JEE PYQ: Differential Equation Question 36
Question 36 - 2019 (09 Apr Shift 2)
If $\cos x\frac{dy}{dx} - y\sin x = 6x$, $(0 < x < \frac{\pi}{2})$ and $y\left(\frac{\pi}{3}\right) = 0$, then $y\left(\frac{\pi}{6}\right)$ is equal to:
(1) $\frac{\pi^2}{2\sqrt{3}}$
(2) $-\frac{\pi^2}{2}$
(3) $-\frac{\pi^2}{2\sqrt{3}}$
(4) $-\frac{\pi^2}{4\sqrt{3}}$
Show Answer
Answer: (3)
Solution
$\frac{dy}{dx} - y\tan x = 6x\sec x$. Actually: $\frac{d}{dx}(y\cos x) = 6x$. Integrate: $y\cos x = 3x^2 + C$. At $x = \frac{\pi}{3}$: $0 = \frac{3\pi^2}{9} + C$, so $C = -\frac{\pi^2}{3}$. At $x = \frac{\pi}{6}$: $y\cdot\frac{\sqrt{3}}{2} = \frac{\pi^2}{12} - \frac{\pi^2}{3} = -\frac{\pi^2}{4}$. So $y = -\frac{\pi^2}{2\sqrt{3}}$.
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