JEE Main 2024 (Online) 6th April Morning ShiftQuestion 1
Let $y=y(x)$ be the solution of the differential equation $\left(1+x^{2}\right)\frac{dy}{dx}+y=e^{\tan^{-1}x}$, $y(1)=0$. Then $y(0)$ is
Differential Equation
Solution
Concept. Divide by $1+x^2$: linear in $y$ with IF $e^{\tan^{-1}x}$. Formulas used. (1) $\int\dfrac{e^{2\tan^{-1}x}}{1+x^2}\,dx=\frac12e^{2\tan^{-1}x}$Step 1 - Linear form. $$y'+\frac{y}{1+x^2}=\frac{e^{\tan^{-1}x}}{1+x^2},\qquad\text{IF}=e^{\tan^{-1}x}.$$ Step 2 - Solve. By (1), $y\,e^{\tan^{-1}x}=\frac12e^{2\tan^{-1}x}+C$. Step 3 - Constant. $y(1)=0$: $0=\frac12e^{\pi/2}+C$, so $C=-\frac12e^{\pi/2}$. Step 4 - At $x=0$. $y(0)=\frac12-\frac12e^{\pi/2}=\frac12\left(1-e^{\pi/2}\right)$. Answer: (C)