JEE Main 2022 · 24 June · morning · Single correct · +4 / −1

If x = x(y) is the solution of the differential equation
ydx/dy = 2x + y³(y + 1)eʸ, x(1) = 0; then x(e) is equal to :

  1. (a) e³(eᵉ - 1)
  2. (b) eᵉ(e³ - 1)
  3. (c) e²(eᵉ + 1)
  4. (d) eᵉ(e² - 1)
Answer and solution

Answer (a) e³(eᵉ - 1)

Solution

Maths sir · 4 steps

Treat x as the function and y as the variable: this is a linear equation in x with integrating factor 1/y².

  1. Standard linear form. Divide by y to get dx/dy - 2/yx = y²(y+1)eʸ. This is linear in x, with P(y) = -2/y and Q(y) = y²(y+1)eʸ.
  2. Integrating factor. IF = e^(∫ -2/y dy) = e⁻²ˡⁿʸ = 1/y². Multiplying through, the left side becomes d/dy(x/y²) and the right side becomes (y+1)eʸ.

    x/y² = ∫ (y+1)eʸ dy = yeʸ + C

  3. Use x(1) = 0. The integral works because d/dy(yeʸ) = (y+1)eʸ. Putting y = 1, x = 0 gives 0 = e + C, so C = -e and x = y²(yeʸ - e).
  4. So. At y = e: x(e) = e²(e · eᵉ - e) = e³(eᵉ - 1), which is option (a).

The idea to keep. When the equation has dx/dy and x appears only to the first power, make it linear in x. Also check whether the right side has the form (f + f')eʸ; then it integrates to feʸ in one line.

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.