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JEE Main 2000 Mathematics question on Limits Continuity and Differentiability

If x2+y2=1x^2 + y^2 = 1, then
  1. A.
    yy′′−2(y′)2+1=0yy'' - 2(y')^2 + 1 = 0
  2. B.
    yy′′+(y′)2+1=0yy'' + (y')^2 + 1 = 0
  3. C.
    yy′′−(y′)2−1=0yy'' - (y')^2 - 1 = 0
  4. D.
    yy′′+2(y′)2+1=0yy'' + 2(y')^2 + 1 = 0
Show answer and solution

Correct option: B

Solution

Differentiating x² + y² = 1 gives 2x + 2yy' = 0, so x + yy' = 0.

Differentiating again: 1 + (y')² + yy'' = 0, which rearranges to yy'' + (y')² + 1 = 0.

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