transient response of simple digitalized RC low-pass filter
If you assume a zero initial condition (i.e., if the system is switched on at $n=0$, then $y[-1]=0$), then the system is fully described by its impulse response, i.e. its response to the unit impulse $x[n]=\delta[n]$ (which is zero everywhere, except at $n=0$, where it is $1$).

Write the system equation as

$$y[n]=b_0x[n]+b_1x[n-1]-a_1y[n-1]\tag{1}$$

In your example $b_0=b_1$. If you apply $x[n]=\delta[n]$, you get the following response:

$$\begin{align}y[0]&amp;=b_0\\y[1]&amp;=b_1-a_1b_0\\
y[2]&amp;=-a_1(b_1-a_1b_0)\\
y[3]&amp;=a_1^2(b_1-a_1b_0)\\\vdots\end{align}$$

So we get for the impulse response

$$h[n]=\begin{cases}0,&amp;n&lt;0\\b_0,&amp;n=0\\
(-a_1)^{n-1}(b_1-a_1b_0),&amp;n&gt;0\end{cases}\tag{2}$$

The general input-output relation is described by the convolution of the input signal and the impulse response:

$$y[n]=\sum_{k=0}^{\infty}h[k]x[n-k]\tag{3}$$

E.g., the step response can be easily computed from the impulse response:

$$a[n]=\sum_{k=0}^{n}h[k]\tag{4}$$