JEE Main 2020 · 7 January · evening · Single correct · +4 / −1

Let y = y(x) be the solution curve of the differential equation,
( y² - x )dy/dx = 1, satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is :

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

Answer (d) 2 - e

Solution

Maths sir · 4 steps

Do not solve for y. Flip it to dx/dy and the equation becomes linear in x.

  1. Flip the derivative. Write the equation as dx/dy = y² - x, that is dx/dy + x = y². This is linear in x, with y as the variable.
  2. Integrating factor. The coefficient of x is 1, so the integrating factor is e^(∫ 1 dy) = eʸ. Then x eʸ = ∫ y² eʸ dy, and integrating by parts twice gives eʸ(y² - 2y + 2) + C.

    x = y² - 2y + 2 + C e⁻ʸ

  3. Use y(0) = 1. Put x = 0, y = 1: 0 = 1 - 2 + 2 + C e⁻¹, so C = -e. The curve is x = y² - 2y + 2 - e¹⁻ʸ.
  4. So. On the x-axis y = 0, so x = 0 - 0 + 2 - e¹ = 2 - e. The answer is (d).

The idea to keep. If an equation looks messy as dy/dx, try dx/dy. Whenever x appears only to the first power, it becomes a linear equation in 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.