What is the solution for y^'=(1-2t)y^2,y(0)=-1/6 ?

The solution for y^'=(1-2t)y^2,y(0)=-1/6 is y=-1/(t-t^2+6)

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y^'=(1-2t)y^2,y(0)=-1/6

y^{^{'}} = (1 - 2 t) y^{2}, y (0) = - \frac{1}{6}

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

Solution steps

y^{'} (t) = (1 - 2 t) y^{2}

Solve separable ODE:

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

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

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What is the solution for y^'=(1-2t)y^2,y(0)=-1/6 ?
The solution for y^'=(1-2t)y^2,y(0)=-1/6 is y=-1/(t-t^2+6)