JEE Main 2020 · 2 September · evening · Single correct · +4 / −1

If a curve y = f(x), passing through the point (1, 2), is the solution of the differential equation,
2x²dy= (2xy + y²)dx, then f( 1/2 ) is equal to :

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

Answer (b) 1/(1 + log ₑ2)

Solution

Maths sir · 5 steps

Every term is degree 2 in x and y, so put y = vx and the equation separates at once.

  1. Spot homogeneous. Rewrite it as dy/dx = (2xy + y²)/2x². Every term on top and bottom has degree 2, so the substitution y = vx will work.
  2. Substitute y = vx. Then dy/dx = v + xdv/dx, and the right side becomes v + v²/2. The v cancels on both sides, leaving xdv/dx = v²/2.
  3. Separate and integrate. (2 dv)/v² = dx/x gives -2/v = logₑ x + C. Putting back v = y/x gives -2x/y = logₑ x + C.
  4. Use the point. At (1, 2): -2/2 = 0 + C, so C = -1. The curve is therefore:

    y = 2x/(1 - logₑ x)

  5. So. At x = 1/2, logₑ 1/2 = -logₑ 2, so f(1/2) = 1/(1 + logₑ 2), which is option (b).

The idea to keep. If every term in a first-order equation has the same total degree, put y = vx. Find C from the given point before substituting the new x.

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.