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

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

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

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

\int - cos (x) sin (n x) d x

\frac{\partial}{\partial x} (3 x cos (x y))

g (x) = x (x^{3} - x)

\int (\frac{d y}{1}) d x

\int \frac{2 x ^{2} + 3 x + 3}{( x + 1 ) ^{3}} d x

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