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:

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

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

d er i v a t i v e f (x) = \frac{π}{3}

(x, y) \to (0, 0) lim (e^{x y})

\frac{d}{d x} ((17 - (13 - 2 x^{4})^{7})^{3})

\frac{\partial}{\partial x} (2 x^{2} ln (3 x y^{2} + 4 y))

\frac{d}{d x} (x e^{- \frac{x}{2}})

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