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

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

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(y-yx^2)y^'=(y+1)^2

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

y = - 1 - \frac{1}{W ( - \frac{x - 1}{c _{1} x + 1} )}

Solution steps

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

Solve separable ODE:

y = - 1 - \frac{1}{W ( - \frac{x - 1}{c _{1} x + 1} )}

y = - 1 - \frac{1}{W ( - \frac{x - 1}{c _{1} x + 1} )}

\frac{d}{d x} (y cos (x y))

5 x^{2}, x^{2} + 4

x^{2} y^{^{'}} + x y = 8

\frac{\partial}{\partial x} (4 x y + 3)

\int x cos (\frac{π x}{2}) d x

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