# Construction of the General Solution of a System of Equations Using the Method of Undetermined Coefficients

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the general solution of the linear system of equations

$\frac{{dx}}{{dt}} = x - y,\; \frac{{dy}}{{dt}} = x + 3y.$

### Example 2

Find the general solution of the system

$\frac{{dx}}{{dt}} = - 2x - 3y - 5z,\;\ \frac{{dy}}{{dt}} = x + 4y + z,\; \frac{{dz}}{{dt}} = 2x + 5z.$

### Example 3

Find the general solution of the system of differential equations

$\frac{{dx}}{{dt}} = - 6x + 5y,\; \frac{{dy}}{{dt}} = - 2x - y + 5z,\; \frac{{dz}}{{dt}} = x - 3y + 4z.$

### 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.