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MHT CET202210 Aug 2022Evening ShiftMathematicsDifferential EquationsActual

The particular solution of y x d y ~d x = 1+y^2 1+x^2 when x=2, y=1 is

Options

  1. A(1+y^2 )=2 (1+x^2 )
  2. B2 (1+y^2 )=5 (1+x^2 )
  3. C2 (1+y^2 )= (1+x^2 )
  4. D5 (1+y^2 )=2 (1+x^2 )

Correct answer

D. 5 (1+y^2 )=2 (1+x^2 )

Step-by-step solution

aligned & y x d y ~d x = 1+y^2 1+x^2 & y 1+y^2 ~d y= x 1+x^2 ~d x & 1 2 2 y ~d y 1+y^2 = 1 2 2 x 1+x^2 ~d x & 1 2 |1+y^2 |= 1 2 |1+x^2 |+ 1 2 c & 1+y^2 1+x^2 = c & 1+y^2 1+x^2 =C aligned Putting x=2, y=1 we get C= 2 5 aligned & 1+y^2 1+x^2 = 2 5 & 5 (1+y^2 )=2 (1+x^2 ) aligned

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