The equation of the curve whose slope is y-1 x^2+x and which passes through the point (1, 0) is

Step-by-step solution

The slope of the curve is given by dy dx . dy dx = y-1 x^2+x Separating the variables, we get: dy y-1 = dx x(x+1) Integrating both sides: dy y-1 = ( 1 x - 1 x+1 ) dx |y-1| = |x| - |x+1| + |C| |y-1| = | Cx x+1 | y-1 = Cx x+1 Since the curve passes through the point (1, 0) , we substitute x = 1 and y = 0 : 0 - 1 = C(1) 1+1 -1 = C 2 C = -2 Substituting C = -2 back into the equation: y-1 = -2x x+1 (y-1)(x+1) = -2x xy + y - x - 1 = -2x xy + x + y - 1 = 0 Answer: xy + x + y - 1 = 0