What is the solution for y^'=x^2-2x^2y ?

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

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y^'=x^2-2x^2y

y^{^{'}} = x^{2} - 2 x^{2} y

y = \frac{e ^{\frac{- 2 x ^{3} + c _{1}}{3}}}{2} + \frac{1}{2}

Solution steps

y^{'} (x) = x^{2} - 2 x^{2} y

Solve separable ODE:

y = \frac{e ^{\frac{- 2 x ^{3} + c _{1}}{3}}}{2} + \frac{1}{2}

y = \frac{e ^{\frac{- 2 x ^{3} + c _{1}}{3}}}{2} + \frac{1}{2}

1 - y^{2} (d x) - 1 + x^{2} (d y) = 0

1 = \frac{5}{3} y + \frac{d y}{d x}

\frac{d y}{d t} + 5 t y = e^{t} y

x \cdot d y - 2 y \cdot d x = 0

\frac{d y}{d x} = \frac{4 x 1 + y ^{2}}{y}

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