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

\int (sin (x))^{3} (cos (x))^{3} d x

e^{a r c s i n (x)}

\frac{d y}{d x} = - \frac{1}{2 y}

\int_{- 1}^{1} - 2 y^{2} + 2 d y

y^{^{''}} + 18 y^{^{'}} + 80 y = 0

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}