# RC step response equation derivation

So this isn't really a big question, but I really need to get something clarified from the derivation of the RC step response equation, which goes like this: $$V(t)=V_S+(V_0-V_S)e^{-\frac{t}{RC}}$$

Given an RC circuit, one can, using KVL, deduce the following: $$V_S-V_R-V(t)=0$$ Where $V_S$ - the supply voltage; $V_R$ - the resistor voltage and $V(t)$ - capacitor voltage.

Using Ohm's Law and capacitor i-v characteristics, $V_R=RC\frac{dV}{dt}$ $$V_S-RC\frac{dV}{dt}-V(t)=0$$ $$\frac{1}{V(t)-V_S}\frac{dV}{dt}=-\frac{1}{RC}$$

And then, we take the integral of both sides with respect to $t$ from $0$ to another independent time variable $t$.

$$\int_0^t{\frac{1}{V(t)-V_S}\frac{dV}{dt}dt}=\int_0^t{-\frac{1}{RC}dt}$$ Until you end up with $$V(t)=V_S+(V_0-V_S)e^{-\frac{t}{RC}}$$ Where $V_0=V(0)$.

But, what wasn't clear to me was why we integrated from zero to $t$? I mean you can do this too: $$\int{\frac{1}{V(t)-V_S}\frac{dV}{dt}dt}=\int{-\frac{1}{RC}dt}$$

Then end up with $$V(t)=V_S+e^{-\frac{t}{RC}}$$ But we know that this cannot be the step response. It is a solution for the differential equation shown above (I'm familiar with differential equations having multiple solutions), but how did we know that only one actually works as the step response?

$$\int{\frac{1}{V(t)-V_S}\frac{dV}{dt}dt}=X+\int{-\frac{1}{RC}dt}$$