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

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

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

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

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

Solution steps

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

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}

\frac{( 5 + 5 s )}{( s ^{2} + 25 )}

\frac{s ^{2}}{s ^{2} + 4 s + 5}

4 - x^{2}, 14 - 7 x

x^{3} - 3 x^{2} + 3 x - 3

(7 x^{\frac{1}{5}}) - (6 x^{\frac{4}{7}})

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