Let $x=x(y)$ be the solution of the differential equation $\mathrm{y}=\left(\mathrm{x}-\mathrm{y} \frac{\mathrm{dx}}{\mathrm{dy}}\right) \sin \left(\frac{\mathrm{x}}{\mathrm{y}}\right), \mathrm{y}>0$ and $\mathrm{x}(1)=\frac{\pi}{2}$. Then $\cos (\mathrm{x}(2))$ is equal to :

(B) $2\left(\log _{e} 2\right)^{2}-1$

(A) $1-2\left(\log _{e} 2\right)^{2}$

(C) $2\left(\log _{e} 2\right)-1$

(D) $1-2\left(\log _{\mathrm{e}} 2\right)$

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Differential Equations

Let

x = x (y)

be the solution of the differential equation

y = (x - y \frac{dx}{dy}) sin (\frac{x}{y}), y > 0

and

x (1) = \frac{π}{2}

.

Then

cos (x (2))

is equal to :

Answer: (B)

Starting from

y = (x - y \frac{d x}{d y}) sin (\frac{x}{y})

divide both sides by

y

1 = (\frac{x}{y} - \frac{d x}{d y}) sin (\frac{x}{y})

Let

v = \frac{x}{y}

, so

x = v y

and

\frac{d x}{d y} = v + y \frac{d v}{d y}

. Then

1 = (v - v - y \frac{d v}{d y}) sin v = - y \frac{d v}{d y} sin v

\frac{d y}{y} = - sin v d v

Integrating:

ln y = cos v + C = cos (\frac{x}{y}) + C

Using

x (1) = \frac{π}{2}

ln 1 = cos (\frac{π}{2}) + C \Rightarrow 0 = 0 + C \Rightarrow C = 0

.

So

ln y = cos (\frac{x}{y})

. At

y = 2

ln 2 = cos (\frac{x}{2})

Using

cos x = 2 cos^{2} (\frac{x}{2}) - 1

cos (x (2)) = 2 (ln 2)^{2} - 1

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option