Question

In Problems, 1-8 use the method of undetermined coefficients to solve the given nonhomogeneous system.

\({\rm{6}}{\rm{. }}{{\bf{X}}^\prime } = \left( {\begin{array}{*{20}{l}}{ - 1}&5\\{ - 1}&1\end{array}} \right){\bf{X}} + \left( {\begin{array}{*{20}{c}}{\sin t}\\{ - 2\cos t}\end{array}} \right)\)

Short answer

The method of undetermined coefficients to solve the nonhomogeneous system of\({X^\prime } = \left( {\begin{array}{*{20}{l}}{ - 1}&5\\{ - 1}&1\end{array}} \right)X + \left( {\begin{array}{*{20}{c}}{\sin t}\\{ - 2\cos t}\end{array}} \right)\) is \({\rm{ }}{\bf{X}}(t) = {c_1}\left( {\begin{array}{*{20}{c}}{\cos 2t + \sin 2t}\\{\cos 2t}\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}{\sin 2t - 2\cos 2t}\\{\sin 2t}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - \frac{1}{3}}\\{\frac{1}{3}}\end{array}} \right)\sin t + \left( {\begin{array}{*{20}{c}}{ - 3}\\{ - \frac{2}{3}}\end{array}} \right)\cos t\)

Step-by-step solution

Determine the Nonhomogeneous Linear system:

A nonhomogeneous system has a homogeneous system associated with it, which can be obtained by replacing the constant term in each equation with zero.

Find the characteristic equation of the coefficient matrix: 

We have

\({X^\prime } = \left( {\begin{array}{*{20}{l}}{ - 1}&5\\{ - 1}&1\end{array}} \right)X + \left( {\begin{array}{*{20}{c}}{\sin t}\\{ - 2\cos t}\end{array}} \right)\)

Where

\(A = \left( {\begin{array}{*{20}{l}}{ - 1}&5\\{ - 1}&1\end{array}} \right)\)

Now, finding the characteristic equation of the coefficient matrix

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

\(( - 1 - \lambda )(1 - \lambda ) - ( - 5) = 0\)

\( - 1 + \lambda - \lambda + {\lambda ^2} + 5 = 0\)

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

\(\lambda = \pm 2i\)

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

find the method of undetermined coefficients:

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

\((A - 2iI\mid 0) = \left( {\begin{array}{*{20}{c}}{ - 1 - 2i}&5&0\\{ - 1}&{1 - 2i}&0\end{array}} \right)\)

Apply row operation \({R_2} - \frac{1}{{1 + 2i}}{R_1} \to {R_2}:\)

\( = \left( {\begin{array}{*{20}{c}}{ - 1 - 2i}&5&0\\0&0&0\end{array}} \right)\)

here we have a single equation,

\( - (1 + 2i){k_1} + 5{k_2} = 0\;\;\; \to \;\;\; - (1 + 2i){k_1}{\rm{ }} = - 5{k_2}\)

\({k_1}{\rm{ }} = (1 - 2i){k_2}\)

choosing\({k_2} = 1\)yields\({k_1} = 1 - 2i\).

This gives an eigenvector:

\({K_{\bf{1}}} = \left( {\begin{array}{*{20}{c}}{1 - 2i}\\1\end{array}} \right) = \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right) + i\left( {\begin{array}{*{20}{r}}{ - 2}\\0\end{array}} \right)\)

where

\({B_1} = \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right)\;\;\;{\rm{ and }}\;\;\;{B_2} = \left( {\begin{array}{*{20}{r}}{ - 2}\\0\end{array}} \right)\)

also,

\(\lambda = \alpha + \beta i\;\;\; \to \;\;\;{\lambda _1} = 0 + 2i\)

where\(\alpha = 0\)and\(\beta = 2.\)

\({X_1}{\rm{ }} = \left[ {{B_1}\cos \beta t + {B_2}\sin \beta t} \right]{e^{\alpha t}}\)

\( = \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right)\cos 2t - \left( {\begin{array}{*{20}{r}}{ - 2}\\0\end{array}} \right)\sin 2t = \left( {\begin{array}{*{20}{c}}{\cos 2t + \sin 2t}\\{\cos 2t}\end{array}} \right)\)

And

\({X_{\bf{2}}}{\rm{ }} = \left[ {{B_2}\cos \beta t + {B_{\bf{1}}}\sin \beta t} \right]{e^{\alpha t}}\)

\( = \left( {\begin{array}{*{20}{r}}{ - 2}\\0\end{array}} \right)\cos 2t + \left( {\begin{array}{*{20}{l}}1\\1\end{array}} \right)\sin 2t = \left( {\begin{array}{*{20}{c}}{\sin 2t - 2\cos 2t}\\{\sin 2t}\end{array}} \right)\)

therefore,

\({{\bf{X}}_{\bf{c}}}{\rm{ }} = {c_1}{{\bf{X}}_{\bf{1}}} + {c_2}{{\bf{X}}_{\bf{2}}}\)

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

Find the particular solution of the system:

Since\({\bf{F}}(t) = \left( {\begin{array}{*{20}{c}}{\sin t}\\{ - 2\cos t}\end{array}} \right),\)we shall try to find a particular solution of the system that possesses the same form:

\({{\bf{X}}_{\bf{p}}} = \left( {\begin{array}{*{20}{l}}{{a_1}}\\{{b_1}}\end{array}} \right)\sin t + \left( {\begin{array}{*{20}{l}}{{a_2}}\\{{b_2}}\end{array}} \right)\cos t\)

differentiating,

\({\bf{X}}_{\bf{p}}^\prime = \left( {\begin{array}{*{20}{l}}{{a_1}}\\{{b_1}}\end{array}} \right)\cos t - \left( {\begin{array}{*{20}{l}}{{a_2}}\\{{b_2}}\end{array}} \right)\sin t\)

Substituting into the given system yields

\(\left( {\begin{array}{*{20}{l}}{{a_1}}\\{{b_1}}\end{array}} \right)\cos t - \left( {\begin{array}{*{20}{l}}{{a_2}}\\{{b_2}}\end{array}} \right)\sin t = \left( {\begin{array}{*{20}{l}}{ - 1}&5\\{ - 1}&1\end{array}} \right)\left[ {\left( {\begin{array}{*{20}{l}}{{a_1}}\\{{b_1}}\end{array}} \right)\sin t + \left( {\begin{array}{*{20}{l}}{{a_2}}\\{{b_2}}\end{array}} \right)\cos t} \right] + \left( {\begin{array}{*{20}{c}}{\sin t}\\{ - 2\cos t}\end{array}} \right)\)

\(\left( {\begin{array}{*{20}{l}}{{a_1}}\\{{b_1}}\end{array}} \right)\cos t - \left( {\begin{array}{*{20}{l}}{{a_2}}\\{{b_2}}\end{array}} \right)\sin t = \left( {\begin{array}{*{20}{c}}{ - {a_1} + 5{b_1}}\\{ - {a_1} + {b_1}}\end{array}} \right)\sin t + \left( {\begin{array}{*{20}{c}}{ - {a_2} + 5{b_2}}\\{ - {a_2} + {b_2}}\end{array}} \right)\cos t + \left( {\begin{array}{*{20}{c}}{\sin t}\\{ - 2\cos t}\end{array}} \right)\)

Compare the terms on each side of the equation we obtained above, and equate the similar terms, where we will obtain

\({a_1} = - {a_2} + 5{b_2}\) \((1)\)

\({b_1} = - {a_2} + {b_2} - 2\) \((2)\)

\( - {a_2}{\rm{ }} = - {a_1} + 5{b_1} + 1\) \((3)\)

\( - {b_2} = - {a_1} + {b_1}{\rm{ }}\) \((4)\)

Step 5: Simplify and substituting the values:

from equation (4), we have

\({b_1} = {a_1} - {b_2}\)

we can substitute this into equation (2), which will become,

\({a_1} - {b_2}{\rm{ }} = - {a_2} + {b_2} - 2\)

\({a_1} = - {a_2} + 2{b_2} - 2\) \((5)\)

subtracting equation (5) from (1), we get

\({a_1} - {a_1}{\rm{ }} = - {a_2} + {a_2} + 5{b_2} - 2{b_2} - 2\)

\(0{\rm{ }} = 0 + 3{b_2} + 2\)

\({b_2}{\rm{ }} = - \frac{2}{3}\)

we substitute \({a_1}\) and \({b_1}\) from equations (1) and (2) into equation (3),

\( - {a_2} = - \left( { - {a_2} + 5{b_2}} \right) + 5\left( { - {a_2} + {b_2} - 2} \right) + 1\)

simplify and substitute the value of\({b_2}\),

\( - {a_2}{\rm{ }} = {a_2} - 5\left( { - \frac{2}{3}} \right) - 5{a_2} + 5\left( { - \frac{2}{3}} \right) - 10 + 1\)

\( - {a_2} - {a_2} + 5{a_2}{\rm{ }} = - 9\)

\(3{a_2}{\rm{ }} = - 9\)

\({a_2} = - 3\)

to find \({a_1}\) we can simply substitute this value of \({a_2}\) that we obtained along with the value of \({b_2}\) into equation (1), where,

\({a_1} = 3 + 5\left( { - \frac{2}{3}} \right)\;\;\; \to \;\;\;{a_1} = - \frac{1}{3}\)

and for\({b_1}\), we substitute the values of \({a_2}\) and \({b_2}\) into\((2)\), where

\({b_1} = 3 - \frac{2}{3} - 2\;\;\; \to \;\;\;{b_1} = \frac{1}{3}\)

Find the solution for Nonhomogeneous Linear system:

we plug these values into our particular solution, where

\({{\bf{X}}_{\bf{p}}} = \left( {\begin{array}{*{20}{l}}{{a_1}}\\{{b_1}}\end{array}} \right)\sin t + \left( {\begin{array}{*{20}{l}}{{a_2}}\\{{b_2}}\end{array}} \right)\cos t = \left( {\begin{array}{*{20}{c}}{ - \frac{1}{3}}\\{\frac{1}{3}}\end{array}} \right)\sin t + \left( {\begin{array}{*{20}{c}}{ - 3}\\{ - \frac{2}{3}}\end{array}} \right)\cos t\)

And

we finally conclude that the general solution of the system is

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

\( = {c_1}\left( {\begin{array}{*{20}{c}}{\cos 2t + \sin 2t}\\{\cos 2t}\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}{\sin 2t - 2\cos 2t}\\{\sin 2t}\end{array}} \right) + \left( {\begin{array}{*{20}{c}}{ - \frac{1}{3}}\\{\frac{1}{3}}\end{array}} \right)\sin t + \left( {\begin{array}{*{20}{c}}{ - 3}\\3\end{array}} \right)\cos t\)