Zero State and Zero Input Responses from Steady State Response
Some books prefer different notations, but I will assume that zero-state response is the one due to pure input $x[n] = \cos(0.2 \pi n) u[n]$ under system zero-initial conditions, while the zero-input response is the one with $x[n] = 0$ and initial state being $y[-1] = 2$.

Now, first the zero-input ($x[n] = 0$ ) response, purely due to the nonzero initial condition ($y[-1] = 2$), of the system can be found by a simple recursion of the causal system as:

$$
\begin{align}
y[0] &amp;= 0.9 \cdot 2 \\
y[1] &amp;= 0.9^2 \cdot 2 \\
y[2] &amp;= 0.9^3 \cdot 2 \\
... &amp;= ... \\
y[n] &amp;= 2 \cdot 0.9^{n+1} u[n] \\
\end{align}
$$

which yields for the zero input response (for $n \geq 0$) as

$$\boxed{ y_{zi}[n] = 2 \cdot (0.9)^{n+1} u[n] }$$

Now we shall find the zero-state response, assuming zero initial conditions. This will be lengthy because of the input form $x[n] = 20 \cos(0.2 \pi n) u[n]$, we can follow the following approach. 

Let $X(w)$ be the Fourier transform of the input, and $H(w)$ be the Freuqency response of the system described by the above differential equation, then the output $y[n]$ will have the Fourier transform given by $$Y(\omega) = H(\omega) X(\omega).$$

Now $$H(\omega) = \frac{0.1}{1- 0.9 e^{-j \omega} } $$ and you have to find out what $X(w)$ is, then multiply them to find out $Y(\omega)$ and perform an inverse Fourier transform, after a simple partial fraction expansion. Denoting this solution as $y_{zs}[n]$, then your total solution is

$$y[n] = y_{zi}[n] + y_{zs}[n]$$

Now, try this on your own and see if you can complete the solution... Then write a matlab program to check your result.