Constant-coefficient Riccati equation ($y^{\prime }+(y-a)(y-b)=0,y=z^{\prime }/z$)

The linearization of a Riccati equation turns this equation into $z^{\prime \prime }-(a+b)z^{\prime }+abz=0$. For distinct real roots, every nonzero linearizing solution is $z=A{e}^{ax}+B{e}^{bx}$ and $y=(aA{e}^{ax}+bB{e}^{bx})/(A{e}^{ax}+B{e}^{bx})$ on intervals with nonzero denominator. The constants $y=a,b$ arise when one coefficient vanishes. With $A\ne 0$, the relative coefficient $B/A$ can have either sign; replacing it by a fixed-sign real exponential would omit a real family of solutions.

For $a=b$, the solutions are $z=(A+Bx)e^{ax}$ and $y=a+B/(A+Bx)$. The equilibrium $y=a$ occurs at $B=0$. Zeros of a nontrivial $z$ give poles of a Riccati solution from zeros of its linearizing solution, so the corresponding $y$ is defined separately between them. Multiplying both coefficients by the same nonzero scalar changes $z$ but not $y$.