Is initializing a digital filter's output with no "momentum" a non-trivial task?
If you want a stable linear time-invariant system to output constant $y$, it must have received input $x$ that is the ratio of the constant output and the zero frequency response of the system $H(1):$

$$y = H(1)x\quad \Leftrightarrow \quad x = \frac{y}{H(1)}$$

Or, you'd want to change the state of the system so that it reflects that situation. The system must satisfy the condition $H(1) \ne 0.$

For a composite filter that consists of a serially connected cascade of second-order sections, you can start with the first section with input $x$ and calculate sequentially using $y = H(1)x$ (recycling the variable names to mean the inputs and outputs of a single section) the constant intermediate output of each section. Or you could start from the other end and use $x = y/H(1).$ Knowing both the input $x$ and output $y$, use the signal flow diagram of each section to calculate its state variables, starting with the summation point with dependency only to known constants. Because all oscillation has settled due to constant input and stability of the filter, all signals are constants rather than functions of time, and the delays in the flow diagram have become identities.

Figure 1. Transposed direct form II biquad signal flow diagram. (CC BY-SA 3.0 by Fcorthay)

Given constant input $x$, the constant output $y$ of a biquad section decipted in Fig. 1 is:

$$y = H(1)x = \frac{b_0+b_11^{-1}+b_21^{-2}}{1+a_11^{-1}+a_21^{-2}}x = \frac{b_0 + b_1 + b_2}{1 + a_1 + a_2}x$$

As for the state variables, the top sum $m_0,$ the middle sum $m_1,$ and the bottom sum $m_2$ are calculated as (start calculation from the bottom):

$$\begin{align}
m_0 &amp;= b_0x + m_1\\
m_1 &amp;= b_1x - a_1y + m_2\\
m_2 &amp;= b_2x - a_2y\\
\end{align}$$

The top sum $m_0$ should equal $y$, which we can verify:

$$\begin{align}m_0 &amp;= b_0x + b_1x - a_1y + m_2\\
&amp;= b_0x + b_1x - a_1y + b_2x - a_2y\\
&amp;= b_0x + b_1x - a_1\frac{b_0 + b_1 + b_2}{1 + a_1 + a_2}x + b_2x - a_2\frac{b_0 + b_1 + b_2}{1 + a_1 + a_2}x\\
&amp;= \frac{b_0 + b_1 + b_2}{1 + a_1 + a_2}x\\
&amp;= y
\end{align}$$