What is the solution for y^'=(3x^2)/(2y-4) ?

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

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y^'=(3x^2)/(2y-4)

y^{^{'}} = \frac{3 x ^{2}}{2 y - 4}

y = 2 + x^{3} + c_{1} + 4, y = 2 - x^{3} + c_{1} + 4

Solution steps

y^{'} (x) = \frac{3 x ^{2}}{2 y - 4}

Solve separable ODE:

y = 2 + x^{3} + c_{1} + 4, y = 2 - x^{3} + c_{1} + 4

y = 2 + x^{3} + c_{1} + 4, y = 2 - x^{3} + c_{1} + 4

y^{^{'}} + 100 y = 5

\frac{d y}{d x} + 6 x^{2} y = 5 x^{2}, y (0) = 2

\frac{d y}{d x} = sin (x) (1 - cos (x)) (y^{2} - 4)

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

(x^{2} + 1) tan (y) y^{^{'}} = x

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