JEE Main 2020 · 4 September · morning · Single correct · +4 / −1

Let y = y(x) be the solution of the differential equation,
xy'- y = x²(xcosx + sinx), x > 0. if y (π) = π then
y''( π/2 ) + y( π/2 ) is equal to :

  1. (a) 1 + π/2
  2. (b) 2 + π/2 + (π ²)/4
  3. (c) 2 + π/2
  4. (d) 1 + π/2 + (π ²)/4
Answer and solution

Answer (c) 2 + π/2

Solution

Maths sir · 4 steps

Divide by x² and the left side becomes the derivative of y/x, so one integration finishes it.

  1. Spot the quotient. Divide the equation by x². The left side (xy' - y)/x² is exactly d/dx(y/x). The right side x cos x + sin x is exactly d/dx(x sin x).

    d/dx(y/x) = d/dx(x sin x) ⇒ y/x = x sin x + C

  2. Fix the constant. Put x = π and y = π: 1 = π sin π + C = C. So C = 1 and y = x²sin x + x.
  3. Differentiate twice. y' = 2x sin x + x²cos x + 1, and y'' = 2 sin x + 4x cos x - x²sin x. At x = π/2 we have sin x = 1 and cos x = 0, so y'' = 2 - π²/4 and y = π²/4 + π/2.
  4. So. Add them. The π²/4 terms cancel, which leaves y'' + y = 2 + π/2, option (c).

The idea to keep. Whenever you see xy' - y, think of d/dx(y/x) after dividing by x². Before you integrate, check whether the right side is also an exact derivative.

Answer from the NTA final answer key. Solution by PrepWiser.

Study Differential Equations

Asked about 200 times in JEE Main since 2020. This question is from 2020.