Editing Matrix exponential (section)

=== Inhomogeneous case generalization:  variation of parameters ===
For the inhomogeneous case, we can use [[integrating factor]]s (a method akin to [[variation of parameters]]). We seek a particular solution of the form {{math|1='''y'''<sub>p</sub>(''t'') = exp(''tA'') '''z'''(''t'')}},
<math display="block">\begin{align}
  \mathbf{y}_p'(t)
    & = \left(e^{tA}\right)'\mathbf{z}(t) + e^{tA}\mathbf{z}'(t) \\[6pt]
    & = Ae^{tA}\mathbf{z}(t) + e^{tA}\mathbf{z}'(t) \\[6pt]
    & = A\mathbf{y}_p(t) + e^{tA}\mathbf{z}'(t)~.
\end{align}</math>

For {{math|'''y'''<sub>''p''</sub>}} to be a solution,
<math display="block">\begin{align}
  e^{tA}\mathbf{z}'(t) &= \mathbf{b}(t) \\[6pt]
        \mathbf{z}'(t) &= \left(e^{tA}\right)^{-1}\mathbf{b}(t) \\[6pt]
         \mathbf{z}(t) &= \int_0^t e^{-uA}\mathbf{b}(u)\,du + \mathbf{c} ~.
\end{align}</math>

Thus,
<math display="block">\begin{align}
  \mathbf{y}_p(t)
    & = e^{tA}\int_0^t e^{-uA}\mathbf{b}(u)\,du + e^{tA}\mathbf{c} \\
    & = \int_0^t e^{(t - u)A}\mathbf{b}(u)\,du + e^{tA}\mathbf{c}~,
\end{align}</math>
where {{math|'''''c'''''}} is determined by the initial conditions of the problem.

More precisely, consider the equation
<math display="block">Y' - A\ Y = F(t)</math>

with the initial condition {{math|1=''Y''(''t''<sub>0</sub>) = ''Y''<sub>0</sub>}}, where
* {{mvar|A}} is an {{mvar|n}} by {{mvar|n}} complex matrix,
* {{mvar|F}} is a continuous function from some open interval {{mvar|I}} to {{math|'''C'''<sup>''n''</sup>}},
* <math>t_0</math> is a point of {{mvar|I}}, and
* <math>Y_0</math> is a vector of {{math|'''C'''<sup>''n''</sup>}}.

Left-multiplying the above displayed equality by {{math|''e<sup>−tA</sup>''}} yields
<math display="block">Y(t) = e^{(t - t_0)A}\ Y_0 + \int_{t_0}^t e^{(t - x)A}\ F(x)\ dx ~.</math>

We claim that the solution to the equation
<math display="block">P(d/dt)\ y = f(t)</math>

with the initial conditions <math>y^{(k)}(t_0) = y_k</math> for {{math|0 ≤ ''k'' < ''n''}} is
<math display="block">y(t) = \sum_{k=0}^{n-1}\ y_k\ s_k(t - t_0) + \int_{t_0}^t s_{n-1}(t - x)\ f(x)\ dx ~,</math>

where the notation is as follows:
* <math>P\in\mathbb{C}[X]</math> is a monic polynomial of degree {{math|''n'' > 0}},
* {{mvar|f}} is a continuous complex valued function defined on some open interval {{mvar|I}},
* <math>t_0</math> is a point of {{mvar|I}},
* <math>y_k</math> is a complex number, and

{{math|''s<sub>k</sub>''(''t'')}}  is the coefficient of <math>X^k</math> in the polynomial denoted by <math>S_t\in\mathbb{C}[X]</math> in Subsection [[matrix exponential#Evaluation by Laurent series|Evaluation by Laurent series]] above.

To justify this claim, we transform our order {{mvar|n}} scalar equation into an order one vector equation by the usual [[Ordinary differential equation#Reduction to a first-order system|reduction to a first order system]]. Our vector equation takes the form
<math display="block">\frac{dY}{dt} - A\ Y = F(t),\quad Y(t_0) = Y_0,</math>
where {{mvar|A}} is the [[transpose]] [[companion matrix]] of {{mvar|P}}. We solve this equation as explained above, computing the matrix exponentials by the observation made in Subsection [[matrix exponential#Evaluation by implementation of Sylvester's formula|Evaluation by implementation of Sylvester's formula]] above.

In the case {{mvar|n}} = 2 we get the following statement. The solution to
<math display="block">
  y'' - (\alpha + \beta)\ y' + \alpha\,\beta\ y = f(t),\quad
  y(t_0) = y_0,\quad
  y'(t_0) = y_1
</math>

is
<math display="block">y(t) = y_0\ s_0(t - t_0) + y_1\ s_1(t - t_0) + \int_{t_0}^t s_1(t - x)\,f(x)\ dx,</math>

where the functions {{math|''s''<sub>0</sub>}}  and {{math|''s''<sub>1</sub>}} are as in Subsection [[matrix exponential#Evaluation by Laurent series|Evaluation by Laurent series]] above.