QJEEM21Ma9e9b: JEE Main 2021 Mathematics - Class 12 JEEM21Ma9e9b Solution: Let y = y(x) be a solution curve of the differential equation , …

Question

Verified Solution • 2025–26 Session moderate

Let y = y(x) be a solution curve of the differential equation

(y + 1){\tan ^2}x\,dx + \tan x\,dy + y\,dx = 0

,

x \in \left( {0,{\pi \over 2}} \right)

. If

\mathop {\lim }\limits_{x \to 0 + } xy(x) = 1

, then the value of

y\left( {{\pi \over 4}} \right)

is :

Accepted Answer

(y + 1){\tan ^2}x\,dx + \tan x\,dy + y\,dx = 0

or

{{dy} \over {dx}} + {{{{\sec }^2}x} \over {\tan x}}.y = - \tan x

IF = {e^{\int {{{{{\sec }^2}x} \over {\tan x}}dx} }} = {e^{\ln \tan x}} = \tan x

\therefore

y\tan x = - \int {{{\tan }^2}x\,dx}

or

y\tan x = - \tan x + x + C

or

y = - 1 + {x \over {\tan x}} + {C \over {\tan x}}

or

\mathop {\lim }\limits_{x \to 0} xy = - x + {{{x^2}} \over {\tan x}} + {{Cx} \over {\tan x}} = 1

or C = 1

y(x) = \cot x + x\cot x - 1

y\left( {{\pi \over 4}} \right) = {\pi \over 4}

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