What is the solution for (2y+2yx)(dy)/(dx)=1-y^2 ?

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

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(2y+2yx)(dy)/(dx)=1-y^2

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

y = \frac{- c _{1} + x + 1}{x + 1}, y = - \frac{- c _{1} + x + 1}{x + 1}

Solution steps

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

Solve separable ODE:

y = \frac{- c _{1} + x + 1}{x + 1}, y = - \frac{- c _{1} + x + 1}{x + 1}

y = \frac{- c _{1} + x + 1}{x + 1}, y = - \frac{- c _{1} + x + 1}{x + 1}

x \to 1 - lim (2 x - 2 - 2)

\frac{d}{d x} (x ln ∣ x ∣)

\frac{\partial}{\partial x} (x e^{2 y})

y^{^{''}} - 5 y^{^{'}} = 0

\frac{d}{d x} (\frac{sin ( 2 x )}{1 + cos ( 2 x )})

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