What is the solution for 1/(2x)y^'= x/(3y) ?

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

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1/(2x)y^'= x/(3y)

\frac{1}{2 x} y^{^{'}} = \frac{x}{3 y}

y = \frac{c _{1} + 4 x ^{3}}{3}, y = - \frac{c _{1} + 4 x ^{3}}{3}

Solution steps

\frac{1}{2 x} y^{'} (x) = \frac{x}{3 y}

Solve separable ODE:

y = \frac{c _{1} + 4 x ^{3}}{3}, y = - \frac{c _{1} + 4 x ^{3}}{3}

y = \frac{c _{1} + 4 x ^{3}}{3}, y = - \frac{c _{1} + 4 x ^{3}}{3}

\frac{d y}{d x} = (y^{2} - 9)

\frac{\partial}{\partial x} (e^{- 5 t} cos (π x))

x \to - 1 lim (- 1)

\int 2 x (x^{2} + 13)^{7} d x

(6, 1), (4, 3)

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