Question

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

\(16.{{\bf{X}}^\prime } = \left( {\begin{array}{*{20}{r}}2&{ - 1}\\4&2\end{array}} \right){\bf{X}} + \left( {\begin{array}{*{20}{c}}{\sin 2t}\\{2\cos 2t}\end{array}} \right){e^{2t}}\)

Short answer

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

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

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

Where

\(A = \left( {\begin{array}{*{20}{r}}2&{ - 1}\\4&2\end{array}} \right)\)

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

\(\det (A - \lambda I) = 0\;\;\; \to \left| {\begin{array}{*{20}{c}}{2 - \lambda }&{ - 1}\\4&{2 - \lambda }\end{array}} \right| = 0\)

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

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

\({\lambda ^2} - 4\lambda + 8 = 0\)

Where,

\(\lambda {\rm{ }} = \frac{{4 \pm \sqrt {16 - 4(1)(8)} }}{2}\)

\( = \frac{{4 \pm 4i}}{2}\)

\( = 2 \pm 2i\)

So our eigenvalues are\({\lambda _1} = 2 + 2i\), and\({\lambda _2} = \overline {{\lambda _1}} = 2 - 2i\).

Determine the Complementary function

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

\((A - (2 + 2i)I\mid 0){\rm{ }} = \left( {\begin{array}{*{20}{c}}{2 - (2 + 2i)}&{ - 1}&0\\4&{2 - (2 + 2i)}&0\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}{ - 2i}&{ - 1}&0\\4&{ - 2i}&0\end{array}} \right)\)

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

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

So here we have a single equation,

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

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

This gives an eigenvector:

\(K = \left( {\begin{array}{*{20}{l}}i\\2\end{array}} \right) = \left( {\begin{array}{*{20}{l}}0\\2\end{array}} \right) + i\left( {\begin{array}{*{20}{l}}1\\0\end{array}} \right)\)

And column vectors:

\({B_1} = \left( {\begin{array}{*{20}{l}}0\\2\end{array}} \right),\)And \({B_2} = \left( {\begin{array}{*{20}{l}}1\\0\end{array}} \right)\)

Also,

\(\lambda = \alpha + \beta i\;\;\; \to \;\;\;\lambda = 2 + 2i\)

Where \(\alpha = 2\) and \(\beta = 2\)

Therefore,

\({X_{\bf{1}}} = \left[ {{B_1}\cos \beta t - {B_2}\sin \beta t} \right]{e^{\alpha t}}\)

\( = \left[ {\left( {\begin{array}{*{20}{l}}0\\2\end{array}} \right)\cos 2t - \left( {\begin{array}{*{20}{l}}1\\0\end{array}} \right)\sin 2t} \right]{e^{2t}}\)

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

And

\({X_{\bf{2}}} = \left[ {{B_2}\cos \beta t + {B_1}\sin \beta t} \right]{e^{\alpha t}}\)

\( = \left[ {\left( {\begin{array}{*{20}{l}}1\\0\end{array}} \right)\cos 2t + \left( {\begin{array}{*{20}{l}}0\\2\end{array}} \right)\sin 2t} \right]{e^{2t}}\)

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

Hence, the complementary function is

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

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,

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

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

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

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

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

\( = - 2{e^{4t}}\left( {{{\sin }^2}2t + {{\cos }^2}2t} \right)\)

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

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

So now,

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

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

Obtaining the particular solution.

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

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

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

Use trigonometric identity\({\cos ^2}t - {\sin ^2}t = \cos 2t\),

\( = \left( {\begin{array}{*{20}{c}}{ - {e^{2t}}\sin 2t}&{{e^{2t}}\cos 2t}\\{2{e^{2t}}\cos 2t}&{2{e^{2t}}\sin 2t}\end{array}} \right)\smallint \left( {\begin{array}{*{20}{c}}{\cos 4t}\\{2\sin 2t\cos 2t}\end{array}} \right)dt\)

To integrate

\(\int {\cos } 4tdt\)

Use the method of substitution:

\(u = 4t\;\;\; \to \;\;\;dt = \frac{1}{4}du\)

Where,

\(\int {\cos } 4tdt = \frac{1}{4}\)

\(\int {\cos } udu = \frac{{\sin u}}{4}\)

Undo substitution, \( = \frac{{\sin 4t}}{4}\)

And to integrate

\(\int 2 \sin 2t\cos 2tdt\)

We will also use the method of substitution:

\(u = \sin 2t\;\;\; \to \;\;\;dt = \frac{1}{{2\cos 2t}}du\)

Where,

\(\int 2 \sin 2t\cos 2tdt{\rm{ }} = \int 2 u\cos 2t\frac{1}{{2\cos 2t}}du\)

\( = \int u du\)

\( = \frac{{{u^2}}}{2}\)

Undo substitution,

\( = \frac{{{{\sin }^2}2l}}{2}\)

\( = - \frac{{\cos 4l}}{4}\)

We now can go back to obtaining the particular solution,

\({X_{\bf{p}}} = \left( {\begin{array}{*{20}{c}}{ - {e^{2t}}\sin 2t}&{{e^{2t}}\cos 2t}\\{2{e^{2t}}\cos 2t}&{2{e^{2t}}\sin 2t}\end{array}} \right)\smallint \left( {\begin{array}{*{20}{c}}{\cos 4t}\\{2\sin 2t\cos 2t}\end{array}} \right)dt\)

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

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

Finally, the general solution of the system is

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

\( = {c_1}\left( {\begin{array}{*{20}{c}}{ - \sin 2t}\\{2\cos 2t}\end{array}} \right){e^{2t}} + {c_2}\left( {\begin{array}{*{20}{c}}{\cos 2t}\\{2\sin 2t}\end{array}} \right){e^{2t}} + \left( {\begin{array}{*{20}{c}}{ - \frac{{\sin 2t\sin 4t}}{4} - \frac{{\cos 2t\cos 4t}}{4}}\\{\frac{{\cos 2t\sin 4t}}{2} - \frac{{\sin 2t\cos 4t}}{2}}\end{array}} \right){e^{2t}}\)

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