What is the solution for (6x^2y^2-4y^4)dx+(2x^3y-4xy^3)dy=0 ?

The solution for (6x^2y^2-4y^4)dx+(2x^3y-4xy^3)dy=0 is

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(6x^2y^2-4y^4)dx+(2x^3y-4xy^3)dy=0

(6 x^{2} y^{2} - 4 y^{4}) d x + (2 x^{3} y - 4 x y^{3}) d y = 0

y = - \frac{- x ^{4} + x ^{8} - c _{1}}{2 x ^{2}}, y = - - \frac{- x ^{4} + x ^{8} - c _{1}}{2 x ^{2}}, y = \frac{x ^{4} + x ^{8} - c _{1}}{2 x}, y = - \frac{x ^{4} + x ^{8} - c _{1}}{2 x}

Solution steps

(6 x^{2} y^{2} - 4 y^{4}) d x + (2 x^{3} y - 4 x y^{3}) d y = 0

Solve exact ODE:

y = - \frac{- x ^{4} + x ^{8} - c _{1}}{2 x ^{2}}, y = - - \frac{- x ^{4} + x ^{8} - c _{1}}{2 x ^{2}}, y = \frac{x ^{4} + x ^{8} - c _{1}}{2 x}, y = - \frac{x ^{4} + x ^{8} - c _{1}}{2 x}

y = - \frac{- x ^{4} + x ^{8} - c _{1}}{2 x ^{2}}, y = - - \frac{- x ^{4} + x ^{8} - c _{1}}{2 x ^{2}}, y = \frac{x ^{4} + x ^{8} - c _{1}}{2 x}, y = - \frac{x ^{4} + x ^{8} - c _{1}}{2 x}

\int cos (π x - nπ x) d x

y = (2 t - 1) (3 t - 4)^{- 1}

\int \frac{1 + cos ( 2 θ )}{2} d θ

\frac{\partial}{\partial s} (4 s t)

\frac{d y}{d x} = y - 2 x

What is the solution for (6x^2y^2-4y^4)dx+(2x^3y-4xy^3)dy=0 ?
The solution for (6x^2y^2-4y^4)dx+(2x^3y-4xy^3)dy=0 is