Question

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

\(7.{{\bf{X}}^\prime } = \left( {\begin{array}{*{20}{l}}1&1&1\\0&2&3\\0&0&5\end{array}} \right)X + \left( {\begin{array}{*{20}{r}}1\\{ - 1}\\2\end{array}} \right){e^{4t}}\)

Short answer

The method of undetermined coefficients to solve the nonhomogeneous system of

\({{\bf{X}}^\prime } = \left( {\begin{array}{*{20}{l}}1&1&1\\0&2&3\\0&0&5\end{array}} \right){\bf{X}} + \left( {\begin{array}{*{20}{r}}1\\{ - 1}\\2\end{array}} \right){e^{4t}}\) is \(X(t) = {c_1}\left( {\begin{array}{*{20}{l}}1\\0\\0\end{array}} \right){e^t} + {c_2}\left( {\begin{array}{*{20}{l}}1\\1\\0\end{array}} \right){e^{2t}} + {c_3}\left( {\begin{array}{*{20}{l}}1\\2\\2\end{array}} \right){e^{5t}} + \left( {\begin{array}{*{20}{r}}{ - \frac{7}{2}}\\{ - \frac{3}{2}}\\{ - 2}\end{array}} \right){e^{4t}}\)

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

\({{\bf{X}}^\prime } = \left( {\begin{array}{*{20}{l}}1&1&1\\0&2&3\\0&0&5\end{array}} \right){\bf{X}} + \left( {\begin{array}{*{20}{r}}1\\{ - 1}\\2\end{array}} \right){e^{4t}}\)

where

\({\bf{A}} = \left( {\begin{array}{*{20}{l}}1&1&1\\0&2&3\\0&0&5\end{array}} \right)\)

Now, finding the characteristic equation of the coefficient matrix

\(\det (A - \lambda I) = 0\;\;\; \to \left| {\begin{array}{*{20}{c}}{1 - \lambda }&1&1\\0&{2 - \lambda }&3\\0&0&{5 - \lambda }\end{array}} \right| = 0\)

\((1 - \lambda )\left| {\begin{array}{*{20}{c}}{2 - \lambda }&3\\0&{5 - \lambda }\end{array}} \right| = 0\)

\((1 - \lambda )(2 - \lambda )(5 - \lambda ) = 0\)

so our eigenvalues are\({\lambda _1} = 1,{\lambda _2} = 2\), and \({\lambda _3} = 5\)

find the method of undetermined coefficients:

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

\((A - I\mid 0) = \left( {\begin{array}{*{20}{c}}{1 - 1}&1&1&0\\0&{2 - 1}&3&0\\0&0&{5 - 1}&0\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}0&1&1&0\\0&1&3&0\\0&0&4&0\end{array}} \right)\)

we see that

\(4{k_3} = 0\;\;\; \to \;\;\;{k_3} = 0\)

And

\({k_2} = 0\)

as for\({k_1}\), we can choose an arbitrary value, say\({k_1} = 1\). This gives an eigenvector and a corresponding solution vector:

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

For\({\lambda _ - }2 = 2:\)

\((A - 2I\mid 0) = \left( {\begin{array}{*{20}{c}}{1 - 2}&1&1&0\\0&{2 - 2}&3&0\\0&0&{5 - 2}&0\end{array}} \right)\)

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

\(3{k_3} = 0\;\;\; \to \;\;\;{k_3} = 0\)

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

choosing \({k_2} = 1\) yields\({k_1} = 1\). This gives an eigenvector and a corresponding solution vector:

\({K_2} = \left( {\begin{array}{*{20}{l}}1\\1\\0\end{array}} \right),\;\;\;{X_2} = \left( {\begin{array}{*{20}{l}}1\\1\\0\end{array}} \right){e^{2t}}\)

Step 4: Simplify and substituting the values:

For\({\lambda _3} = 5:\)

\((A - 5I\mid 0) = \left( {\begin{array}{*{20}{c}}{1 - 5}&1&1&0&0\\0&2&{ - 5}&3&0\\0&0&{5 - 5}&0&0\end{array}} \right)\)

\( = \left( {\begin{array}{*{20}{c}}{ - 4}&1&1&0\\0&{ - 3}&3&0\\0&0&0&0\end{array}} \right)\)

from the first row,

\( - 4{k_1} + {k_2} + {k_3} = 0\)

from the second row we have\({k_2} = {k_3}\), we can substitute that into the equation,

\( - 4{k_1} + {k_3} + {k_3}{\rm{ }} = 0\)

\( - 4{k_1}{\rm{ }} = - 2{k_3}\)

\({k_1}{\rm{ }} = \frac{1}{2}{k_3}\)

choosing \({k_3} = 2\) yields, \({k_1} = 1\) and \({k_2} = 2\). This gives an eigenvector and a corresponding solution vector:

\({K_3} = \left( {\begin{array}{*{20}{l}}1\\2\\2\end{array}} \right),\;\;\;{X_3} = \left( {\begin{array}{*{20}{l}}1\\2\\2\end{array}} \right){e^{5t}}\)

Therefore,

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

Find the particular solution of the system:

Since\(F(t) = \left( {\begin{array}{*{20}{r}}1\\{ - 1}\\2\end{array}} \right){e^{4t}}\), we shall try to find a particular solution of the system that possesses the same form:

\({X_p} = \left( {\begin{array}{*{20}{l}}{{a_1}}\\{{a_2}}\\{{a_3}}\end{array}} \right){e^{4t}}\)

differentiating,

\(X_{\bf{p}}^\prime = \left( {\begin{array}{*{20}{l}}{{a_1}}\\{{a_2}}\\{{a_3}}\end{array}} \right)4{e^{4t}}\)

Substituting into the given system Yields

\(\left( {\begin{array}{*{20}{l}}{{a_1}}\\{{a_2}}\\{{a_3}}\end{array}} \right)4{e^{4t}}{\rm{ }} = \left( {\begin{array}{*{20}{l}}1&1&1\\0&2&3\\0&0&5\end{array}} \right)\left[ {\left( {\begin{array}{*{20}{l}}{{a_1}}\\{{a_2}}\\{{a_3}}\end{array}} \right){e^{4t}}} \right] + \left( {\begin{array}{*{20}{r}}1\\{ - 1}\\2\end{array}} \right){e^{4t}}\)

\(\left( {\begin{array}{*{20}{l}}{4{a_1}}\\{4{a_2}}\\{4{a_3}}\end{array}} \right){e^{4t}} = \left( {\begin{array}{*{20}{c}}{{a_1} + {a_2} + {a_3}}\\{2{a_2} + 3{a_3}}\\{5{a_3}}\end{array}} \right){e^{4t}} + \left( {\begin{array}{*{20}{r}}1\\{ - 1}\\2\end{array}} \right){e^{4t}}\)

\(\left( {\begin{array}{*{20}{l}}{4{a_1}}\\{4{a_2}}\\{4{a_3}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{a_1} + {a_2} + {a_3}}\\{2{a_2} + 3{a_3}}\\{5{a_3}}\end{array}} \right) + \left( {\begin{array}{*{20}{r}}1\\{ - 1}\\2\end{array}} \right)\)

Find the solution for Nonhomogeneous Linear system:

here we have

\(4{a_1} = {a_1} + {a_2} + {a_3} + 1\) \((1)\)

\(3{a_1} - {a_2} - {a_3} = 1\) \((2)\)

\(4{a_2} = 2{a_2} + 3{a_3} - 1\) (3)

\(2{a_2} - 3{a_3}{\rm{ }} = - 1\) \((4)\)

\(4{a_3} = 5{a_3} + 2\) \((5)\)

\({a_3} = 2\; - \;\) \((6)\)

from the third equation we get\({a_3} = - 2\), we can substitute this value into \((2)\) and solve for\({a_2}\),

\(2{a_2} - 3( - 2) = - 1 \to 2{a_2}{\rm{ }} = - 7\)

\({a_2}{\rm{ }} = - \frac{7}{2}\)

finally, we can substitute the value of \({a_2}\) and \({a_3}\) into (1) and solve for \({a_1},\)

\(3{a_2} + \frac{7}{2} + 2 = 1\;\;\; \to \;\)\(3{a_2} = - \frac{9}{2}\)

\({a_2} = - \frac{3}{2}\)

we plug these values into our particular solution, where

\({X_p} = \left( {\begin{array}{*{20}{c}}{{a_1}}\\{{a_2}}\\{{a_3}}\end{array}} \right){e^{4t}} = {X_p} = \left( {\begin{array}{*{20}{r}}{ - \frac{7}{2}}\\{ - \frac{3}{2}}\\{ - 2}\end{array}} \right){e^{4t}}\)

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

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

\( = {c_1}\left( {\begin{array}{*{20}{l}}1\\0\\0\end{array}} \right){e^t} + {c_2}\left( {\begin{array}{*{20}{l}}1\\1\\0\end{array}} \right){e^{2t}} + {c_3}\left( {\begin{array}{*{20}{l}}1\\2\\2\end{array}} \right){e^{5t}} + \left( {\begin{array}{*{20}{r}}{ - \frac{7}{2}}\\{ - \frac{3}{2}}\\{ - 2}\end{array}} \right){e^{4t}}\)