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

The solution for 2x-6y^2(dy)/(dx)+4=0 is y=\sqrt[3]{(x^2)/2+2x+c_{1}}

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

2 x - 6 y^{2} \frac{d y}{d x} + 4 = 0

y = 3 \frac{x ^{2}}{2} + 2 x + c_{1}

Solution steps

2 x - 6 y^{2} \frac{d y}{d x} + 4 = 0

Solve separable ODE:

y = 3 \frac{x ^{2}}{2} + 2 x + c_{1}

y = 3 \frac{x ^{2}}{2} + 2 x + c_{1}

\frac{d y}{d x} = \frac{x}{e ^{- y} ( 1 + x ^{2} ) ^{4}}

(3 x^{2} + 2)^{2} y^{^{'}} - x y^{2} = 0, y (0) = \frac{3}{2}

\frac{x d x}{1 + x ^{2}} = \frac{d y}{y}

\frac{d y}{d x} = 2 x y, y (1) = 5

x^{^{'}} = 20 t - \frac{4 x}{30} t

What is the solution for 2x-6y^2(dy)/(dx)+4=0 ?
The solution for 2x-6y^2(dy)/(dx)+4=0 is y=\sqrt[3]{(x^2)/2+2x+c_{1}}