Question

$\frac{dy}{dx}+\sqrt{\left(\frac{1-y^2}{1-x^2}\right)}=0.$
Solve the equation.

• A. ${\sin }^{-1}y+{\sin }^{-1}x=c$
• B. ${\sin }^{-1}y+{\sin }^{-1}x=-c$
• C. $-{\sin }^{-1}y+{\sin }^{-1}x=c$
• D. ${\sin }^{-1}y-{\sin }^{-1}x=c$

Answer 1

Answer: (A), (B)

The correct options are
A ${\sin }^{-1}y+{\sin }^{-1}x=c$
B ${\sin }^{-1}y+{\sin }^{-1}x=-c$

$\frac{dy}{dx}+\sqrt{\left(\frac{1-y^2}{1-x^2}\right)}=0.$
$\Rightarrow \frac{dy}{\sqrt{1-y^2}}=-\frac{dx}{\sqrt{1-x^2}}$
Integrating we get,
$\Rightarrow {\sin }^{-1}y=-{\sin }^{-1}x+c\Rightarrow {\sin }^{-1}x+{\sin }^{-1}y+c$