JEE Main 2023 · 15 April · morning · Single correct · +4 / −1

Let x=x(y) be the solution of the differential equation
2(y+2) log ₑ(y+2) d x+(x+4-2 log ₑ(y+2)) d y=0, y>-1
with x(e⁴-2)=1. Then x(e⁹-2) is equal to :

  1. (a) 4/9
  2. (b) 32/9
  3. (c) 10/3
  4. (d) 3
Answer and solution

Answer (b) 32/9

Solution

Maths sir · 5 steps

Read it as dx/dy: the equation is linear in x, and √ln(y+2) is the integrating factor.

  1. Write dx/dy. Divide by dy and let L = logₑ(y+2). The equation becomes dx/dy = -(x + 4 - 2L)/2(y+2)L, and x appears only to the first power.

    dx/dy + 1/2(y+2)L x = (L-2)/(y+2)L

  2. Integrating factor. Put t = L, so dt = dy/(y+2). Then ∫ dy/2(y+2)L = 1/2 ln t, and the integrating factor is e^(1/2 ln t) = √t.
  3. Integrate. Multiply through by √t: x√t = ∫ (t-2)/√t dt = 2/3t^(3/2) - 4√t + C.
  4. Find C. At y = e⁴ - 2, t = 4 and x = 1. So 2 = 16/3 - 8 + C, which gives C = 14/3.
  5. So. At y = e⁹ - 2, t = 9. Then 3x = 18 - 12 + 14/3 = 32/3, so x = 32/9, option (b).

The idea to keep. If the equation is not linear in y, check whether it is linear in x. Then use dx/dy + P(y)x = Q(y), and substitute for a repeated logarithm so the integrals become easy.

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 2023.