Let us consider a curve, y = f (x) passing through the point (- 2, 2) and the slope of the tangent to the curve at any point ( x, f ( x ) ) is given by f ( x ) + x f ^ ' ( x ) = x ^ 2 . Then

Correct answer: A. x ^ 3 - 3xf ( x ) - 4 = 0 y + x(dy)/(dx) = x ^ 2 (dy)/(dx) + (y)/(x) = x I . F . = e ^ ∫ (1)/(x)dx = x Solution of differential equation, y · x = ∫ x · xdx xy = x ^ 3 3 + C Passes through (- 2, 2) , So, - 12 = - 8 + 3C C = (- 4)/(3) ∴ 3xy = x ^ 3 - 4 i.e 3x · f(x) = x ^ 3 - 4 x ^ 3 - 3xf(x) - 4 = 0

JEE Main 202127 AugustMorning Shift

Exam: JEE Main
Year: 2021
Date: 27 August
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2021 · 27 August · Morning Shift · Mathematics · Differential Equations

Let us consider a curve, y = f (x) passing through the point (- 2,...

Let us consider a curve,

y = f \left(x\right) passing through

the

point \left(- 2, 2\right) and the slope of the tangent to the curve at any point

( x, f ( x ) )

is given by

f ( x ) + x{f}^{'} ( x ) = {x}^{2} .

Then

Options

A ${x}^{3} - 3xf ( x ) - 4 = 0$
B ${x}^{2} + 2xf ( x ) - 12 = 0$
C ${x}^{3} + xf ( x ) + 12 = 0$
D ${x}^{2} + 2xf ( x ) + 4 = 0$

Correct answer

A. ${x}^{3} - 3xf ( x ) - 4 = 0$

Step-by-step solution

y + x\dfrac{dy}{dx} = {x}^{2} $$ \dfrac{dy}{dx} + \dfrac{y}{x} = x $$ I . F . = {e}^{∫ \dfrac{1}{x}dx} = x $Solution of differential equation,$ y \cdot x = ∫ x \cdot xdx $$ xy = \dfrac{{x}^{3}}{3} + C $Passes through \left(- 2, 2\right)$ , So,

$- 12 = - 8 + 3C$

$C = \dfrac{- 4}{3}$

$∴ 3xy = {x}^{3} - 4$

i.e $3x \cdot f\left(x\right) = {x}^{3} - 4$

${x}^{3} - 3xf\left(x\right) - 4 = 0$

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