What is the solution for y^'=y^3-6x^2y^3,y(1)=-1/2 ?

The solution for y^'=y^3-6x^2y^3,y(1)=-1/2 is y=-sqrt(-1/(2(x-2x^3-1)))

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y^'=y^3-6x^2y^3,y(1)=-1/2

y^{^{'}} = y^{3} - 6 x^{2} y^{3}, y (1) = - \frac{1}{2}

y = - - \frac{1}{2 ( x - 2 x ^{3} - 1 )}

Solution steps

y^{'} (x) = y^{3} - 6 x^{2} y^{3}

Solve separable ODE:

y = - - \frac{1}{2 ( x - 2 x ^{3} - 1 )}

y = - - \frac{1}{2 ( x - 2 x ^{3} - 1 )}

x \to 1 lim (\frac{x ^{2} + 1}{x + 1})

f (x) = 80640 (2 x + 5)^{2}

f (x) = x^{2} - x

2 x x

y = x + tan (x), (π, π)

What is the solution for y^'=y^3-6x^2y^3,y(1)=-1/2 ?
The solution for y^'=y^3-6x^2y^3,y(1)=-1/2 is y=-sqrt(-1/(2(x-2x^3-1)))