What is the solution for (dy)/(dx)=(y(2x^3-y^3))/(x(2x^3-3y^3)) ?

The solution for (dy)/(dx)=(y(2x^3-y^3))/(x(2x^3-3y^3)) is y=-\sqrt[3]{2/(3\W(-(2c_{1)x^2)/3)}}x

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(dy)/(dx)=(y(2x^3-y^3))/(x(2x^3-3y^3))

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

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

Solution steps

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

Solve using substitution:

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

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

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What is the solution for (dy)/(dx)=(y(2x^3-y^3))/(x(2x^3-3y^3)) ?
The solution for (dy)/(dx)=(y(2x^3-y^3))/(x(2x^3-3y^3)) is y=-\sqrt[3]{2/(3\W(-(2c_{1)x^2)/3)}}x