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

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

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

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

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

Solution steps

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

Solve exact ODE:

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

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

\int - 2 x^{2} + 6 x d x

(cos^{2} (s) - 11 s^{2})^{2}

\frac{d}{d x} (- 3 cot (x) csc (x))

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

x \to 4 + lim (\frac{3 + x}{4 - x})

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