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?
Best Answer
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.