Question

\[\sqrt{1 - y^2} dx + \sqrt{1 - x^2} dx = 0\]

Answer 1

\[\sqrt{1 - y^2}dx + \sqrt{1 - x^2}dy = 0\]
\[ \Rightarrow \sqrt{1 - y^2}dx = - \sqrt{1 - x^2}dy\]
\[ \Rightarrow - \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} = \frac{dy}{dx}\]
\[ \Rightarrow \sqrt{1 - x^2}\frac{dy}{dx} + \sqrt{1 - y^2} = 0\]

In this differential equation, the order of the highest order derivative is 1 and its power is 1. So, it is a differential equation of order 1 and degree 1.

It is a non-linear equation, as the exponent of dependent variable \[\left( y \right)\] is more than 1 (on expanding \[\sqrt{1 - y^2}\] binomially).