= Constant-coefficient Riccati equation
{title2=$y'+(y-a)(y-b)=0,\quad y=z'/z$}
The <linearization of a Riccati equation> turns this equation into $z''-(a+b)z'+abz=0$. For distinct real roots, every nonzero linearizing solution is $z=Ae^{ax}+Be^{bx}$ and $y=(aAe^{ax}+bBe^{bx})/(Ae^{ax}+Be^{bx})$ on intervals with nonzero denominator. The constants $y=a,b$ arise when one coefficient vanishes. With $A\ne0$, 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 $y$.
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