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

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

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

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

\frac{d}{d x} (4 (2 x - 5)^{- \frac{1}{2}})

\frac{\partial}{\partial x} (2 y sin (x))

\int_{0}^{3} e^{x} sin (x) d x

f (x) = \frac{x ^{3}}{3} + \frac{1}{4 x}

\int e^{x^{3}} x^{5} d x

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