What is the solution for y^'= x/(1+2y) ?

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

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y^'= x/(1+2y)

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

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

Solution steps

y^{'} (x) = \frac{x}{1 + 2 y}

Solve separable ODE:

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

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

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