Exercise
Consider the system $ \dot x = Ax+bu$ with $$ A = \begin{bmatrix} -6 & 2 \\ -6 & 1 \end{bmatrix},~~~ b= \begin{bmatrix} 1 \\ 0 \end{bmatrix},~~~ c=\begin{bmatrix} 1 & 1 \end{bmatrix} $$
Calculate the output response with initial values $x(0)= \begin{bmatrix} 2 & 1 \end{bmatrix}^T $ and a step input $u(t) = 2σ(t)$
Solution
The state response with the given input is $$ x(t) = x_0(t) + \int_0^t e^{A(t-\tau)}bu(\tau)d\tau $$ The output response with the given input is, with $y_0(t)$ from above as shown below. The matrix $e^A(t−τ)$ is partitioned as $$e^{A(t-\tau)}= \begin{bmatrix} \Phi_{11} & \Phi_{12} \\ \Phi_{21} & \Phi_{22} \end{bmatrix} $$ $$y(t) = y_0(t) +c \int_0^t e^{A(t-\tau)} bu(\tau)d\tau$$ $$=y_0(t)+2 \int_0^t \begin{bmatrix} 1 & 1 \end{bmatrix} \begin{bmatrix} \Phi_{11} & \Phi_{12} \\ \Phi_{21} & \Phi_{22} \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} d\tau $$ $$=y_0(t)+2 \int_0^t (\Phi_{11} +\Phi_{21}) d\tau $$ $$=y_0(t)+2 \int_0^t (\alpha_0(t-\tau)-6\alpha_1(t-\tau)-6\alpha_1(t-\tau))d\tau$$ $$=y_0(t)+2 \int_0^t (10e^{-(t-\tau)}-9e^{-2(t-\tau)})d\tau $$
$$ y(t) = y_0(t)-\frac{20}{3} (e^{-3t}-1)+\frac{18}{2}(e-^{-2t}-1)$$ $$(\int_0^t e^{g(t-\tau)}d\tau = \frac{1}{g}(e^{gt}-1)) $$
Questions
From where does the $c$ in front of $\int_0^t e^{A(t-\tau)} bu(\tau)d\tau$ come?
Why is a $2$ in front of the next $\int_0^t$?
How is $(\Phi_{11} +\Phi_{21})$ converted into $(\alpha_0(t-\tau)-6\alpha_1(t-\tau)-6\alpha_1(t-\tau))$?
$y(t)$ is defined as $c\,x(t)$. So by just multiplying both terms of $x(t)$ by $c$ you get $y(t)$. Remember that $y_0(t)=c\,x_0(t)$.
After this you also need to use the substitution for $u(t)$ in the integral, which is two times the step function. Two two is a constant so can be taken outside of the integral. Since the step function always one on the interval of the integral (assuming $t\geq0$), so that term also vanishes since multiplying by one does not change anything.
Your last question is related to the pervious question you asked on this site (however I never use those intermediate variables myself, but I am not familiar with you method you are following).