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

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

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y^'= 3/x (y-y^{2/3})

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

u^{^{'}} = 5 e^{8 x - H (x)}

y^{^{'}} = 9 y

\frac{d y}{d x} = \frac{( 1 + y ^{2} )}{( 1 + x ^{2} )}

y^{^{'}} = 2 y - 1, y (0) = \frac{3}{2}

\frac{d y}{d x} = \frac{- 1}{x ^{2} y ^{2}}, y (1) = 2

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