Question

The solution of the equation $\dfrac{{dy}}{{dx}} + \sqrt {\dfrac{{1 - {y^2}}}{{1 - {x^2}}}} = 0$ is

1. $x\sqrt {1 - {y^2}} - y\sqrt {1 - {x^2}} = c$
2. $x\sqrt {1 - {y^2}} + y\sqrt {1 - {x^2}} = c$
3. $x\sqrt {1 + {y^2}} + y\sqrt {1 + {x^2}} = c$
4. None of these

Answer 1

Hint: Here, we will separate the variables $x$ and $y$ in the given equation. Use the formula of integration $\int {\dfrac{1}{{\sqrt {1 - {x^2}} }}dx} = {\sin ^{ - 1}}x + c$ to find the solution of the equation.

Complete step-by-step answer:
Given equation is $\dfrac{{dy}}{{dx}} + \sqrt {\dfrac{{1 - {y^2}}}{{1 - {x^2}}}} = 0$.

We can also write this differential equation.

$$\Rightarrow \dfrac{{dy}}{{dx}} = - \sqrt {\dfrac{{1 - {y^2}}}{{1 - {x^2}}}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\sqrt {1 - {y^2}} }}{{\sqrt {1 - {x^2}} }} \\\$$

We know that the easiest way to solve the above integration is by using the method of separating variables.

Separating the variables $x$ and $y$ in the above equation and integrating it, we get

$\int {\dfrac{1}{{\sqrt {1 - {y^2}} }}} dy = - \int {\dfrac{1}{{\sqrt {1 - {x^2}} }}} dx$

Using the formula of integration $\int {\dfrac{1}{{\sqrt {1 - {x^2}} }}dx} = {\sin ^{ - 1}}x + c$ in the above equation, we get

$${\sin ^{ - 1}}y = - {\sin ^{ - 1}}x + c \\\ y = \sin \left( { - {{\sin }^{ - 1}}x + c} \right) \\\$$

Thus, the solution of the equation $\dfrac{{dy}}{{dx}} + \sqrt {\dfrac{{1 - {y^2}}}{{1 - {x^2}}}} = 0$ is $y = \sin \left( { - {{\sin }^{ - 1}}x + c} \right)$.

Hence, option D is correct.

Note: In these types of questions, we will require some knowledge of integration to find the solution of the given equation. First, we will rewrite the given equation by separating variable and then integrate it. We will then use the formula of integration $\int {\dfrac{1}{{\sqrt {1 - {x^2}} }}dx} = {\sin ^{ - 1}}x + c$ to make the integration easier to solve.