What is the solution for xyy^'+4x^2+y^2=0 ?

The solution for xyy^'+4x^2+y^2=0 is y=(sqrt(-2x^4+c_{1)})/x ,y=-(sqrt(-2x^4+c_{1)})/x

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xyy^'+4x^2+y^2=0

x y y^{^{'}} + 4 x^{2} + y^{2} = 0

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

Solution steps

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

Solve Bernoulli ODE:

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

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

What is the solution for xyy^'+4x^2+y^2=0 ?
The solution for xyy^'+4x^2+y^2=0 is y=(sqrt(-2x^4+c_{1)})/x ,y=-(sqrt(-2x^4+c_{1)})/x