Strictly speaking, an LTI system (characterized by an LCCDE) can have a zero-state response, but not a zero-input response. The latter requires nonzero initial conditions which conflicts with the requirement that an LTI system's LCCDE should have zero initial conditions, a.k.a. initial-rest.
However, this problem can be understood in the sense that the given LCCDE represents an LTI system under initial-rest assumptions, yet we still want to compute its response due to nonzero initial conditions as well, in which case the LCCDE actually represents a non LTI system.
Then you can proceed as follows:
Lets use one-sided (unilateral) Z-transform to account for the initial conditions:
$$ X(z) = \sum_{n=0}^{\infty} x[n] z^{-n} \tag{1} $$
and its useful property:
$$ x[n-1] \longleftrightarrow z^{-1}X(z) + x[-1] \tag{2}$$
and for general shift of k:
$$ x[n-k] \longleftrightarrow z^{-k} X(z) + \sum_{m=0}^{k-1} z^{-m}x[m-k] \tag{3} $$
Given an N-th order LCCDE:
$$ \sum_{k=0}^{N} a_k y[n-k] = \sum_{k=0}^{M} b_k x[n-k] \tag{4} $$
Apply Eq.3 to Eq.4, assuming $x[n]=0, n< 0$:
$$ \sum_{k=0}^{N} a_k \left( z^{-k} Y(z) + \sum_{m=0}^{k-1} z^{-m} y[m-k] \right) = \sum_{k=0}^{M} b_k z^{-k} X(z) \tag{5} $$
$$ Y(z) \left( \sum_{k=0}^{N} a_k z^{-k} \right) + \sum_{k=0}^{N} a_k \sum_{m=0}^{k-1} z^{-m} y[m-k] = X(z) \left(\sum_{k=0}^{M} b_k z^{-k} \right) \tag{6} $$
Denote the double sum $\sum_{k=0}^{N} a_k \sum_{m=0}^{k-1} z^{-m} y[m-k]$ as $Y_{ic}(z)$ the initial-conditions part of output, then Eq.6 becomes:
$$ Y(z) \left( \sum_{k=0}^{N} a_k z^{-k} \right) + Y_{ic}(z) = X(z) \left(\sum_{k=0}^{M} b_k z^{-k} \right) \tag{7} $$
Then express $Y(z)$ as:
$$ Y(z) = \frac{ \sum_{k=0}^{M} b_k z^{-k}}{\sum_{k=0}^{N} a_k z^{-k}} X(z) - \frac{ Y_{ic}(z) }{\sum_{k=0}^{N} a_k z^{-k} } \tag{8} $$
Denoting $\frac{ \sum_{k=0}^{M} b_k z^{-k}}{\sum_{k=0}^{N} a_k z^{-k}}$ as $H(z)$; the Transfer Function of the associated LTI system, we have:
$$ Y(z) = H(z) X(z) - \frac{ Y_{ic}(z) }{\sum_{k=0}^{N} a_k z^{-k} } \tag{9} $$
Eq.9 expresses the LCCDE output $Y(z)$ in terms of the input $X(z)$ and the initial-conditions $Y_{ic}(z)$.
The first term $H(z) X(z)$ is the zero-state response (or forced response) $Y_{zs}(z)$ corresponding to the input $X(z)$, and the second term is the zero-input response (or natural response) $Y_{zi}(z)$ corresponding to the initial-conditions $Y_{ic}(z)$ which is typically a transient response due to initial conditions.
Hence we have:
$$Y(z) = Y_{zs}(z) + Y_{zi}(z) \tag{10}$$
The zero-state response, $Y_{zs}(z) = H(z)X(z)$ corresponds to the LTI system that's characterized by the LCCDE, because it assumes zero initial conditions;i.e., $Y_{IC}(z) = 0$, and then the LCCDE output is given by :
$$ Y(z) = H(z)X(s) \longleftrightarrow y[n] = h[n] \star x[n] \tag{11} $$
Eq.11 displays the equivalence between the zero-state response of an LCCDE and the convolution output of the corresponding LTI system with impulse response $h[n]$.
For a suddenly applied sinudoidal $x[n] = K \cos(\omega_0 n)u[n]$, the zero-state response will include two terms: a switching transient part that corresponds to the sudden application of the input; i.e. $x[n] = 0 , n < 0$ , and a sinusoidal steady-state part that remains persistent after switching transients (and initial-condition transients as well, if that also exists) decay, and which corresponds to the response of the LTI system as if the input were $x[n] = \cos(\omega_0 n)$.
When applied to your LCCDE, you get the following:
$$ y[n] + a y[n-1] = b_0 x[n] ~~~,~~~ y[-1]=2 \tag{12} $$
and
$$x[n] = K \cos(\omega_0 n)u[n] \leftrightarrow \frac{K/2}{ 1 - e^{j \omega_0} z^{-1} } + \frac{K/2}{ 1 - e^{-j \omega_0} z^{-1} } \tag{13}$$
where $a=-0.9 ,~ b_0 = 0.1 ,~ \omega_0 = 0.2 \pi$ , and $K=20$.
Then the total output becomes:
$$ Y(z) = \frac{b_0}{1 + a z^{-1}} \left( \frac{K / 2} {1 - e^{j \omega_0} z^{-1}} + \frac{K / 2} {1 - e^{-j \omega_0} z^{-1}} + \right) - \frac{ a y[-1]}{1 + a z^{-1}} \tag{14}$$
Where the term in the paranthesis corresponds to the zero-state output, and the remaing last term is the zero-input response.
Apply partial fraction expansion, PFE, to the zero-state response part:
$$Y(z) = \frac{A_1}{1 - e^{j\omega_0} z^{-1}} + \frac{A_2}{1 - e^{-j\omega_0} z^{-1}} + \frac{B_1}{1 + a z^{-1}} + \frac{B_2}{1 + a z^{-1}} - \frac{a y[-1]}{1 + a z^{-1}}\tag{15} $$
It will turn out that $A_2 = A_1^*$ and $B_2 = B_1^*$, and we have
$$A = \frac{b_0 K/2}{1 + a e^{-j \omega_0}} ~~ ,~~ B = \frac{b_0 K/2}{1 + a^{-1} e^{j\omega_0}} $$.
Finally in the time-domain, the following is obtained for the total output:
$$y[n] = 2|A| \cos(\omega_0 n + \phi_A) + 2 \mathcal{Re}\{B\} (-a)^n - a y[-1]( -a)^n \tag{16} $$
For the given values it happens to be:
$$y[n] = 3.3626 \cos(0.2\pi n - 1.0960) u[n] + 0.4629 (0.9)^n u[n] + 1.8 (0.9)^n u[n] \tag{17} $$
Finally we can denote the parts of total output as follows:
the sinusoidal steady-state response:
$$y_{ss}[n] = 3.3626 \cos(0.2\pi n - 1.0960) u[n] $$
the input switching transient response:
$$y_{is}[n] = 0.4629 (0.9)^n u[n] $$
the zero-state response:
$$y_{zs}[n] = y_{ss}[n] + y_{is}[n] $$
the zero-input (transient) response:
$$ y_{zi}[n] = 1.8 (0.9)^n u[n] $$
