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

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

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y^'=(3y^2+4x)/(2xy)

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

\int \frac{x ^{2}}{- 4 x ^{2} - 12 x - 5} d x

\frac{d}{d x} (f (x) e^{\frac{1}{f ( x )}})

f^{^{'}} (x) = \frac{( 5 x ^{2} - 9 x + 8 )}{7 x + 6}

\frac{\partial}{\partial x} (e^{- y})

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