JEE PYQ: Differential Equation Question 30
Question 30 - 2020 (08 Jan Shift 1)
Let $y = y(x)$ be a solution of the differential equation $\sqrt{1-x^2}\frac{dy}{dx} + \sqrt{1-y^2} = 0$, $|x| < 1$. If $y\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{2}$, then $y\left(\frac{-1}{\sqrt{2}}\right)$ is equal to:
(1) $\frac{\sqrt{3}}{2}$
(2) $-\frac{1}{\sqrt{2}}$
(3) $\frac{1}{\sqrt{2}}$
(4) $-\frac{\sqrt{3}}{2}$
Show Answer
Answer: (3)
Solution
$\frac{dy}{\sqrt{1-y^2}} = -\frac{dx}{\sqrt{1-x^2}}$. Integrating: $\sin^{-1}y + \sin^{-1}x = c$. At $x = \frac{1}{2}$, $y = \frac{\sqrt{3}}{2}$: $c = \frac{\pi}{3} + \frac{\pi}{6} = \frac{\pi}{2}$. So $\sin^{-1}y = \frac{\pi}{2} - \sin^{-1}x = \cos^{-1}x$. At $x = -\frac{1}{\sqrt{2}}$: $y = \sin\left(\cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)\right) = \sin\frac{3\pi}{4} = \frac{1}{\sqrt{2}}$.
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