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}

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