Let y = y(x) be the solution of the differential equation (dy)/(dx) = 2x (x + y) ^ 3 - x(x + y) - 1, y(0) = 1 . Then, ( 1 2 + y( 1 2 )) ^ 2 equals:

Correct answer: D. 1 2 - e Given: (dy)/(dx) = 2x (x + y) ^ 3 - x(x + y) - 1 Let, x + y = t ⇒ 1 + (dy)/(dx) = (dt)/(dx) ⇒ (dy)/(dx) = (dt)/(dx) - 1 ⇒ (dt)/(dx) - 1 = 2x t ^ 3 - xt - 1 ⇒ (dt)/(dx) = 2x t ^ 3 - xt ⇒ (dt)/(dx) + xt = 2x t ^ 3 ⇒ 1 t ^ 3 (dt)/(dx) + x t ^ 2 = 2x . . . (i) Putting, 1 t ^ 2 = u ⇒ - 2 t ^ 3 (dt)/(dx) = (du)/(dx) ⇒ 1 t ^ 3 (dt)/(dx) = (- 1)/(2)(du)/(dx) Putting this value in equation (i) ⇒ (- 1)/(2)(du)/(dx) + xu = 2x ⇒ (du)/(dx) - 2xu = - 4x Integrating factor, IF = e ^ ∫ - 2xdx ⇒ IF = e ^ - x ^ 2 So, solution of the given differential equation is, u × e ^ - x ^ 2 = ∫ e ^ - x ^ 2 (- 4x)dx ⇒ e ^ - x ^ 2 t ^ 2 = ∫ e ^ - x ^ 2 (- 4x)dx Putting, - x ^ 2 = z ⇒ - 2xdx = dz ⇒ e ^ - x ^ 2 (x + y) ^ 2 = ∫ 2 e ^ z dz ⇒ e ^ - x ^ 2 (x + y) ^ 2 = 2 e ^ z + C ⇒ e ^ - x ^ 2 (x + y) ^ 2 = 2 e ^ - x ^ 2 + C ⇒ 1 (x + y) ^ 2 = 2 + C e ^ x ^ 2 It is given that, at x = 0 , y = 1 . ⇒ 1 = 2 + C ⇒ C = - 1 ⇒ (x + y) ^ 2 = 1 2 - e ^ x ^ 2 Putting, x = 1 2 ⇒ (y + 1 2 ) ^ 2 = 1 2 - e ^ (1)/(2) ⇒ (y( 1 2 ) + 1 2 ) ^ 2 = ( 1 2 - e )

JEE Main 20241 FebruaryMorning Shift

Exam: JEE Main
Year: 2024
Date: 1 February
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2024 · 1 February · Morning Shift · Mathematics · Differential Equations

Let y = y(x) be the solution of the differential equation (dy)/(dx)...

Let $y = y\left(x\right)$ be the solution of the differential equation $\dfrac{dy}{dx} = 2x{\left(x + y\right)}^{3} - x\left(x + y\right) - 1, y\left(0\right) = 1$ . Then, ${\left(\dfrac{1}{\sqrt{2}} + y\left(\dfrac{1}{\sqrt{2}}\right)\right)}^{2}$ equals:

Options

A $\dfrac{4}{4 + \sqrt{e}}$
B $\dfrac{3}{3 - \sqrt{e}}$
C $\dfrac{2}{1 + \sqrt{e}}$
D $\dfrac{1}{2 - \sqrt{e}}$

Correct answer

D. $\dfrac{1}{2 - \sqrt{e}}$

Step-by-step solution

Given: $\dfrac{dy}{dx} = 2x{\left(x + y\right)}^{3} - x\left(x + y\right) - 1$

Let, $x + y = t$

$⇒ 1 + \dfrac{dy}{dx} = \dfrac{dt}{dx}$

$⇒ \dfrac{dy}{dx} = \dfrac{dt}{dx} - 1$

$⇒ \dfrac{dt}{dx} - 1 = 2x{t}^{3} - xt - 1$

$⇒ \dfrac{dt}{dx} = 2x{t}^{3} - xt$

$⇒ \dfrac{dt}{dx} + xt = 2x{t}^{3}$

$⇒ \dfrac{1}{{t}^{3}}\dfrac{dt}{dx} + \dfrac{x}{{t}^{2}} = 2x . . . \left(i\right)$

Putting, $\dfrac{1}{{t}^{2}} = u$

$⇒ \dfrac{- 2}{{t}^{3}}\dfrac{dt}{dx} = \dfrac{du}{dx}$

$⇒ \dfrac{1}{{t}^{3}}\dfrac{dt}{dx} = \dfrac{- 1}{2}\dfrac{du}{dx}$

Putting this value in equation \left(i\right) $$ ⇒ \dfrac{- 1}{2}\dfrac{du}{dx} + xu = 2x $$ ⇒ \dfrac{du}{dx} - 2xu = - 4x $Integrating factor,$ IF = {e}^{∫ - 2xdx} $$ ⇒ IF = {e}^{- {x}^{2}} $So, solution of the given differential equation is,$ u \times {e}^{- {x}^{2}} = ∫ {e}^{- {x}^{2}}\left(- 4x\right)dx $$ ⇒ \dfrac{{e}^{- {x}^{2}}}{{t}^{2}} = ∫ {e}^{- {x}^{2}}\left(- 4x\right)dx $Putting,$ - {x}^{2} = z $$ ⇒ - 2xdx = dz $$ ⇒ \dfrac{{e}^{- {x}^{2}}}{{\left(x + y\right)}^{2}} = ∫ 2{e}^{z}dz $$ ⇒ \dfrac{{e}^{- {x}^{2}}}{{\left(x + y\right)}^{2}} = 2{e}^{z} + C $$ ⇒ \dfrac{{e}^{- {x}^{2}}}{{\left(x + y\right)}^{2}} = 2{e}^{- {x}^{2}} + C $$ ⇒ \dfrac{1}{{\left(x + y\right)}^{2}} = 2 + C{e}^{{x}^{2}} $It is given that, at$ x = 0 $,$ y = 1 $.$ ⇒ 1 = 2 + C $$ ⇒ C = - 1 $$ ⇒ {\left(x + y\right)}^{2} = \dfrac{1}{2 - {e}^{{x}^{2}}} $Putting,$ x = \dfrac{1}{\sqrt{2}} $$ ⇒ {\left(y + \dfrac{1}{\sqrt{2}}\right)}^{2} = \dfrac{1}{2 - {e}^{\dfrac{1}{2}}} $$ ⇒ {\left(y\left(\dfrac{1}{\sqrt{2}}\right) + \dfrac{1}{\sqrt{2}}\right)}^{2} = \left(\dfrac{1}{2 - \sqrt{e}}\right)$

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