What is the solution for y^'=(3y^2+9)/((2xy)) ?

The solution for y^'=(3y^2+9)/((2xy)) is y=sqrt(c_{1)x^3-3},y=-sqrt(c_{1)x^3-3}

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y^'=(3y^2+9)/((2xy))

y^{^{'}} = \frac{3 y ^{2} + 9}{( 2 x y )}

y = c_{1} x^{3} - 3, y = - c_{1} x^{3} - 3

Solution steps

y^{'} (x) = \frac{3 y ^{2} + 9}{( 2 x y )}

Solve separable ODE:

y = c_{1} x^{3} - 3, y = - c_{1} x^{3} - 3

y = c_{1} x^{3} - 3, y = - c_{1} x^{3} - 3

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

\frac{d y}{d x} = \frac{( ln ( x ) ) ^{6}}{x y ^{2}}, y (1) = 4

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

\frac{d}{d x} ((y)^{2}) + \frac{d y}{d x} - 2 y = 0

y^{^{'}} + x y = 5 x y^{2}

What is the solution for y^'=(3y^2+9)/((2xy)) ?
The solution for y^'=(3y^2+9)/((2xy)) is y=sqrt(c_{1)x^3-3},y=-sqrt(c_{1)x^3-3}