The charge, \(q\), on a capacitor in an \(L C R\) series circuit satisfies the second-order differential equation $$ L \frac{\mathrm{d}^{2} q}{\mathrm{~d} t^{2}}+R \frac{\mathrm{d} q}{\mathrm{~d} t}+\frac{1}{C} q=E $$ where \(L, R, C\) and \(E\) are constants. Show that if \(2 L=C R^{2}\) the general solution of this equation is $$ \begin{aligned} &q= \\ &\mathrm{e}^{-t /(C R)}\left(A \cos \frac{1}{C R} t+B \sin \frac{1}{C R} t\right)+C E \end{aligned} $$ If \(i=\frac{\mathrm{d} q}{\mathrm{~d} t}=0\) and \(q=0\) when \(t=0\) show that the current in the circuit is $$ i=\frac{2 E}{R} \mathrm{e}^{-t /(C R)} \sin \frac{1}{C R} t $$

Question: Prove that for a second-order differential equation representing the charge, q, on a capacitor in an LCR series circuit given by \(L \frac{d^2 q}{dt^2} + R \frac{dq}{dt} + \frac{1}{C}q = E\) and the condition \(2L = CR^2\), the general solution is given by \(q(t) = CE + Ae^{-t/(CR)} \cos(\frac{1}{CR}t) + B e^{-t/(CR)}\sin(\frac{1}{CR}t)\), and that the current in the circuit is given by \(i(t) = \frac{2E}{R}e^{-t/(CR)}\sin(\frac{1}{CR}t)\) when \(i=\frac{dq}{dt}=0\) and \(q=0\) initially. Answer: By analyzing the given equation and applying the given condition, we derived the general solution of the equation as \(q(t) = CE + Ae^{-t/(CR)} \cos(\frac{1}{CR}t) + B e^{-t/(CR)}\sin(\frac{1}{CR}t)\). Then, we applied the initial conditions for the charge and current, and proved that the current in the circuit is given by \(i(t) = \frac{2E}{R}e^{-t/(CR)}\sin(\frac{1}{CR}t)\) under these given conditions.