What is the solution for (dy)/(dx)=xy^2-y^2+x-1 ?

The solution for (dy)/(dx)=xy^2-y^2+x-1 is y=tan((x^2)/2-x+c_{1})

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(dy)/(dx)=xy^2-y^2+x-1

\frac{d y}{d x} = x y^{2} - y^{2} + x - 1

y = tan (\frac{x ^{2}}{2} - x + c_{1})

Solution steps

\frac{d y}{d x} = x y^{2} - y^{2} + x - 1

Solve separable ODE:

y = tan (\frac{x ^{2}}{2} - x + c_{1})

y = tan (\frac{x ^{2}}{2} - x + c_{1})

14 y^{6} \frac{d y}{d x} = 8 x^{7}

7 x y \frac{d y}{d x} = 4, y > 0, x > 0

y^{^{'}} = \frac{ln ( x ) ( y ^{2} + 1 )}{x y}

\frac{d y}{d x} = x y, y (- 1) = \frac{25}{16}

(1 + e^{x}) e^{y} y^{^{'}} = y^{- 1}

What is the solution for (dy)/(dx)=xy^2-y^2+x-1 ?
The solution for (dy)/(dx)=xy^2-y^2+x-1 is y=tan((x^2)/2-x+c_{1})