Question

Solve the differential equation:

$\frac{dy}{dx}+\sqrt{\frac{1-y^2}{(1+y{)}^2}}=0$.

• A. $\to {\sin }^{-1}y+x-\sqrt{1-y^2}=C$
• B. $x-\frac{\sqrt{1-y^2}}{2}=c$
• C. $x+\frac{\sqrt{1-y}}{2}=c$
• D. $x-\frac{\sqrt{1-y}}{2}=c$

Answer 1

The correct option is A $\to {\sin }^{-1}y+x-\sqrt{1-y^2}=C$

$\frac{dy}{dx}=-\frac{\sqrt{1-y^2}}{(1+y)}$

$\to \frac{(1+y)dy}{\sqrt{1-y^2}}=-dx$

$\to \frac{dy}{\sqrt{1-y^2}}+\frac{ydy}{\sqrt{1-y^2}}=-dx$

On integrating, we get

$\to {\sin }^{-1}y-\sqrt{1-y^2}=-x+C$

$\to {\sin }^{-1}y+x-\sqrt{1-y^2}=C$