Question

Solve the differential equation:

$\sqrt{1-x^2}dy+\sqrt{1-y^2}dx=0$

Answer 1

Given the differential equation,

$\sqrt{1-x^2}dy+\sqrt{1-y^2}dx=0$

or, $\frac{dx}{\sqrt{1-x^2}}+\frac{dy}{\sqrt{1-y^2}}=0$

Now integrating we get,

${\sin }^{-1}x+{\sin }^{-1}y={\sin }^{-1}c$ [ Where $c$ is integrating constant]

or, ${\sin }^{-1}(x\sqrt{1-y^2}+y{\sin }^{-1}\sqrt{1-x^2})={\sin }^{-1}c$

or, $x\sqrt{1-y^2}+y\sqrt{1-x^2}=c$.