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

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

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

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

y = \frac{2 x ^{2}}{c _{1} + x ^{4}}, y = - \frac{2 x ^{2}}{c _{1} + x ^{4}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{2 x ^{2}}{c _{1} + x ^{4}}, y = - \frac{2 x ^{2}}{c _{1} + x ^{4}}

y = \frac{2 x ^{2}}{c _{1} + x ^{4}}, y = - \frac{2 x ^{2}}{c _{1} + x ^{4}}

\frac{d}{d t} (\frac{1}{t ^{2}})

\int \frac{5 x - 1}{( 2 x + 1 ) ( x - 1 )} d x

f (x) = arctan (3 x)

\frac{d y}{d t} = k y (100 - y)

\frac{d}{d x} (lo g_{4} ((x^{2} + 2 x)^{3}))

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