If the solution curve $y=y(x)$ of the differential equation $y^{2} d x+\left(x^{2}-x y+y^{2}\right) d y=0$, which passes through the point $(1,1)$ and intersects the line $\mathrm{y}=\sqrt{3} \mathrm{x}$ at the point $(\alpha, \sqrt{3} \alpha)$, then value of $\log _{e}(\sqrt{3} \alpha)$ is equal to

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Differential Equations

If the solution curve

y = y (x)

of the differential equation

y^{2} d x + (x^{2} - x y + y^{2}) d y = 0

, which passes through the point

(1, 1)

and intersects the line

y = 3 x

at the point

(α, 3 α)

, then value of

lo g_{e} (3 α)

is equal to

Answer: (C)

Writing

x = v y

so that

d x = v d y + y d v

and substituting into

y^{2} d x + (x^{2} - x y + y^{2}) d y = 0

, the common factor

y^{2}

cancels and the equation collapses to

y d v + (v^{2} + 1) d y = 0

, a separable form

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

. Integrating gives

tan^{- 1} (\frac{x}{y}) = - ln y + C

. The curve passes through

(1, 1)

, where

\frac{x}{y} = 1

, so

tan^{- 1} 1 = - ln 1 + C

and

C = \frac{π}{4}

. At the intersection with

y = 3 x

the point is

(α, 3 α)

, for which

\frac{x}{y} = \frac{1}{3}

and

tan^{- 1} \frac{1}{3} = \frac{π}{6}

. Hence

\frac{π}{6} = - ln (3 α) + \frac{π}{4}

, which rearranges to

ln (3 α) = \frac{π}{4} - \frac{π}{6} = \frac{π}{12}

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Single correct option