Current in charged RC circuit
Capacitor is charged up to 10 V when the switch is closed.

When the switch opens, the circuit consists of one capacitor (the source, with initial voltage U = 10V) and two resistors (R1 and R2) connected in series. 

So

\$V_C + V_{R1} + V_{R2} = 0 \$

Since the current is the same

\$I_C = C\frac{dV_C}{dt} = I_{R1} = I_{R2} = I \$

With this info, it's easy to continue the analysis.



EDIT

You can prove that after the switch opens (t=0) capacitor's voltage is

\$V_C(t) = V_C(0) \cdot e^{-\frac{t}{(R_1+R_2)C}} \$

and from this find current' s equation