What is the solution for y^'=6x(y-1)^{2/3} ?

The solution for y^'=6x(y-1)^{2/3} is y=x^6+c_{1}x^4+(c_{1}^2x^2)/3+(c_{1}^3)/(27)+1

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y^'=6x(y-1)^{2/3}

y^{^{'}} = 6 x (y - 1)^{\frac{2}{3}}

y = x^{6} + c_{1} x^{4} + \frac{c _{1}^{2} x ^{2}}{3} + \frac{c _{1}^{3}}{27} + 1

Solution steps

y^{'} (x) = 6 x (y - 1)^{\frac{2}{3}}

Solve separable ODE:

y = x^{6} + c_{1} x^{4} + \frac{c _{1}^{2} x ^{2}}{3} + \frac{c _{1}^{3}}{27} + 1

y = x^{6} + c_{1} x^{4} + \frac{c _{1}^{2} x ^{2}}{3} + \frac{c _{1}^{3}}{27} + 1

k = 0 \sum \infty \frac{k}{2 ^{k}} x^{k}

y^{^{''}} + 214 y^{^{'}} + 14 y = 0

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

\frac{\partial}{\partial x} (e^{3 x} cos (2 x))

\int \frac{1}{x ^{4}} d x

What is the solution for y^'=6x(y-1)^{2/3} ?
The solution for y^'=6x(y-1)^{2/3} is y=x^6+c_{1}x^4+(c_{1}^2x^2)/3+(c_{1}^3)/(27)+1