Is there any reference with some example, about how to solve a "riccati" equation in this (below) form :$$y'(x)+a(x)y^2(x)+b(x)y(x)+c(x)=0$$ by Kovacic's algorithm?
Or can anybody help me to understand how Kovacic's algorithm works?
(In Kovacic's 1985 article, the algorithm is described only for the case of $y''=ry$ with some complicated examples. I am searching for some easier examples. I am new to this study.)
Thanks in advance.
In order to use Kovacic's result, we shall transform your equation into precisely the one he studies. All the coefficients will be functions of $x$, so for an easier typing I shall omit the argument. Keep in mind that Kovacic's algorithm applies to equations the coefficients of which are rational functions of a complex variable.
1) First, note that your equation is not linear, so Kovacic's algorithm cannot be applied to it. Fortunately, if $a \ne 0$, the usual change of unknown $w = -a y$ transforms your equation into another Riccati equation:
$$ w' = w^2 + (\frac {a'} a - b) w + a c .$$
2) The above is still not linear, so we next perform the usual change of unknown $w = - \frac {v'} v$, obtaining
$$v'' + (b - \frac {a'} a) v' + ac = 0 .$$
3) Let $f = b - \frac {a'} a$. In order to get rid of the term containing $v'$, make the last substitution $v = \Bbb e ^{\frac f 2} u$, obtaining
$$u'' = -(\frac {(f')^2} 4 + ac) u .$$
Let $r = -(\frac {(f')^2} 4 + ac) = \frac {a'' a - a^2 b' - (a')^2 - 4 a^3 c} {4 a^2}$ in order for your equation to become $u'' = r u$.
Now you may apply Kovacic's algorithm if $r$ is a fraction of polynomial functions of a complex variable. Unfortunately, there is nothing to be done in the general case if you do not provide the concrete expressions of $a,b,c$.