JEE Main 2019 (Online) 9th April Evening ShiftQuestion 1
If $\cos x\frac{dy}{dx}-y\sin x=6x$, $\left(0<x<\frac\pi2\right)$ and $y\left(\frac\pi3\right)=0$, then $y\left(\frac\pi6\right)$ is equal to
Differential Equation
Solution
Concept. $\cos x\,y'-y\sin x$ is the derivative of $y\cos x$. Formulas used. (1) $(y\cos x)'=y'\cos x-y\sin x$Step 1 - Integrate. By (1), $(y\cos x)'=6x$, so $y\cos x=3x^2+c$. Step 2 - Use $y\left(\frac\pi3\right)=0$. $0=\frac{\pi^2}{3}+c$, so $c=-\frac{\pi^2}{3}$. Step 3 - At $x=\frac\pi6$. $$\frac{\sqrt3}{2}\,y=\frac{\pi^2}{12}-\frac{\pi^2}{3}=-\frac{\pi^2}{4}\ \Longrightarrow\ y=-\frac{\pi^2}{2\sqrt3}.$$ Answer: (D)