\(\displaystyle L\frac{d^2 q}{dt^2} + R\frac{dq}{dt} + \frac{1}{C}q = v\)
We continue from here.
\(\displaystyle L\frac{d^2 q}{dt^2} + R\frac{dq}{dt} + \frac{1}{C}q = v\)
\(\displaystyle \frac{d^2 q}{dt^2} + \frac{R}{L}\frac{dq}{dt} + \frac{1}{LC}q = \frac{v}{L}\)
\(\displaystyle \frac{d^2 q}{dt^2} + 2\frac{dq}{dt} + 16q = 2\)
\(\displaystyle r = \frac{-2 \pm \sqrt{4 - 4(1)(16)}}{2(1)}\)
\(\displaystyle r = -1 \pm i\sqrt{15}\)
\(\displaystyle q(t) = c_1e^{-t}\cos \sqrt{15}t + c_2e^{-t}\sin \sqrt{15}t + \frac{1}{8}\)
\(\displaystyle i(0) = 0\) also means \(\displaystyle q(0) = 0\), then
\(\displaystyle c_1 = -\frac{1}{8}\)
\(\displaystyle q(t) = -\frac{1}{8}e^{-t}\cos \sqrt{15}t + c_2e^{-t}\sin \sqrt{15}t + \frac{1}{8}\)
\(\displaystyle i(t) = \frac{dq}{dt} = \frac{1}{8}e^{-t}\bigg[\left(\sqrt{15} - 8c_2\right)\sin\sqrt{15}t + \left(8\sqrt{15}c_2 + 1\right)\cos\sqrt{15}t\bigg]\)
Apply \(\displaystyle i(0) = 0\).
\(\displaystyle 0 = \frac{1}{8}e^{-t}\bigg[\left(8\sqrt{15}c_2 + 1\right)\bigg]\)
\(\displaystyle c_2 = -\frac{1}{8\sqrt{15}}\)
\(\displaystyle i(t) = \frac{1}{8}e^{-t}\bigg[\left(\sqrt{15} + \frac{1}{\sqrt{15}}\right)\sin\sqrt{15}t\bigg]\)
Simplify.
\(\displaystyle i(t) = \frac{2}{\sqrt{15}}e^{-t}\sin\sqrt{15}t\)
