JEE Main 2020 · 4 September · evening · Single correct · +4 / −1

The solution of the differential equation
dy/dx - (y + 3x)/(log ₑ( y + 3x )) + 3 = 0 is:
(where c is a constant of integration)

  1. (a) x - 1/2( log ₑ( y + 3x ) )² = C
  2. (b) y + 3x - 1/2( log ₑx )² = C
  3. (c) x – logₑ(y+3x) = C
  4. (d) x – 2logₑ(y+3x) = C
Answer and solution

Answer (a) x - 1/2( log ₑ( y + 3x ) )² = C

Solution

Maths sir · 4 steps

The bracket y + 3x appears twice, so call it v; the +3 then becomes part of dv/dx.

  1. Substitute. Put v = y + 3x. Then dv/dx = dy/dx + 3, which is exactly the dy/dx and the +3 already in the equation.
  2. Rewrite. Move the middle term across: dy/dx + 3 = v/(logₑ v). The left side is dv/dx, so dv/dx = v/(logₑ v), and the variables separate.

    ∫ (logₑ v)/v dv = ∫ dx ⇒ 1/2(logₑ v)² = x + C

  3. Integrate. Put t = logₑ v, so dt = dv/v. The left integral becomes ∫ t dt = t²/2.
  4. So. Put v = y + 3x back in and collect the constant: x - 1/2(logₑ(y + 3x))² = C, which is option (a).

The idea to keep. When one expression like ax + by keeps repeating, make it the new variable. Its derivative usually swallows the extra constant term, and the equation separates.

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 2020.