JEE PYQ: Differential Equation Question 24
Question 24 - 2020 (05 Sep Shift 2)
Let $y = y(x)$ be the solution of the differential equation $\cos x\frac{dy}{dx} + 2y\sin x = \sin 2x$, $x \in \left(0, \frac{\pi}{2}\right)$. If $y(\pi/3) = 0$, then $y(\pi/4)$ is equal to:
(1) $2 - \sqrt{2}$
(2) $2 + \sqrt{2}$
(3) $\sqrt{2} - 2$
(4) $\frac{1}{\sqrt{2}} - 1$
Show Answer
Answer: (3)
Solution
$\frac{dy}{dx} + 2y\tan x = 2\sin x$. I.F. $= \sec^2 x$. Solution: $y\sec^2 x = 2\sec x + C$. At $x = \frac{\pi}{3}$: $0 \cdot 4 = 4 + C$, so $C = -4$. At $x = \frac{\pi}{4}$: $2y = 2\sqrt{2} - 4$, $y = \sqrt{2} - 2$.
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