What is the solution for y^'-8/x y=(y^4)/(x^{17)},y(1)=1 ?

The solution for y^'-8/x y=(y^4)/(x^{17)},y(1)=1 is y=(2x^8)/(\sqrt[3]{-3x^8+11)}

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y^'-8/x y=(y^4)/(x^{17)},y(1)=1

y^{^{'}} - \frac{8}{x} y = \frac{y ^{4}}{x ^{17}}, y (1) = 1

Solution steps

y^{'} (x) - \frac{8}{x} y = \frac{y ^{4}}{x ^{17}}

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x \to 3 + lim (- \frac{2}{x ^{2} - 9})

\int 5 e^{- 0.4 x} d x

f (x) = 7 sec (x), at x = \frac{π}{6}

x y^{^{'}} + y = x

x \to a lim (\frac{1}{0})

What is the solution for y^'-8/x y=(y^4)/(x^{17)},y(1)=1 ?
The solution for y^'-8/x y=(y^4)/(x^{17)},y(1)=1 is y=(2x^8)/(\sqrt[3]{-3x^8+11)}