Let y=y(x) be the solution of the differential equation x (d y)/(d x)- 2 y=x^ 3 (2-x^ 3 ) ^ 2 y, x ≠ 0 . If y(2)=0 , then (y(1)) is equal to

Correct answer: D. (7)/(4) Divide the equation by ^2 y : x ^2 y (dy)/(dx) - 2 y = x^3(2-x^3) . Let v = y , so ^2 y (dy)/(dx) = (dv)/(dx) . The equation becomes x(dv)/(dx) - 2v = x^3(2-x^3) . Dividing by x : (dv)/(dx) - (2v)/(x) = x^2(2-x^3) . Using integrating factor μ = x^ -2 : (d)/(dx)(x^ -2 v) = 2 - x^3 . Integrating: x^ -2 v = 2x - (x^4)/(4) + C . From y(2) = 0 , we have 0 = 0 = 16 - 16 + 4C , giving C = 0 . Therefore y = 2x^3 - (x^6)/(4) . At x = 1 : (y(1)) = 2 - (1)/(4) = (7)/(4) .

JEE Main 202628 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 28 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2026 · 28 January · Morning Shift · Mathematics · Differential Equations

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

Let

y=y(x)

be the solution of the differential equation

x \frac{d y}{d x}-\sin 2 y=x^{3}\left(2-x^{3}\right) \cos ^{2} y, x \neq 0

. If

y(2)=0

, then

\tan (y(1))

is equal to

Options

A $-\frac{3}{4}$
B $\frac{3}{4}$
C $-\frac{7}{4}$
D $\frac{7}{4}$

Correct answer

D. $\frac{7}{4}$

Step-by-step solution

Divide the equation by

\cos^2 y

x\sec^2 y \frac{dy}{dx} - 2\tan y = x^3(2-x^3)

.
Let

v = \tan y

, so

\sec^2 y \frac{dy}{dx} = \frac{dv}{dx}

.
The equation becomes

x\frac{dv}{dx} - 2v = x^3(2-x^3)

.
Dividing by

x

\frac{dv}{dx} - \frac{2v}{x} = x^2(2-x^3)

.
Using integrating factor

\mu = x^{-2}

\frac{d}{dx}(x^{-2}v) = 2 - x^3

.
Integrating:

x^{-2}v = 2x - \frac{x^4}{4} + C

.
From

y(2) = 0

, we have

\tan 0 = 0 = 16 - 16 + 4C

, giving

C = 0

.
Therefore

\tan y = 2x^3 - \frac{x^6}{4}

.
At

x = 1

\tan(y(1)) = 2 - \frac{1}{4} = \frac{7}{4}

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