Answer
$i=\frac{E_0C}{1-aRC}(\frac{1}{RC}e^{\frac{-t}{RC}}-ae^{-at})$
Work Step by Step
Using RC Circuit, we get:
$$\frac{dq}{dt}+\frac{q}{RC}=\frac{E}{R}$$
Since $E=E_0e^{-at}$, we get:
$$\frac{dq}{dt}+\frac{q}{RC}=\frac{E_0}{R}e^{-at}$$
The integrating factor is:
$$I=e^{\frac{1}{RC}dt}=e^{\frac{t}{RC}}$$
Multiplying by integrating factor:
$$e^{\frac{t}{RC}}\frac{dq}{dt}+e^{\frac{t}{RC}}\frac{q}{RC}=e^{\frac{t}{RC}}\frac{E}{R}$$
$$\frac{d}{dt}(qe^{\frac{t}{RC}})=e^{\frac{1}{RC-a}t}\frac{E_0}{R}$$
Integrating both sides:
$$qe^{\frac{t}{RC}}=c+e^{\frac{1}{RC-a}t}\frac{E_0C}{1-aRC}$$
$$q=ce^{\frac{-t}{RC}}+\frac{E_0C}{1-aRC}e^{-at}$$
Since $q(0)=0$, we have:
$$0=c+\frac{E_0C}{1-aRC}$$
Solve for $c$:
$$ \rightarrow c=-\frac{E_0C}{1-aRC}$$
Thus, the current flowing in an RC circuit is:
$$i=\frac{E_0C}{1-aRC}(\frac{1}{RC}e^{\frac{-t}{RC}}-ae^{-at})$$
