Question

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

\(15.{X^\prime } = \left( {\begin{array}{*{20}{c}}3&{ - 5}\\{\frac{3}{4}}&{ - 1}\end{array}} \right)X + \left( {\begin{array}{*{20}{r}}1\\{ - 1}\end{array}} \right){e^{t/2}}\)

Short answer

The general solution of the system for \({X^\prime } = \left( {\begin{array}{*{20}{c}}3&{ - 5}\\{\frac{3}{4}}&{ - 1}\end{array}} \right)X + \left( {\begin{array}{*{20}{r}}1\\{ - 1}\end{array}} \right){e^{t/2}}\) is \(X(t) = {c_1}\left( {\begin{array}{*{20}{c}}{10}\\3\end{array}} \right){e^{1.5t}} + {c_2}\left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right){e^{0.5t}} + \left( {\begin{array}{*{20}{c}}{ - \frac{{15}}{2}}\\{ - \frac{9}{4}}\end{array}} \right){e^{0.5t}} + \left( {\begin{array}{*{20}{c}}{ - \frac{{13}}{2}}\\{ - \frac{{13}}{4}}\end{array}} \right)t{e^{0.5t}}\)

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

\({X^\prime } = \left( {\begin{array}{*{20}{c}}3&{ - 5}\\{3/4}&{ - 1}\end{array}} \right)X + \left( {\begin{array}{*{20}{r}}1\\{ - 1}\end{array}} \right){e^{\frac{t}{2}}}\)

Where

\(A = \left( {\begin{array}{*{20}{c}}3&{ - 5}\\{3/4}&{ - 1}\end{array}} \right)\)

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

\(\det (A - \lambda I) = 0\;\;\; \to \left| {\begin{array}{*{20}{c}}{3 - \lambda }&{ - 5}\\{3/4}&{ - 1 - \lambda }\end{array}} \right| = 0\)

\((3 - \lambda )( - 1 - \lambda ) + \frac{{15}}{4} = 0\)

\( - 3 - 3\lambda + \lambda + {\lambda ^2} + \frac{{15}}{4} = 0\)

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

\(\left( {\lambda - \frac{3}{2}} \right)\left( {\lambda - \frac{1}{2}} \right) = 0\)

So our eigenvalues are\({\lambda _1} = \frac{3}{2}\), and \({\lambda _2} = \frac{1}{2}\)

Determine the Eigen vector and the corresponding solution vector

For \({\lambda _1} = 1.5\) :

\((A - 1.5I\mid 0) = \left( {\begin{array}{*{20}{c}}{3 - 1.5}&{ - 5}&0\\{0.75}&{ - 1 - 1.5}&0\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}{1.5}&{ - 5}&0\\{0.75}&{ - 2.5}&0\end{array}} \right)\)

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

\( = \left( {\begin{array}{*{20}{c}}{1.5}&{ - 5}&0\\0&0&0\end{array}} \right)\)

So here we have a single equation,

\(1.5{k_1} - 5{k_2} = 0\;\;\; \to \;\;\;{k_1} = \frac{{10}}{3}{k_2}\)

Choosing \({k_2} = 3\)yields\({k_1} = 10\).

This gives an eigenvector and a corresponding solution vector:

\({K_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{10}\\3\end{array}} \right),\;\;\;{X_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{10}\\3\end{array}} \right){e^{1.5t}} = \left( {\begin{array}{*{20}{c}}{10{e^{1.5t}}}\\{3{e^{1.5t}}}\end{array}} \right)\)

For \({\lambda _2} = 0.5:\)

\((A - 0.5I\mid 0) = \left( {\begin{array}{*{20}{c}}{3 - 0.5}&{ - 5}&0\\{0.75}&{ - 1 - 0.5}&0\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}{2.5}&{ - 5}&0\\{0.75}&{ - 1.5}&0\end{array}} \right)\)

Apply row operation\(\frac{{10}}{3}{R_2} - {R_1} \to {R_2}\):

\( = \left( {\begin{array}{*{20}{c}}{2.5}&{ - 5}&0\\0&0&0\end{array}} \right)\)

So here we have a single equation,

\(2.5{k_1} - 5{k_2} = 0\;\;\; \to \;\;\;{k_1} = 2{k_2}\)

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

This gives an eigenvector and a corresponding solution vector:

\({K_2} = \left( {\begin{array}{*{20}{l}}2\\1\end{array}} \right),\;\;\;{X_2} = \left( {\begin{array}{*{20}{l}}2\\1\end{array}} \right){e^{0.5t}} = \left( {\begin{array}{*{20}{c}}{2{e^{0.5t}}}\\{{e^{0.5t}}}\end{array}} \right)\)

Hence, the complementary function is

\({X_{\bf{c}}} = {c_1}\left( {\begin{array}{*{20}{c}}{10}\\3\end{array}} \right){e^{1.5t}} + {c_2}\left( {\begin{array}{*{20}{l}}2\\1\end{array}} \right){e^{0.5t}}\)

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,

\(\Phi (t) = \left( {\begin{array}{*{20}{c}}{10{e^{1.5t}}}&{2{e^{0.5t}}}\\{3{e^{1.5t}}}&{{e^{0.5t}}}\end{array}} \right)\)

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

\(|\Phi (t)|{\rm{ }} = \left| {\begin{array}{*{20}{c}}{10{e^{1.5t}}}&{2{e^{0.5t}}}\\{3{e^{1.5t}}}&{{e^{0.5t}}}\end{array}} \right|\)

\( = \left( {10{e^{1.5t}}} \right)\left( {{e^{0.5t}}} \right) - \left( {2{e^{0.5t}}} \right)\left( {3{e^{1.5t}}} \right)\)

\( = 10{e^2} - 6{e^2}\)

\( = 4{e^2} \ne 0\)

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

So now, \({\Phi ^{ - 1}}(t) = \frac{1}{{4{e^2}}}\left( {\begin{array}{*{20}{c}}{{e^{0.5t}}}&{ - 2{e^{0.5t}}}\\{ - 3{e^{1.5t}}}&{10{e^{1.5t}}}\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}{\frac{1}{4}{e^{ - 1.5t}}}&{ - \frac{1}{2}{e^{ - 1.5t}}}\\{ - \frac{3}{4}{e^{ - 0.5t}}}&{\frac{5}{2}{e^{ - 0.5t}}}\end{array}} \right)\)

Obtaining the particular solution,

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

\( = \left( {\begin{array}{*{20}{c}}{10{e^{1.5t}}}&{2{e^{0.5t}}}\\{3{e^{1.5t}}}&{{e^{0.5t}}}\end{array}} \right)/\left( {\begin{array}{*{20}{c}}{\frac{1}{4}{e^{ - 1.5t}}}&{ - \frac{1}{2}{e^{ - 1.5t}}}\\{ - \frac{3}{4}{e^{ - 0.5t}}}&{\frac{5}{2}{e^{ - 0.5t}}}\end{array}} \right)\left( {\begin{array}{*{20}{r}}1\\{ - 1}\end{array}} \right){e^{0.5t}}dt\)

\( = \left( {\begin{array}{*{20}{c}}{10{e^{1.5t}}}&{2{e^{0.5t}}}\\{3{e^{1.5t}}}&{{e^{0.5t}}}\end{array}} \right)/\left( {\begin{array}{*{20}{c}}{\frac{1}{4}{e^{ - t}} + \frac{1}{2}{e^{ - t}}}\\{ - \frac{3}{4} - \frac{5}{2}}\end{array}} \right)dt\)

\( = \left( {\begin{array}{*{20}{c}}{10{e^{1.5t}}}&{2{e^{0.5t}}}\\{3{e^{1.5t}}}&{{e^{0.5t}}}\end{array}} \right)/\left( {\begin{array}{*{20}{c}}{\frac{3}{4}{e^{ - t}}}\\{ - \frac{{13}}{4}}\end{array}} \right)dt\)

\( = \left( {\begin{array}{*{20}{c}}{10{e^{1.5t}}}&{2{e^{0.5t}}}\\{3{e^{1.5t}}}&{{e^{0.5t}}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - \frac{3}{4}{e^{ - t}}}\\{ - \frac{{13}}{4}t}\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}{ - \frac{{15}}{2}{e^{0.5t}} - \frac{{13}}{2}t{e^{0.5t}}}\\{ - \frac{9}{4}{e^{0.5t}} - \frac{{13}}{4}t{e^{0.5t}}}\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}{ - \frac{{15}}{2}}\\{ - \frac{9}{4}}\end{array}} \right){e^{0.5t}} + \left( {\begin{array}{*{20}{c}}{ - \frac{{13}}{2}}\\{ - \frac{{13}}{4}}\end{array}} \right)t{e^{0.5t}}\)

Finally, the general solution of the system is

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

\( = {c_1}\left( {\begin{array}{*{20}{c}}{10}\\3\end{array}} \right){e^{1.5t}} + {c_2}\left( {\begin{array}{*{20}{l}}2\\1\end{array}} \right){e^{0.5t}} + \left( {\begin{array}{*{20}{r}}{ - \frac{{15}}{2}}\\{ - \frac{9}{4}}\end{array}} \right){e^{0.5t}} + \left( {\begin{array}{*{20}{c}}{ - \frac{{13}}{2}}\\{ - \frac{{13}}{4}}\end{array}} \right)t{e^{0.5t}}\)

Therefore, the general solution of the system for \({X^\prime } = \left( {\begin{array}{*{20}{c}}3&{ - 5}\\{\frac{3}{4}}&{ - 1}\end{array}} \right)X + \left( {\begin{array}{*{20}{r}}1\\{ - 1}\end{array}} \right){e^{t/2}}\) is \(X(t) = {c_1}\left( {\begin{array}{*{20}{c}}{10}\\3\end{array}} \right){e^{1.5t}} + {c_2}\left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right){e^{0.5t}} + \left( {\begin{array}{*{20}{c}}{ - \frac{{15}}{2}}\\{ - \frac{9}{4}}\end{array}} \right){e^{0.5t}} + \left( {\begin{array}{*{20}{c}}{ - \frac{{13}}{2}}\\{ - \frac{{13}}{4}}\end{array}} \right)t{e^{0.5t}}\).