What is the solution for y^'=-4xy^3 ?

The solution for y^'=-4xy^3 is y=sqrt(-1/(2(-2x^2+c_{1)))},y=-sqrt(-1/(2(-2x^2+c_{1)))}

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y^'=-4xy^3

y^{^{'}} = - 4 x y^{3}

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

Solution steps

y^{'} (x) = - 4 x y^{3}

Solve separable ODE:

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

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

\int \frac{x - 3}{x ^{3} + 2 x ^{2}} d x

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

\frac{2}{x} - \frac{2}{x ^{3}} + \frac{1}{x ^{4}}

\int \frac{sin ( 2 x )}{1 - cos ( 2 x )} d x

What is the solution for y^'=-4xy^3 ?
The solution for y^'=-4xy^3 is y=sqrt(-1/(2(-2x^2+c_{1)))},y=-sqrt(-1/(2(-2x^2+c_{1)))}