Rational difference equation

First approach

WLoG, the determinant-like quantity ${\displaystyle ad-bc\neq 0}$, and this can be easily seen by noting that, with a division on both the numerator and denominator in Equation (2), you can always set ${\displaystyle c=1}$, in which case, if ${\displaystyle ad-bc=0}$, then the numerator and denominator will cancel away, leaving no difference equation left, being instead so completely reduced as to become exactly the constant ${\displaystyle a}$. Thus, one approach^[7] to developing the transformed variable ${\displaystyle x_{t}}$, is to write

${\displaystyle y_{t+1}=\alpha -{\frac {\beta }{y_{t}}}}$

where ${\displaystyle \alpha =(a+d)/c}$ and ${\displaystyle \beta =(ad-bc)/c^{2}}$ and where ${\displaystyle w_{t}=y_{t}-d/c}$.

Further writing ${\displaystyle y_{t}=x_{t+1}/x_{t}}$ can be shown to yield

${\displaystyle x_{t+2}-\alpha x_{t+1}+\beta x_{t}=0.}$

Second approach

The above approach is already of general applicability. This following approach^[8] gives a first-order difference equation for ${\displaystyle x_{t}}$ instead of a second-order one. Let ${\displaystyle r^{2}=(d-a)^{2}+4bc}$. For the case in which ${\displaystyle r^{2}\geqslant 0}$, every term will be real-valued and this method may be convenient to use. Otherwise, the method still works, but complex numbers will appear, and it might be more convenient to attempt a trigonometric ansatz instead. Substituting ${\displaystyle x_{t}=1/(\eta +w_{t})}$, which implies ${\displaystyle w_{t}=(1-\eta x_{t})/x_{t}}$, into Equation (2), we find that it is always possible to make ${\displaystyle x_{t}}$ evolve according to the simple inhomogeneous first-order linear difference equation

${\displaystyle x_{t+1}=\left({\frac {d-\eta c}{\eta c+a}}\right)\!x_{t}+{\frac {c}{\eta c+a}}.}$

by choosing ${\displaystyle \eta }$ such that ${\displaystyle c\eta ^{2}-(d-a)\eta -b=0}$, and it is clear that this can always be done in either of the two choices ${\displaystyle \eta _{\pm }={\frac {d-a\pm r}{2c}}}$, even when ${\displaystyle r\in \mathbb {C} }$

Third approach

The equation

${\displaystyle w_{t+1}={\frac {aw_{t}+b}{cw_{t}+d}}}$

can also be solved by treating it as a special case of the more general matrix equation

${\displaystyle X_{t+1}=-(E+BX_{t})(C+AX_{t})^{-1},}$

where all of A, B, C, E, and X are n × n matrices (in this case n = 1); the solution of this is^[9]

${\displaystyle X_{t}=N_{t}D_{t}^{-1}}$

where

${\displaystyle {\begin{pmatrix}N_{t}\\D_{t}\end{pmatrix}}={\begin{pmatrix}-B&-E\\A&C\end{pmatrix}}^{t}{\begin{pmatrix}X_{0}\\I\end{pmatrix}}.}$