ExamHabit
Practise free

JEE Main 2018

JEE Main 2018 Mathematics question on Differential Equations

Let y=y(x)y = y(x) be the solution of the differential equation sin⁡xdydx+ycos⁡x=4x\sin x \frac{dy}{dx} + y\cos x = 4x, x∈(0,π)x \in (0, \pi). If y(π2)=0y\left(\frac{\pi}{2}\right) = 0, then y(π6)y\left(\frac{\pi}{6}\right) is equal to
  1. A.
    −493π2\frac{-4}{9\sqrt{3}}\pi^2
  2. B.
    −893π2\frac{-8}{9\sqrt{3}}\pi^2
  3. C.
    −89π2\frac{-8}{9}\pi^2
  4. D.
    −49π2\frac{-4}{9}\pi^2
Show answer and solution

Correct option: C

Solution

The ODE is sin x · dy/dx + y cos x = 4x, which rewrites as d/dx(y sin x) = 4x.

Integrating: y sin x = 2x² + C.

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

So y sin x = 2x² - π²/2.

At x = π/6: sin(π/6) = 1/2, so y(π/6) · (1/2) = 2(π/6)² - π²/2 = 2π²/36 - π²/2 = π²/18 - 9π²/18 = -8π²/18 = -4π²/9.

Therefore y(π/6) = -8π²/9.

[JEE Main 2018]

Practise Differential Equations the way it is asked

A timed set of real JEE Main Differential Equations questions, graded as the exam grades. No sign-up.

Start a Differential Equations test

More: all JEE Main Differential Equations questions · JEE Main 2018 · Mathematics chapter list