JEE Main 2023 (Online) 6th April Morning ShiftQuestion 1
Let $y=y(x)$ be a solution of the differential equation $(x\cos x)dy+(xy\sin x+y\cos x-1)dx=0$, $0<x<\frac{\pi}{2}$. If $\frac{\pi}{3}y\left(\frac{\pi}{3}\right)=\sqrt{3}$, then $\left|\frac{\pi}{6}y''\left(\frac{\pi}{6}\right)+2y'\left(\frac{\pi}{6}\right)\right|$ is equal to $\_\_\_\_\_\_$ .
Differential Equation
Answer$2$
Solution
Concept. IF $x\sec x$ gives $xy=\sin x+\sqrt3\cos x$; then $(xy)''=xy''+2y'$. Formulas used. (1) $\text{IF}=e^{\int(\tan x+\frac1x)dx}=x\sec x$(2) $(xy)''=xy''+2y'$Step 1 - Linear form. $$y'+\left(\tan x+\frac1x\right)y=\frac{1}{x\cos x}.$$ Step 2 - Solve. By (1), $(xy\sec x)'=\sec^2x$, so $xy\sec x=\tan x+C$. Step 3 - Constant. At $\frac\pi3$: $\sqrt3\cdot2=\sqrt3+C$, so $C=\sqrt3$ and $$xy=\sin x+\sqrt3\cos x.$$ Step 4 - Second derivative. By (2), $xy''+2y'=-\sin x-\sqrt3\cos x$. At $\frac\pi6$: $-\frac12-\frac32=-2$. The absolute value is $2$. Answer: 2