What is the solution for y^'-8/x y=(y^4)/(x^{13)} ?

The solution for y^'-8/x y=(y^4)/(x^{13)} is y=(\sqrt[3]{4}x^8)/(\sqrt[3]{-x^{12)+c_{1}}}

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y^'-8/x y=(y^4)/(x^{13)}

y^{^{'}} - \frac{8}{x} y = \frac{y ^{4}}{x ^{13}}

Solution steps

y^{'} (x) - \frac{8}{x} y = \frac{y ^{4}}{x ^{13}}

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What is the solution for y^'-8/x y=(y^4)/(x^{13)} ?
The solution for y^'-8/x y=(y^4)/(x^{13)} is y=(\sqrt[3]{4}x^8)/(\sqrt[3]{-x^{12)+c_{1}}}