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

The solution for -3y^'+(3y)/x =xy^4 is y=(\sqrt[3]{5}x)/(\sqrt[3]{x^5+c_{1)}}

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

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

Solution steps

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

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\int \frac{x}{( x ^{2} - 3 ) ^{3}} d x

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What is the solution for -3y^'+(3y)/x =xy^4 ?
The solution for -3y^'+(3y)/x =xy^4 is y=(\sqrt[3]{5}x)/(\sqrt[3]{x^5+c_{1)}}