JEE Main 2021 (Online) 26th August Morning ShiftQuestion 1
Let $y=y(x)$ be a solution curve of the differential equation $(y+1)\tan^{2}x\,dx+\tan x\,dy+y\,dx=0$, $x\in\left(0,\frac{\pi}{2}\right)$. If $\lim_{x\to0+}xy(x)=1$, then the value of $y\left(\frac{\pi}{4}\right)$ is :
Differential Equation
Solution
Concept. Collect the $y$ terms: $y\tan^2x+y=y\sec^2x$. Dividing by $\tan x$ gives a linear equation. Formulas used. (1) $\dfrac{\sec^2x}{\tan x}=\tan x+\cot x$ and $\int(\tan x+\cot x)\,dx=\ln\tan x$(2) $\int\tan^2x\,dx=\tan x-x+C$Step 1 - Linear form. The equation is $\tan x\,y'+y\sec^2x=-\tan^2x$, so $$y'+(\tan x+\cot x)\,y=-\tan x,\qquad\text{IF}=\tan x.$$ Step 2 - Integrate. By (2), $y\tan x=-\int\tan^2x\,dx=x-\tan x+C$, so $$y=-1+\frac{x+C}{\tan x}.$$ Step 3 - The limit. $xy=-x+\frac{x}{\tan x}(x+C)\to0+1\cdot C$ as $x\to0^+$. So $C=1$. Step 4 - Value. $y\left(\frac\pi4\right)=-1+\frac{\frac\pi4+1}{1}=\frac\pi4$. Answer: (D)