Question

In Problems 9 and 10, solve the given initial-value problem.

\(9.{X^\prime } = \left( {\begin{array}{*{20}{r}}{ - 1}&{ - 2}\\3&4\end{array}} \right)X + \left( {\begin{array}{*{20}{l}}3\\3\end{array}} \right),\;\;\;X(0) = \left( {\begin{array}{*{20}{r}}{ - 4}\\5\end{array}} \right)\)

Short answer

The initial value \({X^\prime } = \left( {\begin{array}{*{20}{r}}{ - 1}&{ - 2}\\3&4\end{array}} \right)X + \left( {\begin{array}{*{20}{l}}3\\3\end{array}} \right),\;\;\;X(0) = \left( {\begin{array}{*{20}{r}}{ - 4}\\5\end{array}} \right)\) is \(X(t) = 4\left( {\begin{array}{*{20}{r}}{ - 2}\\3\end{array}} \right){e^{2t}} + 13\left( {\begin{array}{*{20}{r}}1\\{ - 1}\end{array}} \right){e^t} + \left( {\begin{array}{*{20}{r}}{ - 9}\\6\end{array}} \right)\)

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

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

where

\(A = \left( {\begin{array}{*{20}{r}}{ - 1}&{ - 2}\\3&4\end{array}} \right)\)

Now,

\(det(A - \lambda I) = 0\;\;\; \to \left| {\begin{array}{*{20}{c}}{ - 1 - \lambda }&{ - 2}\\3&{4 - \lambda }\end{array}} \right| = 0\)

\(( - 1 - \lambda )(4 - \lambda ) + 6 = 0\)

\( - 4 + \lambda - 4\lambda + {\lambda ^2} + 6 = 0\)

\(\begin{array}{*{20}{r}}{{\lambda ^2} - 3\lambda + 2 = 0}\\{(\lambda - 2)(\lambda - 1) = 0}\end{array}\)

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

Find the eigenvector and corresponding solution vector:

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

\((A - 2I\mid 0) = \left( {\begin{array}{*{20}{c}}{ - 1 - 2}&{ - 2}&0\\3&{4 - 2}&0\end{array}} \right)\)

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

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

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

so here we have,

\( - 3{k_1} - 2{k_2} = 0\;\;\; \to \;\;\;{k_1} = - \frac{2}{3}{k_2}\)

choosing\({k_2} = 3\)yields\({k_1} = - 2\).

This gives an eigenvector and a corresponding solution vector:

\({K_1} = \left( {\begin{array}{*{20}{r}}{ - 2}\\3\end{array}} \right),\;\;\;{X_1} = \left( {\begin{array}{*{20}{r}}{ - 2}\\3\end{array}} \right){e^{2t}}\)

For\({\lambda _2} = 1:\)

\((A - I\mid 0) = \left( {\begin{array}{*{20}{c}}{ - 1 - 1}&{ - 2}&0\\3&{4 - 1}&0\end{array}} \right)\)

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

Apply row operation\(\frac{2}{3}{R_2} + {R_1} \to {R_2}\):

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

so here we have,

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

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

This gives an eigenvector and a corresponding solution vector:

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

therefore,

\({X_c} = {c_1}\left( {\begin{array}{*{20}{r}}{ - 2}\\3\end{array}} \right){e^{2t}} + {c_2}\left( {\begin{array}{*{20}{r}}1\\{ - 1}\end{array}} \right){e^t}\)

Find the particular solution of the system:

Since \(F(t) = \left( {\begin{array}{*{20}{l}}3\\3\end{array}} \right)\) we shall try to find a particular solution to the system that possesses the same form:

\({X_p} = \left( {\begin{array}{*{20}{l}}{{a_1}}\\{{b_1}}\end{array}} \right)\)

differentiating,

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

Substituting into the given system yields

\(\left( {\begin{array}{*{20}{l}}0\\0\end{array}} \right) = \left( {\begin{array}{*{20}{r}}{ - 1}&{ - 2}\\3&4\end{array}} \right)\left( {\begin{array}{*{20}{l}}{{a_1}}\\{{b_1}}\end{array}} \right) + \left( {\begin{array}{*{20}{l}}3\\3\end{array}} \right)\)

\(\left( {\begin{array}{*{20}{l}}0\\0\end{array}} \right) = \left( {\begin{array}{*{20}{r}}{ - {a_1} - 2{b_1}}\\{3{a_1} + 4{b_1}}\end{array}} \right) + \left( {\begin{array}{*{20}{l}}3\\3\end{array}} \right)\)

Simplify and substitute the values:

where,

\( - {a_1} - 2{b_1} + 3 = 0\) \(\left( 1 \right)\)

\(3{a_1} + 4{b_1} + 3 = 0\) \(\left( 2 \right)\)

multiply \(\;\left( 1 \right)\) by\(3\) , the equation will become

\( - 3{a_1} - 6{b_1} + 9 = 0\) \(\left( 3 \right)\)

adding \({\rm{\backslash begin}}aligned4{b_1} - 6{b_1} + 3 + 9 = 0 \to 2{b_1} = 12\) to \((2)\),

\(4{b_1} - 6{b_1} + 3 + 9 = 0 \to 2{b_1} = 12\)

\({b_1} = 6\)

substituting this value \({b_1}\) that we obtained into any of the equations above,

\( - {a_1} - 2(6) + 3 = 0\;\;\; \to \;\;\; - {a_1} - 9 = 0\)

\({a_1} = - 9\)

the particular solution is then

\({X_{\bf{p}}} = \left( {\begin{array}{*{20}{l}}{{a_1}}\\{{b_1}}\end{array}} \right) = \left( {\begin{array}{*{20}{r}}{ - 9}\\6\end{array}} \right)\)

and the general solution is

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

\( = {c_1}\left( {\begin{array}{*{20}{r}}{ - 2}\\3\end{array}} \right){e^{2t}} + {c_2}\left( {\begin{array}{*{20}{r}}1\\{ - 1}\end{array}} \right){e^t} + \left( {\begin{array}{*{20}{r}}{ - 9}\\6\end{array}} \right)\)

Find the initial value for the Nonhomogeneous Linear system:

applying the initial condition

\(X(0) = \left( {\begin{array}{*{20}{r}}{ - 4}\\5\end{array}} \right)\)

\(\left( {\begin{array}{*{20}{r}}{ - 4}\\5\end{array}} \right) = {c_1}\left( {\begin{array}{*{20}{r}}{ - 2}\\3\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{r}}1\\{ - 1}\end{array}} \right) + \left( {\begin{array}{*{20}{r}}{ - 9}\\6\end{array}} \right)\)

where,

\( - 4 = - 2{c_1} + {c_2} - 9\;\;\; \to \;\;\;{c_2} = 5 + 2{c_1}\) \((4)\)

we can substitute it into the second equation,

\(5 = 3{c_1} - {c_2} + 6 \to 5 = 3{c_1} - \left( {5 + 2{c_1}} \right) + 6\)

\( - 1 = 3{c_1} - 5 - 2{c_1}\)

\({c_1} = 4\)

and,

\({c_2} = 5 + 2(4) = 13\)

Therefore,

\(X(t) = 4\left( {\begin{array}{*{20}{r}}{ - 2}\\3\end{array}} \right){e^{2t}} + 13\left( {\begin{array}{*{20}{r}}1\\{ - 1}\end{array}} \right){e^t} + \left( {\begin{array}{*{20}{r}}{ - 9}\\6\end{array}} \right)\)