Let the solution curve $y =y(x)$ of the differential equation, $$\left[\frac{x}{\sqrt{x^{2}-y^{2}}}+e^{\frac{y}{x}}\right] x \frac{d y}{d x}=x+\left[\frac{x}{\sqrt{x^{2}-y^{2}}}+e^{\frac{y}{x}}\right] y$$ pass through the points $(1,0)$ and $(2 \alpha, \alpha), \alpha>0$. Then $\alpha$ is equal to

(A) $\frac{1}{2} \exp \left(\frac{\pi}{6}+\sqrt{\mathrm{e}}-1\right)$

(B) $\frac{1}{2} \exp \left(\frac{\pi}{3}+\sqrt{\mathrm{e}}-1\right)$

(C) $\exp \left(\frac{\pi}{6}+\sqrt{\mathrm{e}}+1\right)$

(D) $2 \exp \left(\frac{\pi}{3}+\sqrt{\mathrm{e}}-1\right)$

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Differential Equations

Let the solution curve

y = y (x)

of the differential equation,

[\frac{x}{x ^{2} - y ^{2}} + e^{\frac{y}{x}}] x \frac{d y}{d x} = x + [\frac{x}{x ^{2} - y ^{2}} + e^{\frac{y}{x}}] y

pass through the points

(1, 0)

and

(2 α, α), α > 0

.
Then

α

is equal to

Answer: (A)

Rearranging the differential equation:

e^{y / x} (x d y - y d x) + \frac{x}{x ^{2} - y ^{2}} (x d y - y d x) = x d x

Dividing by

x^{2}

e^{y / x} d (\frac{y}{x}) + \frac{1}{1 - ( y / x ) ^{2}} d (\frac{y}{x}) = \frac{d x}{x}

Integrating:

e^{y / x} + sin^{- 1} (\frac{y}{x}) = ln x + c

Using

(1, 0)

e^{0} + 0 = 0 + c

, so

c = 1

.

Using

(2 α, α)

e^{1/2} + sin^{- 1} (1/2) = ln (2 α) + 1

e + \frac{π}{6} = ln (2 α) + 1

ln (2 α) = e + \frac{π}{6} - 1

α = \frac{1}{2} exp (\frac{π}{6} + e - 1)

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Single correct option