JEE Main 2019 (Online) 8th April Morning ShiftQuestion 1
Let $y=y(x)$ be the solution of the differential equation $\left(x^{2}+1\right)^{2}\frac{dy}{dx}+2x\left(x^{2}+1\right)y=1$ such that $y(0)=0$. If $\sqrt a\,y(1)=\frac{\pi}{32}$, then the value of $a$ is
Differential Equation
Solution
Concept. Divide by $x^2+1$; the left side becomes $\big((x^2+1)y\big)'$. Formulas used. (1) $\big((x^2+1)y\big)'=(x^2+1)y'+2xy$Step 1 - Rewrite. Dividing by $x^2+1$ and using (1): $$\big((x^2+1)y\big)'=\frac{1}{x^2+1}.$$ Step 2 - Integrate. $(x^2+1)y=\tan^{-1}x+C$; $y(0)=0$ gives $C=0$. Step 3 - $y(1)$. $2y(1)=\frac\pi4$, so $y(1)=\frac\pi8$. Step 4 - Find $a$. $\sqrt a\cdot\frac\pi8=\frac{\pi}{32}$, so $\sqrt a=\frac14$ and $a=\frac1{16}$. Answer: (A)