Answer
The general solution of the given differential equation is
$$
V(t)=e^{-\frac{2t}{3}}[c_{1}\sin(\frac{\sqrt{5}t}{3})+c_{2}\cos(\frac{\sqrt{5}t}{3})].
$$
Work Step by Step
$$
3\frac{d^{2} V}{d t^{2}}+4 \frac{d V}{d t}+3 V=0
$$
The given equation is a homogeneous linear equation. So the characteristic equation of the differential equation is
$$
3r^{2}+4r+3 =0
$$
whose roots are
$$
r_{1,\:2}=\frac{-4\pm \sqrt{4^2-4\cdot \:3\cdot \:3}}{2\cdot \:3}=-\frac{2}{3} \pm \frac{\sqrt{5}}{3} i
$$
Therefore, the general solution of the given differential equation is
$$
V(t)=e^{-\frac{2t}{3}}[c_{1}\sin(\frac{\sqrt{5}t}{3})+c_{2}\cos(\frac{\sqrt{5}t}{3})]
$$
