If I was given a differential equation of the form $$y'(t) + a(t)y(t) = x(t)$$ how would I be able to decide its linearity, time-invariance, and causality?

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    $\begingroup$ well, it ain't time-invariant if $a(t)$ changes. if $a(t)=a$ is constant, it is time-invariant. $\endgroup$ – robert bristow-johnson Nov 14 '18 at 3:19
  • $\begingroup$ Thank you! Appreciate it. Apologies for the formatting in my question. $\endgroup$ – Benny Nov 14 '18 at 3:21
  • $\begingroup$ i haven't been picking on you about formatting. sometimes it's easier just to do it. $\endgroup$ – robert bristow-johnson Nov 14 '18 at 3:23
  • $\begingroup$ usually we use the terms "linearity", "time-invariancy", and "causality" to describe properties of a system with an input or inputs and an output or outputs. that system may have an input/output relationship defined by a differential equations such as you have shown. you might want to be explicit about what signal is the input and what is the output. $\endgroup$ – robert bristow-johnson Nov 14 '18 at 3:27
  • $\begingroup$ without stating the initial conditions as initial rest you cannot determine whether the equation signifies an LTI systen or not. $\endgroup$ – Fat32 Nov 19 '18 at 11:05

If the input $x(t)$ and the output $y(t)$ and their derivatives occur linearly in the equation, i.e., not in the form $f\left(x^{(n)}(t)\right)$ or $f\left(y^{(n)}(t)\right)$, where $f(\cdot)$ is a non-linear function, then the differential equation can describe a linear system.

Time-invariance is determined by the factors with which $x(t)$ and $y(t)$ and their derivatives are multiplied. If they are constant, and, consequently, not functions of $t$, then the equation can describe a time-invariant system. If that is the case, and if the equation is also linear, then it is called Linear Constant-Coefficient Differential Equation. The general form of such an equation is


Given only the differential equation, you cannot draw any conclusions concerning causality. Taking your example with $a(t)=a$ being constant (for time-invariance), you can verify that the given differential equation describes two different systems. One is a causal system with impulse response $h(t)=e^{-at}u(t)$, and the other one is anti-causal with impulse response $h(t)=-e^{-at}u(-t)$.

  • $\begingroup$ for your first paragraph; that describe a linear differential equation but necessarily a linear system, unless initial conditions are all zero (initial rest). Specifically if $y(0)=1$ for then the output will not be linear in $x$. That's also true for time-invariance; initial conditions should be consistent. Assuming initial rest is an easy cover up... $\endgroup$ – Fat32 Nov 19 '18 at 11:12
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    $\begingroup$ @Fat32: yes, I agree. I've had many (fruitless) discussions about this, and I wanted to avoid it this time. This is why I said "can describe a linear system". $\endgroup$ – Matt L. Nov 19 '18 at 11:16
  • $\begingroup$ And I've added a second "can" in the other paragraph on time-invariance. $\endgroup$ – Matt L. Nov 19 '18 at 11:18
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    $\begingroup$ Anyway, Benny doesn't seem to care anymore ... :) $\endgroup$ – Matt L. Nov 19 '18 at 11:20

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