Let y = y(x) be a solution curve of the differential equation (y + 1){ an ^2}x\,dx + an x\,dy + y\,dx = 0, x \in \left( {0,{\pi \over 2}} ight). If \mathop {\lim }\limits_{x o 0 + } xy(x) = 1, then the value of y\left( {{\pi \over 4}} ight) is :

Question

Let y = y(x) be a solution curve of the differential equation (y + 1){	an ^2}x\,dx + 	an x\,dy + y\,dx = 0, x \in \left( {0,{\pi \over 2}} ight). If \mathop {\lim }\limits_{x 	o 0 + } xy(x) = 1, then the value of y\left( {{\pi \over 4}} ight) is :

Accepted Answer

Answer: D. {\pi \over 4} (y + 1){	an ^2}x\,dx + 	an x\,dy + y\,dx = 0 or {{dy} \over {dx}} + {{{{\sec }^2}x} \over {	an x}}.y = - 	an x IF = {e^{\int {{{{{\sec }^2}x} \over {	an x}}dx} }} = {e^{\ln 	an x}} = 	an x 	herefore y	an x = - \int {{{	an }^2}x\,dx} or y	an x = - 	an x + x + C or y = - 1 + {x \over {	an x}} + {C \over {	an x}} or \mathop {\lim }\limits_{x 	o 0} xy = - x + {{{x^2}} \over {	an x}} + {{Cx} \over {	an x}} = 1 or C = 1 y(x) = \cot x + x\cot x - 1 y\left( {{\pi \over 4}} ight) = {\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 :

Answer

Explanation

More JEE Math questions