JEE Main 2021 · 27 July · morning · Single correct · +4 / −1

Let y = y(x) be solution of the differential equation
log ( dy/dx ) = 3x + 4y, with y(0) = 0.
If y( - 2/3log ₑ2 ) = α log ₑ2, then the value of α is equal to :

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

Answer (a) - 1/4

Solution

Maths sir · 4 steps

Take the exponential of both sides first: e³ˣ⁺⁴ʸ splits into e³ˣe⁴ʸ, and the variables separate.

  1. Remove the log. The log here is natural, so dy/dx = e³ˣ⁺⁴ʸ = e³ˣ e⁴ʸ. Move e⁴ʸ to the left to get e⁻⁴ʸ dy = e³ˣ dx.
  2. Integrate both sides. This gives -1/4e⁻⁴ʸ = 1/3e³ˣ + C. Put x = 0, y = 0: -1/4 = 1/3 + C, so C = -7/12.

    e⁻⁴ʸ = (7 - 4e³ˣ)/3

  3. Put in the point. At x = -2/3 logₑ 2 we have 3x = -2 logₑ 2, so e³ˣ = 1/4. Then e⁻⁴ʸ = (7 - 1)/3 = 2.
  4. So. -4y = logₑ 2, which gives y = -1/4 logₑ 2. So α = -1/4, option (a).

The idea to keep. When the right side is eᵃˣ⁺ᵇʸ, write it as eᵃˣeᵇʸ and separate. Fix the constant with the initial condition before you substitute the new point.

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.