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

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

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

y^{^{'}} = \frac{x - y}{x + 2 y}

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

Solution steps

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

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{d ^{2}}{d x ^{2}} (\frac{x}{4 - 6 x})

t^{2} + 2 t + 50

\frac{d}{d x} (t (a + b t))

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{1}{4 + e ^{x}}

x \to 2 + lim (e^{(\frac{3}{2 - x})})

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