Linear Nonhomogeneous Systems of Differential Equations with Constant Coefficients
Solved Problems
Example 1.
Solve the system of equations by elimination:
\[x' = x + 2y + {e^{ - 2t}},\; y' = 4x - y.\]
Solution.
We differentiate the first equation and substitute into it the derivative \(y'\) from the second equation:
\[x^{\prime\prime} = x' + 2y' - 2{e^{ - 2t}},\;\; \Rightarrow x^{\prime\prime} = x' + 2\left( {4x - y} \right) - 2{e^{ - 2t}},\;\; \Rightarrow x^{\prime\prime} = x' + 8x - 2y - 2{e^{ - 2t}}.\]
We can solve the first equation for \(2y\) and substitute in the last equation:
\[2y = x' - x - {e^{ - 2t}},\;\; \Rightarrow x^{\prime\prime} = x' + 8x - \left( {x' - x - {e^{ - 2t}}} \right) - 2{e^{ - 2t}},\;\; \Rightarrow x^{\prime\prime} = x' + 8x - x' + x + {e^{ - 2t}} - 2{e^{ - 2t}},\;\; \Rightarrow x^{\prime\prime} - 9x = - {e^{ - 2t}}.\]
We have a second order linear homogeneous equation for the function \(x\left( t \right).\)
First solve the homogeneous equation
\[x^{\prime\prime} - 9x = 0.\]
The roots of the auxiliary equation are
\[{\lambda ^2} - 9 = 0,\;\; \Rightarrow {\lambda _{1,2}} = \pm 3.\]
Then the solution of the homogeneous equation for \(x\left( t \right)\) is as follows:
\[{x_0}\left( t \right) = {C_1}{e^{3t}} + {C_2}{e^{ - 3t}},\]
where \({C_1},\) \({C_2}\) are arbitrary numbers.
By virtue of the form of the inhomogeneous term in the equation for \(x\left( t \right),\) we seek a particular solution \({x_1}\left( t \right)\) in the form
\[{x_1}\left( t \right) = A{e^{ - 2t}}.\]
Substituting this trial solution into the nonhomogeneous differential equation, we determine the coefficient \(A:\)
\[{\left( { - 2} \right)^2}A{e^{ - 2t}} - 9A{e^{ - 2t}} = - {e^{ - 2t}},\;\; \Rightarrow 4A - 9A = - 1,\;\; \Rightarrow 5A = 1,\;\; \Rightarrow A = \frac{1}{5}.\]
Hence, a particular solution \({x_1}\left( t \right)\) is given by
\[{x_1}\left( t \right) = \frac{1}{5}{e^{ - 2t}}.\]
Accordingly, the general solution for the function \(x\left( t \right)\) is written as
\[x\left( t \right) = {x_0}\left( t \right) + {x_1}\left( t \right) = {C_1}{e^{3t}} + {C_2}{e^{ - 3t}} + \frac{1}{5}{e^{ - 2t}}.\]
It remains to find the function \(y\left( t \right).\) We calculate the derivative \(x'\left( t \right)\) and substitute it into the first equation of the original system:
\[x'\left( t \right) = 3{C_1}{e^{3t}} - 3{C_2}{e^{ - 3t}} - \frac{2}{5}{e^{ - 2t}},\;\; \Rightarrow 3{C_1}{e^{3t}} - 3{C_2}{e^{ - 3t}} - \frac{2}{5}{e^{ - 2t}} = {C_1}{e^{3t}} + {C_2}{e^{ - 3t}} + \frac{1}{5}{e^{ - 2t}} + 2y + {e^{ - 2t}},\;\; \Rightarrow 2y = 2{C_1}{e^{3t}} - 4{C_2}{e^{ - 3t}} - \frac{8}{5}{e^{ - 2t}},\;\; \Rightarrow y\left( t \right) = {C_1}{e^{3t}} - 2{C_2}{e^{ - 3t}} - \frac{4}{{5}}{e^{ - 2t}}.\]
The final answer is written as follows:
\[\left\{ \begin{array}{l} {x\left( t \right) = {C_1}{e^{3t}} + {C_2}{e^{ - 3t}} + \frac{1}{5}{e^{ - 2t}} }\\ {y\left( t \right) = {C_1}{e^{3t}} - 2{C_2}{e^{ - 3t}} - \frac{4}{{5}}{e^{ - 2t}} }\end{array} \right..\]
Example 2.
Solve the system of equations by the method of undetermined coefficients:
\[\frac{{dx}}{{dt}} = 2x + y,\; \frac{{dy}}{{dt}} = 3y + t{e^t}.\]
Solution.
We write this system in matrix form:
\[\mathbf{X}'\left( t \right) = A\mathbf{X}\left( t \right) + \mathbf{f}\left( t \right),\;\;
\text{where}\;\;\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
{x\left( t \right)}\\
{y\left( t \right)}
\end{array}} \right],\;\;
A = \left[ {\begin{array}{*{20}{c}}
2&1\\
0&3
\end{array}} \right],\;\;
\mathbf{f}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
0\\
{t{e^t}}
\end{array}} \right].\]
We first find the solution of the homogeneous system. Calculate the eigenvalues of the matrix \(A:\)
\[\det \left( {A - \lambda I} \right) = \left| {\begin{array}{*{20}{c}}
{2 - \lambda }&1\\
0&{3 - \lambda }
\end{array}} \right| = 0,\;\; \Rightarrow
\left( {2 - \lambda } \right)\left( {3 - \lambda } \right) = 0,\;\; \Rightarrow
{\lambda _1} = 2,\;{\lambda _2} = 3.\]
Determine the eigenvector \({\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}}} \right)^T}\) for the value \({\lambda _1} = 2:\)
\[\left( {A - {\lambda _1}I} \right){\mathbf{V}_1} = \mathbf{0},\;\; \Rightarrow
\left[ {\begin{array}{*{20}{c}}
{2 - 2}&1\\
0&{3 - 2}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
\left[ {\begin{array}{*{20}{c}}
0&1\\
0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
0 \cdot {V_{11}} + 1 \cdot {V_{21}} = 0.\]
It is seen that \({V_{21}} = 0,\) and the coordinate \({V_{11}}\) can be arbitrary. For simplicity, we choose \({V_{11}} = 1.\) Therefore, \({\mathbf{V}_1} = {\left( {1,0} \right)^T}.\)
Similarly, we find the eigenvector \({\mathbf{V}_2} = {\left( {{V_{12}},{V_{22}}} \right)^T},\) corresponding to the eigenvalue \({\lambda _2} = 3:\)
\[\left( {A - {\lambda _2}I} \right){\mathbf{V}_2} = \mathbf{0},\;\; \Rightarrow
\left[ {\begin{array}{*{20}{c}}
{2 - 3}&1\\
0&{3 - 3}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{12}}}\\
{{V_{22}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
\left[ {\begin{array}{*{20}{r}}
{-1}&1\\
0&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{12}}}\\
{{V_{22}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
-{V_{12}} + {V_{22}} = 0.\]
Putting \({V_{22}} = t,\) we have: \({V_{12}} = {V_{22}} = t.\) Then
\[{\mathbf{V}_2} = \left[ {\begin{array}{*{20}{c}}
{{V_{12}}}\\
{{V_{22}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
t\\
t
\end{array}} \right]
= t\left[ {\begin{array}{*{20}{c}}
1\\
1
\end{array}} \right]
\sim \left[ {\begin{array}{*{20}{c}}
1\\
1
\end{array}} \right].\]
Thus, the general solution of the homogeneous system is given by
\[{\mathbf{X}_0}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
x\\
y
\end{array}} \right]
= {C_1}{e^{{\lambda _1}t}}{\mathbf{V}_1} + {C_2}{e^{{\lambda _2}t}}{\mathbf{V}_2}
= {C_1}{e^{2t}}\left[ {\begin{array}{*{20}{c}}
1\\
0
\end{array}} \right] + {C_2}{e^{3t}}\left[ {\begin{array}{*{20}{c}}
1\\
1
\end{array}} \right].\]
Now we find a particular solution \({\mathbf{X}_1}\left( t \right).\) The inhomogeneous term has the form of a quasi-polynomial \({\mathbf{P}_1}\left( t \right){e^t}.\) The degree of the exponential function is \(\alpha = 1.\) Because it does not coincide with any of the eigenvalues \({\lambda _1} = 2,\) \({\lambda _2} = 3,\) then we find a particular solution in a form similar to \(\mathbf{f}\left( t \right),\) that is assume that
\[{\mathbf{X}_1}\left( t \right) = \left( {\begin{array}{*{20}{c}}
{{x_1}\left( t \right)}\\
{{y_1}\left( t \right)}
\end{array}} \right) = {\mathbf{P}_1}\left( t \right){e^t},\;\;
\text{where}\;\; {\mathbf{P}_1}\left( t \right) = {\mathbf{A}_0} + {\mathbf{A}_1}t.\]
We find the unknown vectors \({\mathbf{A}_0},{\mathbf{A}_1}\) using the method of undetermined coefficients. Let
\[{\mathbf{A}_0} = \left[ {\begin{array}{*{20}{c}}
{{a_0}}\\
{{b_0}}
\end{array}} \right],\;\; {\mathbf{A}_1} = \left[ {\begin{array}{*{20}{c}}
{{a_1}}\\
{{b_1}}
\end{array}} \right].\]
Therefore, the particular solution can be written as
\[{\mathbf{X}_1}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
{{x_1}\left( t \right)}\\
{{y_1}\left( t \right)}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{\left( {{a_0} + {a_1}t} \right){e^t}}\\
{\left( {{b_0} + {b_1}t} \right){e^t}}
\end{array}} \right].\]
Substitute \({\mathbf{X}_1}\left( t \right)\) into the original nonhomogeneous equation:
\[{\mathbf{X'}_1}\left( t \right) = A{\mathbf{X}_1}\left( t \right) + \mathbf{f}\left( t \right),\;\; \Rightarrow
\left[ {\begin{array}{*{20}{c}}
{{a_1}{e^t} + \left( {{a_0} + {a_1}t} \right){e^t}}\\
{{b_1}{e^t} + \left( {{b_0} + {b_1}t} \right){e^t}}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
2&1\\
0&3
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{\left( {{a_0} + {a_1}t} \right){e^t}}\\
{\left( {{b_0} + {b_1}t} \right){e^t}}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
0\\
{t{e^t}}
\end{array}} \right],\;\Rightarrow
\left\{ {\begin{array}{*{20}{l}}
{{\left( {{a_1} + {a_0} + {a_1}t} \right){e^t} = \left( {2{a_0} + 2{a_1}t} \right){e^t} + \left( {{b_0} + {b_1}t} \right){e^t}} }\\
{{\left( {{b_1} + {b_0} + {b_1}t} \right){e^t} = \left( {3{b_0} + 3{b_1}t} \right){e^t} + t{e^t}} }
\end{array}} \right..\]
Divide both sides of each equation by \({{e^t}}:\)
\[\left\{ {\begin{array}{*{20}{l}}
{{{a_1} + {a_0} + {a_1}t }={ 2{a_0} + 2{a_1}t + {b_0} + {b_1}t} }\\
{{{b_1} + {b_0} + {b_1}t }={ 3{b_0} + 3{b_1}t + t}}
\end{array}} \right..\]
By equating the coefficients of like terms, we obtain the following system of equations:
\[\left\{ \begin{array}{l}
{a_1} + {a_0} = 2{a_0} + {b_0}\\
{a_1} = 2{a_1} + {b_1}\\
{b_1} + {b_0} = 3{b_0}\\
{b_1} = 3{b_1} + 1
\end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l}
{a_1} = {a_0} + {b_0}\\
{a_1} + {b_1} = 0\\
{b_1} = 2{b_0}\\
2{b_1} + 1 = 0
\end{array} \right..\]
Solve the system and find the unknown coefficients \({a_0},{a_1},{b_0},{b_1}:\)
\[{b_1} = - \frac{1}{2},\;\; {b_0} = \frac{{{b_1}}}{2} = - \frac{1}{4},\;\; {a_1} = - {b_1} = \frac{1}{2},\;\; {a_0} = {a_1} - {b_0} = \frac{1}{2} - \left( { - \frac{1}{4}} \right) = \frac{3}{4}.\]
Thus, a particular solution \({\mathbf{X}_1}\left( t \right)\) can be written as
\[{\mathbf{X}_1}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
{{x_1}\left( t \right)}\\
{{y_1}\left( t \right)}
\end{array}} \right]
= {\mathbf{P}_1}\left( t \right){e^t}
= \left[ {\begin{array}{*{20}{c}}
{{a_0} + {a_1}t}\\
{{b_0} + {b_1}t}
\end{array}} \right]{e^t}
= \left[ {\begin{array}{*{20}{c}}
{\frac{3}{4} + \frac{1}{2}t}\\
{ - \frac{1}{4} - \frac{1}{2}t}
\end{array}} \right]{e^t}
= \frac{1}{4}{e^t}\left[ {\begin{array}{*{20}{c}}
{3 + 2t}\\
{ - 1 - 2t}
\end{array}} \right].\]
Then the general solution of the original nonhomogeneous system is given by the following formula:
\[\mathbf{X}\left( t \right) = {\mathbf{X}_0}\left( t \right) + {\mathbf{X}_1}\left( t \right)
= {C_1}{e^{2t}}\left[ {\begin{array}{*{20}{c}}
1\\
0
\end{array}} \right]
+ {C_2}{e^{3t}}\left[ {\begin{array}{*{20}{c}}
1\\
1
\end{array}} \right]
+ \frac{1}{4}{e^t}\left[ {\begin{array}{*{20}{c}}
{3 + 2t}\\
{ - 1 - 2t}
\end{array}} \right].\]
Example 3.
Solve the system of equations by the method of undetermined coefficients:
\[\frac{{dx}}{{dt}} = x + {e^t},\; \frac{{dy}}{{dt}} = x + y - {e^t}.\]
Solution.
We calculate the eigenvalues \({\lambda _i}\) of the matrix \(A\) and construct the general solution of the associated homogeneous system:
\[A = \left[ {\begin{array}{*{20}{c}}
1&0\\
1&1
\end{array}} \right],\;\; \Rightarrow
\det \left( {A - \lambda I} \right) = \left| {\begin{array}{*{20}{c}}
{1 - \lambda }&0\\
1&{1 - \lambda }
\end{array}} \right| = 0,\;\; \Rightarrow
{\left( {1 - \lambda } \right)^2} = 0,\;\; \Rightarrow
{\lambda _1} = 1.\]
So, here we have one eigenvalue \({\lambda _1} = 1\) of multiplicity \({k_1} = 2.\) Find the eigenvector \({\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}}} \right)^T}\) for \({\lambda _1} = 1:\)
\[\left( {A - {\lambda _1}I} \right){\mathbf{V}_1} = \mathbf{0},\;\; \Rightarrow
\left[ {\begin{array}{*{20}{c}}
{1 - 1}&0\\
1&{1 - 1}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
\left[ {\begin{array}{*{20}{c}}
0&0\\
1&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
1 \cdot {V_{11}} + 0 \cdot {V_{21}} = 0.\]
It is clear that \({V_{11}} = 0,\) and the coordinate \({V_{21}}\) can be arbitrary. By setting \({V_{21}} = 1,\) we obtain \({\mathbf{V}_1} = {\left( {0,1} \right)^T}.\)
We calculate the second linearly independent vector \({\mathbf{V}_2} = {\left( {{V_{12}},{V_{22}}} \right)^T}\) as a generalized eigenvector, connected to \({\mathbf{V}_1}:\)
\[\left( {A - {\lambda _1}I} \right){\mathbf{V}_2} = {\mathbf{V}_1},\;\; \Rightarrow
\left[ {\begin{array}{*{20}{c}}
0&0\\
1&0
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{12}}}\\
{{V_{22}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
0\\
1
\end{array}} \right],\;\; \Rightarrow
1 \cdot {V_{12}} + 0 \cdot {V_{22}} = 1.\]
Here \({V_{12}} = 1,\) and \({V_{22}}\) can be chosen arbitrarily, for example, \({V_{22}} = 0.\) Then \({\mathbf{V}_2} = {\left( {1,0} \right)^T}.\)
The general solution of the homogeneous system will be expressed by the formula
\[{\mathbf{X}_0}\left( t \right) = {C_1}{e^t}\left[ {\begin{array}{*{20}{c}}
0\\
1
\end{array}} \right]
+ {C_2}{e^t}\left( {t\left[ {\begin{array}{*{20}{c}}
0\\
1
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
1\\
0
\end{array}} \right]} \right).\]
where \({C_1},{C_2}\) are arbitrary constants.
As you can see, the length of the Jordan chain for \({\lambda _1} = 1\) is equal to \(2.\)
We now turn to finding a particular solution \({\mathbf{X}_1}\left( t \right)\) of the nonhomogeneous equation. The inhomogeneous terms in each equation contain the exponential function \({e^t},\) which coincides with the exponential function in the solution of the homogeneous equation. This means that we have the resonance case.
Therefore, the particular solution \({\mathbf{X}_1}\left( t \right)\) should be sought in the form of a vector quasi-polynomial
\[{\mathbf{X}_1}\left( t \right) = {\mathbf{P}_{m + k}}\left( t \right){e^t},\]
where \(m = 0\) (\(m\) denotes the degree of the vector polynomial \(\mathbf{f}\left( t \right)\)) and \(k = 2\) (\(k\) is the length of the Jordan chain for the resonant eigenvalue \({\lambda _1} = 1\)).
So, in this case, we select the second degree polynomial:
\[{\mathbf{X}_1}\left( t \right) = {\mathbf{P}_2}\left( t \right){e^t} = \left( {{\mathbf{A}_0} + {\mathbf{A}_1}t + {\mathbf{A}_2}{t^2}} \right){e^t}.\]
The vector coefficients \({\mathbf{A}_0},{\mathbf{A}_1},{\mathbf{A}_2}\) can be found by direct substitution of the function \({\mathbf{X}_1}\left( t \right)\) into the nonhomogeneous system. Let
\[{\mathbf{A}_0} = \left[ {\begin{array}{*{20}{c}}
{{a_0}}\\
{{b_0}}
\end{array}} \right],\;\;
{\mathbf{A}_1} = \left[ {\begin{array}{*{20}{c}}
{{a_1}}\\
{{b_1}}
\end{array}} \right],\;\;
{\mathbf{A}_2} = \left[ {\begin{array}{*{20}{c}}
{{a_2}}\\
{{b_2}}
\end{array}} \right].\]
Then
\[{\mathbf{X}_1}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
{{x_1}\left( t \right)}\\
{{y_1}\left( t \right)}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{\left( {{a_0} + {a_1}t + {a_2}{t^2}} \right){e^t}}\\
{\left( {{b_0} + {b_1}t + {b_2}{t^2}} \right){e^t}}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{{a_0} + {a_1}t + {a_2}{t^2}}\\
{{b_0} + {b_1}t + {b_2}{t^2}}
\end{array}} \right]{e^t}.\]
The derivative is given by
\[{\mathbf{X'}_1}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
{{x'_1}\left( t \right)}\\
{{y'_1}\left( t \right)}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{{a_0} + {a_1}t + {a_2}{t^2}}\\
{{b_0} + {b_1}t + {b_2}{t^2}}
\end{array}} \right]{e^t} + \left[ {\begin{array}{*{20}{c}}
{{a_1} + 2{a_2}t}\\
{{b_1} + 2{b_2}t}
\end{array}} \right]{e^t}
= \left[ {\begin{array}{*{20}{c}}
{a_0} + {a_1} + \left( {{a_1} + 2{a_2}} \right)t + {a_2}{t^2}\\
{b_0} + {b_1} + \left( {{b_1} + 2{b_2}} \right)t + {b_2}{t^2}
\end{array}} \right]{e^t}.\]
After the substitution in the original system, we obtain:
\[\left( {{a_0} + {a_1} + \left( {{a_1} + 2{a_2}} \right)t + {a_2}{t^2}} \right){e^t} = \left( {{a_0} + {a_1}t + {a_2}{t^2}} \right){e^t} + {e^t},\]
\[\left( {{b_0} + {b_1} + \left( {{b_1} + 2{b_2}} \right)t + {b_2}{t^2}} \right){e^t}
= \left( {{a_0} + {a_1}t + {a_2}{t^2}} \right){e^t}
+ \left( {{b_0} + {b_1}t + {b_2}{t^2}} \right){e^t} - {e^t}.\]
Dividing both sides of each equation by \({e^t}\) and equating the coefficients of the terms with equal powers of \(t,\) we obtain the following system for the unknown coefficients:
\[\left\{ \begin{array}{l}
{a_0} + {a_1} + \left( {{a_1} + 2{a_2}} \right)t + \cancel{{a_2}{t^2}} = {a_0} + {a_1}t + \cancel{{a_2}{t^2}} + 1\\
{b_0} + {b_1} + \left( {{b_1} + 2{b_2}} \right)t + \cancel{{b_2}{t^2}} = {a_0} + {a_1}t + {a_2}{t^2} + {b_0} + {b_1}t + \cancel{{b_2}{t^2}} - 1
\end{array} \right.,\]
\[\left\{ \begin{array}{l}
\cancel{a_0} + {a_1} = \cancel{a_0} + 1\\
\cancel{a_1} + 2{a_2} = \cancel{a_1}\\
\cancel{a_2} = \cancel{a_2}\\
\cancel{b_0} + {b_1} = {a_0} + \cancel{b_0} - 1\\
\cancel{b_1} + 2{b_2} = {a_1} + \cancel{b_1}\\
\cancel{b_2} = {a_2} + \cancel{b_2}
\end{array} \right.,\;\; \Rightarrow
\left\{ {\begin{array}{*{20}{l}}
{{a_1} = 1}\\
{{a_2} = 0}\\
{{b_1} = {a_0} - 1}\\
{2{b_2} = {a_1}}
\end{array}} \right..\]
There are only four independent equations here. The numbers \({a_0}\) and \({b_0}\) can be chosen arbitrarily, for example, \({a_0} = 0,\) \({b_0} = 0.\) As a result, the coefficients have the following values:
\[{a_0} = 0,\;\; {a_1} = 1,\;\; {a_2} = 0,\;\; {b_0} = 0,\;\; {b_1} = - 1,\;\; {b_2} = \frac{1}{2}.\]
Thus, a particular solution \({\mathbf{X}_1}\left( t \right)\) is given by
\[{\mathbf{X}_1}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
{{x_1}\left( t \right)}\\
{{y_1}\left( t \right)}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{{a_0} + {a_1}t + {a_2}{t^2}}\\
{{b_0} + {b_1}t + {b_2}{t^2}}
\end{array}} \right]{e^t}
= \left[ {\begin{array}{*{20}{c}}
t\\
{\frac{1}{2}{t^2} - t}
\end{array}} \right]{e^t}.\]
Finally the general solution of the nonhomogeneous system is written as
\[
\mathbf{X}\left( t \right) = {\mathbf{X}_0}\left( t \right) + {\mathbf{X}_1}\left( t \right)
= {C_1}{e^t}\left[ {\begin{array}{*{20}{c}}
0\\
1
\end{array}} \right] + {C_2}{e^t}\left( {t\left[ {\begin{array}{*{20}{c}}
0\\
1
\end{array}} \right] }
+ {\left[ {\begin{array}{*{20}{c}}
1\\
0
\end{array}} \right]} \right) + {e^t}\left[ {\begin{array}{*{20}{c}}
t\\
{\frac{1}{2}{t^2} - t}
\end{array}} \right].\]
Example 4.
Solve the system of equations by the method of undetermined coefficients:
\[\frac{{dx}}{{dt}} = - y,\; \frac{{dy}}{{dt}} = x + \cos t.\]
Solution.
We first construct the general solution of the homogeneous system. The eigenvalues of the matrix \(A\) are
\[A = \left[ {\begin{array}{*{20}{r}}
0&{ - 1}\\
1&0
\end{array}} \right],\;\; \Rightarrow
\det \left( {A - \lambda I} \right) = \left| {\begin{array}{*{20}{r}}
{ - \lambda }&{ - 1}\\
1&{ - \lambda }
\end{array}} \right| = 0,\;\; \Rightarrow
{\lambda ^2} + 1 = 0,\;\; \Rightarrow
{\lambda ^2} = - 1,\;\; \Rightarrow
{\lambda _{1,2}} = \pm i.\]
In this case, we have a pair of complex conjugate eigenvalues of multiplicity \(1.\) In accordance with the general theory, we find the complex solution, for example, for the eigenvalue \({\lambda _1} = + i,\) and then separate the real and imaginary parts, which form a fundamental system of solutions.
Determine the eigenvector \({\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}}} \right)^T}\) for the eigenvalue \({\lambda _1} = + i:\)
\[\left( {A - {\lambda _1}I} \right){\mathbf{V}_1} = \mathbf{0},\;\; \Rightarrow
\left[ {\begin{array}{*{20}{c}}
{ - i}&{ - 1}\\
1&{ - i}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
{V_{11}} - i{V_{21}} = 0.\]
Let \({V_{21}} = t.\) Then \({V_{11}} = i{V_{21}} = it.\) Hence,
\[{\mathbf{V}_1} = \left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{it}\\
t
\end{array}} \right]
= t\left[ {\begin{array}{*{20}{c}}
i\\
1
\end{array}} \right]
\sim \left[ {\begin{array}{*{20}{c}}
i\\
1
\end{array}} \right].\]
The eigenvalue \({\lambda _1}\) and the eigenvector \({\mathbf{V}_1}\) form a complex solution of the type
\[{\mathbf{Z}_1}\left( t \right) = {e^{{\lambda _1}t}}{\mathbf{V}_1}
= {e^{it}}\left[ {\begin{array}{*{20}{c}}
i\\
1
\end{array}} \right]
= \left( {\cos t + i\sin t} \right)\left[ {\begin{array}{*{20}{c}}
i\\
1
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{i\cos t - \sin t}\\
{\cos t + i\sin t}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{ - \sin t}\\
{\cos t}
\end{array}} \right] + i\left[ {\begin{array}{*{20}{c}}
{\cos t}\\
{\sin t}
\end{array}} \right]
= \text{Re}\left[ {{\mathbf{Z}_1}\left( t \right)} \right] + \text{Im}\left[ {{\mathbf{Z}_1}\left( t \right)} \right].\]
The general solution of the homogeneous system is written as
\[{\mathbf{X}_0}\left( t \right) = {C_1}\left[ {\begin{array}{*{20}{c}}
{ - \sin t}\\
{\cos t}
\end{array}} \right] + {C_2}\left[ {\begin{array}{*{20}{c}}
{\cos t}\\
{\sin t}
\end{array}} \right],\]
where \({C_1},{C_2}\) are arbitrary numbers.
We now find a particular solution \({\mathbf{X}_1}\left( t \right)\) of the nonhomogeneous system. Here we meet again with the resonance case, since the inhomogeneous term
\[\mathbf{f}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
0\\
{\cos t}
\end{array}} \right]\]
is described by the complex number \(\gamma = \alpha + \beta i \) \(= + i\) and coincides with the eigenvalue \({\lambda _1} = +i\) of the matrix \(A.\) Hence, we seek the particular solution \({\mathbf{X}_1}\left( t \right)\) in the form
\[{\mathbf{X}_1}\left( t \right) = \left( {{\mathbf{A}_0} + {\mathbf{A}_1}t} \right)\cos t
+ \left( {{\mathbf{B}_0} + {\mathbf{B}_1}t} \right)\sin t.\]
Let the vectors \({\mathbf{A}_0},\) \({\mathbf{A}_1},\) \({\mathbf{B}_0},\) \({\mathbf{B}_1}\) have the following coordinates:
\[{\mathbf{A}_0} = \left[ {\begin{array}{*{20}{c}}
{{a_0}}\\
{{b_0}}
\end{array}} \right],\;\; {\mathbf{A}_1} = \left[ {\begin{array}{*{20}{c}}
{{a_1}}\\
{{b_1}}
\end{array}} \right],\;\;
{\mathbf{B}_0} = \left[ {\begin{array}{*{20}{c}}
{{c_0}}\\
{{d_0}}
\end{array}} \right],\;\; {\mathbf{B}_1} = \left[ {\begin{array}{*{20}{c}}
{{c_1}}\\
{{d_1}}
\end{array}} \right].\]
Then the components \({x_1}\left( t \right),{y_1}\left( t \right)\) of the vector \({\mathbf{X}_1}\left( t \right)\) can be written as
\[{x_1}\left( t \right) = \left( {{a_0} + {a_1}t} \right)\cos t + \left( {{c_0} + {c_1}t} \right)\sin t,\]
\[{y_1}\left( t \right) = \left( {{b_0} + {b_1}t} \right)\cos t + \left( {{d_0} + {d_1}t} \right)\sin t.\]
The derivatives of these functions are given by
\[{x'_1}\left( t \right) = {a_1}\cos t - \left( {{a_0} + {a_1}t} \right)\sin t + {c_1}\sin t + \left( {{c_0} + {c_1}t} \right)\cos t,\]
\[{y'_1}\left( t \right) = {b_1}\cos t - \left( {{b_0} + {b_1}t} \right)\sin t
+ {d_1}\sin t + \left( {{d_0} + {d_1}t} \right)\cos t.\]
We substitute these expressions into the nonhomogeneous system:
\[\color{blue}{{a_1}\cos t} - \color{red}{{a_0}\sin t} - \color{magenta}{{a_1}t\sin t} + \color{red}{{c_1}\sin t}
+ \color{blue}{{c_0}\cos t} + \color{green}{{c_1}t\cos t}
= - \color{blue}{{b_0}\cos t} - \color{green}{{b_1}t\cos t}
- \color{red}{{d_0}\sin t} - \color{magenta}{{d_1}t\sin t},\]
\[\color{blue}{{b_1}\cos t} - \color{red}{{b_0}\sin t} - \color{magenta}{{b_1}t\sin t} + \color{red}{{d_1}\sin t}
+ \color{blue}{{d_0}\cos t} + \color{green}{{d_1}t\cos t}
= \color{blue}{{a_0}\cos t} + \color{green}{{a_1}t\cos t}
+ \color{red}{{c_0}\sin t} + \color{magenta}{{c_1}t\sin t}
+ \color{blue}{\cos t},\]
By equating the coefficients of like functions on the left and right sides, we obtain an algebraic system for the unknown coefficients:
\[\left\{ \begin{array}{l}
{a_1} + {c_0} = - {b_0}\\
- {a_0} + {c_1} = - {d_0}\\
- {a_1} = - {d_1}\\
{c_1} = - {b_1}\\
{b_1} + {d_0} = {a_0} + 1\\
- {b_0} + {d_1} = {c_0}\\
{d_1} = {a_1}\\
- {b_1} = {c_1}
\end{array} \right..\]
A part of the equations in this system is dependent on the others. Therefore, some coefficients can be chosen arbitrarily (for example, we can set them equal to zero). As a result, we get the following set of numbers:
\[{a_0} = 0,\;\; {b_0} = 0,\;\;
{c_0} = 0,\;\; {d_0} = \frac{1}{2},\;\;
{a_1} = 0,\;\; {b_1} = \frac{1}{2},\;\;
{c_1} = - \frac{1}{2},\;\; {d_1} = 0.\]
Thus, a particular solution \({\mathbf{X}_1}\left( t \right)\) has the form:
\[{\mathbf{X}_1}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
{{x_1}\left( t \right)}\\
{{y_1}\left( t \right)}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{{\left( {{a_0} + {a_1}t} \right)\cos t + \left( {{c_0} + {c_1}t} \right)\sin t}}\\
{{\left( {{b_0} + {b_1}t} \right)\cos t + \left( {{d_0} + {d_1}t} \right)\sin t}}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{ - \frac{1}{2}t\sin t}\\
{\frac{1}{2}t\cos t + \frac{1}{2}\sin t}
\end{array}} \right].\]
The general solution of the original system is written as
\[\mathbf{X}\left( t \right) = {\mathbf{X}_0}\left( t \right) + {\mathbf{X}_1}\left( t \right)
= {C_1}\left[ {\begin{array}{*{20}{c}}
{ - \sin t}\\
{\cos t}
\end{array}} \right] + {C_2}\left[ {\begin{array}{*{20}{c}}
{\cos t}\\
{\sin t}
\end{array}} \right]
+ \left[ {\begin{array}{*{20}{c}}
{ - \frac{1}{2}t\sin t}\\
{\frac{1}{2}t\cos t + \frac{1}{2}\sin t}
\end{array}} \right].\]
Example 5.
Solve the system of equations by variation of parameters:
\[\frac{{dx}}{{dt}} = y + \frac{1}{{\cos t}},\; \frac{{dy}}{{dt}} = - x.\]
Solution.
We first construct the general solution of the homogeneous system. Calculate the eigenvalues:
\[A = \left[ {\begin{array}{*{20}{r}}
0&1\\
{ - 1}&0
\end{array}} \right],\;\; \Rightarrow
\det \left( {A - \lambda I} \right) = \left| {\begin{array}{*{20}{r}}
{-\lambda} &1\\
{ - 1}&{ - \lambda }
\end{array}} \right| = 0,\;\; \Rightarrow
{\lambda ^2} + 1 = 0,\;\; \Rightarrow
{\lambda ^2} = - 1,\;\; \Rightarrow
{\lambda _{1,2}} = \pm i.\]
Find the complex eigenvector \({\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}}} \right)^T}\) for the eigenvalue \({\lambda _1} = + i:\)
\[\left( {A - {\lambda _1}I} \right){\mathbf{V}_1} = \mathbf{0},\;\; \Rightarrow
\left[ {\begin{array}{*{20}{r}}
{ - i}&1\\
{ - 1}&{ - i}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
- i{V_{11}} + {V_{21}} = 0.\]
We set \({V_{11}} = t.\) Then \({V_{21}} = i{V_{11}} = it.\) Hence,
\[{\mathbf{V}_1} = \left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
t\\
{it}
\end{array}} \right]
= t\left[ {\begin{array}{*{20}{c}}
1\\
i
\end{array}} \right]
\sim \left[ {\begin{array}{*{20}{c}}
1\\
i
\end{array}} \right].\]
The eigenvalue \({\lambda _1}\) and the eigenvector \({\mathbf{V}_1}\) correspond to a solution of the form
\[{\mathbf{Z}_1}\left( t \right) = {e^{{\lambda _1}t}}{\mathbf{V}_1} = {e^{it}}\left[ {\begin{array}{*{20}{c}}
1\\
i
\end{array}} \right]
= \left( {\cos t + i\sin t} \right)\left[ {\begin{array}{*{20}{c}}
1\\
i
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{\cos t + i\sin t}\\
{i\cos t - \sin t}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
{\cos t}\\
{ - \sin t}
\end{array}} \right] + i\left[ {\begin{array}{*{20}{c}}
{\sin t}\\
{\cos t}
\end{array}} \right].\]
The real and imaginary parts in the last expression form a fundamental system of solutions:
\[{\mathbf{X}_0}\left( t \right) = {C_1}\left[ {\begin{array}{*{20}{c}}
{\cos t}\\
{ - \sin t}
\end{array}} \right] + {C_2}\left[ {\begin{array}{*{20}{c}}
{\sin t}\\
{\cos t}
\end{array}} \right],\]
where \({C_1},{C_2}\) are arbitrary constants.
Write down the solution separately for each coordinate:
\[\left\{ \begin{array}{l}
{x_0}\left( t \right) = {C_1}\cos t + {C_2}\sin t\\
{y_0}\left( t \right) = - {C_1}\sin t + {C_2}\cos t
\end{array} \right..\]
Now consider the nonhomogeneous system. In accordance with the method of variation of constants, we assume that \({C_1},{C_2}\) are functions of the variable \(t:\)
\[\left\{ \begin{array}{l}
{x\left( t \right) = {C_1}\left( t \right)\cos t + {C_2}\left( t \right)\sin t}\\
{y\left( t \right) = - {C_1}\left( t \right)\sin t + {C_2}\left( t \right)\cos t}
\end{array} \right..\]
Substituting these expressions in the original nonhomogeneous system yields:
\[{C'_1}\cos t - \cancel{{C_1}\sin t} + {C'_2}\sin t + \cancel{{C_2}\cos t} = - \cancel{{C_1}\sin t} + \cancel{{C_2}\cos t} + \frac{1}{{\cos t}},\]
\[- {C'_1}\sin t - \cancel{{C_1}\cos t} + {C'_2}\cos t - \cancel{{C_2}\sin t} = - \cancel{{C_1}\cos t} - \cancel{{C_2}\sin t},\]
\[\Rightarrow
\left\{ \begin{array}{l}
{C'_1}\cos t + {C'_2}\sin t = \frac{1}{{\cos t}}\\
- {C'_1}\sin t + {C'_2}\cos t = 0
\end{array} \right..\]
We solve the resulting system and find the functions \({C_1}\left( t \right),\) \({C_2}\left( t \right).\) This is conveniently done by using Cramer's rule:
\[\Delta = \left| {\begin{array}{*{20}{r}}
{\cos t}&{\sin t}\\
{ - \sin t}&{\cos t}
\end{array}} \right| = {\cos ^2}t + {\sin ^2}t = 1,\;\;
{\Delta _1} = \left| {\begin{array}{*{20}{c}}
{\frac{1}{{\cos t}}}&{\sin t}\\
0&{\cos t}
\end{array}} \right| = \frac{1}{{\cos t}} \cdot \cos t - 0 = 1,\;\;
{\Delta _2} = \left| {\begin{array}{*{20}{r}}
{\cos t}&{\frac{1}{{\cos t}}}\\
{ - \sin t}&0
\end{array}} \right| = 0 + \frac{1}{{\cos t}} \cdot \sin t = \tan t.\]
Hence, we obtain:
\[{C'_1} = \frac{{{\Delta _1}}}{\Delta } = \frac{1}{1} = 1,\;\; {C'_2} = \frac{{{\Delta _2}}}{\Delta } = \frac{{\tan t}}{1} = \tan t.\]
Integrating, we find:
\[{C_1}\left( t \right) = \int {1dt} = t + {A_1},\;\;
{C_2}\left( t \right) = \int {\tan tdt}
= \int {\frac{{\sin t}}{{\cos t}}dt}
= - \int {\frac{{d\left( {\cos t} \right)}}{{\cos t}}dt}
= - \ln \left| {\cos t} \right| + {A_2},\]
where \({A_1},{A_2}\) are constants of integration.
As a result, we get the following expressions for \(x\left( t \right)\) and \(y\left( t \right):\)
\[x\left( t \right) = {C_1}\left( t \right)\cos t + {C_2}\left( t \right)\sin t = \left( {t + {A_1}} \right)\cos t + \left( { - \ln \left| {\cos t} \right| + {A_2}} \right)\sin t = {A_1}\cos t + {A_2}\sin t + t\cos t - \sin t\ln \left| {\cos t} \right|,\]
\[y\left( t \right) = - {C_1}\left( t \right)\sin t + {C_2}\left( t \right)\cos t = - \left( {t + {A_1}} \right)\sin t + \left( { - \ln \left| {\cos t} \right| + {A_2}} \right)\cos t = - {A_1}\sin t + {A_2}\cos t - t\sin t - \cos t\ln \left| {\cos t} \right|.\]
The first two terms with the coefficients \({A_1},{A_2}\) in every expression describe the solution of the homogeneous system. The remaining terms are due to the inhomogeneous part. The final answer can be expressed as
\[\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
{x\left( t \right)}\\
{y\left( t \right)}
\end{array}} \right]
= {A_1}\left[ {\begin{array}{*{20}{c}}
{\cos t}\\
{ - \sin t}
\end{array}} \right]
+ {A_2}\left[ {\begin{array}{*{20}{c}}
{\sin t}\\
{\cos t}
\end{array}} \right]
+ \left[ {\begin{array}{*{20}{c}}
{t\cos t - \sin t\ln \left| {\cos t} \right|}\\
{ - t\sin t - \cos t\ln \left| {\cos t} \right|}
\end{array}} \right].\]
Example 6.
Solve the linear nonhomogeneous system by variation of constants:
\[\frac{{dx}}{{dt}} = 2x - y + {e^{2t}},\;\frac{{dy}}{{dt}} = 6x - 3y + {e^t} + 1.\]
Solution.
We begin with the construction of the general solution of the homogeneous system. Calculate the eigenvalues of the matrix \(A\) and the corresponding eigenvectors.
\[\det \left( {A - \lambda I} \right) = \left| {\begin{array}{*{20}{c}}
2&{ - 1}\\
6&{ - 3}
\end{array}} \right| = 0,\;\; \Rightarrow
\left( {2 - \lambda } \right)\left( { - 3 - \lambda } \right) + 6 = 0,\;\; \Rightarrow
\left( {\lambda - 2} \right)\left( {\lambda + 3} \right) + 6 = 0,\;\; \Rightarrow
{\lambda ^2} - 2\lambda + 3\lambda - \cancel{6} + \cancel{6} = 0,\;\; \Rightarrow
{\lambda ^2} + \lambda = 0,\;\; \Rightarrow
\lambda \left( {\lambda + 1} \right) = 0,\;\; \Rightarrow
{\lambda _1} = 0,\;{\lambda _2} = - 1.\]
The eigenvector \({\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}}} \right)^T}\) for the eigenvalue \({\lambda _1} = 0\) is equal to
\[\left( {A - {\lambda _1}I} \right){\mathbf{V}_1} = \mathbf{0},\;\; \Rightarrow
\left[ {\begin{array}{*{20}{c}}
2&{ - 1}\\
6&{ - 3}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
2{V_{11}} - {V_{21}} = 0,\;\; \Rightarrow
{V_{11}} = t,\;\;
{V_{21}} = 2{V_{11}} = 2t,\;\; \Rightarrow
{\mathbf{V}_1} = \left[ {\begin{array}{*{20}{c}}
{{V_{11}}}\\
{{V_{21}}}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
t\\
{2t}
\end{array}} \right]
= t\left[ {\begin{array}{*{20}{c}}
1\\
2
\end{array}} \right]
\sim \left[ {\begin{array}{*{20}{c}}
1\\
2
\end{array}} \right].\]
Similarly we find the eigenvector \({\mathbf{V}_2} = {\left( {{V_{12}},{V_{22}}} \right)^T}\) associated with the eigenvalue \({\lambda _2} = -1:\)
\[\left( {A - {\lambda _2}I} \right){\mathbf{V}_2} = \mathbf{0},\;\; \Rightarrow
\left[ {\begin{array}{*{20}{c}}
3&{ - 1}\\
6&{ - 2}
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
{{V_{12}}}\\
{{V_{22}}}
\end{array}} \right] = \mathbf{0},\;\; \Rightarrow
3{V_{12}} - {V_{22}} = 0,\;\; \Rightarrow
{V_{12}} = t,\;\;
{V_{22}} = 3{V_{12}} = 3t,\;\; \Rightarrow
{\mathbf{V}_2} = \left[ {\begin{array}{*{20}{c}}
{{V_{12}}}\\
{{V_{22}}}
\end{array}} \right]
= \left[ {\begin{array}{*{20}{c}}
t\\
{3t}
\end{array}} \right]
= t\left[ {\begin{array}{*{20}{c}}
1\\
3
\end{array}} \right]
\sim \left[ {\begin{array}{*{20}{c}}
1\\
3
\end{array}} \right].\]
Hence, the general solution of the homogeneous system is given by
\[{\mathbf{X}_0}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
{{x_0}\left( t \right)}\\
{{y_0}\left( t \right)}
\end{array}} \right]
= {C_1}\left[ {\begin{array}{*{20}{c}}
1\\
2
\end{array}} \right] + {C_2}{e^{ - t}}\left[ {\begin{array}{*{20}{c}}
1\\
3
\end{array}} \right],\]
where \({C_1},{C_2}\) are constant numbers.
Consider the original inhomogeneous system and find its solution by variation of parameters. We replace the constants \({C_1},{C_2}\) with the functions \({C_1}\left( t \right),\) \({C_2}\left( t \right),\) i.e. we will seek the solution in the form
\[\mathbf{X}\left( t \right) = {C_1}\left( t \right)\left[ {\begin{array}{*{20}{c}}
1\\
2
\end{array}} \right] + {C_2}\left( t \right){e^{ - t}}\left[ {\begin{array}{*{20}{c}}
1\\
3
\end{array}} \right]\]
or
\[\left\{ \begin{array}{l}
x\left( t \right) = {C_1}\left( t \right) + {C_2}\left( t \right){e^{ - t}}\\
y\left( t \right) = 2{C_1}\left( t \right) + 3{C_2}\left( t \right){e^{ - t}}
\end{array} \right..\]
The derivatives of these functions are
\[\left\{ \begin{array}{l}
{x'\left( t \right) = {C'_1} + {C'_2}{e^{ - t}} - {C_2}{e^{ - t}} }\\
{y'\left( t \right) = 2{C'_1} + 3{C'_2}{e^{ - t}} - 3{C_2}{e^{ - t}} }
\end{array} \right..\]
Next, we substitute these expressions into the nonhomogeneous system:
\[{C'_1} + {C'_2}{e^{ - t}} - \cancel{{C_2}{e^{ - t}}} = \cancel{2{C_1}} + \cancel{2{C_2}{e^{ - t}}} - \cancel{2{C_1}} - \cancel{3{C_2}{e^{ - t}}} + {e^{2t}} ,\]
\[2{C'_1} + 3{C'_2}{e^{ - t}} - \cancel{3{C_2}{e^{ - t}}} = \cancel{6{C_1}} + \cancel{6{C_2}{e^{ - t}}} - \cancel{6{C_1}} - \cancel{9{C_2}{e^{ - t}}} + {e^t} + 1,\]
\[\Rightarrow
\left\{ {\begin{array}{*{20}{l}}
{{C'_1} + {C'_2}{e^{ - t}} = {e^{2t}}}\\
{2{C'_1} + 3{C'_2}{e^{ - t}} = {e^t} + 1}
\end{array}} \right..\]
Solving the resulting system of algebraic equations, we find the derivatives \({C'_1},{C'_2}\) and then the functions \({C_1}\left( t \right), {C_2}\left( t \right):\)
\[\Delta = \left| {\begin{array}{*{20}{c}} 1&{{e^{ - t}}}\\ 2&{3{e^{ - t}}} \end{array}} \right| = 3{e^{ - t}} - 2{e^{ - t} = e^{ - t}},\]
\[{\Delta _1} = \left| {\begin{array}{*{20}{c}} {{e^{2t}}}&{{e^{ - t}}}\\ {{e^t} + 1}&{3{e^{ - t}}} \end{array}} \right| = 3{e^{2t}}{e^{ - t}} - {e^{ - t}}\left( {{e^t} + 1} \right) = 3{e^t} - {e^{ - t}} - 1,\]
\[{\Delta _2} = \left| {\begin{array}{*{20}{c}}
1&{{e^{2t}}}\\
2&{{e^t} + 1}
\end{array}} \right|
= {e^t} - 2{e^{2t}} + 1,\]
\[\Rightarrow {C'_1} = \frac{{{\Delta _1}}}{\Delta } = \frac{{3{e^t} - {e^{ - t}} - 1}}{{{e^{ - t}}}} = 3{e^{2t}} - {e^t} - 1,\]
\[\Rightarrow {C'_2} = \frac{{{\Delta _2}}}{\Delta } = \frac{{{e^t} - 2{e^{2t}} + 1}}{{{e^{ - t}}}} = {e^{2t}} - 2{e^{3t}} + {e^t}.\]
Integrating, we obtain:
\[{C_1}\left( t \right) = \int {\left( {3{e^{2t}} + {e^t} - 1} \right)dt} = \frac{3}{2}{e^{2t}} - {e^t} - t + {A_1},\]
\[{C_2}\left( t \right) = \int {\left( {{e^{2t}} - 2{e^{3t}} + {e^t}} \right)dt} = \frac{1}{2}{e^{2t}} - \frac{2}{3}{e^{3t}} + {e^t} + {A_2}.\]
The functions \(x\left( t \right),y\left( t \right)\) will have the following form:
\[ x\left( t \right) = {C_1}\left( t \right) + {C_2}\left( t \right){e^{ - t}} = \left( {\frac{3}{2}{e^{2t}} - {e^t} - t + {A_1}} \right) + \left( {\frac{1}{2}{e^{2t}} - \frac{2}{3}{e^{3t}} + {e^t} + {A_2}} \right){e^{ - t}} = {A_1} + {A_2}{e^{ - t}} + \frac{5}{6}{e^{2t}} - \frac{1}{2}{e^t} - t + 1,\]
\[y\left( t \right) = 2{C_1}\left( t \right) + 3{C_2}\left( t \right){e^{ - t}} = 2\left( {\frac{3}{2}{e^{2t}} - {e^t} - t + {A_1}} \right) + 3\left( {\frac{1}{2}{e^{2t}} - \frac{2}{3}{e^{3t}} + {e^t} + {A_2}} \right){e^{ - t}} = 2{A_1} + 3{A_2}{e^{ - t}} + {e^{2t}} - \frac{1}{2}{e^t} - 2t + 3.\]
The final answer can be represented in the form:
\[\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}} {x\left( t \right)}\\ {y\left( t \right)} \end{array}} \right] = {A_1}\left[ {\begin{array}{*{20}{c}} 1\\ 2 \end{array}} \right] + {A_2}{e^{ - t}}\left[ {\begin{array}{*{20}{c}} 1\\ 3 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {\frac{5}{6}{e^{2t}} - \frac{1}{2}{e^t} - t + 1}\\ {{e^{2t}} - \frac{1}{2}{e^t} - 2t + 3} \end{array}} \right].\]
Note that the inhomogeneous part in this problem consists of quasi-polynomials. Therefore, the solution to the system of equations can also be obtained using the method of undetermined coefficients and the principle of superposition.
