 Linearization of a Riccati equation

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 $a-c^{\prime }/c$.