Let y = y (x) be the solution of the differential equation x(1 - x ^ 2 )(dy)/(dx) + (3 x ^ 2 y - y - 4 x ^ 3 ) = 0, x > 1 with y(2) = - 2 . Then y(3) is equal to

Correct answer: A. - 18 Given x(1 - x ^ 2 )(dy)/(dx) + (3 x ^ 2 y - y - 4 x ^ 3 ) = 0 (x - x ^ 3 )(dy)/(dx) + (3 x ^ 2 - 1)y = 4 x ^ 3 (dy)/(dx) + (3 x ^ 2 - 1) (x - x ^ 3 ) y = 4 x ^ 3 (x - x ^ 3 ) This is a linear differential equation of the form (dy)/(dx) + Py = Q Here IF = e ^ ∫ Pdx = e ^ ∫ 3 x ^ 2 - 1 x - x ^ 3 dx Let x - x ^ 3 = t ⇒ IF = e ^ ∫ - (dt)/(t) = e ^ - t = (1)/(t) ∴ IF = 1 x - x ^ 3 So the general solution of the differential equation will be y × IF = ∫ Q × IFdx y( 1 x - x ^ 3 ) = ∫ 4 x ^ 3 x - x ^ 3 × 1 (x - x ^ 3 ) dx = ∫ 4 x ^ 3 (x - x ^ 3 ) ^ 2 dx = ∫ 4x (1 - x ^ 2 ) ^ 2 dx Now let I = ∫ 4x (1 - x ^ 2 ) ^ 2 dx Substituting 1 - x ^ 2 = z I = 2 ∫ - dz z ^ 2 = - 2(- (1)/(z)) + c = 2 1 - x ^ 2 + c Hence the solution of the differential equation becomes y x - x ^ 3 = 2 1 - x ^ 2 + c At x = 2, y = - 2 (- 2)/(2 - 8) = (2)/(1 - 4) + c (1)/(3) = (- 2)/(3) + c ∴ c = 1 Hence the particular solution will be y x - x ^ 3 = 2 1 - x ^ 2 + 1 Put x = 3 (y)/(3 - 27) = (2)/(1 - 9) + 1 (y)/(- 24) = - (1)/(4) + 1 - (y)/(24) = (3)/(4) y = (3)/(4)(- 24) = - 18

JEE Main 202228 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 28 June
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2022 · 28 June · Morning Shift · Mathematics · Differential Equations

Let y = y (x) be the solution of the differential equation x(1 - x...

Let

y = y

\left(x\right)

be the solution of the differential equation

x\left(1 - {x}^{2}\right)\dfrac{dy}{dx} + \left(3{x}^{2}y - y - 4{x}^{3}\right) = 0, x > 1

with

y\left(2\right) = - 2

. Then

y\left(3\right) is equal to

Options

A $- 18$
B $- 12$
C $- 6$
D $- 3$

Correct answer

A. $- 18$

Step-by-step solution

Given $x\left(1 - {x}^{2}\right)\dfrac{dy}{dx} + \left(3{x}^{2}y - y - 4{x}^{3}\right) = 0$

\left(x - {x}^{3}\right)\dfrac{dy}{dx} + \left(3{x}^{2} - 1\right)y = 4{x}^{3} $$ \dfrac{dy}{dx} + \dfrac{\left(3{x}^{2} - 1\right)}{\left(x - {x}^{3}\right)}y = \dfrac{4{x}^{3}}{\left(x - {x}^{3}\right)} $This is a linear differential equation of the form$ \dfrac{dy}{dx} + Py = Q $Here$ IF = {e}^{∫ Pdx} = {e}^{∫ \dfrac{3{x}^{2} - 1}{x - {x}^{3}}dx} $Let$ x - {x}^{3} = t ⇒ IF = {e}^{∫ - \dfrac{dt}{t}} $$ = {e}^{- \ln t} = \dfrac{1}{t} $$ ∴ IF = \dfrac{1}{x - {x}^{3}} $So the general solution of the differential equation will be $ y \times IF = ∫ Q \times IFdx $$ y\left(\dfrac{1}{x - {x}^{3}}\right) = ∫ \dfrac{4{x}^{3}}{x - {x}^{3}} \times \dfrac{1}{\left(x - {x}^{3}\right)}dx $$ = ∫ \dfrac{4{x}^{3}}{{\left(x - {x}^{3}\right)}^{2}}dx $$ = ∫ \dfrac{4x}{{\left(1 - {x}^{2}\right)}^{2}}dx $Now let$ I = ∫ \dfrac{4x}{{\left(1 - {x}^{2}\right)}^{2}}dx $Substituting$ 1 - {x}^{2} = z $$ I = 2 ∫ \dfrac{- dz}{{z}^{2}} $$ = - 2\left(- \dfrac{1}{z}\right) + c = \dfrac{2}{1 - {x}^{2}} + c $Hence the solution of the differential equation becomes $ \dfrac{y}{x - {x}^{3}} = \dfrac{2}{1 - {x}^{2}} + c $At$ x = 2, y = - 2 $$ \dfrac{- 2}{2 - 8} = \dfrac{2}{1 - 4} + c $$ \dfrac{1}{3} = \dfrac{- 2}{3} + c $$ ∴ c = 1 $Hence the particular solution will be$ \dfrac{y}{x - {x}^{3}} = \dfrac{2}{1 - {x}^{2}} + 1 $Put$ x = 3 $$ \dfrac{y}{3 - 27} = \dfrac{2}{1 - 9} + 1 $$ \dfrac{y}{- 24} = - \dfrac{1}{4} + 1 $$ - \dfrac{y}{24} = \dfrac{3}{4} $$ y = \dfrac{3}{4}\left(- 24\right) = - 18$

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