The equation of the curve satisfying the eqution `(xy-x^(2)) (dy)/(dx) = y^(2)` and passing through the point `(-1,1)` is

Options

A. `y=(logy-1)x`
B. `y=(log y+1)x`
C. `x= (log x-1)y`
D. `x = (log x+1)y`

Answer

b

Explanation

We have , `(xy-x^(2)) (dy)/(dx) = y^(2)`
` rArr y^(2) (dx)/(dy) = xy -x^(2)`
` rArr 1/(x^(2)) (dx)/(dy) - 1/x . 1/y = -1/(y^(2))`
Put `1/x = v`
` rArr -1/(x^(2)) (dx)/(dy) = (dv)/(dy)`
` (dv)/(dy) + v/y = 1/(y^(2))`, which is linear
` :. IF - e^(int1/ydy) = e^(logy) = y`
` :. " The solution " vy = int 1/(y^(2) ) . y dy + C`
` rArr y/x = log y + C rArr y = x ( log y + C)`
This passes through the point `(-1),1)`
` :. 1 = -1 ( log 1 +C) " i.e " C =-1`
Thus , the equation of the curve is y = x `(log y -1)`