Let y = y (x) be the solution of the differential equation (x - x ^ 3 )dy = (y + y x ^ 2 - 3 x ^ 4 )dx, x > 2 If y(3) = 3, then y(4) is equal to:

Correct answer: B. 12 (x - x ^ 3 )dy = (y + y x ^ 2 - 3 x ^ 4 )dx ⇒ xdy - ydx = (y x ^ 2 - 3 x ^ 4 )dx + x ^ 3 dy ⇒ xdy - ydx x ^ 2 = ( ydx + xdy ) - 3 x ^ 2 dx ⇒ d((y)/(x)) = d(xy) - d( x ^ 3 ) Integrate ⇒ (y)/(x) = xy - x ^ 3 + c given f(3) = 3 ⇒ (3)/(3) = 3 × 3 - 3 ^ 3 + c ⇒ c = 19 ∴ (y)/(x) = xy - x ^ 3 + 19 at x = 4, (y)/(4) = 4y - 64 + 19 15y = 4 × 45 ⇒ y = 12

JEE Main 202127 JulyEvening Shift

Exam: JEE Main
Year: 2021
Date: 27 July
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2021 · 27 July · Evening Shift · Mathematics · Differential Equations

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

Let

y = y \left(x\right) be

the

solution of the differential equation \left(x - {x}^{3}\right)dy = \left(y + y{x}^{2} - 3{x}^{4}\right)dx, x > 2

y\left(3\right) = 3,

then

y\left(4\right) is equal to:

Options

A $4$
B $12$
C $8$
D $16$

Correct answer

B. $12$

Step-by-step solution

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

$⇒ xdy - ydx = \left(y{x}^{2} - 3{x}^{4}\right)dx + {x}^{3}dy$

$⇒ \dfrac{xdy - ydx}{{x}^{2}} = ( ydx + xdy ) - 3{x}^{2}dx$

$⇒ d\left(\dfrac{y}{x}\right) = d\left(xy\right) - d\left({x}^{3}\right)$

Integrate

$⇒ \dfrac{y}{x} = xy - {x}^{3} + c$

given $f\left(3\right) = 3$

$⇒ \dfrac{3}{3} = 3 \times 3 - {3}^{3} + c$

$⇒ c = 19$

$∴ \dfrac{y}{x} = xy - {x}^{3} + 19$

at $x = 4, \dfrac{y}{4} = 4y - 64 + 19$

$15y = 4 \times 45$

$⇒ y = 12$

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