"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 :"

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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

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

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

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 :

Solution

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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 :

Solution

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