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

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

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y^'=-2xy^3

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

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

Solution steps

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

Solve separable ODE:

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

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

x \to \frac{3 π}{2} lim (sec (x))

x \to 0 lim (\frac{x ^{3}}{ln ∣ x ∣})

y^{^{'}} = y (x y^{3} + 5)

\int \frac{e ^{u}}{( 7 - e ^{u} ) ^{2}} d u

(x^{2} + 1) d y = (x^{3} - 2 x y + x) d x

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