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

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

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2xy*y^'=-(x^2+y^2)

2 \cdot x \cdot y \cdot y^{^{'}} = - (x^{2} + y^{2})

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

Solution steps

2 \cdot x \cdot y \cdot y^{'} (x) = - (x^{2} + y^{2})

Solve Bernoulli ODE:

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

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

\frac{d}{d x} (\frac{x ^{2}}{( x ^{2} + 5 ) ^{3}})

9 y^{7} + 4 x^{3} = 8 y + 5 x, (1, 1)

\frac{\partial}{\partial x} (cos^{5} (x^{7} y^{5}))

x y^{^{'}} + (1 - x) y = 0

\frac{d N}{d t} = N (1 - 0.0005 N)

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