What is the solution for y^'-4/x y=(y^3)/(x^5),y(1)=1 ?

The solution for y^'-4/x y=(y^3)/(x^5),y(1)=1 is y=(sqrt(2)x^4)/(sqrt(-x^4+3))

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y^'-4/x y=(y^3)/(x^5),y(1)=1

y^{^{'}} - \frac{4}{x} y = \frac{y ^{3}}{x ^{5}}, y (1) = 1

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

Solution steps

y^{'} (x) - \frac{4}{x} y = \frac{y ^{3}}{x ^{5}}

Solve Bernoulli ODE:

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

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

(1.1) (2.5)

n = 1 \sum \infty \frac{3 e ^{n}}{n ^{5}}

\frac{\partial}{\partial x} ((1 - 3 x y) e^{- x y})

\int (5^{x} - sec^{2} (x)) d x

\frac{\partial}{\partial y} (sin (3 x + yz))

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