Finding Voltage and Current across a capacitor. Will you check my work?
Starting from \$t=0\$, you know that \$V_{C(t=0)}=0\:\textrm{V}\$ and:

$$\begin{align*}
\frac{V_C}{R_1}+\frac{V_C}{R_3}+\frac{\textrm{d}V_C}{\textrm{d}t}C&amp;=\frac{V_0}{R_1}+\frac{0\:\textrm{V}}{R_3}\\\\
\frac{\textrm{d}V_C}{\textrm{d}t}+V_C\left(\frac{1}{R_1}+\frac{1}{R_3}\right)\frac{1}{C}&amp;=\frac{V_0}{R_1 \:C}\\\\
\hline\\
V_C^{'} + P_t\: V_C &amp;= Q_t,\\\\&amp; &amp; P_t&amp;=\frac{1}{C\left(R_1\:\mid\mid\: R_3\right)}\\\\
&amp; &amp; Q_t &amp;=\frac{V_0}{R_1 \:C}
\end{align*}$$

Which is in standard form for the obvious solution method used for 1st order linear differential equations.



The final solution (now that you've posted your own writing on this) is:

$$\begin{align*}
V_{C}&amp;=V_0\frac{R_3}{R_1+R_3}\left(1- e^{-\frac{-t}{C\cdot\left(R_1\:\mid\mid\:R_3\right)}}\right)\\\\
&amp;=8\left(1-e^{\frac{-t}{2}}\right)
\end{align*}$$

Just as you wrote.