Let $y=y(x)$ be a solution of the differential equation, $\sqrt{1-x^{2}} \frac{d y}{d x}+\sqrt{1-y^{2}}=0,|x|<1$. If $\mathrm{y}\left(\frac{1}{2}\right)=\frac{\sqrt{3}}{2}$, then $\mathrm{y}\left(\frac{1}{\sqrt{2}}\right)$ is equal to:

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Differential Equations

Let

y = y (x)

be a solution of the differential equation,

1 - x^{2} \frac{d y}{d x} + 1 - y^{2} = 0, ∣ x ∣ < 1

. If

y (\frac{1}{2}) = \frac{3}{2}

, then

y (\frac{1}{2})

is equal to:

Answer: (C)

Separating the equation gives

\frac{d y}{1 - y ^{2}} + \frac{d x}{1 - x ^{2}} = 0

, which integrates to

sin^{- 1} y + sin^{- 1} x = c

. The initial condition

y (\frac{1}{2}) = \frac{3}{2}

gives

sin^{- 1} \frac{3}{2} + sin^{- 1} \frac{1}{2} = \frac{π}{3} + \frac{π}{6} = \frac{π}{2}

, so

c = \frac{π}{2}

and

sin^{- 1} y = \frac{π}{2} - sin^{- 1} x = cos^{- 1} x

. Hence

y = sin (cos^{- 1} x) = 1 - x^{2}

, and at

x = \frac{1}{2}

this gives

y = 1 - \frac{1}{2} = \frac{1}{2}

JEE Main · 2020 · 08 Jan 2020 · Shift 1 · Single correct option