What is the solution for y^'=(64xy)^{1/3} ?

The solution for y^'=(64xy)^{1/3} is

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y^'=(64xy)^{1/3}

y^{^{'}} = (64 x y)^{\frac{1}{3}}

y = \frac{2 2 27 x ^{4} + 27 c _{1} x ^{\frac{8}{3}} + 9 c _{1}^{2} x ^{\frac{4}{3}} + c _{1}^{3}}{3 3}, y = - \frac{2 2 27 x ^{4} + 27 c _{1} x ^{\frac{8}{3}} + 9 c _{1}^{2} x ^{\frac{4}{3}} + c _{1}^{3}}{3 3}

Solution steps

y^{'} (x) = (64 x y)^{\frac{1}{3}}

Solve separable ODE:

y = \frac{2 2 27 x ^{4} + 27 c _{1} x ^{\frac{8}{3}} + 9 c _{1}^{2} x ^{\frac{4}{3}} + c _{1}^{3}}{3 3}, y = - \frac{2 2 27 x ^{4} + 27 c _{1} x ^{\frac{8}{3}} + 9 c _{1}^{2} x ^{\frac{4}{3}} + c _{1}^{3}}{3 3}

y = \frac{2 2 27 x ^{4} + 27 c _{1} x ^{\frac{8}{3}} + 9 c _{1}^{2} x ^{\frac{4}{3}} + c _{1}^{3}}{3 3}, y = - \frac{2 2 27 x ^{4} + 27 c _{1} x ^{\frac{8}{3}} + 9 c _{1}^{2} x ^{\frac{4}{3}} + c _{1}^{3}}{3 3}

\int_{1}^{\infty} \frac{1}{y ^{2}} d y

y = \frac{x}{2}

\frac{9}{x ^{3}}

\int \frac{( 3 x ^{2} + 7 x + 3 )}{x ( x + 1 ) ^{2}} d x

\frac{\partial}{\partial x} (\frac{sin ( y )}{x ^{2} + 1})

What is the solution for y^'=(64xy)^{1/3} ?
The solution for y^'=(64xy)^{1/3} is