Simply speaking my goal is to integrate the differential Riccati equation
$$x'(t)=a(t)x^2+b(t)x+c(t)$$
I know that this is impossible to express the solution by the means of the elementary functions. However if one consider the linear equation
$$x'(t)=x(t)$$
one cannot solve this equation by quadratures in the set of the rational functions. One needs to introduce the exponential function to solve this equation. Moreover, the solution for all linear equations
$$x'(t)=a(t)x(t)+b(t)$$ can be expressed by the exponential functions, which is obvious.
Therefore I am seeking for the Riccati function $Ri(t)$, which solves the Riccati equation
$$Ri'=a_0(t)Ri^2+b_0(t)Ri+c_0(t)$$
such that any other solution to
$$x'(t)=a(t)x^2+b(t)x+c(t)$$
can be express by this function. Am I trying to reinvent the wheel? Is it known that there is no Riccati function?
You can as always set $x=-\frac{u'}{au}$ so that $$ -\frac{u''}{au}+\frac{u'^2}{au^2}+\frac{a'u'}{a^2u}=\frac{u'^2}{au^2}-\frac{bu'}{au}+c \\\iff\\ 0=u''-\left(\frac{a'}{a}+b\right)u'+ac\,u $$ where all results on linear ODE with non-constant coefficients apply. Especially that you can span the solution space with two basis solutions $u_1,u_2$ so that the general solution of the Riccati equation has the parametrization $$ y=\frac{c_1u_1'+c_2u_2'}{a(c_1u_1+c_2u_2)} $$ which has really only one parameter, $\frac{c_1}{c_2}$ or $\frac{c_2}{c_1}$ or $(c_1,c_2)$ on some arc that crosses all lines through the origin exactly once, like a semi-circle $c_2=\sqrt{1-c_1^2}$, $c_1\in(-1,1]$.