What is the solution for -3y^'+(3y)/x =x^4y^4 ?

The solution for -3y^'+(3y)/x =x^4y^4 is y=(2x)/(\sqrt[3]{x^8+c_{1)}}

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-3y^'+(3y)/x =x^4y^4

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

Solution steps

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

Solve Bernoulli ODE:

\int \frac{4}{2 x - 1} d x

\frac{d}{d x} (2 x e^{3 x})

x \to 6 lim (\frac{3}{x - 6})

\frac{d}{d x} (\frac{( x ^{5} - 5 )}{( x + 1 )})

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

What is the solution for -3y^'+(3y)/x =x^4y^4 ?
The solution for -3y^'+(3y)/x =x^4y^4 is y=(2x)/(\sqrt[3]{x^8+c_{1)}}