JEE Main 2022 · 28 June · morning · Single correct · +4 / −1

Let the solution curve y = y(x) of the differential equation
[ x/√(x² - y²) + e^(y/x) ]xdy/dx = x + [ x/√(x² - y²) + e^(y/x) ]y
pass through the points (1, 0) and (2α, α), α > 0. Then α is equal to

  1. (a) 1/2 exp ( π/6 + √e - 1 )
  2. (b) 1/2 exp ( π/6 + e - 1 )
  3. (c) exp ( π/6 + √e + 1 )
  4. (d) 2 exp ( π/3 + √e - 1 )
Answer and solution

Answer (a) 1/2 exp ( π/6 + √e - 1 )

Solution

Maths sir · 5 steps

Pull out the common bracket: what remains is x dy - y dx, which is exactly x² dv when y = vx.

  1. Group the bracket. Call the bracket f. Move the fy term to the left side to get f(xdy/dx - y) = x, which is f (x dy - y dx) = x dx.
  2. Substitute y = vx. Then x dy - y dx = x² dv. For x > 0, x/√(x² - y²) = 1/√(1 - v²) and e^(y/x) = eᵛ. Divide both sides by x² and the variables separate.

    (1/√(1 - v²) + eᵛ)dv = dx/x ⇒ sin⁻¹ (y/x) + e^(y/x) = ln x + C

  3. Find C. At (1, 0): sin⁻¹ 0 + e⁰ = ln 1 + C, so C = 1.
  4. Use the second point. At (2α, α), y/x = 1/2. This gives π/6 + √e = ln(2α) + 1, so 2α = exp(π/6 + √e - 1).
  5. So. α = 1/2 exp(π/6 + √e - 1), which is option (a).

The idea to keep. When a homogeneous equation has the combination x dy - y dx hidden in it, use y = vx to turn it into x² dv. The equation then separates in one line.

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.