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

The solution for y^'=(2x)/((1+2y)) is y=(-1+sqrt(1-4(-c_{1)-x^2)})/2 ,y=(-1-sqrt(1-4(-c_{1)-x^2)})/2

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

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

y = \frac{- 1 + 1 - 4 ( - c _{1} - x ^{2} )}{2}, y = \frac{- 1 - 1 - 4 ( - c _{1} - x ^{2} )}{2}

Solution steps

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

Solve separable ODE:

y = \frac{- 1 + 1 - 4 ( - c _{1} - x ^{2} )}{2}, y = \frac{- 1 - 1 - 4 ( - c _{1} - x ^{2} )}{2}

y = \frac{- 1 + 1 - 4 ( - c _{1} - x ^{2} )}{2}, y = \frac{- 1 - 1 - 4 ( - c _{1} - x ^{2} )}{2}

\int \frac{x}{x ^{4} - 6 x ^{2} + 5} d x

\frac{d}{d x} (lo g_{3} (x^{2} + 1))

\frac{d}{d x} ((x + \frac{1}{x})^{2})

\frac{d}{d t} ((t - 2)^{2})

\frac{\partial}{\partial x} (y x + y)

What is the solution for y^'=(2x)/((1+2y)) ?
The solution for y^'=(2x)/((1+2y)) is y=(-1+sqrt(1-4(-c_{1)-x^2)})/2 ,y=(-1-sqrt(1-4(-c_{1)-x^2)})/2