JEE Main 2022 · 25 June · evening · Single correct · +4 / −1

If y = y(x) is the solution of the differential equation
2x²dy/dx - 2xy + 3y² = 0 such that y(e) = e/3, then y(1) is equal to :

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

Answer (b) 2/3

Solution

Maths sir · 5 steps

Every term has 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 - 3y²)/2x². The right side depends only on y/x, so substitute y = vx and dy/dx = v + xdv/dx.
  2. Separate variables. Substituting gives 2x²(v + xdv/dx) - 2x² v + 3x²v² = 0. The v terms cancel, leaving 2xdv/dx = -3v², that is (2 dv)/v² = -(3 dx)/x.
  3. Integrate. Integrating gives -2/v = -3 ln x - C. Multiply by -1 and put back v = y/x.

    2x/y = 3 ln x + C

  4. Find C. At x = e, y = e/3, so the left side is 2e/(e/3) = 6. Since ln e = 1, this gives 6 = 3 + C, so C = 3.
  5. So. At x = 1, ln 1 = 0, so 2/y = 3 and y(1) = 2/3, which is option (b).

The idea to keep. If every term in the equation has the same total degree in x and y, put y = vx. The equation separates into v and x. Find the constant from the given condition before you plug in 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 2022.