In my field, electrical engineering, we frequently study linear time-invariant systems of the following form:

$$ a_n\frac{d^ny}{dt^n} + a_{n-1}\frac{d^{n-1}y}{dt^{n-1}} + \ldots + a_1\frac{dy}{dt} + a_0y(t) = b_m\frac{d^nx}{dt^n} + b_{m-1}\frac{d^{n-1}x}{dt^{n-1}} + \ldots + b_1\frac{dx}{dt} + b_0x(t), $$ or, using operator notation, $$ Ly = Rx, $$ where $x(t)$ is some input function (signal), and $y(t)$ is the system response. Using the Laplace transform, one can find the solution $y(t)$ for a generic input $x(t)$, assuming the system is relaxed, using the convolution integral: $$ y(t) = \int h(\tau) x(t-\tau) d\tau, $$ where $h(t)$ is the response of the system to a unit impulse, $\delta(t)$. This formulation is ubiquitous in engineering literature. The reason we don't simply just compute the derivatives on the RHS is that we want to study the behavior of the system independent of what input is being applied.

However, in physics and applied mathematics literature, I usually see the following form: $$ Ly = x, $$ for which the solution is found using the Green's function: $$ y(t) = \int G(\tau) x(t-\tau) d\tau, $$ which appears to be a special case when $R = I$.

How do physicists deal with the case when $R \ne I$?

Another related question: Is is possible to rewrite any linear, constant coefficient differential equation: $$ Ly = Rx $$ in this form?: $$ R^{-1}Ly = x. $$ If so, that could alleviate this concern.

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    $\begingroup$ Final comment in question seems accurate to me, though it is no longer a differential equation, then, but an integro-differential equation. $\endgroup$ – Sean E. Lake Oct 19 '17 at 16:40
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    $\begingroup$ I think I'm saying the same thing, but what is wrong with inserting Rx after the Green's function G in the integral? $\endgroup$ – Cosmas Zachos Oct 19 '17 at 18:10
  • $\begingroup$ Isn't the $L \ y = x$ form just a linear response while the $L \ y = R \ x$ form is a nonlinear response? $\endgroup$ – honeste_vivere Oct 19 '17 at 18:54
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    $\begingroup$ @honeste_vivere, it's not a nonlinear response since both $L$ and $R$ are differential operators, e.g., $$R = \sum_n r_n \frac{d^n}{dt^n}$$ Taking the Laplace transform of both sides gives the transfer function: $$H(s) \equiv \frac{Y(s)}{X(s)} = \frac{R(s)}{L(s)}$$ i.e., the poles come from the $L$ operator while the zeros come from the $R$ operator. $\endgroup$ – Alfred Centauri Oct 19 '17 at 19:45
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    $\begingroup$ Green's functions are more general since time (or shift) invariance isn't required. For example, a time varying system has a Green's function $G(t,\tau)$ where, in general, $G(t - \tau, 0) \ne G(t,\tau)$, i.e., the response to $\delta(t - \tau)$ is not the time shifted response to $\delta(t)$. $\endgroup$ – Alfred Centauri Oct 19 '17 at 20:01

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