Understanding equation for capacitor-voltage
\$A_1\$ is the voltage that will be across the C if the actual circuit would remain like it is for \$t \rightarrow \infty\$


It looks the exponential components lacks a sign. \$e^{+xt}\$ goes to \$\infty\$ when \$t \rightarrow \infty\$.
But with a minus sign, yes, your understanding is correct. $$ \lim_{x \to \infty} u_c = \lim_{x \to \infty} (A_1+A_2\cdot e^{\color{red}{-}\frac{t}{\tau}}) = A_1 $$

\$A_1\$ is the steady state value.




  \$A_2\$ = actual voltage over the C - \$A_1\$


\$u_c\$ is the actual voltage over the C. So, substituting this in your statement:
$$A_2 = u_c - A_1 $$
But, \$u_c\$ has been defined as
$$ u_c = A_1+A_2\cdot e^{\color{red}{-}\frac{t}{\tau}} $$
Substiting \$u_c\$ yields

$$u_c - A_1 = (A_1+A_2\cdot e^{-\frac{t}{\tau}})-A_1 = A_2\cdot e^{-\frac{t}{\tau}} \color{red}{\neq} A_2 $$

So, that quoted statement is not correct.

 
\$A_2\$ is the factor to solve the boundary condition of the ODE (at \$t=0\$).