Current in \$R{_1}\$ of resistor divider with capacitor
My first instinct would be to first apply Thevenin to \$V\$, \$R_1\$, and \$R_2\$ in order to reduce this to a simple series RC problem, whose solution is:

$$\begin{align*}
R_{th} &amp;= \frac{R_1\cdot R_2}{R_1+R_2} = 5\:\textrm{k}\Omega\\
 \\
V_{th} &amp;= V\cdot\frac{R_2}{R_1+R_2} = \frac{1}{2}\:V \\
 \\
\tau&amp;= R_{th} \cdot C_1 = 6\:\textrm{ms}\\
 \\
V_x=V_{R_1}&amp;=V_{th}\cdot\left(1-e^{-\cfrac{t}{\tau}}\right)=\frac{1}{2}\:V\cdot\left(1-e^{-\cfrac{t}{\tau}}\right)
\end{align*}$$

But that's not a general approach. It just works in this case.



Nodal analysis is way, way more general and would be, setting the bottom node arbitrarily to \$0\:\textrm{V}\$:

$$\begin{align*}
\frac{V_x}{R_1} + \frac{V_x}{R_2} + C_1\frac{\textrm{d} V_x}{\textrm{d} t}&amp;= \frac{V}{R_1} + \frac{0\:\textrm{V}}{R_2} \\
 \\
V_x\cdot\left(\frac{1}{R_1} + \frac{1}{R_2}\right) + C_1\frac{\textrm{d} V_x}{\textrm{d} t}&amp;= \frac{V}{R_1} \\
 \\
\frac{\textrm{d} V_x}{\textrm{d} t}+\frac{1}{C_1\frac{R_1\cdot R_2}{R_1+R_2}}V_x&amp;= \frac{V}{R_1 C_1} \\
 \\
\frac{\textrm{d} V_x}{\textrm{d} t}+\frac{1}{C_1 R_{th}}V_x&amp;= \frac{V_{th}}{C_1 R_{th}}
\end{align*}$$

Which is in a very familiar first-order ODE form. You can apply Laplace to that.

But just using the usual ODE solution method for first-order, gives:

$$\begin{align*}
P_t = \frac{1}{C_1 R_{th}},~~Q_t &amp;= \frac{V_{th}}{C_1 R_{th}},~~\therefore~~ \frac{\textrm{d} V_x}{\textrm{d} t}+P_t V_x= Q_t\\
 \\
\mu &amp;= e^{\int P_t\:\textrm{d} t} = e^{\left[\cfrac{t}{C_1 R_{th}}\right]}\\
 \\
V_x &amp;= \frac{1}{\mu}\int \mu\: Q_t~  \:\textrm{d} t \\
 \\
 &amp;= e^{\left[\cfrac{-t}{C_1 R_{th}}\right]}\int e^{\left[\cfrac{t}{C_1 R_{th}}\right]}\: \frac{V_{th}}{C_1 R_{th}}~  \:\textrm{d} t \\
 \\
 &amp;= e^{\left[\cfrac{-t}{C_1 R_{th}}\right]}\cdot \left(\frac{V_{th}}{C_1 R_{th}}\right)\cdot \int e^{\left[\cfrac{t}{C_1 R_{th}}\right]}\:~  \:\textrm{d} t \\
 \\
 &amp;= e^{\left[\cfrac{-t}{C_1 R_{th}}\right]}\cdot \left(\frac{V_{th}}{C_1 R_{th}}\right)\cdot \left(C_1 R_{th}\: e^{\left[\cfrac{t}{C_1 R_{th}}\right]}+C_0\right) \\
 \\
 &amp;= V_{th}\:e^{\left[\cfrac{-t}{C_1 R_{th}}\right]}\cdot \left(e^{\left[\cfrac{t}{C_1 R_{th}}\right]}+C_0\right) \\
 \\
 &amp;= V_{th}\left(1 + C_0\:e^{\left[\cfrac{-t}{C_1 R_{th}}\right]}\right)
\end{align*}$$

Using the initial condition that \$V_x\left(t=0\right) = 0\:\textrm{V}\$, this results in \$C_0=-1\$, so:

$$\begin{align*}
V_x &amp;= V_{th}\left(1 -e^{\left[\cfrac{-t}{C_1 R_{th}}\right]}\right) = \frac{1}{2}\:V\cdot\left(1-e^{-\cfrac{t}{\tau}}\right)
\end{align*}$$

And now you can see why the Thevenin approach would be chosen first.



Or use Laplace applied to the ODE mentioned earlier. Using the known initial condition, you get \$Y_s = \frac{Q_t}{s^2+P_t s}\$, which solves out as \$y_t=\frac{Q_t}{P_t}\left(1-e^{-P_t t}\right)=V_{th}\left(1-e^{\frac{-t}{\tau}}\right)=\frac{1}{2} V\left(1-e^{\frac{-t}{\tau}}\right)\$.



Of course, knowing \$V_x\$ it is trivial to then answer your title question regarding the current in \$R_1\$ over time.