If m and n are order and degree of the equation $\left(\frac{d^2 y}{d x^2}\right)^5+4 . \frac{\left(\frac{d^2 y}{d x^2}\right)^3}{\frac{d^3 y}{d x^3}}+\frac{d^3 y}{d x^3}=x^2-1$, then

m = 3, n = 2

The given differential equation can be written as

$\left(\frac{d^2 y}{d x^2}\right)^5\left(\frac{d^3 y}{d x^3}\right)+4 . \left(\frac{d^2 y}{d x^2}\right)^3+\left(\frac{d^3 y}{d x^3}\right)^2=\left(x^2-1\right) \frac{d^3 y}{d x^3}$

⇒ m = 3, n = 2

Hence (2) is the correct answer.