1. The problem is to find the characteristic equation of the differential equation $$2 \frac{d^3X}{dt^3} + 4 \frac{dX}{dt} - 3X = 0$$. 2. For linear differential equations with constant coefficients, the characteristic equation is found by substituting $X = e^{rt}$, which transforms derivatives as follows: $$\frac{dX}{dt} = r e^{rt}, \quad \frac{d^3X}{dt^3} = r^3 e^{rt}$$. 3. Substitute these into the differential equation: $$2 r^3 e^{rt} + 4 r e^{rt} - 3 e^{rt} = 0$$. 4. Factor out $e^{rt}$ (which is never zero): $$e^{rt} (2 r^3 + 4 r - 3) = 0$$. 5. The characteristic equation is therefore: $$2 r^3 + 4 r - 3 = 0$$. This cubic equation in $r$ determines the behavior of the solutions to the differential equation.