What is the solution for y^'+x^{-1}y= 2/3 x^4y^4 ?

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

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y^'+x^{-1}y= 2/3 x^4y^4

y^{^{'}} + x^{- 1} y = \frac{2}{3} x^{4} y^{4}

Solution steps

y^{'} (x) + x^{- 1} y = \frac{2}{3} x^{4} y^{4}

Solve Bernoulli ODE:

y^{^{''}} + 2 y^{^{'}} + y = 12 e^{- t}

\frac{\partial}{\partial x} (x e^{x} + e^{x})

\frac{\partial}{\partial y} (x - 2 y)

y^{^{''}} - 14 y^{^{'}} + 50 y = 0, y (0) = 10, y (\frac{π}{2}) = 3

f (x) = \frac{e ^{x} x - e ^{x}}{x ^{2}}

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