What is the solution for y^'=y^3-6x^2y^3 ?

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

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y^'=y^3-6x^2y^3

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

y = - \frac{1}{2 ( x - 2 x ^{3} + c _{1} )}, y = - - \frac{1}{2 ( x - 2 x ^{3} + c _{1} )}

Solution steps

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

Solve separable ODE:

y = - \frac{1}{2 ( x - 2 x ^{3} + c _{1} )}, y = - - \frac{1}{2 ( x - 2 x ^{3} + c _{1} )}

y = - \frac{1}{2 ( x - 2 x ^{3} + c _{1} )}, y = - - \frac{1}{2 ( x - 2 x ^{3} + c _{1} )}

\int e^{5 θ} sin (6 θ) d θ

1 0^{x}

\frac{d y}{d x} - 5 y = 6 e^{8 x}

\frac{d}{d x} (e^{\frac{1}{x ^{2}}} + \frac{1}{e ^{x^{2}}})

\int \frac{x ^{2}}{4} - 4 x + 17 d x

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