JEE Main 2021 · 17 March · morning · Single correct · +4 / −1

Which of the following is true for y(x) that satisfies the differential equation
dy/dx = xy - 1 + x - y; y(0) = 0 :

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

Answer (b) y(1) = e -1/2 - 1

Solution

Maths sir · 4 steps

Group the right side in pairs and it factors into (y+1)(x-1), so the variables separate at once.

  1. Factor the right side. Pair the terms: xy + x - y - 1 = x(y+1) - (y+1) = (y+1)(x-1). So dy/dx = (y+1)(x-1).
  2. Separate and integrate. Divide by y+1 and integrate both sides: ∫ dy/(y+1) = ∫ (x-1) dx, which gives ln|y+1| = x²/2 - x + C.
  3. Use y(0) = 0. At x = 0, y = 0, so ln 1 = 0 + C and C = 0. Near x = 0 we have y + 1 > 0, so the modulus can be dropped.

    y = e^(x²/2 - x) - 1

  4. So. At x = 1 the exponent is 1/2 - 1 = -1/2, so y(1) = e^(-1/2) - 1. That is option (b).

The idea to keep. When dy/dx has four terms like xy + x - y - 1, try grouping them in pairs. They often factor into (function of x) times (function of y), and then you can separate.

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.