Let y = y(x) be a solution curve of the differential equation [(y + 1) ^2x,dx + x,dy + y,d…

Question

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

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

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

Solutions · 1