Electronic – How would a Laplace transform be useful in analyzing this circuit

Elaborating on Andy aka's answer ...

You need the transform pair

$$\frac{\omega_0}{(s+\alpha)^2+\omega_0^2}\Longleftrightarrow e^{-\alpha t}\sin(\omega_0t)u(t)$$

with

$$\omega_0^2=\frac{1}{LC}-\frac{R^2}{4L^2}$$

and

$$\alpha=\frac{R}{2L}$$

Of course you'll also get some multiplicative constant, but that's straightforward.

Taking a particular V(s) and then performing the inverse LT will give you i(t), but that will be in transcendental form, and not a differential equation. You can obtain a DE from your equation using the property: \$sX(s)\rightarrow\frac{dx(t)}{dt}\$. The resultant DE is not particularly user-friendly, though!

***Added in response to comment:

Write your original equation in TF form and add the two fractions:

\$\frac{I(s)}{V(s)}= \frac{As^3 +Bs}{Cs^4 +Ds^2+1}\$

Cross multiply:

\$Cs^4 I(s) + Ds^2 I(s) + I(s)=As^3 V(s)+Bs V(s)\$

Inverse LT:

\$C\frac{d^4I(t)}{dt^4}+D\frac{d^2I(t)}{dt^2}+I(t)= A\frac{d^3V(t)}{dt^3}+B\frac{dV(t)}{dt}\$