Time invariance of y(t) = y(t−4) + x(t−4)

A system is time-invariant if a time shift of the input causes the same time shift of the output: if $x(t) \to y(t)$, then $x(t-t_0) \to y(t-t_0)$.

Given: $y(t) = y(t-4) + x(t-4)$ (a recursive system; assume it starts at rest).

Step 1: apply a delayed input $x_1(t) = x(t-t_0)$. Its output $y_1(t)$ satisfies

$$y_1(t) = y_1(t-4) + x_1(t-4) = y_1(t-4) + x(t-t_0-4)$$

Step 2: shift the original equation by $t_0$ (replace $t$ by $t-t_0$):

$$y(t-t_0) = y(t-t_0-4) + x(t-t_0-4)$$

Step 3: compare. The delayed output $y(t-t_0)$ satisfies exactly the same equation, driven by the same input $x(t-t_0-4)$, as $y_1(t)$. With the same (rest) initial condition, the solutions are equal:

$$y_1(t) = y(t-t_0)$$

The reason is that the coefficients (both equal to 1) and the delay (4 s) are constants; time $t$ does not appear explicitly in the equation.

Answer: the system $y(t) = y(t-4) + x(t-4)$ is time-invariant.