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}{d x} (x^{3} y^{5} - 2 x^{2} y + x)

\int (x - 2) (x + 2) d x

f (x) = 4 tan (x), at x = \frac{π}{6}

\frac{d}{d x} (ln (x^{3}) + 2 x)

\frac{d}{d x} (cos (x) sin^{2} (x))

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)