Let $y(x)$ be the solution of the differential equation $2 x^{2} d y+\left(e^{y}-2 x\right) d x=0, x>0$. If $y(e)=1$, then $y(1)$ is equal to:

(C) $\log _{\mathrm{e}}(2 \mathrm{e})$

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Differential Equations

Let

y (x)

be the solution of the differential equation

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

. If

y (e) = 1

, then

y (1)

is equal to:

Answer: (D)

Rearranging,

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

. Multiplying by

\frac{e ^{- y}}{2 x ^{2}}

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

Put

v = e^{- y}

, so that

\frac{d v}{d x} = - e^{- y} \frac{d y}{d x}

. The equation becomes linear in

v

\frac{d v}{d x} + \frac{v}{x} = \frac{1}{2 x ^{2}}

with integrating factor

x

. Then

\frac{d}{d x} (xv) = \frac{1}{2 x}

, so

xv = \frac{1}{2} ln x + C

, that is

e^{- y} = \frac{ln x}{2 x} + \frac{C}{x}

The condition

y (e) = 1

gives

e^{- 1} = \frac{1}{2 e} + \frac{C}{e}

, so

C = \frac{1}{2}

. Putting

x = 1

, where

ln x = 0

, gives

e^{- y (1)} = \frac{1}{2}

, so

y (1) = lo g_{e} 2

JEE Main · 2021 · 26 Aug 2021 · Shift 2 · Single correct option