Variance of Autoregressive models, AR(1)

We have 
$$y_i =\phi_0+ \phi_1 y_{i-1} + \epsilon_i.$$
Then we use the formula for the variance of the sum of three random variables https://www.probabilitycourse.com/chapter6/6_1_2_sums_random_variables.php
\[Var(x_1+x_2+x_3)=Var(x_1)+Var(x_2)+Var(x_3)+2Cov(x_1,x_2)+2Cov(x_1,x_3)+2Cov(x_2,x_3).     (*)\]
In this case we have $x_1=\varphi_0$, $x_2=\varphi_1 y_{i-1}$, and $x_3=\epsilon_i$.
Using (*) we get 

$$Var(y_t)=var(\varphi_0+ \phi_1 y_{i-1} + \epsilon_i.)$$
\[=Var(\varphi_0)+Var(\phi_1 y_{i-1})+Var(\epsilon_i)+2Cov(\phi_0,\phi_1 y_{i-1})+2Cov(\phi_0,\epsilon_i)+2Cov(\phi_1 y_{i-1},\epsilon_i)  (1).\]
Now note that since $\phi_0$ is constant it has zero variance $var (\phi_0)=0.$ Also $\phi_0$, $y_{i-1}$, and $\epsilon_i$ are independent with respect to each other, and hence
\[ Cov(\phi_0,\phi_1 y_{i-1})=Cov(\phi_0,\epsilon_i)=Cov(\phi_1 y_{i-1},\epsilon_i)=0. \]
Indeed a constant random variable is independent of any rndom variable. Finally note that
\[Var(\phi_1 y_{i-1}) =\phi_1^2 Var(y_{i-1}) \]
as in general $var (c x)=c^2var(x)$ for any constant $c$ and any random variable $x$. Substituting all this in the formula (1) we get 
\[=Var(\varphi_0)+Var(\phi_1 y_{i-1})+Var(\epsilon_i)+2Cov(\phi_0,\phi_1 y_{i-1})+2Cov(\phi_0,\epsilon_i)+2Cov(\phi_1 y_{i-1},\epsilon_i) \]
\[=0+\phi_1^2Var( y_{i-1})+Var(\epsilon_i)+0+0+0  \]
\[=\phi_1^2Var( y_{i-1})+\sigma^2\]
as $Var(\epsilon_i)=\sigma^2$.

This took me about 30 minutes to answer. Please offer higher bounties in the future :)