}\\{5{e^{ - t}}}\end{array}} \right)dt\)

\( = \left( {\begin{array}{*{20}{l}}1&{3{e^t}}\\1&{2{e^t}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 11t}\\{ - 5{e^{ - t}}}\end{array}} \right)\)

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

\( = \left( {\begin{array}{*{20}{l}}{ - 11}\\{ - 11}\end{array}} \right)t + \left( {\begin{array}{*{20}{l}}{ - 15}\\{ - 10}\end{array}} \right)\)

Therefore, the general solution of the system is

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

\( = {c_1}\left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{l}}3\\2\end{array}} \right){e^t} + \left( {\begin{array}{*{20}{l}}{ - 11}\\{ - 11}\end{array}} \right)t + \left( {\begin{array}{*{20}{l}}{ - 15}\\{ - 10}\end{array}} \right)\)

Therefore, the general solution of the system for \({X^\prime } = \left( {\begin{array}{*{20}{l}}3&{ - 3}\\2&{ - 2}\end{array}} \right)X + \left( {\begin{array}{*{20}{r}}4\\{ - 1}\end{array}} \right)\) is \(X = {c_1}\left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{l}}3\\2\end{array}} \right){e^t} + \left( {\begin{array}{*{20}{l}}{ - 11}\\{ - 11}\end{array}} \right)t + \left( {\begin{array}{*{20}{l}}{ - 15}\\{ - 10}\end{array}} \right)\)