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

Let the equation of a curve passing through the point (0,1) be given b y=intx^2e^(x^3)dx . If the equation of the curve is written in the form x=f(y) , then f(y) is

A curve C passes through (2,0) and the slope at (x , y) as ((x+1)^2+(y-3))/(x+1)dot Find the equation of the curve.

If the slope of the tangent at (x,y) to a curve passing through the point (2,1) is (x^(2)+y^(2))/(2xy) , then the equation of the curve is-

If the gradient of the tangent at any point (x,y) of a curve which passes through the point (1, (pi)/(4)) is {(y)/(x)-sin^(2)((y)/(x))} , then the equation of the curve is-

The gradient (siope) of a curve at any point (x,y) is (x^(2)-4)/(x^(2)) . If the curve passes through the point (2 , 7) , then the equation of the curve is ,

The gradient (slope) of a curve at any point (x,y) is (x^(2) -4)/(x^(2)) . If the curve passes through the point (2,7) , then the equation of the curve is. . . .

A curve y=f(x) passes through point P(1,1). The normal to the curve at P is a (y-1)+(x-1)=0. If the slope of the tangent at any point on the curve is proportional to the ordinate of the point,then the equation of the curve is

The slope of the tangent at (x,y) to a curve passing through a point (2,1) is (x^(2)+y^(2))/(2xy) then the equation of the curve is

A curve y=f(x) passes through the point P(1,1) . The normal to the curve at P is a(y-1)+(x-1)=0 . If the slope of the tangent at any point on the curve is proportional to the ordinate of the point. Determine the equation of the curve. Also obtain the area bounded by the y-axis, the curve and the normal to the curve at P .