What is the solution for y^'-9/x y=(y^4)/(x^{12)} ?

The solution for y^'-9/x y=(y^4)/(x^{12)} is y=(2\sqrt[3]{2}x^9)/(\sqrt[3]{-3x^{16)+c_{1}}}

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y^'-9/x y=(y^4)/(x^{12)}

y^{^{'}} - \frac{9}{x} y = \frac{y ^{4}}{x ^{12}}

Solution steps

y^{'} (x) - \frac{9}{x} y = \frac{y ^{4}}{x ^{12}}

Solve Bernoulli ODE:

\frac{d y}{d t} = - y^{2} t - y^{2} t^{2}

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

x \to 2 - lim (2 - \frac{1}{x - 2})

\frac{d y}{d x} = \frac{4 x ^{2} + 2 x}{y ^{2}}

f (x) = \frac{1}{x ^{2}}, 1, at x = 1

What is the solution for y^'-9/x y=(y^4)/(x^{12)} ?
The solution for y^'-9/x y=(y^4)/(x^{12)} is y=(2\sqrt[3]{2}x^9)/(\sqrt[3]{-3x^{16)+c_{1}}}