Question

The solution of $\frac{dy}{dx}=\sqrt{1-x^2-y^2+x^2{y}^2}$ is:
where c is an arbitrary constant

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

Answer 1

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

Solution:

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

$=\sqrt{(1-x^2)-y^2(1-x^2)}$

$=\sqrt{(1-x^2)(1-y^2)}$

$=\left(\sqrt{1-x^2}\right)\left(\sqrt{1-y^2}\right)$

or, $\frac{dy}{\sqrt{1-y^2}}=\left(\sqrt{1-x^2}\right)dx$

On integrating both sides, we get

$\int$$\frac{dy}{\sqrt{1-y^2}}=\int \left(\sqrt{1-x^2}\right)dx$

or, ${\sin }^{-1}y=\frac{1}{2}[x\sqrt{1-x^2}+{\sin }^{-1}x]+A$

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

Hence, C is the correct option.