Attempting to use the trapezoidal rule to form a difference equation representing a circuit
The Laplace transform of the original differential equation is
$$Cs V_o(s) = V_i(s) \left( \frac{1}{R_1} \right) - V_o(s) \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$$
$$R_1 R_2 C s V_o(s) = R_2 V_i(s) - (R_1 + R_2) V_o(s).$$
In the Laplace domain, the ratio $V_o(s)/V_i(s)$ is the transfer function for the circuit,
$$H(s) = \frac{ V_o(s) }{ V_i(s) } = \frac{ R_2 }{ R_1 + R_2 + R_1 R_1 C s }
= \frac{K}{1 + R_p C s},$$
where $K = R_2 / (R_1 + R_2)$ is the voltage divider ratio for $R_1$ and $R_2$
in a series connection, and $R_p = R_1 R_2 / (R_1 + R_2)$ is the equivalent
resistant for $R_1$ and $R_2$ in a parallel connection.
We can obtain the $z$-domain transfer function for an equivalent digital filter
using the bilinear transformation,
$$s = \frac{2}{T} \left( \frac{z-1}{z+1} \right)$$.
In this transformation, the constant is set to $2/T$, which will produce a digital filter that is an approximation to the analog filter that is equivalent to approximating the convolution integral by the trapezoidal rule.  I assume this is what you are looking for.
By making this substitution, the transfer function for the digital filter is
$$H(z) = \frac{K}{1 + R_p C \left( \dfrac{2}{T} \right) 
\left( \dfrac{z-1}{z+1} \right)}
= \dfrac{ KT (z+1) }{ T(z+1) + 2 R_p C (z-1) }$$.
This can be further simplified to
$$H(z) = \frac{V_o(z)}{V_i(z)} =
\frac{ \left( \dfrac{KT}{T + 2 R_p C} \right) \left( 1 - z^{-1} \right) }
{1 + \left( \dfrac{T - 2 R_p C}{T + 2 R_p C} \right) z^{-1}}.$$
Taking the inverse $z$-transform of this equation, we obtain a
difference equation for the digital filter,
$$Y(z) \left[ 1 + \left( \dfrac{T - 2 R_p C}{T + 2 R_p C} \right) z^{-1} \right]
= X(z) \left[ \left( \dfrac{KT}{T + 2 R_p C} \right) \left( 1 - z^{-1} \right) \right].$$
$$y[n] = \left( \dfrac{KT}{T + 2 R_p C} \right) \left( x[n] - x[n-1] \right)
- \left( \dfrac{T - 2 R_p C}{T + 2 R_p C} \right) y[n-1].$$
You should be able to use this equation to recursively compute $y[n]$.