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

The solution for -2y^'+(4y)/x =x^4y^3 is y=(3x^2)/(sqrt(x^9+c_{1))},y=-(3x^2)/(sqrt(x^9+c_{1))}

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

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

y = \frac{3 x ^{2}}{x ^{9} + c _{1}}, y = - \frac{3 x ^{2}}{x ^{9} + c _{1}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{3 x ^{2}}{x ^{9} + c _{1}}, y = - \frac{3 x ^{2}}{x ^{9} + c _{1}}

y = \frac{3 x ^{2}}{x ^{9} + c _{1}}, y = - \frac{3 x ^{2}}{x ^{9} + c _{1}}

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

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

\int \frac{1}{2 x - 1} d x

\frac{d}{d x} (\frac{5 1 - x ^{3}}{6})

\int 13 arctan (x) d x

What is the solution for -2y^'+(4y)/x =x^4y^3 ?
The solution for -2y^'+(4y)/x =x^4y^3 is y=(3x^2)/(sqrt(x^9+c_{1))},y=-(3x^2)/(sqrt(x^9+c_{1))}