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

The solution for y^'-4/x y=(y^3)/(x^3),y(1)=1 is y=(sqrt(3)x^4)/(sqrt(-x^6+4))

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

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

y = \frac{3 x ^{4}}{- x ^{6} + 4}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{3 x ^{4}}{- x ^{6} + 4}

y = \frac{3 x ^{4}}{- x ^{6} + 4}

\frac{d}{d x} (\frac{7}{10} x^{2} + 4 x^{3})

\frac{d y}{d x} - y = - \frac{x}{2}

x \to 0 lim (1 - \frac{1}{x ^{2}})

x \to 3 lim (\frac{( x ^{2} - 2 x - 3 )}{x ^{2} - 9})

y^{^{'}} (x^{2} + 2) + 4 x (y^{2} + 2 y + 1) = 0

What is the solution for y^'-4/x y=(y^3)/(x^3),y(1)=1 ?
The solution for y^'-4/x y=(y^3)/(x^3),y(1)=1 is y=(sqrt(3)x^4)/(sqrt(-x^6+4))