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AP EAMCET202419 May 2024Evening ShiftMathematicsDifferential EquationsActual

If x d y+ (y+y^2 x ) d x=0 and y=1 at x=1 . then

Options

  1. Ay= x 1+ x
  2. By= 1+ x x
  3. Cy=x(1+ x)
  4. Dy= 1 x(1+ x)

Correct answer

D. y= 1 x(1+ x)

Step-by-step solution

Since, x d y+ (y+y^2 x ) d x=0 d y d x + y x +y^2=0...(i) Let y= 1 t d y d x =- 1 t^2 d t d x aligned & - 1 t^2 d t d x + 1 t x + 1 t^2 =0 d t d x - t x =1 & I.F. =e^ - 1 x d x =e^ - x = 1 x aligned So, solution is given by t x = 1 x d x= x+c 1 y x = x+c Since, y(1)=1 1 1 = 1+c c=1 . 1 y x = x+1 y= 1 x(1+ x) .

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