JEE Main 2020 · 2 September · morning · Single correct · +4 / −1

Let y = y(x) be the solution of the differential equation,
(2 + sin x)/(y + 1).dy/dx = - cos x, y > 0,y(0) = 1.
If y(π) = a and dy/dx at x = π is b, then the ordered pair (a, b) is equal to :

  1. (a) (2, 1)
  2. (b) ( 2,3/2 )
  3. (c) (1, -1)
  4. (d) (1, 1)
Answer and solution

Answer (d) (1, 1)

Solution

Maths sir · 5 steps

Put all the y terms with dy and all the x terms with dx, then integrate both sides as logs.

  1. Separate the variables. Move y + 1 to the left and 2 + sin x to the right. Each side now has only one variable.

    ∫ dy/(y+1) = -∫ (cos x)/(2+sin x) dx

  2. Integrate. On the right, the top is the derivative of the bottom. So ln(y+1) = -ln(2+sin x) + C, which gives (y+1)(2+sin x) = k for some constant k.
  3. Use y(0) = 1. At x = 0: (1+1)(2+0) = k, so k = 4. The solution is (y+1)(2+sin x) = 4.
  4. Find a and b. At x = π, sin π = 0, so y + 1 = 2 and a = 1. From the equation, dy/dx = -((y+1)cos x)/(2+sin x). With cos π = -1 this gives b = -(2 × (-1))/2 = 1.
  5. So. (a, b) = (1, 1), which is option (d).

The idea to keep. When the numerator is the derivative of the denominator, the integral is a log. Combine the logs into one product equal to a constant before you put in the initial condition.

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.