What is the solution for y^'=(1-2x)y^2,y(0)=(-1)/6 ?

The solution for y^'=(1-2x)y^2,y(0)=(-1)/6 is y=-1/(x-x^2+6)

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y^'=(1-2x)y^2,y(0)=(-1)/6

y^{^{'}} = (1 - 2 x) y^{2}, y (0) = \frac{- 1}{6}

y = - \frac{1}{x - x ^{2} + 6}

Solution steps

y^{'} (x) = (1 - 2 x) y^{2}

Solve separable ODE:

y = - \frac{1}{x - x ^{2} + 6}

y = - \frac{1}{x - x ^{2} + 6}

\frac{d y}{d x} - y^{5} = 0, y (0) = 1

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

(1 + x^{2}) \frac{d y}{d x} + 2 x y = - x, x > 0

(x + 2) \frac{d y}{d x} = y

x^{3} y^{^{'}} - (8 x^{2} - 5) y = 0

What is the solution for y^'=(1-2x)y^2,y(0)=(-1)/6 ?
The solution for y^'=(1-2x)y^2,y(0)=(-1)/6 is y=-1/(x-x^2+6)