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

The solution for (dy)/(dx)=1-y+x^2-yx^2 is y=e^{-(x^3)/3-x+c_{1}}+1

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

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

y = e^{- \frac{x ^{3}}{3} - x + c_{1}} + 1

Solution steps

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

Solve separable ODE:

y = e^{- \frac{x ^{3}}{3} - x + c_{1}} + 1

y = e^{- \frac{x ^{3}}{3} - x + c_{1}} + 1

\int_{0}^{1} (x^{2} + 1) e^{- x} d x

n = 0 \sum \infty \frac{2}{5 ^{n}}

\frac{d}{d x} (csc (2 x))

\frac{d y}{d x} = - \frac{x}{4 y}

\frac{\partial}{\partial y} (8 x y^{2} z^{3})

What is the solution for (dy)/(dx)=1-y+x^2-yx^2 ?
The solution for (dy)/(dx)=1-y+x^2-yx^2 is y=e^{-(x^3)/3-x+c_{1}}+1