What is the solution for -3y^'+(4y)/x =(y^4)/(x^2) ?

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

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-3y^'+(4y)/x =(y^4)/(x^2)

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

Solution steps

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

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