JEE Main 2024 (Online) 30th January Morning ShiftQuestion 1
Let $y=y(x)$ be the solution of the differential equation $\sec x\,dy+\{2(1-x)\tan x+x(2-x)\}dx=0$ such that $y(0)=2$. Then $y(2)$ is equal to :
Differential Equation
Solution
Concept. After multiplying by $\cos x$, the right side is the derivative of $x(2-x)\sin x$. Formulas used. (1) $\frac{d}{dx}\big[x(2-x)\sin x\big]=(2-2x)\sin x+x(2-x)\cos x$Step 1 - Rewrite. Multiply by $\cos x$: $$\frac{dy}{dx}=-\big[2(1-x)\sin x+x(2-x)\cos x\big].$$ Step 2 - Integrate. By (1), $y=-x(2-x)\sin x+C$. Step 3 - Constant. $y(0)=2$ gives $C=2$. Step 4 - At $x=2$. $x(2-x)=0$, so $y(2)=2$. Answer: (B)