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}

f (x) = 6 sin (x), (\frac{π}{6}, 3)

\frac{\partial}{\partial u ( v )} (\frac{e ^{v}}{u ( v ) + v ^{8}})

\frac{d y}{d x}, 9 cos (x y) = 4 x + 5 y

\int_{- 2}^{2} y^{2} - 2 - y d y

\frac{\partial}{\partial x} (\frac{( x - 3 y )}{x + 3 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