JEE Main · 2019 · 08 Apr 2019 · Shift 1 · Differential Equations

Divide through by $(x^2+1{)}^2$ to reach the standard linear form $\frac{dy}{dx}+\frac{2x}{x^2+1}y=\frac{1}{(x^2+1{)}^2}$. The integrating factor is $e^{\int \frac{2x}{x^2+1}dx}=x^2+1$, and multiplying through makes the left side the derivative of $y(x^2+1)$, so $y(x^2+1)=\int \frac{dx}{x^2+1}={\tan }^{-1}x+c$. The condition $y(0)=0$ forces $c=0$, giving $y=\frac{{\tan }^{-1}x}{x^2+1}$ and $y(1)=\frac{\pi /4}{2}=\frac{\pi }{8}$. Then $\sqrt{a}\cdot \frac{\pi }{8}=\frac{\pi }{32}$ gives $\sqrt{a}=\frac{1}{4}$, so $a=\frac{1}{16}$.