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

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

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

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

y = \frac{- x ^{2} + x ^{4} - 4 ( x ^{3} - c _{1} )}{2}, y = - \frac{- x ^{2} + x ^{4} - 4 ( x ^{3} - c _{1} )}{2}, y = \frac{- x ^{2} - x ^{4} - 4 ( x ^{3} - c _{1} )}{2}, y = - \frac{- x ^{2} - x ^{4} - 4 ( x ^{3} - c _{1} )}{2}

Solution steps

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

Solve exact ODE:

y = \frac{- x ^{2} + x ^{4} - 4 ( x ^{3} - c _{1} )}{2}, y = - \frac{- x ^{2} + x ^{4} - 4 ( x ^{3} - c _{1} )}{2}, y = \frac{- x ^{2} - x ^{4} - 4 ( x ^{3} - c _{1} )}{2}, y = - \frac{- x ^{2} - x ^{4} - 4 ( x ^{3} - c _{1} )}{2}

y = \frac{- x ^{2} + x ^{4} - 4 ( x ^{3} - c _{1} )}{2}, y = - \frac{- x ^{2} + x ^{4} - 4 ( x ^{3} - c _{1} )}{2}, y = \frac{- x ^{2} - x ^{4} - 4 ( x ^{3} - c _{1} )}{2}, y = - \frac{- x ^{2} - x ^{4} - 4 ( x ^{3} - c _{1} )}{2}

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