If $m$ and $n$ are the order and degree of the differential 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

$\left(\frac{d^3 y}{d x^3}\right)^2-\left(x^2-1\right) \frac{d^3 y}{d x^3}+\left(\frac{d^3 y}{d x^3}\right)\left(\frac{d^2 y}{d x^2}\right)^5+4\left(\frac{d^2 y}{d x^2}\right)^3=0$

Clearly, its order is 3 and degree is 2.

In differential equation, power of derivative terms cannot be fractional/negative)

only whole numbers allowed

∴  m = 3 and n = 2