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

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

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

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

\frac{d}{d x} (x - ln (x + 1))

\int cos^{7} (9 - x) sin (9 - x) d x

v^{'} - 2 = 2 v^{2}

\frac{d}{d x} ((2 x + 7)^{3})

\frac{4}{s} (\frac{1}{6 + 5 s})

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