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

The solution for y^'-2/x y=(y^3)/(x^2),y(1)=1 is y=(sqrt(3)x^2)/(sqrt(-2x^3+5))

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

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

y = \frac{3 x ^{2}}{- 2 x ^{3} + 5}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{3 x ^{2}}{- 2 x ^{3} + 5}

y = \frac{3 x ^{2}}{- 2 x ^{3} + 5}

\frac{d}{d s} (\frac{1}{s ^{2} - 1})

\int \frac{csc ( θ )}{csc ( θ ) - sin ( θ )} d θ

\int 5 e^{- 2 t} sin (t) d t

\int (x + 2)^{2} (2 - x)^{6} d x

x^{3} + 3 x^{2} - 9 x

What is the solution for y^'-2/x y=(y^3)/(x^2),y(1)=1 ?
The solution for y^'-2/x y=(y^3)/(x^2),y(1)=1 is y=(sqrt(3)x^2)/(sqrt(-2x^3+5))