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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?

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A math error.

If two functions are equal, it doesn't mean that ALL their integral functions are equal. You must add an integration constant (=an unknown number X which can be real or complex) to the equation, where two indefinite integrals are equal. That constant can be solved from boundary condition V=Vo when t=0.

So, continue from this:

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

Definite integrals declare different thing. They say that the integrals grow as much during the period from 0 to t and they are zero, if t=0.

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