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
- A(1+y^2 )=2 (1+x^2 )
- B2 (1+y^2 )=5 (1+x^2 )
- C2 (1+y^2 )= (1+x^2 )
- 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