What is the solution for y^'-3/x y=(y^4)/(x^2) ?

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

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y^'-3/x y=(y^4)/(x^2)

y^{^{'}} - \frac{3}{x} y = \frac{y ^{4}}{x ^{2}}

y = \frac{2 x ^{3}}{3 - 3 x ^{8} + c _{1}}

Solution steps

y^{'} (x) - \frac{3}{x} y = \frac{y ^{4}}{x ^{2}}

Solve Bernoulli ODE:

y = \frac{2 x ^{3}}{3 - 3 x ^{8} + c _{1}}

y = \frac{2 x ^{3}}{3 - 3 x ^{8} + c _{1}}

d er i v a t i v e \frac{x ^{2} + 1}{x}

\int sin (x) cos (n) x d x

y^{^{'}} = \frac{t}{1 + t ^{2}} \cdot e^{y}

\frac{d}{d r} (\frac{1}{r ^{2}})

\frac{d}{d n} ((n + 1)^{n})

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