Q89 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
The particular solution of $\dfrac{d y}{d x}+\sqrt{\dfrac{1-y^2}{1-x^2}}=0$, when $x=0, y=\dfrac{1}{2}$ is
- A$\sin ^{-1} x+\sin ^{-1} y=\dfrac{\pi}{2}$
- B$\sin ^{-1} x+\sin ^{-1} y=\dfrac{\pi}{3}$
- C$\sin ^{-1} x-\sin ^{-1} y=\dfrac{\pi}{2}$
- Correct answerD$\sin ^{-1} x+\sin ^{-1} y=\dfrac{\pi}{6}$
Solution
The given differential equation is $\dfrac{dy}{dx} + \sqrt{\dfrac{1-y^2}{1-x^2}} = 0$.
Separating the variables, we have $\dfrac{dy}{\sqrt{1-y^2}} = -\dfrac{dx}{\sqrt{1-x^2}}$.
Integrating both sides, we get $\int \dfrac{dy}{\sqrt{1-y^2}} = -\int \dfrac{dx}{\sqrt{1-x^2}}$.
This yields $\sin^{-1} y = -\sin^{-1} x + C$, which simplifies to $\sin^{-1} x + \sin^{-1} y = C$.
Given the initial condition $x=0$ and $y=\dfrac{1}{2}$, we substitute these values into the equation:
$\sin^{-1}(0) + \sin^{-1}\left(\dfrac{1}{2}\right) = C$.
$0 + \dfrac{\pi}{6} = C$, so $C = \dfrac{\pi}{6}$.
Therefore, the particular solution is $\sin^{-1} x + \sin^{-1} y = \dfrac{\pi}{6}$.
Answer: $\sin ^{-1} x+\sin ^{-1} y=\dfrac{\pi}{6}$
Separating the variables, we have $\dfrac{dy}{\sqrt{1-y^2}} = -\dfrac{dx}{\sqrt{1-x^2}}$.
Integrating both sides, we get $\int \dfrac{dy}{\sqrt{1-y^2}} = -\int \dfrac{dx}{\sqrt{1-x^2}}$.
This yields $\sin^{-1} y = -\sin^{-1} x + C$, which simplifies to $\sin^{-1} x + \sin^{-1} y = C$.
Given the initial condition $x=0$ and $y=\dfrac{1}{2}$, we substitute these values into the equation:
$\sin^{-1}(0) + \sin^{-1}\left(\dfrac{1}{2}\right) = C$.
$0 + \dfrac{\pi}{6} = C$, so $C = \dfrac{\pi}{6}$.
Therefore, the particular solution is $\sin^{-1} x + \sin^{-1} y = \dfrac{\pi}{6}$.
Answer: $\sin ^{-1} x+\sin ^{-1} y=\dfrac{\pi}{6}$
Hence, the correct answer is option: 'D'.