KCL on source-free RC circuit - confused about current direction
In your solution, you have written the differential equation for a charging capacitor. Clearly, with your nominated current direction and assuming the top plate of \$\small C\$ is initially positively charged, the circuit represents a discharging capacitor where the rate of change of capacitor voltage is negative. An appropriate solution is presented below.

The voltage across the capacitor, \$ v\$, for \$\small t \ge 0\$ is the difference between the initial voltage, \$\small V_0\$, and the voltage that has 'leaked' away due to the current flow, thus: $$v= V_0-\dfrac{1}{C}\int i\:dt$$

Differentiating: $$\dfrac{dv}{dt}=-\dfrac{i}{C}$$

But \$i\$ is the current flowing through \$R\$; thus \$i=\dfrac{v}{R}\$, and substituting in the above equation:

$$\dfrac{dv}{dt}=-\dfrac{v}{RC}$$

Solving the differential equation gives:

$$v= V_0e^{-t/RC}$$