The equation of the curve satisfying the eqution `(xy-x^(2)) (dy)/(dx) = y^(2)` and passing through the point `(-1,1)` isA. `y=(logy-1)x`B. `y=(log y+1)x`C. `x= (log x-1)y`D. `x = (log x+1)y`

The equation of the curve satisfying the eqution `(xy-x^(2)) (dy)/(dx)

1.	The equation of the curve satisfying the eqution `(xy-x^(2)) (dy)/(dx) = y^(2)` and passing through the point `(-1,1)` isA. `y=(logy-1)x`B. `y=(log y+1)x`C. `x= (log x-1)y`D. `x = (log x+1)y`
Answer» Correct Answer - b 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)`

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The equation of the curve satisfying the eqution `(xy-x^(2)) (dy)/(dx) = y^(2)` and passing through the point `(-1,1)` isA. `y=(logy-1)x`B. `y=(log y+1)x`C. `x= (log x-1)y`D. `x = (log x+1)y`

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