RLC (5-10-2020)

$\textcolor{#20A900}{Fun Problem.}$ I will be very happy if you continue this series.
The basic equations are $V_{s}-IR-L\frac{dI}{dT}-\frac{q}{c}=0$
After substituting values $10-\frac{I}{4}-\frac{dI}{dT}-q=0$
As we know $I=\frac{dq}{dT}=\dot{q}$ After substituting this $10-0.25\dot{q}-\ddot{q}-q=0$ Solving this double differential equation leads to $\large q(t) =c_{1}e^{-0.125t}sin(0.992157t) +c_{2}e^{-0.125t}cos(0.992157) +10$ where $c_{1}$ and $c_{2}$ are arbitrary constant.
Using this $q(0) =0$ $\dot{q}(0)=0$ we can find the value of $c_{1}$ and $c_{2}$ easily.
After that we will reach $\large q(t) =-1.25988125e^{-0.125t}sin(0.992157t) -10e^{-0.125t}cos(0.992157) +10$ Differentiate $q(t)$ to get $I(t)$ $I(t) =10.0791e^{-0.125t}sin(0.992157t)$ Heat dissipated $H = \int_{0}^{∞} (I(t))^{2}Rdt$ $\textcolor{#3D99F6}{\boxed{H=50.0005}}$