If $\sin \left(\frac{y}{x}\right)=\log _{e}|x|+\frac{\alpha}{2}$ is the solution of the differential equation $x \cos \left(\frac{y}{x}\right) \frac{d y}{d x}=y \cos \left(\frac{y}{x}\right)+x$ and $y(1)=\frac{\pi}{3}$, then $\alpha^{2}$ is equal to

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Differential Equations

sin (\frac{y}{x}) = lo g_{e} ∣ x ∣ + \frac{α}{2}

is the solution of the differential equation

x cos (\frac{y}{x}) \frac{d y}{d x} = y cos (\frac{y}{x}) + x

and

y (1) = \frac{π}{3}

, then

α^{2}

is equal to

Answer: (A)

x cos (\frac{y}{x}) \frac{d y}{d x} = y cos (\frac{y}{x}) + x

.

Put

y = v x

, so

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

x cos v (v + x \frac{d v}{d x}) = v x cos v + x

xv cos v + x^{2} cos v \frac{d v}{d x} = v x cos v + x

x^{2} cos v \frac{d v}{d x} = x \Rightarrow cos v d v = \frac{d x}{x}

.

Integrating:

sin v = ln ∣ x ∣ + c

, i.e.

sin (\frac{y}{x}) = ln ∣ x ∣ + c

.

Using

y (1) = \frac{π}{3}

sin \frac{π}{3} = ln 1 + c \Rightarrow c = \frac{3}{2}

.

Comparing with

sin (\frac{y}{x}) = ln ∣ x ∣ + \frac{α}{2}

\frac{α}{2} = \frac{3}{2} \Rightarrow α = 3 \Rightarrow α^{2} = 3

JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Single correct option

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