Does this equation have a closed form
$$y[n] = x[n] + (1-Q) y[n-1]$$

where $y[-1] = 0$ and otherwise $n \ge 0$.

this is a difference equation with both a "natural response" (when the input is always zero, but you might have initial conditions) and a "forced response".  a.k.a. the "transient response" and the "steady-state response" the math guys might call these the "homogeneous solution" and the "particular solution".

this is LTI so then the exponential function is an eigenfunction for the natural response.

for the natural solution, try 
$$ y_\text{n}[n] = y_\text{n}[0] a^n $$

then
$$\begin{align}
 y[n] &amp;= x[n] + (1-Q) y[n-1] \\
 y_\text{n}[0] a^n &amp;= 0 + (1-Q) y_\text{n}[0] a^{n-1} \\
               a^n &amp;= (1-Q) a^{n-1} \\
                   &amp;= (1-Q) a^n a^{-1} \\
               1   &amp;= (1-Q) a^{-1} \\
\end{align}$$

so i guess $a = 1-Q$.

for steady state, you posit a solution that is like the input plus all of its finite differences.  since the input is constant and all finite differences of a discrete constant function are zero, then the steady state output is a constant.  you sum the scaled steady state with the scaled transient response for the complete response.

$$\begin{align}
 y[n] &amp;= x[n] + (1-Q) y[n-1] \\
 C + y_\text{n}[0] a^n &amp;= D + (1-Q)(C + y_\text{n}[0] a^{n-1}) \\
 C + y_\text{n}[0] (1-Q)^n &amp;= D + (1-Q)(C + y_\text{n}[0] (1-Q)^{n-1}) \\
                           &amp;= D + (1-Q)C + y_\text{n}[0] (1-Q)^n) \\
                      C &amp;= D + (1-Q)C \\
\end{align}$$

i guess that means that $QC = D$ and $C = \tfrac{D}{Q}$.

now you add your natural response to your forced response.

$$ y[n] =  y_\text{n}[n] + C $$

lastly, for your initial condition

$$\begin{align}
 y[n] &amp;= x[n] + (1-Q) y[n-1] \\
 y[0] &amp;= x[0] + (1-Q) y[-1] \\
 C + y_\text{n}[0] (1-Q)^0 &amp;= D + (1-Q)\cdot 0 \\
 C + y_\text{n}[0] &amp;= D  \\
\end{align}$$

i guess that means that $ y_\text{n}[0] = D-C = D - \tfrac{D}{Q} $.

$$\begin{align}
  y[n] &amp;= y_\text{n}[n] + C \\
       &amp;= y_\text{n}[0]a^n + C \\
       &amp;= (D - \tfrac{D}{Q})(1-Q)^n + \tfrac{D}{Q} \\
\end{align}$$

for $n \ge 0$.