Question

Degree of the differential equation $\frac{d^{2}y}{d{x}^2}=\sqrt[3]{31+(\frac{dy}{d}{)}^4}$ is ___

Answer 1

This may look look like an easy question and some of us may think that the answer is 1. $\frac{d^{2}y}{d{x}^2}$ is the highest derivative and the power of that is 1. This is wrong because we are not considering the last part of the definition of degree when we proceed this way. Definition of degree goes like this- Degree of a differential equation is defined as the highest power of highest derivative in it after differential equation is cleared of radicals and fractions so far as the derivatives are concerned. Last part of this says that the derivatives should be free of radicals. $\frac{dy}{dx}$ is inside the radical symbol(cube root). We will first remove the radical symbol by cubing both the sides.
So we get $(\frac{d^{2}y}{d{x}^2}{)}^3=1+(\frac{dy}{dx}{)}^4$
There are no radical symbols in this equation. So the degree will be power of $\frac{d^{2}y}{d{x}^2}$ which is 3.