What is the solution for (dy)/(dx)= x/((1+2y)),y(-1)=0 ?

The solution for (dy)/(dx)= x/((1+2y)),y(-1)=0 is y=(-1+sqrt(2x^2-1))/2

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(dy)/(dx)= x/((1+2y)),y(-1)=0

\frac{d y}{d x} = \frac{x}{( 1 + 2 y )}, y (- 1) = 0

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

Solution steps

\frac{d y}{d 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}

f (x) = 9 + 5 x^{2} - 2 x^{3}, at x = a

\int \frac{1}{9 - 36 x ^{2}} d x

\frac{d}{d x} (\frac{4 x}{x + 2})

f (x) = 64 - x^{2}

\frac{\partial}{\partial x} (x y cos (5 yz))

What is the solution for (dy)/(dx)= x/((1+2y)),y(-1)=0 ?
The solution for (dy)/(dx)= x/((1+2y)),y(-1)=0 is y=(-1+sqrt(2x^2-1))/2