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JEE Main 2019 · 9 Apr · Shift II

JEE Main 2019 Mathematics question on Differential Equations

If cos⁡xdydx−ysin⁡x=6x\cos x \dfrac{dy}{dx} - y \sin x = 6x, (0<x<π2)\left(0 < x < \dfrac{\pi}{2}\right) and y(π3)=0y\left(\dfrac{\pi}{3}\right) = 0, then y(π6)y\left(\dfrac{\pi}{6}\right) is equal to
  1. A.
    π223\dfrac{\pi^2}{2\sqrt{3}}
  2. B.
    −π223-\dfrac{\pi^2}{2\sqrt{3}}
  3. C.
    −π243-\dfrac{\pi^2}{4\sqrt{3}}
  4. D.
    −π22-\dfrac{\pi^2}{2}
Show answer and solution

Correct option: B

Solution

Divide by cos x: dy/dx - y tan x = 6x sec x.

Integrating factor: e^(∫-tan x dx) = e^(ln cos x) = cos x.

Multiply through: d/dx(y cos x) = 6x, so y cos x = 3x² + C.

Apply y(π/3) = 0: 0·(1/2) = 3(π/3)² + C → C = -π²/3.

So y cos x = 3x² - π²/3.

At x = π/6: y·(√3/2) = 3(π²/36) - π²/3 = π²/12 - π²/3 = -π²/4.

Thus y = (-π²/4)·(2/√3) = -π²/(2√3).

[JEE Main 2019 · 9 Apr · Shift II]

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