Question:
Consider the RLC circuit shown in Figure, with $ = 110 \Omega, = 1 H, = 0.001 F$, and a battery supplying $_0 = 90 V$. Initially there is no current in the circuit and no charge on the capacitor. At time $ = 0$ the switch is closed and left closed for 1 second. After time $ = 1$ it is opened and left open thereafter. Find the resulting current in the circuit. $$(/)+ +(1/) \int_{0}^{t} (\tau)\tau = e().$$
Needs help in this regard.
My try:
$di/dt + 110i+1000 \int_{0}^{t} i (\tau) d\tau=90$
I know how to tackle this. But my question is that how to get rid of integral. Should I replace $i(\tau)$ by $dq/d\tau(=q'(\tau))$ and to apply FFTC.
The more usual approach is simply to differentiate the entire equation to: $$L\,\frac{d^2i}{dt^2}+r\,\frac{di}{dt}+\frac1C\,i=\frac{de(t)}{dt}.$$ You would use the IC's in the course of solving this DE - I would recommend using Laplace Transforms: the IC's get used quite naturally in that process.
Alternatively, if you use the Laplace Transform technique, you can simply transform the integral term as-is.