Question

${\left(\frac{d^{3}y}{{dx}^3}\right)}^{2/3}+4-3\frac{d^{2}y}{{dx}^2}+5\frac{dy}{dx}=0$

Answer 1

Given,

${\left(\frac{d^{3}y}{d{x}^3}\right)}^{\frac{2}{3}}+4-3\frac{d^{2}y}{d{x}^2}+5\frac{dy}{dx}=0$

${\left(\frac{d^{3}y}{d{x}^3}\right)}^{\frac{2}{3}}=3\frac{d^{2}y}{d{x}^2}-5\frac{dy}{dx}-4$

cubing on both sides, we get,

${\left(\frac{d^{3}y}{d{x}^3}\right)}^2={(3\frac{d^{2}y}{d{x}^2}-5\frac{dy}{dx}-4)}^3$

$Order=3$

$Degree=2$