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

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

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

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

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

Solution steps

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

Solve separable ODE:

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

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

x, x - 2, [0, 4]

\frac{d ^{3}}{d x ^{3}} (3 t^{3} - \frac{2}{t ^{2}})

\int (x^{3} + x + \frac{1}{x ^{3}}) d x

\int_{0}^{\frac{π}{2}} - 10 sin^{2} (3 x) d x

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

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