JEE Main 2019 (Online) 10th April Evening ShiftQuestion 1
Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}+y\tan x=2x+x^{2}\tan x$, $x\in\left(-\frac\pi2,\frac\pi2\right)$, such that $y(0)=1$. Then
Differential Equation
Solution
Concept. Linear with $P=\tan x$, IF $=\sec x$; the right side becomes an exact derivative. Formulas used. (1) $\left(x^2\sec x\right)'=2x\sec x+x^2\sec x\tan x$Step 1 - Integrate. Multiply by $\sec x$; by (1), $$(y\sec x)'=2x\sec x+x^2\sec x\tan x=\left(x^2\sec x\right)'.$$ So $y\sec x=x^2\sec x+C$, i.e. $y=x^2+C\cos x$. Step 2 - Use $y(0)=1$. $C=1$, so $y=x^2+\cos x$ and $y'=2x-\sin x$. Step 3 - Check (A). $$y'\left(\tfrac\pi4\right)-y'\left(-\tfrac\pi4\right)=\left(\tfrac\pi2-\tfrac1{\sqrt2}\right)-\left(-\tfrac\pi2+\tfrac1{\sqrt2}\right)=\pi-\sqrt2.$$ (The sums in (B), (D) are $0$, and the difference in (C) is $0$.) Answer: (A)