Is $y(t) = y(t-4)+x(t-4)$ time invariant or not?

I want to check the time invariability of this recursively defined function $$y(t) = y(t-4)+x(t-4)$$

We can check time invariability of functions expressed in terms of x(t), but I couldn't find anything for such recursively defined function. In the book "Signals and System by A. Nagoorkani", we can test for time invariance as follow:

1. Delay the input signal by m units of time and determine the response of the system for this delayed input signal. Let this response be y1(t).
2. Delay the response of the system for unshifted input by m unit of time. Let this delayed response by y2(t).
3. Check whether y1(t) = y2(t). If they are equal then the system is time invariant. Otherwise the system is time variant.

Following these steps, the delayed version of input signal is $$x((t-m)-4)$$ But what is its response (y1)? Is it

$$y(t-4) + x((t-m)-4)$$ $$\text{or}$$ $$y((t-m)-4) + x((t-m)-4)?$$

In other cases, we just shift the function $$x(t)$$ and leave all other terms as it is but the problem here is, the shifting of $$x(t)$$ may affect the $$y(t)$$ and so $$y(t-4)$$ as well. So, should I shift that term as well or keep it as it is?

Let's re-write your difference equation: $$y(t) - y(t-4) = x(t-4)$$
• Delaying the input gives the following difference equation for the output $$y_1(t)$$:$$y_1(t) - y_1(t-4) = x(t-m-4)$$
• A delayed response to an input $$x(t)$$ gives $$y(t-m) - y(t-m-4) = x(t-m-4)$$
From there you can see that $$y_1(t)$$ and $$y(t-m)$$ satisfy the same difference equation. The system is therefore indeed time-invariant.