What is the solution for y^'= x/(1+2y),y(-1)=0 ?

The solution for y^'= x/(1+2y),y(-1)=0 is y=(-1+sqrt(2x^2-1))/2

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y^'= x/(1+2y),y(-1)=0

y^{^{'}} = \frac{x}{1 + 2 y}, y (- 1) = 0

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

Solution steps

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

Solve separable ODE:

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

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What is the solution for y^'= x/(1+2y),y(-1)=0 ?
The solution for y^'= x/(1+2y),y(-1)=0 is y=(-1+sqrt(2x^2-1))/2