Linearization of a Riccati equation ($y=x\prime /(cx)$)

For constant nonzero $c$, putting $y=x^{\prime }/(cx)$ in $y^{\prime }+c{y}^2+a(t)y+b(t)=0$ cancels the quadratic derivative terms and gives $x^{\prime \prime }+a(t)x^{\prime }+cb(t)x=0$. The logarithmic derivative is defined on intervals where $x\ne 0$, and multiplying $x$ by a nonzero scalar does not change $y$. For variable $c(t)\ne 0$, the derivative coefficient becomes $a-c^{\prime }/c$.

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 $y$.

At a regular point of a second-order linear ordinary differential equation, a nontrivial solution cannot have both $x$ and $x^{\prime }$ zero, by initial-value uniqueness. Its zeros are therefore simple. Under the linearization of a Riccati equation, such a zero $t_{\ast }$ produces $y=x^{\prime }/(cx)\sim 1/[c(t-t_{\ast })]$. Zeros separate the intervals on which the Riccati equation solution is finite.