The degree of the differential equation y_2^3/2-y_1^1/2-4=0 is:

$(\frac{d^2y}{dx^2})^{3/2}-(\frac{dy}{dx})^{1/2}-4=0⇒(\frac{d^2y}{dx^2})^3=16+(\frac{dy}{dx})+8(\frac{dv}{dx})^{1/2}$

$⇒[(\frac{d^2y}{dx^2})^3-16-(\frac{dy}{dx})]^2=64(\frac{dy}{dx})$

⇒ degree = 6 and order = 2