What is the solution for y^'-6/x y=(y^4)/(x^{12)} ?

The solution for y^'-6/x y=(y^4)/(x^{12)} is y=(\sqrt[3]{7}x^6)/(\sqrt[3]{-3x^7+c_{1)}}

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y^'-6/x y=(y^4)/(x^{12)}

y^{^{'}} - \frac{6}{x} y = \frac{y ^{4}}{x ^{12}}

y = \frac{3 7 x ^{6}}{3 - 3 x ^{7} + c _{1}}

Solution steps

y^{'} (x) - \frac{6}{x} y = \frac{y ^{4}}{x ^{12}}

Solve Bernoulli ODE:

y = \frac{3 7 x ^{6}}{3 - 3 x ^{7} + c _{1}}

y = \frac{3 7 x ^{6}}{3 - 3 x ^{7} + c _{1}}

\int \frac{4 ln ( 4 x ^{4} )}{x ^{4}} d x

\frac{d}{d x} (\frac{4 x}{2 x ^{2} + 1})

(x + 2)^{2} + (10 - 3 x)^{2}

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

(1, 0), (0, 1)

What is the solution for y^'-6/x y=(y^4)/(x^{12)} ?
The solution for y^'-6/x y=(y^4)/(x^{12)} is y=(\sqrt[3]{7}x^6)/(\sqrt[3]{-3x^7+c_{1)}}