The equation of a curve passing through (1,0) and having slope of tangent at any point ( x , y ) of the curve as y -1 x^2+x is

Step-by-step solution

The slope of the tangent at any point (x, y) is given by dy dx = y-1 x^2+x . Separating variables yields dy y-1 = dx x(x+1) . Integrating both sides gives dy y-1 = dx x(x+1) . The left side integrates to |y-1| + C₁ . For the right side, decompose into partial fractions: 1 x(x+1) = 1 x - 1 x+1 . Integrating gives |x| - |x+1| + C₂ = | x x+1 | + C₂ . Combining results, |y-1| = | x x+1 | + C . Using the point (1, 0) to find C : |0-1| = | 1 2 | + C , so 0 = 1 2 + C and C = 2 . Substituting back: |y-1| = | x x+1 | + 2 =