JEE Main 2021 · 18 March · evening · Single correct · +4 / −1

Let y = y(x) be the solution of the differential equation
dy/dx = (y + 1)( (y + 1)e^(x²/2) - x ), 0 < x < 2.1, with y(2) = 0. Then the value of dy/dx at x = 1 is equal to :

  1. (a) e^(5/2)/(1 + e²)²
  2. (b) 5e^(1/2)/(e² + 1)²
  3. (c) - 2e²/(1 + e²)²
  4. (d) (- e^(3/2))/(e² + 1)²
Answer and solution

Answer (d) (- e^(3/2))/(e² + 1)²

Solution

Maths sir · 4 steps

Put u = y + 1, then v = 1/u: the Bernoulli equation turns linear and e^(-x²/2) finishes it.

  1. Shift and flip. Let u = y + 1. Then du/dx = u² e^(x²/2) - xu, which is a Bernoulli equation. Divide by u² and put v = 1/u, so dv/dx = -1/u²du/dx.

    dv/dx - xv = -e^(x²/2) ⇒ d/dx(v e^(-x²/2)) = -1

  2. Integrate and fit. Integrating gives v e^(-x²/2) = C - x. At x = 2, y = 0, so u = 1 and v = 1. This gives e⁻² = C - 2, so C = 2 + e⁻².
  3. Evaluate at one. At x = 1, v e^(-1/2) = 1 + e⁻² = (e² + 1)/e², so v = (e² + 1)/e^(3/2). Then y + 1 = u = e^(3/2)/(e² + 1).
  4. So. dy/dx = u(u e^(1/2) - 1) = u(e²/(e² + 1) - 1) = -u/(e² + 1) = -e^(3/2)/(e² + 1)². That is option (d).

The idea to keep. If the right side has a u² term next to a u term, it is Bernoulli. Put v = 1/u, it becomes linear, and the integrating factor does the rest.

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.