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

The solution for x^2+y^2+2xyy^'=0 is y=sqrt(-(x^2)/3+(c_{1))/x},y=-sqrt(-(x^2)/3+(c_{1))/x}

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

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

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

Solution steps

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

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}

\int (x + 1) e^{- 2 x} d x

f (x) = \frac{x ^{2} - 2 x - 3}{x}

\frac{\partial}{\partial x} (3 x^{2} - 12 y)

\frac{d}{d x} (\frac{18}{x ^{2} + 2})

\frac{4}{x - 4} + \frac{27}{2 x + 2}

What is the solution for x^2+y^2+2xyy^'=0 ?
The solution for x^2+y^2+2xyy^'=0 is y=sqrt(-(x^2)/3+(c_{1))/x},y=-sqrt(-(x^2)/3+(c_{1))/x}