What is the solution for y^'-4/x y= 1/(x^3)y^4 ?

The solution for y^'-4/x y= 1/(x^3)y^4 is y=(\sqrt[3]{10}x^4)/(\sqrt[3]{-3x^{10)+c_{1}}}

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y^'-4/x y= 1/(x^3)y^4

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

Solution steps

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

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What is the solution for y^'-4/x y= 1/(x^3)y^4 ?
The solution for y^'-4/x y= 1/(x^3)y^4 is y=(\sqrt[3]{10}x^4)/(\sqrt[3]{-3x^{10)+c_{1}}}