Consider a linear and time-invariant system with differential equation $Q(D)y(t) = P(D)x(t)$ where $Q(D)$ and $P(D)$ are differential operator polynomials.
We define the zero-state response as the solution to this differential equation with all initial conditions set to $0$
$$Q(D)y_{zs}(t) = P(D)x(t) \tag{1}.$$
We define the zero-input response as the the solution to this differential equation with non-zero initial conditions
$$Q(D)y_{zi}(t) = 0 \tag{2}.$$
The decomposition property for linear systems states that the solution to $(1)$ plus the solution to $(2)$ is the same as the solution to
$$Q(D)[y_{zs}(t)+y_{zi}(t)] = P(D)x(t) $$
with non-zero initial conditions.
My question is why is this true? How can one show that the decomposition holds for any linear system?
Clearly, $y_{zs} + y_{zi}$ satisfies the initial conditions, so we just need to show that it also satisfies the differential equation: $$ Q(D)[y_{zs}(t) + y_{zi}(t)] = Q(D) y_{zs}(t) + Q(D)y_{zi}(t) = P(D)x(t) + 0 = P(D) x(t) $$
By uniqueness of solution, if $y_{zs}+y_{zi}$ satisfies the differential equation and the initial conditions, we must have that $y = y_{zs}+y_{zi}$.