A curve passes through the point (0,1) and the gradient at (x,y) on it is `y(xy-1)`. The equation of the curve is

Options

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

Answer

a

Explanation

We have , `(dy)/(dx) = y (xy -1)`
` rArr dy = xy^(2)dx - y dx`
` rArr y dx + dy = xy^(2) dx rArr (y" "dx+dy)/(y^(2)) = xdx`
` rArr (ye^(-x) dx + e^(-x) dy)/(y^(2)) = xe^(-x) dx`
` :. -d ((e^(-x))/y) = xe^(-x) dx`
On integrating , we get
` - (e^(-x)/y ) = - xe^(-x) - e^(-x)-C`
` rArr 1/y = x+1 + Ce^(x) " " ` ...(i)
Since , it passes through (0,1)
Therefore , `1 = 1+C rArr C = 0 `
On putting C = 0 in Eq . (i) , we get
` 1/y = x+1 rArr y (x+1) =1`