Let $y=y(x)$ be a solution curve of the differential equation $(y+1) \tan ^{2} x d x+\tan x d y+y d x=0, x \in\left(0, \frac{\pi}{2}\right)$. If $\lim _{x \rightarrow 0^{+}} x y(x)=1$, then the value of $y\left(\frac{\pi}{4}\right)$ is:

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Differential Equations

Let

y = y (x)

be a solution curve of the differential equation

(y + 1) tan^{2} x d x + tan x d y + y d x = 0, x \in (0, \frac{π}{2})

. If

lim_{x \to 0^{+}} x y (x) = 1

, then the value of

y (\frac{π}{4})

is:

Answer: (A)

Collecting the terms in

y

tan x d y + [y (tan^{2} x + 1) + tan^{2} x] d x = 0 \Rightarrow \frac{d y}{d x} + y \frac{sec ^{2} x}{tan x} = - tan x .

The integrating factor is

e^{\int s e c^{2} x / t a n x d x} = tan x

, so

\frac{d}{d x} (y tan x) = - tan^{2} x = 1 - sec^{2} x \Rightarrow y tan x = x - tan x + C .

Then

x y = \frac{x}{tan x} (x - tan x + C)

, and as

x \to 0^{+}

one has

\frac{x}{tan x} \to 1

and

x - tan x \to 0

, so

lim_{x \to 0^{+}} x y = C

. The given condition therefore fixes

C = 1

.
At

x = \frac{π}{4}

, where

tan x = 1

y (\frac{π}{4}) = \frac{π}{4} - 1 + 1 = \frac{π}{4} .

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option