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}

\int \frac{1}{x ^{2}} ln (x) d x

\frac{d}{d y} (\frac{ln ( y )}{2 y} + \frac{1}{y})

x \to - 2 lim (- 10)

x y^{^{'}} + y = y^{2}, y (1) = - 9

\frac{d}{d x} (cosh (3 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