What is the solution for y^'= 1/x y-(x^2)/3 y^4,y(1)=1 ?

The solution for y^'= 1/x y-(x^2)/3 y^4,y(1)=1 is y=(\sqrt[3]{6}x)/(\sqrt[3]{x^6+5)}

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y^'= 1/x y-(x^2)/3 y^4,y(1)=1

y^{^{'}} = \frac{1}{x} y - \frac{x ^{2}}{3} y^{4}, y (1) = 1

y = \frac{3 6 x}{3 x ^{6} + 5}

Solution steps

y^{'} (x) = \frac{1}{x} y - \frac{x ^{2}}{3} y^{4}

Solve Bernoulli ODE:

y = \frac{3 6 x}{3 x ^{6} + 5}

y = \frac{3 6 x}{3 x ^{6} + 5}

f (x) = (9 x + 3)^{3}

cos (sin (tan (π x)))

\frac{d ^{3} y}{d x ^{3}} + 4 \frac{d y}{d x} = 0

\frac{d y}{d x} = x + 2 y, y (0) = 4

\int_{0}^{8} x^{2} e^{- 3 x} d x

What is the solution for y^'= 1/x y-(x^2)/3 y^4,y(1)=1 ?
The solution for y^'= 1/x y-(x^2)/3 y^4,y(1)=1 is y=(\sqrt[3]{6}x)/(\sqrt[3]{x^6+5)}