I'm trying to find a general solution to this Riccati equation:
$f'(x) = -a f(x)^2 + g(x)$
where $a$ is a constant, $g$ is a smooth real-valued function of the real independent variable $x$, and $f$ is the unknown real-valued function of $x$.
I know that the general Riccati equation can't be solved in closed form for arbitrary coefficient functions, but the one I have involves one constant coefficient $a$ and one arbitrary function $g$. Does that change anything?
More specifically, is this equation solvable in closed form (or at all)? The author of this simulation claims that Riccati solutions are ratios of Bessel functions... but that seems a bit too easy because if that were true, some mathematician would have probably worked that into a general form by now since these things have been around for 300 years:
https://demonstrations.wolfram.com/RiccatiDifferentialEquationWithContinuedFractions/
Any help is appreciated, thanks!
$$f'(x) = -a f(x)^2 + g(x)$$ Change of function : $$f(x)=\frac{y'(x)}{a\:y(x)}\quad\implies\quad f'(x)=-\frac{(y')^2}{a\:y^2}+\frac{y''}{a\:y}$$ $$-\frac{(y')^2}{a\:y^2}+\frac{y''}{a\:y} = -a \left(\frac{y'}{a\:y} \right)^2 + g(x)$$ $$\frac{y''}{a\:y}=g(x)$$ $$y''(x)-ag(x)y(x)=0$$ This is a second order linear ODE. This ODE can be of various kind depending on the function $g(x)$.
Of course for some functions $f(x)$ the result involves Bessel functions. In this case $f(x)=\frac{y'(x)}{a\:y(x)}$ is a ratio of Bessel functions.
For other kind of functions $g(x)$ the solution of the above second order linear ODE involves functions which are not of Bessel kind. For example if $g(x)=a+bx+cx^2$ the solution involves parabolic cylinder functions.
Another example : if $g(x)=(b+\frac{c}{x})$ the solution involves confluent hypergeometric functions.
Thus one cannot conclude definitively if the function $g(x)$ is not of a specified kind.