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

The solution for y^'-(2/x y)=((y^3)/(x^3)),y(1)=1 is y=(x^2)/(sqrt(-x^2+2))

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

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

y = \frac{x ^{2}}{- x ^{2} + 2}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{x ^{2}}{- x ^{2} + 2}

y = \frac{x ^{2}}{- x ^{2} + 2}

sin (- 3 t)

\int cos (t) e^{- s t} d t

x \to + 0 lim (\frac{sin ( x ^{2} )}{x})

n = 0 \sum \infty 7^{- n}

y = (4 x^{2} + x) (x - x^{2})

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