Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 2 / 1D / Solution

The direction field has slope $1-y^2$, independent of $x$. Its line segments slope upward for $-1<y<1$, are horizontal at $y=\pm 1$, and slope downward for $|y|>1$. Thus $y=1$ is an attracting equilibrium in forward $x$, while $y=-1$ is repelling. The two constant solutions $y\equiv \pm 1$ must be included separately when separating variables.

The following original sketch shows the slope field and one solution in each requested region. The lower exterior solution ends at a finite pole; it cannot be continued across that pole as a real finite-valued solution.