Let y ( x ) be the solution of the differential equation 2 x ^ 2 dy + ( e ^ y - 2x)dx = 0, x > 0 . If y ( e ) = 1, then y ( 1 ) is equal to:

Correct answer: B. _ e 2 (dy)/(dx) = - e ^ y 2 x ^ 2 + (1)/(x) e ^ - y dy dx = e ^ - y x + - 1 2 x ^ 2 - e ^ - y dy dx + e ^ - y x = 1 2 x ^ 2 Let e ^ - y = t . . . (i) e ^ - y (- 1)(dy)/(dx) = (dt)/(dx) (dt)/(dx) + (t)/(x) = 1 2 x ^ 2 I . F = _ e ^ ∫ (1)/(x)dx = e ^ x = x tx = ∫ 1 2 x ^ 2 · xdx + C Using equation (1) e ^ - y x = (1)/(2)ℓnx + C Given, y(e) = 1 ⇒ e ^ - 1 e = (1)/(2) + C ⇒ C = (1)/(2) e ^ - y x = (1)/(2) ( 1 + ℓnx ) Put x = 1 then y is ⇒ y = ℓn2or _ e 2

JEE Main 202126 AugustEvening Shift

Exam: JEE Main
Year: 2021
Date: 26 August
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2021 · 26 August · Evening Shift · Mathematics · Differential Equations

Let y ( x ) be the solution of the differential equation 2 x ^ 2 dy...

Let

y ( x )

be the solution of the differential equation

2{x}^{2}dy + \left({e}^{y} - 2x\right)dx = 0, x > 0 .

y ( e ) = 1,

then

y ( 1 )

is equal to:

Options

A $\log_{e} ( 2e )$
B $\log_{e}2$
C $2$
D $0$

Correct answer

B. $\log_{e}2$

Step-by-step solution

\dfrac{dy}{dx} = \dfrac{- {e}^{y}}{2{x}^{2}} + \dfrac{1}{x} $$ \dfrac{{e}^{- y}dy}{dx} = \dfrac{{e}^{- y}}{x} + \dfrac{- 1}{2{x}^{2}} $$ \dfrac{- {e}^{- y}dy}{dx} + \dfrac{{e}^{- y}}{x} = \dfrac{1}{2{x}^{2}} $Let$ {e}^{- y} = t . . . \left(i\right) $$ {e}^{- y}\left(- 1\right)\dfrac{dy}{dx} = \dfrac{dt}{dx} $$ \dfrac{dt}{dx} + \dfrac{t}{x} = \dfrac{1}{2{x}^{2}} $$ I . F{=}_{e}^{∫ \dfrac{1}{x}dx} = {e}^{\ln x} = x $$ tx = ∫ \dfrac{1}{2{x}^{2}} \cdot xdx + C $Using equation \left(1\right)$
${e}^{- y}x = \dfrac{1}{2}ℓnx + C$

Given, $y\left(e\right) = 1$

$⇒ {e}^{- 1}e = \dfrac{1}{2} + C ⇒ C = \dfrac{1}{2}$

${e}^{- y}x = \dfrac{1}{2} ( 1 + ℓnx )$

Put $x = 1$ then $y$ is $⇒ y = ℓn2\text{or}\log_{e}2$

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