Construction of the General Solution of a System of Equations Using the Method of Undetermined Coefficients
Solved Problems
Example 1.
Find the general solution of the linear system of equations
\[\frac{{dx}}{{dt}} = x - y,\; \frac{{dy}}{{dt}} = x + 3y.\]
Solution.
We calculate the eigenvalues \({\lambda _i}\) of the matrix \(A\) composed of the coefficients of the equations:
\[
\det \left( {A - \lambda I} \right) = \left| {\begin{array}{*{20}{c}}
{1 - \lambda }&{ - 1}\\
1&{3 - \lambda }
\end{array}} \right| = 0,\;\;\Rightarrow
\left( {1 - \lambda } \right)\left( {3 - \lambda } \right) + 1 = 0,\;\; \Rightarrow
3 - 3\lambda - \lambda + {\lambda ^2} + 1 = 0,\;\; \Rightarrow
{\lambda ^2} - 4\lambda + 4 = 0,\;\; \Rightarrow
{\left( {\lambda - 2} \right)^2} = 0.\]
Therefore, the matrix \(A\) has one eigenvalue \({\lambda _1} = 2\) of multiplicity \({k_1} = 2.\) Find the rank of the matrix \(A - {\lambda _1}I.\) Substituting the value \({\lambda _1} = 2\) into the matrix and performing elementary transformations, we obtain:
\[
\left[ {\begin{array}{*{20}{c}}
{1 - 2}&{ - 1}\\
1&{3 - 2}
\end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}}
{ - 1}&{ - 1}\\
1&1
\end{array}} \right]
\sim \left. {\left[ {\begin{array}{*{20}{c}}
1&1\\
{ - 1}&{ - 1}
\end{array}} \right]} \right|\left. {\begin{array}{*{20}{c}}
{}\\
\small{{R_2} + {R_1}}\normalsize
\end{array}} \right.
\sim \left[ {\begin{array}{*{20}{c}}
1&1\\
0&0
\end{array}} \right]
\sim \left( {\begin{array}{*{20}{c}}
1&1
\end{array}} \right).
\]
Thus, the rank of the matrix \(A - {\lambda _1}I\) is \(1.\) Then we get the geometric multiplicity \({s_1} = 1\) for the number \({\lambda _1} = 2,\) i.e., we have one eigenvector:
\[{s_1} = n - \text{rank}\left( {A - {\lambda _1}I} \right) = 2 - 1 = 1.\]
The general vector solution will be given by
\[
\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
x\\
y
\end{array}} \right] + {\mathbf{P}_{{k_i} - {s_i}}}\left( t \right){e^{{\lambda _i}t}}
= {\mathbf{P}_1}\left( t \right){e^{{\lambda _i}t}}
= \left( {{\mathbf{A}_0} + {\mathbf{A}_1}t} \right){e^{2t}}.
\]
Further, we use the method of undetermined coefficients. Let
\[x = \left( {{a_0} + {a_1}t} \right){e^{2t}},\;\; y = \left( {{b_0} + {b_1}t} \right){e^{2t}}.\]
The derivatives are equal
\[\frac{{dx}}{{dt}} = {a_1}{e^{2t}} + 2\left( {{a_0} + {a_1}t} \right){e^{2t}} = \left( {2{a_0} + {a_1} + 2{a_1}t} \right){e^{2t}},\]
\[\frac{{dy}}{{dt}} = {b_1}{e^{2t}} + 2\left( {{b_0} + {b_1}t} \right){e^{2t}} = \left( {2{b_0} + {b_1} + 2{b_1}t} \right){e^{2t}}.\]
Substitute the functions \(x, y\) and their derivatives in the original system of differential equations:
\[\left\{ \begin{array}{l}
\left( {2{a_0} + {a_1} + 2{a_1}t} \right){e^{2t}} = \left( {{a_0} + {a_1}t} \right){e^{2t}} - \left( {{b_0} + {b_1}t} \right){e^{2t}} \\
\left( {2{b_0} + {b_1} + 2{b_1}t} \right){e^{2t}} = \left( {{a_0} + {a_1}t} \right){e^{2t}} + 3\left( {{b_0} + {b_1}t} \right){e^{2t}}
\end{array} \right..\]
Dividing by \({e^{2t}}\) and equating the coefficients of like terms in the left and right sides, we obtain a system of algebraic equations for the coefficients \({a_0},{a_1},{b_0},{b_1}:\)
\[
\left\{ \begin{array}{l}
2{a_0} + {a_1} = {a_0} - {b_0}\\
2{a_1} = {a_1} - {b_1}\\
2{b_0} + {b_1} = {a_0} + 3{b_0}\\
2{b_1} = {a_1} + 3{b_1}
\end{array} \right.,\;\; \Rightarrow
\left\{ {\begin{array}{*{20}{l}}
{{a_0} + {a_1} + {b_0} = 0}\\
{{a_1} + {b_1} = 0}\\
{{a_0} + {b_0} - {b_1} = 0}\\
{{a_1} + {b_1} = 0}
\end{array}} \right..\]
In this system, there are only two independent equations. We choose as independent coefficients \({a_0} = {C_1}\) and \({a_1} = {C_2}.\) The remaining two numbers \({b_0}\) and \({b_1}\) are expressed in terms of \({C_1}\) and \({C_2}:\)
\[{C_1} + {C_2} + {b_0} = 0,\;\; \Rightarrow {b_0} = - {C_1} - {C_2},\]
\[{C_2} + {b_1} = 0,\;\; \Rightarrow {b_1} = - {C_2}.\]
Thus, the general solution has the form
\[
\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
x\\
y
\end{array}} \right] = {e^{2t}}\left[ {\begin{array}{*{20}{c}}
{{a_0} + {a_1}t}\\
{{b_0} + {b_1}t}
\end{array}} \right]
= {e^{2t}}\left[ {\begin{array}{*{20}{c}}
{{C_1} + {C_2}t}\\
{ - {C_1} - {C_2} - {C_2}t}
\end{array}} \right].\]
It can be conveniently rewritten in vector form:
\[\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right] = {C_1}{e^{2t}}\left[ {\begin{array}{*{20}{c}} 1\\ { - 1} \end{array}} \right] + {C_2}{e^{2t}}\left[ {\begin{array}{*{20}{c}} t\\ { - 1 - t} \end{array}} \right] = {C_1}{e^{2t}}\left[ {\begin{array}{*{20}{c}} 1\\ { - 1} \end{array}} \right] + {C_2}{e^{2t}}\left( {\left[ {\begin{array}{*{20}{c}} 0\\ { - 1} \end{array}} \right] + t\left[ {\begin{array}{*{20}{c}} 1\\ { - 1} \end{array}} \right]} \right).\]
Example 2.
Find the general solution of the system
\[
\frac{{dx}}{{dt}} = - 2x - 3y - 5z,\;\
\frac{{dy}}{{dt}} = x + 4y + z,\;
\frac{{dz}}{{dt}} = 2x + 5z.\]
Solution.
First, we determine the eigenvalues of the system by solving the corresponding characteristic equation:
\[\det \left( {A - \lambda I} \right) = \left| {\begin{array}{*{20}{c}}
{ - 2 - \lambda }&{ - 3}&{ - 5}\\
1&{4 - \lambda }&1\\
2&0&{5 - \lambda }
\end{array}} \right| = 0.\]
Expand the determinant along the third line:
\[
2\left| {\begin{array}{*{20}{c}}
{ - 3}&{ - 5}\\
{4 - \lambda }&1
\end{array}} \right| + \left( {5 - \lambda } \right)\left| {\begin{array}{*{20}{c}}
{ - 2 - \lambda }&{ - 3}\\
1&{4 - \lambda }
\end{array}} \right| = 0,\;\; \Rightarrow
2\left[ { - 3 + 5\left( {4 - \lambda } \right)} \right] + \left( {5 - \lambda } \right) \left[ {\left( { - 2 - \lambda } \right)\left( {4 - \lambda } \right) + 3} \right] = 0,\;\; \Rightarrow
2\left( { - 5\lambda + 17} \right) + \left( {5 - \lambda } \right)\left( {{\lambda ^2} - 2\lambda - 5} \right) = 0,\;\; \Rightarrow
- \color{red}{10\lambda} + \color{green}{34} + \color{blue}{5{\lambda ^2}} - \color{red}{10\lambda} - \color{green}{25} - \color{black}{\lambda ^3} + \color{blue}{2{\lambda ^2}} + \color{red}{5\lambda} = \color{black}0,\;\; \Rightarrow
{\lambda ^3} - \color{blue}{7{\lambda ^2}} + \color{red}{15\lambda} - \color{green}9 = \color{black}0.\]
Note that one of the roots of the cubic equation is the number \(\lambda = 1.\) Taking out the common factor \(\left( {\lambda - 1} \right),\) we obtain:
\[
{\lambda ^3} - {\lambda ^2} - 6{\lambda ^2} + 6\lambda + 9\lambda - 9 = 0,\;\; \Rightarrow
{\lambda ^2}\left( {\lambda - 1} \right) - 6\lambda \left( {\lambda - 1} \right) + 9\left( {\lambda - 1} \right) = 0,\;\; \Rightarrow
\left( {\lambda - 1} \right) \left( {{\lambda ^2} - 6\lambda + 9} \right) = 0,\;\; \Rightarrow
\left( {\lambda - 1} \right){\left( {\lambda - 3} \right)^2} = 0.\]
Thus, the matrix of the system of equations has two eigenvalues: \({\lambda _1} = 1\) of multiplicity \(1\) and \({\lambda _2} = 3\) of multiplicity \(2.\)
Consider the first root \({\lambda _1} = 1\) and find the part of the general solution \({\mathbf{X}_1}\) associated with this eigenvalue. The system of equations for computing the coordinates of the eigenvector \({\mathbf{V}_1}\) is given by
\[
\left[ {\begin{array}{*{20}{c}}
{ - 2 - 1}&{ - 3}&{ - 5}\\
1&{4 - 1}&1\\
2&0&{5 - 1}
\end{array}} \right] \left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}\\
{{V_{31}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
\left[ {\begin{array}{*{20}{c}}
{ - 3}&{ - 3}&{ - 5}\\
1&3&1\\
2&0&4
\end{array}} \right] \left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}\\
{{V_{31}}}
\end{array}} \right] = \mathbf{0}.
\]
Simplify the system:
\[
\left\{ \begin{array}{l}
- 3{V_{11}} - 3{V_{21}} - 5{V_{31}} = 0\\
{V_{11}} + 3{V_{21}} + {V_{31}} = 0\\
2{V_{11}} + 4{V_{31}} = 0
\end{array} \right.,\;\; \Rightarrow
{\left\{ \begin{array}{l}
{V_{11}} + 3{V_{21}} + {V_{31}} = 0\\
- 3{V_{11}} - 3{V_{21}} - 5{V_{31}} = 0\\
2{V_{11}} + 4{V_{31}} = 0
\end{array} \right.},\;\; \Rightarrow
\left\{ \begin{array}{l}
{V_{11}} + 3{V_{21}} + {V_{31}} = 0\\
0 + 6{V_{21}} - 2{V_{31}} = 0\\
0 - 6{V_{21}} + 2{V_{31}} = 0
\end{array} \right.,\;\; \Rightarrow
\left\{ {\begin{array}{*{20}{l}}
{{V_{11}} + 3{V_{21}} + {V_{31}} = 0}\\
{3{V_{21}} - {V_{31}} = 0}
\end{array}} \right..
\]
We choose as an independent variable \({V_{31}} = t.\) The remaining coordinates are expressed in terms of \(t\) as follows:
\[3{V_{21}} = {V_{31}} = t,\;\; \Rightarrow {V_{21}} = \frac{t}{3},\;\; \Rightarrow {V_{11}} = - {V_{31}} - 3{V_{21}} = - t - 3 \cdot \frac{t}{3} = - 2t.\]
Hence, the eigenvector \({\mathbf{V}_1}\) is
\[
{\mathbf{V}_1} = \left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}\\
{{V_{31}}}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{ - 2t}\\
{\frac{t}{3}}\\
t
\end{array}} \right]
\sim t\left[ {\begin{array}{*{20}{c}}
{ - 6}\\
1\\
3
\end{array}} \right]
\sim \left[ {\begin{array}{*{20}{c}}
{ - 6}\\
1\\
3
\end{array}} \right].
\]
Thus, the eigenvalue \({\lambda _1} = 1\) makes the following contribution to the general solution:
\[
{\mathbf{X}_1}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right]
= {C_1}{e^{{\lambda _1}t}}{\mathbf{V}_1}
= {C_1}{e^{{\lambda _1}t}}\left[ {\begin{array}{*{20}{c}}
{ - 6}\\
1\\
3
\end{array}} \right].\]
Now we consider the eigenvalue \({\lambda _2} = 3\) with algebraic multiplicity \({k_2} = 2.\) Find out the rank of the matrix at \({\lambda _2} = 3:\)
\[
\left[ {\begin{array}{*{20}{c}}
{ - 2 - 3}&{ - 3}&{ - 5}\\
1&{4 - 3}&1\\
2&0&{5 - 3}
\end{array}} \right]
\sim \left[ {\begin{array}{*{20}{c}}
{ - 5}&{ - 3}&{ - 5}\\
1&1&1\\
2&0&2
\end{array}} \right]
\sim {\left[ {\begin{array}{*{20}{c}}
1&1&1\\
2&0&2\\
{ - 5}&{ - 3}&{ - 5}
\end{array}} \right]}
\sim \left[ {\begin{array}{*{20}{c}}
1&1&1\\
0&{ - 2}&0\\
0&2&0
\end{array}} \right]
\sim \left[ {\begin{array}{*{20}{c}}
1&1&1\\
0&2&0
\end{array}} \right].
\]
As it can be seen, \(\text{rank}\left( {A - {\lambda _2}I} \right) = 2.\) Hence, the number \({\lambda _2} = 3\) is characterized by the geometric multiplicity \({s_2} = 1\) and has one eigenvector:
\[{s_2} = n - \text{rank}\left( {A - {\lambda _2}I} \right) = 3 - 2 = 1.\]
We seek a solution associated with the eigenvalue \({\lambda _2}\) as a function
\[
{\mathbf{X}_2}\left( t \right) = {\mathbf{P}_{{k_2} - {s_2}}}\left( t \right){e^{{\lambda _2}t}}
= \left( {{\mathbf{A}_0} + {\mathbf{A}_1}t} \right){e^{3t}},\]
where the vector polynomial \({\mathbf{P}_{{k_2} - {s_2}}}\left( t \right)\) has degree \({k_2} - {s_2} = 1.\) By setting
\[
{\mathbf{A}_0} = \left[ {\begin{array}{*{20}{c}}
{{a_0}}\\
{{b_0}}\\
{{d_0}}
\end{array}} \right],\;\; {\mathbf{A}_1} = \left[ {\begin{array}{*{20}{c}}
{{a_1}}\\
{{b_1}}\\
{{d_1}}
\end{array}} \right],\]
we write the formulas for each coordinate of \({\mathbf{X}_2}:\)
\[x = \left( {{a_0} + {a_1}t} \right){e^{3t}},\;\; y = \left( {{b_0} + {b_1}t} \right){e^{3t}},\;\; z = \left( {{d_0} + {d_1}t} \right){e^{3t}}.\]
The derivatives of these functions are
\[\frac{{dx}}{{dt}} = {a_1}{e^{3t}} + 3\left( {{a_0} + {a_1}t} \right){e^{3t}},\;\; \frac{{dy}}{{dt}} = {b_1}{e^{3t}} + 3\left( {{b_0} + {b_1}t} \right){e^{3t}},\;\; \frac{{dz}}{{dt}} = {d_1}{e^{3t}} + 3\left( {{d_0} + {d_1}t} \right){e^{3t}}.\]
Substituting these expressions into the original system and dividing by the factor \({e^{3t}},\) we have:
\[\left\{ \begin{array}{l}
{a_1} + 3\left( {{a_0} + {a_1}t} \right) = - 2\left( {{a_0} + {a_1}t} \right) - 3\left( {{b_0} + {b_1}t} \right) - 5\left( {{d_0} + {d_1}t} \right) \\
{b_1} + 3\left( {{b_0} + {b_1}t} \right) = {a_0} + {a_1}t + 4\left( {{b_0} + {b_1}t} \right) + {d_0} + {d_1}t \\
{d_1} + 3\left( {{d_0} + {d_1}t} \right) = 2\left( {{a_0} + {a_1}t} \right) + 5\left( {{d_0} + {d_1}t} \right)
\end{array} \right..\]
Equating the coefficients of like powers of \(t\) on the left and right sides, we obtain a system of six equations with unknowns \({a_0},{a_1},\) \({b_0},{b_1},\) \({d_0},{d_1}:\)
\[
\left\{ \begin{array}{l}
{a_1} + 3{a_0} = - 2{a_0} - 3{b_0} - 5{d_0}\\
3{a_1} = - 2{a_1} - 3{b_1} - 5{d_1}\\
{b_1} + 3{b_0} = {a_0} + 4{b_0} + {d_0}\\
3{b_1} = {a_1} + 4{b_1} + {d_1}\\
{d_1} + 3{d_0} = 2{a_0} + 5{d_0}\\
3{d_1} = 2{a_1} + 5{d_1}
\end{array} \right.,\;\; \Rightarrow
\left\{ \begin{array}{l}
5{a_0} + {a_1} + 3{b_0} + 5{d_0} = 0\\
5{a_1} + 3{b_1} + 5{d_1} = 0\\
{a_0} + {b_0} - {b_1} + {d_0} = 0\\
{a_1} + {b_1} + {d_1} = 0\\
2{a_0} + 2{d_0} - {d_1} = 0\\
{a_1} + {d_1} = 0
\end{array} \right..
\]
In this system of equations, only two coefficients are independent. This follows from the fact that the eigenvalue \({\lambda_2} = 3\) has algebraic multiplicity \(2\) and, therefore, must have two linearly independent solutions. We choose as free variables \({a_0}\) and \({a_1},\) denoting
\[{a_0} = {C_2},\;\;{a_1} = 2{C_3}.\]
where \({C_2},\) \({C_3}\) are arbitrary numbers, and the factor \(2\) is introduced to get rid of fractions. The remaining coefficients are easily expressed in terms of \({C_2}\) and \({C_3}\) and are represented as
\[{a_0} = {_2},\;\;{b_0} = {C_3},\;\; {d_0} = - {C_3} - {C_2},\;\; {a_1} = 2{C_3},\;\; {b_1} = 0,\;\; {d_1} = - 2{C_3}.\]
Then the part of the general solution associated with the eigenvalue \({\lambda_2} = 3\) can be written as
\[\left\{ \begin{array}{l}
x\left( t \right) = \left( {{a_0} + {a_1}t} \right){e^{{\lambda _2}t}} = \left( {{C_2} + 2{C_3}t} \right){e^{3t}} \\
y\left( t \right) = \left( {{b_0} + {b_1}t} \right){e^{{\lambda _2}t}} = {C_3}{e^{3t}} \\
z\left( t \right) = \left( {{d_0} + {d_1}t} \right){e^{{\lambda _2}t}} = \left( { - {C_3} - {C_2} - 2{C_3}t} \right){e^{3t}}
\end{array} \right..\]
Rewrite the solution in the vector form:
\[{\mathbf{X}_2}\left( t \right) = \left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = {e^{3t}}\left[ {\begin{array}{*{20}{c}} {{C_2} + 2{C_3}t}\\ {{C_3}}\\ { - {C_3} - {C_2} - 2{C_3}t} \end{array}} \right] = {C_2}{e^{3t}}\left[ {\begin{array}{*{20}{c}} 1\\ 0\\ { - 1} \end{array}} \right] + {C_3}{e^{3t}}\left[ {\begin{array}{*{20}{c}} {2t}\\ 1\\ { - 1 - 2t} \end{array}} \right] = {C_2}{e^{3t}}\left[ {\begin{array}{*{20}{c}} 1\\ 0\\ { - 1} \end{array}} \right] + {C_3}{e^{3t}}\left( {\left[ {\begin{array}{*{20}{c}} 0\\ 1\\ { - 1} \end{array}} \right] + t\left[ {\begin{array}{*{20}{c}} 2\\ 0\\ { - 2} \end{array}} \right]} \right). \]
By bringing together all the components, we obtain the general solution of the original system in the form:
\[\mathbf{X}\left( t \right) = {\mathbf{X}_1}\left( t \right) + {\mathbf{X}_2}\left( t \right) = {C_1}{e^t}\left[ {\begin{array}{*{20}{c}} { - 6}\\ 1\\ 3 \end{array}} \right] + {C_2}{e^{3t}}\left[ {\begin{array}{*{20}{c}} 1\\ 0\\ { - 1} \end{array}} \right] + {C_3}{e^{3t}}\left( {\left[ {\begin{array}{*{20}{c}} 0\\ 1\\ { - 1} \end{array}} \right] + t\left[ {\begin{array}{*{20}{c}} 2\\ 0\\ { - 2} \end{array}} \right]} \right). \]
Example 3.
Find the general solution of the system of differential equations
\[
\frac{{dx}}{{dt}} = - 6x + 5y,\;
\frac{{dy}}{{dt}} = - 2x - y + 5z,\;
\frac{{dz}}{{dt}} = x - 3y + 4z.\]
Solution.
We begin by calculating the eigenvalues of the system. Solve the auxiliary equation:
\[\det \left( {A - \lambda I} \right) = \left| {\begin{array}{*{20}{c}}
{ - 6 - \lambda }&5&0\\
{ - 2}&{ - 1 - \lambda }&5\\
1&{ - 3}&{4 - \lambda }
\end{array}} \right| = 0.\]
Expand the determinant along the first row:
\[
\left( { - 6 - \lambda } \right)\left| {\begin{array}{*{20}{c}}
{ - 1 - \lambda }&5\\
{ - 3}&{4 - \lambda }
\end{array}} \right| - 5\left| {\begin{array}{*{20}{c}}
{ - 2}&5\\
1&{4 - \lambda }
\end{array}} \right| = 0,\;\; \Rightarrow
\left( { - 6 - \lambda } \right) \left[ {\left( { - 1 - \lambda } \right)\left( {4 - \lambda } \right) + 15} \right]
- 5\left[ { - 2\left( {4 - \lambda } \right) - 5} \right] = 0,\;\; \Rightarrow
\left( {\lambda + 6} \right)\left( {{\lambda ^2} - 3\lambda + 11} \right) + 5\left( {2\lambda - 13} \right) = 0,\;\; \Rightarrow
{\lambda ^3} + \color{blue}{6{\lambda ^2}} - \color{blue}{3{\lambda ^2}} - \color{red}{18\lambda} + \color{red}{11\lambda} + \color{green}{66} + \color{red}{10\lambda} - \color{green}{65} = \color{black}0,\;\; \Rightarrow
{\lambda ^3} + \color{blue}{3{\lambda ^2}} + \color{red}{3\lambda} + \color{green}1 = \color{black}0,\;\; \Rightarrow
{\left( {\lambda + 1} \right)^3} = 0.
\]
Thus, the matrix has one eigenvalue \({\lambda _1} = - 1\) with algebraic multiplicity \({k_1} = 3.\) We find the rank of the matrix at \({\lambda _1} = - 1\) and geometric multiplicity \({s_1}:\)
\[
\left[ {\begin{array}{*{20}{c}}
{ - 6 + 1}&5&0\\
{ - 2}&{ - 1 + 1}&5\\
1&{ - 3}&{4 + 1}
\end{array}} \right]
\sim \left[ {\begin{array}{*{20}{c}}
{ - 5}&5&0\\
{ - 2}&0&5\\
1&{ - 3}&5
\end{array}} \right]
\sim {\left[ {\begin{array}{*{20}{c}}
{ - 5}&5&0\\
{ - 2}&0&5\\
1&{ - 3}&5
\end{array}} \right]}
\sim \left[ {\begin{array}{*{20}{c}}
1&{ - 3}&5\\
0&{ - 2}&5\\
0&{ - 6}&{15}
\end{array}} \right]
\sim \left[ {\begin{array}{*{20}{c}}
1&{ - 3}&5\\
0&{ - 2}&5
\end{array}} \right].
\]
Hence, \(\text{rank}\left( {A - {\lambda _1}I} \right) = 2.\) Accordingly, the geometric multiplicity (and the number of eigenvectors) for the eigenvalue \({\lambda _1} = - 1\) is
\[{s_1} = n - \text{rank}\left( {A - {\lambda _1}I} \right) = 3 - 2 = 1.\]
With this in mind, we will seek the general solution \(\mathbf{X}\) in the form of a vector function
\[\mathbf{X}\left( t \right) = {\mathbf{P}_{{k_1} - {s_1}}}\left( t \right){e^{{\lambda _1}t}} = \left( {{\mathbf{A}_0} + {\mathbf{A}_1}t + {\mathbf{A}_2}{t^2}} \right){e^{ - t}}.\]
Let the vectors \({\mathbf{A}_0},{\mathbf{A}_1},{\mathbf{A}_2}\) have the coordinates
\[
{\mathbf{A}_0} = \left[ {\begin{array}{*{20}{c}}
{{a_0}}\\
{{b_0}}\\
{{d_0}}
\end{array}} \right],\;\;
{\mathbf{A}_1} = \left[ {\begin{array}{*{20}{c}}
{{a_1}}\\
{{b_1}}\\
{{d_1}}
\end{array}} \right],\;\;
{\mathbf{A}_2} = \left[ {\begin{array}{*{20}{c}}
{{a_2}}\\
{{b_2}}\\
{{d_2}}
\end{array}} \right].
\]
We write the coordinate functions and find their derivatives:
\[x\left( t \right) = \left( {{a_0} + {a_1}t + {a_2}{t^2}} \right){e^{ - t}},\]
\[y\left( t \right) = \left( {{b_0} + {b_1}t + {b_2}{t^2}} \right){e^{ - t}},\]
\[z\left( t \right) = \left( {{d_0} + {d_1}t + {d_2}{t^2}} \right){e^{ - t}},\]
\[
\frac{{dx}}{{dt}} = \left( {{a_1} + 2{a_2}t} \right){e^{ - t}}
- \left( {{a_0} + {a_1}t + {a_2}{t^2}} \right){e^{ - t}},
\]
\[
\frac{{dy}}{{dt}} = \left( {{b_1} + 2{b_2}t} \right){e^{ - t}}
- \left( {{b_0} + {b_1}t + {b_2}{t^2}} \right){e^{ - t}},
\]
\[
\frac{{dz}}{{dt}} = \left( {{d_1} + 2{d_2}t} \right){e^{ - t}}
- \left( {{d_0} + {d_1}t + {d_2}{t^2}} \right){e^{ - t}}.
\]
Substituting into the original system and dividing both sides of each equation by the exponential function \({e^{ - t}},\) we obtain:
\[
{a_1} + 2{a_2}t - {a_0} - {a_1}t - {a_2}{t^2}
= - 6\left( {{a_0} + {a_1}t + {a_2}{t^2}} \right)
+ 5\left( {{b_0} + {b_1}t + {b_2}{t^2}} \right),
\]
\[
{b_1} + 2{b_2}t - {b_0} - {b_1}t - {b_2}{t^2}
= - 2\left( {{a_0} + {a_1}t + {a_2}{t^2}} \right)
- \left( {{b_0} + {b_1}t + {b_2}{t^2}} \right)
+ 5\left( {{d_0} + {d_1}t + {d_2}{t^2}} \right),
\]
\[
{d_1} + 2{d_2}t - {d_0} - {d_1}t - {d_2}{t^2}
= {a_0} + {a_1}t + {a_2}{t^2}
- 3\left( {{b_0} + {b_1}t + {b_2}{t^2}} \right)
+ 4\left( {{d_0} + {d_1}t + {d_2}{t^2}} \right).
\]
Equating terms of equal powers of \(t\) on the left and the right, we get a system of nine equations:
\[\left\{ \begin{array}{l}
{a_1} - {a_0} = - 6{a_0} + 5{b_0}\\
2{a_2} - {a_1} = - 6{a_1} + 5{b_1}\\
- {a_2} = - 6{a_2} + 5{b_2}\\
{b_1} - {b_0} = - 2{a_0} - {b_0} + 5{d_0}\\
2{b_2} - {b_1} = - 2{a_1} - {b_1} + 5{d_1}\\
- {b_2} = - 2{a_2} - {b_2} + 5{d_2}\\
{d_1} - {d_0} = {a_0} - 3{b_0} + 4{d_0}\\
2{d_2} - {d_1} = {a_1} - 3{b_1} + 4{d_1}\\
- {d_2} = {a_2} - 3{b_2} + 4{d_2}
\end{array} \right.,\;\; \Rightarrow
\left\{ \begin{array}{l}
5{a_0} + {a_1} - 5{b_0} = 0\\
5{a_1} + 2{a_2} - 5{b_1} = 0\\
{a_2} - {b_2} = 0\\
2{a_0} + {b_1} - 5{d_0} = 0\\
2{a_1} + 2{b_2} - 5{d_1} = 0\\
2{a_2} - 5{d_2} = 0\\
{a_0} - 3{b_0} + 5{d_0} - {d_1} = 0\\
{a_1} - 3{b_1} + 5{d_1} - 2{d_2} = 0\\
{a_2} - 3{b_2} + 5{d_2} = 0
\end{array} \right..
\]
This system contains only three independent variables. This follows from the fact that the general solution \(\mathbf{X}\) must contain three linearly independent functions. We choose as independent variables
\[{a_0} = {C_1},\;\; {a_1} = {C_2},\;\; {a_2} = {C_3}.\]
The other variables are expressed in terms of \({C_1},\) \({C_2},\) \({C_3}:\)
\[5{b_0} = 5{a_0} + {a_1} = 5{C_1} + {C_2},\;\; {b_0} = {C_1} + \frac{1}{5}{C_2};\]
\[5{b_1} = 5{a_1} + 2{a_2} = 5{C_2} + 2{C_3},\;\; {b_1} = {C_2} + \frac{2}{5}{C_3};\]
\[{b_2} = {a_2} = {C_3};\]
\[5{d_0} = 2{a_0} + {b_1} = 2{C_1} + {C_2} + \frac{2}{5}{C_3},\;\; {d_0} = \frac{2}{5}{C_1} + \frac{1}{5}{C_2} + \frac{2}{{25}}{C_3};\]
\[5{d_1} = 2{a_1} + 2{b_2} = 2{C_2} + 2{C_3},\;\; \Rightarrow {d_1} = \frac{2}{5}{C_2} + \frac{2}{5}{C_3};\]
\[5{d_2} = 2{a_2} = 2{C_3},\;\; \Rightarrow {d_2} = \frac{2}{5}{C_3}.\]
Thus, the general solution can be written as
\[x\left( t \right) = \left( {{a_0} + {a_1}t + {a_2}{t^2}} \right){e^{ - t}} = \left( {{C_1} + {C_2}t + {C_3}{t^2}} \right){e^{ - t}},\]
\[y\left( t \right) = \left( {{b_0} + {b_1}t + {b_2}{t^2}} \right){e^{ - t}} = \left\{ {{C_1} + {\frac{1}{5}{C_2}} + \left( {{C_2} + \frac{2}{5}{C_3}} \right)t + {C_3}{t^2}} \right\} {e^{ - t}},\]
\[z\left( t \right) = \left( {{d_0} + {d_1}t + {d_2}{t^2}} \right){e^{ - t}} = \left\{ {\frac{2}{5}{C_1} + \frac{1}{5}{C_2} + \frac{2}{{25}}{C_3} + \left( {\frac{2}{5}{C_2} + \frac{2}{5}{C_3}} \right) t + \frac{2}{5}{C_3}{t^2}} \right\} {e^{ - t}}.\]
We represent the solution in vector form, identifying explicitly linearly independent vectors:
\[\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right] = {C_1}{e^{ - t}}\left[ {\begin{array}{*{20}{c}}
1\\
1\\
{\frac{2}{5}}
\end{array}} \right] + {C_2}{e^{ - t}}\left[ {\begin{array}{*{20}{c}}
t\\
{\frac{1}{5} + t}\\
{\frac{1}{5} + \frac{2}{5}t}
\end{array}} \right] + {C_3}{e^{ - t}}\left[ {\begin{array}{*{20}{c}}
{{t^2}}\\
{\frac{2}{5}t + {t^2}}\\
{\frac{2}{{25}} + \frac{2}{5}t + \frac{2}{5}{t^2}}
\end{array}} \right].\]
Renormalize the numbers \({C_1},\) \({C_2},\) \({C_3}\) to get rid of the fractional coordinates:
\[{C_1} \to 5{C_1},\;\; {C_2} \to 5{C_2},\;\; {C_3} \to 25{C_3}.\]
Then the answer is written as
\[\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = {C_1}{e^{ - t}}\left[ {\begin{array}{*{20}{c}} 5\\ 5\\ 2 \end{array}} \right] + {C_2}{e^{ - t}}\left[ {\begin{array}{*{20}{c}} {5t}\\ {1 + 5t}\\ {1 + 2t} \end{array}} \right] + {C_3}{e^{ - t}}\left[ {\begin{array}{*{20}{c}} {25{t^2}}\\ {10t + 25{t^2}}\\ {2 + 10t + 10{t^2}} \end{array}} \right] = {C_1}{e^{ - t}}\left[ {\begin{array}{*{20}{c}} 5\\ 5\\ 2 \end{array}} \right] + {C_2}{e^{ - t}}\left( {\left[ {\begin{array}{*{20}{c}} 0\\ 1\\ 1 \end{array}} \right] + t\left[ {\begin{array}{*{20}{c}} 5\\ 5\\ 2 \end{array}} \right]} \right) + {C_3}{e^{ - t}}\left( {\left[ {\begin{array}{*{20}{c}} 0\\ 0\\ 2 \end{array}} \right] + t\left[ {\begin{array}{*{20}{c}} 0\\ {10}\\ {10} \end{array}} \right] + {t^2}\left[ {\begin{array}{*{20}{c}} {25}\\ {25}\\ {10} \end{array}} \right]} \right).\]
Note that the general solution contains \(3\) linearly independent vectors:
\[
{\mathbf{V}_1} = \left[ {\begin{array}{*{20}{c}}
5\\
5\\
2
\end{array}} \right],\;\;
{\mathbf{V}_2} = \left[ {\begin{array}{*{20}{c}}
0\\
0\\
1
\end{array}} \right],\;\;
{\mathbf{V}_3} = \left[ {\begin{array}{*{20}{c}}
0\\
1\\
1
\end{array}} \right].\]
The other vectors are collinear to the specified ones. Among these three vectors, \({\mathbf{V}_1}\) is an ordinary eigenvector, and the vectors \({\mathbf{V}_2},\) \({\mathbf{V}_3}\) are called generalized eigenvectors. The form of the general solution is determined by the structure of the so-called Jordan matrix for the system. This technique is considered in more detail on the page Construction of the General Solution of a System of Differential Equations Using the Jordan Form.
