JEE PYQ: Differential Equation Question 1
Question 1 - 2021 (16 Mar Shift 1)
If $y = y(x)$ is the solution of the differential equation, $\frac{dy}{dx} + 2y\tan x = \sin x$, $y\left(\frac{\pi}{3}\right) = 0$, then the maximum value of the function $y(x)$ over $\mathbb{R}$ is equal to:
(1) 8
(2) $\frac{1}{2}$
(3) $-\frac{15}{4}$
(4) $\frac{1}{8}$
Show Answer
Answer: (4)
Solution
I.F. $= e^{\int 2\tan x,dx} = \sec^2 x$. Solution: $y\sec^2 x = \int \sin x\sec^2 x,dx = \sec x + C$. At $x = \frac{\pi}{3}$, $y = 0$: $0 = 2 + C$, so $C = -2$. Thus $y = \frac{\sec x - 2}{\sec^2 x} = \cos x - 2\cos^2 x$. Setting $\frac{dy}{dx} = 0$: $-\sin x + 4\sin x\cos x = 0 \Rightarrow \cos x = \frac{1}{4}$. Max value $= \frac{1}{4} - 2\cdot\frac{1}{16} = \frac{1}{8}$.
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