JEE Main 2020 · 6 September · evening · Single correct · +4 / −1

If y = ( 2/πx - 1 ) cosec x is the solution of the differential equation,
dy/dx + p( x )y = 2/π cosec x,
0 < x < π/2, then the function p(x) is equal to :

  1. (a) cot x
  2. (b) sec x
  3. (c) tan x
  4. (d) cosec x
Answer and solution

Answer (a) cot x

Solution

Maths sir · 3 steps

Differentiate the given y and spot y itself hiding inside the derivative. Its coefficient is p(x).

  1. Product rule. Write y = (2/πx - 1)cosec x and differentiate. The derivative of cosec x is -cosec x cot x.

    dy/dx = 2/πcosec x - (2/πx - 1)cosec x cot x

  2. Spot y. The second term is (2/πx - 1)cosec x · cot x = y cot x. So dy/dx = 2/πcosec x - y cot x.
  3. So. Rearranging gives dy/dx + (cot x) y = 2/πcosec x. Comparing with the given equation, p(x) = cot x, option (a).

The idea to keep. When the solution is given and a coefficient is missing, do not solve the equation. Differentiate the solution, rewrite the extra term in terms of y, and compare.

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.