JEE Main 2023 · 11 April · morning · Single correct · +4 / −1

Let y=y(x) be a solution curve of the differential equation.
(1-x² y²) d x=y d x+x d y.
If the line x=1 intersects the curve y=y(x) at y=2 and the line x=2 intersects the curve y=y(x) at y=α, then a value of α is :

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

Answer (a) (1+3 e²)/2(3 e²-1)

Solution

Maths sir · 5 steps

Spot that y dx + x dy is d(xy), put t = xy, and the equation separates at once.

  1. Spot the exact form. The right side y dx + x dy is exactly d(xy). Put t = xy, so the equation becomes (1 - t²) dx = dt.

    ∫ dt/(1-t²) = ∫ dx ⇒ 1/2 ln|(1+t)/(1-t)| = x + C

  2. Fix the constant. At x = 1, y = 2, so t = 2 and (1+t)/(1-t) = 3/-1 = -3. Then 1/2 ln 3 = 1 + C, which gives C = 1/2 ln 3 - 1.
  3. Second point. At x = 2, 1/2 ln|(1+t)/(1-t)| = 2 + 1/2 ln 3 - 1, so |(1+t)/(1-t)| = 3e². On this curve the ratio was negative (-3) at the start, so take (1+t)/(1-t) = -3e².
  4. Solve for t. 1 + t = -3e² + 3e² t gives t(3e² - 1) = 1 + 3e², so t = (1+3e²)/(3e²-1). Since t = xy = 2α, we get α = t/2.
  5. So. α = (1+3e²)/2(3e²-1), which is option (a).

The idea to keep. Whenever you see x dy + y dx, think d(xy) and substitute t = xy. When a modulus appears, let the given point decide which sign to keep.

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.