JEE MAIN - Mathematics (2021 - 26th August Morning Shift - No. 2)
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 :
$$ - {\pi \over 4}$$
$${\pi \over 4} - 1$$
$${\pi \over 4} + 1$$
$${\pi \over 4}$$
Explanation
$$(y + 1){\tan ^2}x\,dx + \tan x\,dy + y\,dx = 0$$
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
$$y(x) = \cot x + x\cot x - 1$$
$$y\left( {{\pi \over 4}} \right) = {\pi \over 4}$$
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
$$y(x) = \cot x + x\cot x - 1$$
$$y\left( {{\pi \over 4}} \right) = {\pi \over 4}$$
Comments (0)
