JEE Main 2020 · 6 September · morning · Single correct · +4 / −1

The general solution of the differential equation
√(1 + x² + y² + x²y²) + xydy/dx = 0 is :
(where C is a constant of integration)

  1. (a) √(1 + y²) + √(1 + x²) = 1/2log ₑ( (√(1 + x²) - 1)/(√(1 + x²) + 1) ) + C
  2. (b) √(1 + y²) - √(1 + x²) = 1/2log ₑ( (√(1 + x²) - 1)/(√(1 + x²) + 1) ) + C
  3. (c) √(1 + y²) + √(1 + x²) = 1/2log ₑ( (√(1 + x²) + 1)/(√(1 + x²) - 1) ) + C
  4. (d) √(1 + y²) - √(1 + x²) = 1/2log ₑ( (√(1 + x²) + 1)/(√(1 + x²) - 1) ) + C
Answer and solution

Answer (c) √(1 + y²) + √(1 + x²) = 1/2log ₑ( (√(1 + x²) + 1)/(√(1 + x²) - 1) ) + C

Solution

Maths sir · 4 steps

Factor the root into (1+x²)(1+y²), and the equation separates straight away.

  1. Factor and separate. Group the terms under the root: 1 + x² + y² + x²y² = (1+x²)(1+y²). Divide by x√(1+y²) so that each variable sits on its own side.

    (y dy)/√(1+y²) = -√(1+x²)/x dx

  2. Left side. Put u = 1+y², so y dy = 1/2du. The integral is ∫ du/2√u = √(1+y²).
  3. Right side. Put t = √(1+x²), so t dt = x dx and dx/x = (t dt)/(t²-1). Then ∫ √(1+x²)/xdx = ∫ t²/(t²-1)dt = ∫ (1 + 1/(t²-1))dt = t + 1/2 logₑ(t-1)/(t+1).
  4. So. √(1+y²) = -√(1+x²) - 1/2 logₑ(t-1)/(t+1) + C. Move √(1+x²) to the left and use -logₑ p/q = logₑ q/p. This gives √(1+y²) + √(1+x²) = 1/2 logₑ(√(1+x²)+1)/(√(1+x²)-1) + C, which is option (c).

The idea to keep. When a root holds four terms like 1+x²+y²+x²y², try factorising it into a product. For ∫ √(1+x²)/xdx, put t=√(1+x²) to get a rational integral.

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.