Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 2 / 1D (Differential Equations)

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.

Separate variables and integrate on $|y|<1$:Hence the general solution in this region isEvery such solution increases from values near $-1$ towards $1$, remains strictly between the two equilibria, and is defined for all real $x$.

For $|y|>1$, the separated integral isInverting givesThe upper branch has $x+C>0$, where $y>1$ decreases towards $1$ as $x\to \infty$. The lower branch has $x+C<0$, where $y<-1$ decreases to $-\infty$ as $x\uparrow -C$. These are separate maximal intervals: the pole is not part of a solution. For the domain $x\ge 0$, a lower-branch initial value below $-1$ necessarily reaches that pole in finite positive $x$. The direction field sketch above displays both exterior branches and a central hyperbolic tangent branch.