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

The solution for y^'= 2/(3x)y+(2x)/y is y=sqrt(6x^2+c_{1)x^{4/3}},y=-sqrt(6x^2+c_{1)x^{4/3}}

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

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

y = 6 x^{2} + c_{1} x^{\frac{4}{3}}, y = - 6 x^{2} + c_{1} x^{\frac{4}{3}}

Solution steps

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

Solve Bernoulli ODE:

y = 6 x^{2} + c_{1} x^{\frac{4}{3}}, y = - 6 x^{2} + c_{1} x^{\frac{4}{3}}

y = 6 x^{2} + c_{1} x^{\frac{4}{3}}, y = - 6 x^{2} + c_{1} x^{\frac{4}{3}}

y = x^{2} - 5 x, (7, 14)

x \to - \infty lim (x^{3} + 2)

2 x + y

x \to 6 - lim (\frac{7}{x - 6})

(4 x^{2} + 3 y^{2}) d x - (2 x y) d y = 0

What is the solution for y^'= 2/(3x)y+(2x)/y ?
The solution for y^'= 2/(3x)y+(2x)/y is y=sqrt(6x^2+c_{1)x^{4/3}},y=-sqrt(6x^2+c_{1)x^{4/3}}