Riccati equations for Fibonacci and Lucas generating functions

Lemma 1: $U(a,b;x)$ defined in eq. 2 satisfies eq. 13.

Proof: This follows from the identity

$${{\partial {}}\over{\partial {x}}} {\frac{1}{1-a\,x-b\,x^2}}\ =\ (a+2\,b\,x)\,{\frac{1}{(1-a\,x-b\,x^2)^2}}\\$$ (58)

after multiplication with $(a\, +\, 2\,b\,x)$ and separation of the last term in eq. 13.$\square$.

Corollary 1: For the non-degenerate case $a^2+4\,b \neq 0$ this is a Riccati eq. (also a special Bernoulli eq.) which can be written as shown in eq. 15.
In the degenerate case $U(a;x)\, :=\, U(a,-a^2/4;x)$ satisfies the first order linear differential eq. 27 as well as the second order non-linear differential eq. given as the first of eqs. 29.

Proof: Elementary. $\square$

Note 1: The degenerate case is equivalent to $\lambda_{+}(a,b)\, =\, \lambda_{-}(a,b)$ with the definition of the characteristic roots of the recursion relation eq. 1 given in eq. 4. We may assume that not both, $a$ and $b$, vanish and $x\neq 1/\lambda_{\pm}(a,b) \, =\, -\lambda_{\mp}/b$. In each case $U(a,b;0)\, =\, 1$.

Lemma 2: $V(a,b;x)$ defined in eq. 3 for $a\neq 0$ satisfies eq. 14.

Proof: This follows from the identity

$${{\partial {}}\over{\partial {x}}}\, V(a,b;x) \ =\ \Bigl{(}(2\,{\... ...\,x \ +\ 2\, {\frac{b^2}{a}}\, x^2\,\Bigr{)}\, {\frac{1}{(1-a\,x-b\,x^2)^2}}\ ,$$ (59)

after multiplication with $(a+2\,b\,x)^2$ and separation of the $V^2(a,b;x)$ term shown in eq. 14.$\square$

Corollary 2: For the non-degenerate case $a^2+4\,b \neq 0$ eq. 14 is a Riccati eq. which can be written as shown in eq. 16.
In the degenerate case $V(a;x)\, :=\, V(a,-a^2/4;x)$ satisfies the first order linear differential eq. 28 as well as the first order non-linear differential eq. given as the second of eqs. 29.
In each case $V(a,b,0)\, =\, 1$.

Proof: Elementary. $\square$

Proposition 1 (Solution of eq. 13):

a) For $a^2+4\,b \neq 0$ the solution of the Riccati eq. 13 with $U(a,b;0)\, =\, 1$ is given by eq. 2.

b) For $a^2+4\,b \, =\, 0$ the solution of eqs. 27 is $U(a,-a^2/4;x)$ from eq. 2.

Proof: a) Rewrite this Riccati eq., which is also of the Bernoulli type, $y^{\prime}\, +\, f(x)\,y\, +\, g(x)\, y^2\, =\, 0$ with $f(x)\, =\, 4\,b/(a\,+2\,b\,x)$ and $g(x) \, =\, -(a^2\, +\, 4\, b)/(a\,+2\,b\,x)$, provided $x\neq -a/(2\,b)$, as a first order inhomogeneous linear differential eq. for $z=1/y$: $z^{\prime}\, -\, f(x)\, z\, -\, g(x)\, =\, 0$. Its standard solution is $y^{-1}\, =\, z\, =\, exp(F(x))\, \left[ C\, +\, \int\, dx\,exp(F(x))\, g(x)\right]$ with $F(x)\, :=\, \int\,dx\,f(x)\, =\, ln((a\, +\, 2\,b\,x)^2)$. The integral becomes $(1\, +\, a^2/(4\,b))/(a\, +\, 2\,b\,x)^2$ and $C$ is fixed from the initial value $y^{-1}(0)\, =\, z(0)\, =\, 1$ to $-1/(4\,b)$. The solution $1/(1\, -\, a\,x\, -\, b\,x^2)$ is also correct if $x=-a/(2\,b)$.
b) For $a^2+4\,b \, =\, 0$ and $x\neq 2/a$ the solution of the standard linear differential eq. 27 coincides with eq. 2. For $x \, =\, 2/a$ eq. 27 demands a singular $(ln\, U)^{\prime}$; a requirement which is satisfied by this solution. $\square$

Proposition 2 (Solution of eq. 14):

a) For $a^2+4\,b \neq 0$ the solution of the Riccati eq. 14 with $V(a,b;x)$ is given by eq. 3.

b) For $a^2+4\,b \, =\, 0$ the solution of eqs. 28 is $V(a, -a^2/4;0)\, =\, 1$ from eq. 3.

Proof: a) Similar to the proof of Proposition 1 as standard solution to the inhomogeneous linear differential eq. for $y^{-1}\, =\, z$ where now $f(x)\, =\, 2\,{\frac{b}{a}}/(1\, +\, 2\, {\frac{b}{a}}\, x)$, $g(x)\, =\, -(a\, +\, 4\,{\frac{b}{a}})/(1\, +\, 2\,{\frac{b}{a}}\, x)^2$ and $1\, +\, 2\,{\frac{b}{a}}\,x\, \neq \,0$. The initial value is $y^{-1}(0)\, =\, z(0)\, =\, 1$ which fixes the solution to be $y\, =\, (1\, +\, 2\,{\frac{b}{a}}\, x)/(1\, -\, a\,x\, -\, b\,x^2)$. This is also correct for $1\, +\, 2\,{\frac{b}{a}}\, x\, =\, 0$.
b) For $a^2\, +\, 4\,b\, =\, 0$ and $x\neq 2/a$ the solution of eq. 28 with $V(a;0)\, =\, 1$ coincides with $V(a,-a^2/4;x)\ =\ 1/(1\, -\, a\,x/2)$. For $x \, =\, 2/a$ eq. 28 needs a singular $(ln\,V)^{\prime}$ because $a\neq 0$, and this requirement is indeed fulfilled. $\square$