JEE Main 2022 · 28 June · evening · Single correct · +4 / −1

Let x = x(y) be the solution of the differential equation
2y e^(x/y²)dx + ( y² - 4xe^(x/y²) )dy = 0 such that x(1) = 0. Then, x(e) is equal to :

  1. (a) elog ₑ(2)
  2. (b) - elog ₑ(2)
  3. (c) e²log ₑ(2)
  4. (d) - e²log ₑ(2)
Answer and solution

Answer (d) - e²log ₑ(2)

Solution

Maths sir · 4 steps

The exponent is x/y², so put x = vy² and the messy terms cancel out completely.

  1. Choose the substitution. The same quantity x/y² sits in both exponentials, so let v = x/y², that is x = vy². Then dx = y² dv + 2vy dy.
  2. Substitute and simplify. The first term becomes 2y eᵛ(y² dv + 2vy dy) = 2y³ eᵛ dv + 4vy² eᵛ dy. The second term is (y² - 4vy² eᵛ) dy. The 4vy² eᵛ dy terms cancel, which leaves 2y³ eᵛ dv + y² dy = 0.

    2eᵛ dv = -dy/y ⇒ 2e^(x/y²) = -ln y + C

  3. Use x(1) = 0. At y = 1, x = 0, so 2e⁰ = 0 + C and C = 2. The solution is 2e^(x/y²) = 2 - ln y.
  4. So. At y = e: 2e^(x/e²) = 2 - 1 = 1, so x/e² = ln 1/2 = -ln 2. Hence x(e) = -e² logₑ 2, which is option (d).

The idea to keep. When the same odd combination like x/y² keeps appearing, make it the new variable. A good substitution makes the awkward terms cancel and separates the variables.

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 2022.