Editing Matrix exponential (section)

==== Example (inhomogeneous) ====
Consider now the inhomogeneous system
<math display="block">\begin{matrix}
  x' &=& 2x & - & y & + & z & + & e^{2t} \\
  y' &=&    &   & 3y& - & z & \\
  z' &=& 2x & + & y & + & 3z & + & e^{2t}
\end{matrix} ~.</math>

We again have
<math display="block">A = \left[\begin{array}{rrr}
  2 & -1 &  1 \\
  0 &  3 & -1 \\
  2 &  1 &  3
\end{array}\right] ~,</math>

and
<math display="block">\mathbf{b} = e^{2t}\begin{bmatrix}1 \\0\\1\end{bmatrix}.</math>

From before, we already have the general solution to the homogeneous equation. Since the sum of the homogeneous and particular solutions give the general solution to the inhomogeneous problem, we now only need find the particular solution.

We have, by above,
<math display="block">\begin{align}
  \mathbf{y}_p
  &= e^{tA}\int_0^t e^{(-u)A}\begin{bmatrix}e^{2u} \\0\\e^{2u}\end{bmatrix}\,du+e^{tA}\mathbf{c} \\[6pt]
  &= e^{tA}\int_0^t \begin{bmatrix}
         2e^u - 2ue^{2u} & -2ue^{2u}    & 0 \\
    -2e^u + 2(u+1)e^{2u} & 2(u+1)e^{2u} & 0 \\
                2ue^{2u} & 2ue^{2u}     & 2e^u
  \end{bmatrix}\begin{bmatrix}e^{2u} \\0 \\e^{2u}\end{bmatrix}\,du + e^{tA}\mathbf{c} \\[6pt]
  &= e^{tA}\int_0^t \begin{bmatrix}
     e^{2u}\left( 2e^u - 2ue^{2u}\right) \\
     e^{2u}\left(-2e^u + 2(1 + u)e^{2u}\right) \\
    2e^{3u} + 2ue^{4u}
  \end{bmatrix}\,du + e^{tA}\mathbf{c} \\[6pt]
  &= e^{tA}\begin{bmatrix}
    -{1 \over 24}e^{3t}\left(3e^t(4t - 1) - 16\right) \\
     {1 \over 24}e^{3t}\left(3e^t(4t + 4) - 16\right) \\
     {1 \over 24}e^{3t}\left(3e^t(4t - 1) - 16\right)
  \end{bmatrix} + \begin{bmatrix}
           2e^t - 2te^{2t} &      -2te^{2t} & 0 \\
    -2e^t + 2(t + 1)e^{2t} & 2(t + 1)e^{2t} & 0 \\
                  2te^{2t} &       2te^{2t} & 2e^t
  \end{bmatrix}\begin{bmatrix}c_1 \\c_2 \\c_3\end{bmatrix} ~,
\end{align}</math>
which could be further simplified to get the requisite particular solution determined through variation of parameters.
Note '''c''' = '''y'''<sub>''p''</sub>(0). For more rigor, see the following generalization.