JEE Main 2021 · 27 August · morning · Single correct · +4 / −1

Let y = y(x) be the solution of the differential equation
dy/dx = 2(y + 2 sin x - 5)x - 2 cos x such that y(0) = 7. Then y(π) is equal to :

  1. (a) 2e^(π ²) + 5
  2. (b) e^(π ²) + 5
  3. (c) 3e^(π ²) + 5
  4. (d) 7e^(π ²) + 5
Answer and solution

Answer (a) 2e^(π ²) + 5

Solution

Maths sir · 4 steps

Look inside the bracket: put u = y + 2 sin x - 5 and the -2 cos x term disappears.

  1. Spot the substitution. Let u = y + 2 sin x - 5. Then du/dx = dy/dx + 2 cos x, which is exactly the left side plus the stray 2 cos x.

    du/dx = dy/dx + 2 cos x = 2xu

  2. Separate and integrate. du/u = 2x dx gives ln|u| = x² + c, so u = Ce^(x²).
  3. Use y(0) = 7. At x = 0, u = 7 + 2 sin 0 - 5 = 2, so C = 2. Hence y = 2e^(x²) - 2 sin x + 5.
  4. So. Since sin π = 0, y(π) = 2e^(π²) + 5, which is option (a).

The idea to keep. When a differential equation has a messy bracket next to y, try calling the whole bracket u. If its derivative soaks up the leftover terms, the equation becomes separable.

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 2021.