What I am trying to do is to solve Riccati equation: $$y'=a(x)y^2+b(x)y+c(x)$$ I introduced new function and called it $Ri(x)$, which satisfy the following ODE: $$y'=y^2+x$$
Is it possible that using this function I can solve Riccati eqn in general case by the terms of $Ri$ function?
My motivation is from the first order ODE:
$$y'=a(x)y+b(x)$$
The solution to this one can express by the terms of exponential function which is the solution to the following ODE
$$y'=y$$
The most general approach is to transform the equation into a linear DE or system.
Set $y=\frac{p}{q}$, then $$ p'q-q'p=ap^2+bpq+cq^2 $$ Now separate terms with factor $p$ and terms with factor $q$. There are several solutions for that, for instance $$ (p'-bp-cq)q=(q'+ap)p. $$ With the introduction of the fraction one degree of freedom was gained. Setting $q'=-ap$ adds a restriction that fixes this down to a completely determined system \begin{align} p'&=bp+cq\\ q'&=-ap \end{align} This is now an almost arbitrary linear system, so it appears unlikely that a reduction to a single reference case is possible.