Finding impulse response of an RC circuit from its step response
The answer (How to calculate the impulse response of an RC circuit using time-domain method)  provides a direct time-domain solution of an RC circuit for the impulse reponse $h(t)$. Now this new answer modifies it to solve for the step-response $s(t)$ instead and then computes the impulse response according to :

$$h(t) = s(t)' $$

The differential equation of the first order circuit was derived as 
$$ \boxed{ y'(t) +\frac{1}{RC} y = x'(t) } \tag{1} $$

The step reponse $s(t)$ is defined as the output $y(t)$ of Eq.(1) when the input $x(t)$ is a unit-step function $$ x(t) = u(t) \implies y(t) = s(t)  $$

Let's apply a one stage direct solution to obtain $s(t)$. 

The homogeneous solution is found from
$$ y'(t) +\frac{1}{RC} y = 0  \tag{2} $$

The characteristic equation : $s + \frac{1}{RC} = 0 \implies s = - \frac{1}{RC}$. 

The (causal) homogeneous solution is :

$$y_h(t) = K e^{-t/RC} u(t) \tag{3}$$  

Then, the particular solution $y_p(t)$ will be from the method of undetermined coefficients as follows:

For the particular input $x(t) = u(t)$ we may assume a particular solution as $y_p(t) = A u(t) + B \delta(t)$, then plug this assumed solution into Eq.(1) to find out the coefficients $A$ and $B$. This yields $A=0$ and $B=0$, hence the particular solution is found to be identically zero.

Then since the total solution is:
$$s(t) = y_h(t) + y_p(t) = y_h(t)$$

to find the value of unknown $K$ in Eq.(4) we need an initial condition on the output $y(0^+) =s(0^+) $ ; found from the circuit physics: apply KVL around the loop of voltage source, capacitor and resistor, with the fact that the capacitor voltage is fixed at zero at time $t=0$. Then we find $y(0^+) = s(0^+) = 1$ which yields $K=1$

Hence the step-response is:
$$ \boxed{ s(t) = e^{-t/RC} u(t) } \tag{4}$$ 

Then the impulse response of Eq.(1) is found to be 
$$ h(t) =  s(t)' = \left( e^{-t/RC} u(t) \right)' $$

$$ \boxed{ h(t) = \delta(t) - \frac{1}{RC} e^{-t/RC} u(t) } \tag{5}$$