What is the solution for y^'=((1-x-y))/(x+y) ?

The solution for y^'=((1-x-y))/(x+y) is y=(-2x+sqrt(8x+c_{1)})/2 ,y=(-2x-sqrt(8x+c_{1)})/2

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y^'=((1-x-y))/(x+y)

y^{^{'}} = \frac{( 1 - x - y )}{x + y}

y = \frac{- 2 x + 8 x + c _{1}}{2}, y = \frac{- 2 x - 8 x + c _{1}}{2}

Solution steps

y^{'} (x) = \frac{( 1 - x - y )}{x + y}

Solve exact ODE:

y = \frac{- 2 x + 8 x + c _{1}}{2}, y = \frac{- 2 x - 8 x + c _{1}}{2}

y = \frac{- 2 x + 8 x + c _{1}}{2}, y = \frac{- 2 x - 8 x + c _{1}}{2}

\frac{1}{4} q^{^{''}} + 20 q^{^{'}} + 300 q = 0

\int 2^{X} d X

\frac{2}{x + 1}

\frac{\partial}{\partial x} ((x + 2 y^{3})^{\frac{1}{3}})

\frac{d}{d x} (\frac{( x cos ( x ) + π )}{x - π})

What is the solution for y^'=((1-x-y))/(x+y) ?
The solution for y^'=((1-x-y))/(x+y) is y=(-2x+sqrt(8x+c_{1)})/2 ,y=(-2x-sqrt(8x+c_{1)})/2