Question

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

\(8.{{\bf{X}}^\prime } = \left( {\begin{array}{*{20}{l}}0&0&5\\0&5&0\\5&0&0\end{array}} \right){\bf{X}} + \left( {\begin{array}{*{20}{r}}5\\{ - 10}\\{40}\end{array}} \right)\)

Short answer

The method of undetermined coefficients to solve the nonhomogeneous system of

\({X^\prime } = \left( {\begin{array}{*{20}{l}}0&0&5\\0&5&0\\5&0&0\end{array}} \right)X + \left( {\begin{array}{*{20}{r}}5\\{ - 10}\\{40}\end{array}} \right)\) is \(X(t) = {c_1}\left( {\begin{array}{*{20}{l}}0\\1\\0\end{array}} \right){e^{5t}} + {c_2}\left( {\begin{array}{*{20}{l}}1\\0\\1\end{array}} \right){e^{5t}} + {c_3}\left( {\begin{array}{*{20}{r}}{ - 1}\\0\\1\end{array}} \right){e^{ - 5t}} + \left( {\begin{array}{*{20}{r}}{ - 8}\\2\\{ - 1}\end{array}} \right)\)

Step-by-step solution

Defining 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

\({X^\prime } = \left( {\begin{array}{*{20}{l}}0&0&5\\0&5&0\\5&0&0\end{array}} \right)X + \left( {\begin{array}{*{20}{c}}5\\{ - 10}\\{40}\end{array}} \right)\)

where

\(A = \left( {\begin{array}{*{20}{l}}0&0&5\\0&5&0\\5&0&0\end{array}} \right)\)

Now, finding the characteristic equation of the coefficient matrix

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

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

\(\begin{array}{*{20}{r}}{(5 - \lambda )\left[ {{\lambda ^2} - 25} \right] = 0}\\{(5 - \lambda )(\lambda - 5)(\lambda + 5) = 0}\end{array}\)

so our eigenvalues are \({\lambda _1} = {\lambda _2} = 5,\)and \({\lambda _3} = - 5\)

Find the method of undetermined coefficients:

\( For {\lambda _{ - 1}} = {\lambda _2} = 5:\)

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

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

from the matrix we see that

\( - 5{k_1} + 5{k_3} = 0\;\;\; \to \;\;\;{k_1} = {k_3}\)

as for\({k_2}\), we choose an arbitrary value. Choosing\({k_1} = 0\)yields\({k_3} = 0\), and let\({k_2} = 1\), this gives an eigenvector and a corresponding solution vector:

\({K_1} = \left( {\begin{array}{*{20}{l}}0\\1\\0\end{array}} \right),\;\;\;{X_1} = \left( {\begin{array}{*{20}{l}}0\\1\\0\end{array}} \right){e^{5t}}\)

Choosing\({k_1} = 1\)yields\({k_3} = 1\), and let\({k_2} = 0\), this gives an eigenvector and a corresponding solution vector:

\({K_2} = \left( {\begin{array}{*{20}{l}}1\\0\\1\end{array}} \right),\;\;\;{X_2} = \left( {\begin{array}{*{20}{l}}1\\0\\1\end{array}} \right){e^{5t}}\)

For\({\lambda _ - }3 = - 5:\)

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

\( = \left( {\begin{array}{*{20}{c}}5&0&5&0\\0&{10}&0&0\\5&0&5&0\end{array}} \right)\)

from the matrix we see that\({k_2} = 0\), and

\(5{k_1} + 5{k_3} = 0\;\;\; \to \;\;\;{k_1} = - {k_3}\)

Choosing\({k_3} = 1\)yields\({k_1} = - 1\), this gives an eigenvector and a corresponding solution vector:

\({K_3} = \left( {\begin{array}{*{20}{r}}{ - 1}\\0\\1\end{array}} \right),\;\;\;{X_3} = \left( {\begin{array}{*{20}{r}}{ - 1}\\0\\1\end{array}} \right){e^{ - 5t}}\)

Therefore,

\({X_{\bf{c}}} = {c_1}\left( {\begin{array}{*{20}{l}}0\\1\\0\end{array}} \right){e^{5t}} + {c_2}\left( {\begin{array}{*{20}{l}}1\\0\\1\end{array}} \right){e^{5t}} + {c_3}\left( {\begin{array}{*{20}{r}}{ - 1}\\0\\1\end{array}} \right){e^{ - 5t}}\)

Simplify and substituting the values:

Since \(F(t) = \left( {\begin{array}{*{20}{r}}5\\{ - 10}\\{40}\end{array}} \right)\), we shall try to find a particular solution of the system that possesses the same form:

\({X_p} = \left( {\begin{array}{*{20}{l}}{{a_1}}\\{{a_2}}\\{{a_3}}\end{array}} \right)\)

differentiating,

\(X_{\bf{p}}^\prime = \left( {\begin{array}{*{20}{l}}0\\0\\0\end{array}} \right)\)

Substituting into the given system yields

\(\left( {\begin{array}{*{20}{l}}0\\0\\0\end{array}} \right) = \left( {\begin{array}{*{20}{l}}0&0&5\\0&5&0\\5&0&0\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{a_1}}\\{{a_2}}\\{{a_3}}\end{array}} \right) + \left( {\begin{array}{*{20}{r}}5\\{ - 10}\\{40}\end{array}} \right)\)

\(\left( {\begin{array}{*{20}{l}}0\\0\\0\end{array}} \right) = \left( {\begin{array}{*{20}{l}}{5{a_3}}\\{5{a_2}}\\{5{a_1}}\end{array}} \right) + \left( {\begin{array}{*{20}{r}}5\\{ - 10}\\{40}\end{array}} \right)\)

here we have,

\(5{a_3} + 5 = 0 \to {\rm{ }}{a_3}{\rm{ }} = - 5\)

\({a_3}{\rm{ }} = - 1\)

\(5{a_2} - 10 = 0 \to {\rm{ }}{a_2}{\rm{ }} = 10\)

\({a_2}{\rm{ }} = 2\)

\(5{a_1} + 40 = 0{\rm{ }} \to 5{a_1}{\rm{ }} = - 40\)

\({a_1}{\rm{ }} = - 8\)

Find the solution for Nonhomogeneous Linear system:

we plug these values into our particular solution, where

\({X_p} = \left( {\begin{array}{*{20}{l}}{{a_1}}\\{{a_2}}\\{{a_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{r}}{ - 8}\\2\\{ - 1}\end{array}} \right)\)

and we finally conclude that the general solution of the system is

\(X(t) = {X_c} + {X_p}\)

\( = {c_1}\left( {\begin{array}{*{20}{l}}0\\1\\0\end{array}} \right){e^{5t}} + {c_2}\left( {\begin{array}{*{20}{l}}1\\0\\1\end{array}} \right){e^{5t}} + {c_3}\left( {\begin{array}{*{20}{r}}{ - 1}\\0\\1\end{array}} \right){e^{ - 5t}} + \left( {\begin{array}{*{20}{r}}{ - 8}\\2\\{ - 1}\end{array}} \right)\)