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}

\frac{d}{d x} (cos (x) - 2 sin (x))

x \to 5 - lim (\frac{2}{x - 5})

\frac{d}{d x} (x^{2} y^{3} + x e^{x y})

\int (10 x + 0.5) d x

\frac{d}{d y} (x^{\frac{z}{y}})

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