Solution of Simple Riccati equation

Question

I have a Riccati equation $$ x'(t) + \alpha \, x(t)^2 + \beta(t) =0$$ where $$ \beta(t) = \frac{\beta_0}{1-\mu \beta_0 t}$$ such that $$ \mu =\frac{d\beta(t)}{dt}\frac{1}{\beta(t)^2}=const$$ and $\beta_0$, $\alpha$ and $\mu$ are real. Making the substitution $x(t) = \alpha \, u'(t)/u(t)$ brings the equation into the linear form $$u''(t) + \frac{\beta(t)}{\alpha}u =0 $$ I know that there are standard techniques for finding the general solution of a Riccati equation if a particular solution is known. But how do I find a particular solution to this equation?

Answer 1

Substitute $s = 1 - \mu \beta_0 t$ and let $v(s) = u(t)$ we get, for a constant $c$ the equation. $$ v''(s) - c\frac{1}{s}v(s) = 0 $$

Expand $v$ in a power series $v(s) = \sum_{i=1}^\infty a_i s^i$ and we get the condition,

$$ n(n-1) a_n = c a_{n-1} $$

Hence $$ a_n=\frac{c^{n-1}}{n(n-1)(n-1)(n-2)\cdots 2 \cdot 1} a_1 = \frac{n c^{n-1}}{(n!)^2} $$

This can be expressed by bessel functions