Consider the recurrence relation given by $2P_t + P_{t-1} = 3, P_0 = 0$. Does the graph of $(Q_t, P_t)$ look like a Cobweb where $Q_t = 4-2P_t$?
My calculation tells me that $(Q_t, P_t) = (2 + 2(-0.5)^t, 1-(-0.5)^t)$. The diagram seems to be a mere straight line. Where am I going wrong?
This "solution" will only focus on the difference equations.
Notice that $P_{n}$ is given by the equation $$ P_{n+1} = \frac{1}{2} \, (3 - P_{n}) \hspace{10mm} P_{0} = 0.$$ The first few terms may be obtained and are: $$ P_{n} \in \left\{0,\frac{3}{2},\frac{3}{4},\frac{9}{8},\frac{15}{16},\frac{33}{32},\frac{63}{64},\frac{129}{128},\frac{255}{256},\frac{513}{512},\frac{1023}{1024}, \cdots \right\}. $$ It is seen that the denominators are powers of $2$ and the numerators are of the form $3 \, a_{n}$. This gives $P_{n} = \frac{3 \, a_{n}}{2^n}$ where $a_{n} \in \{0, 1, 1, 3, 5, 11, 21, \cdots \}$. The sequence $a_{n} = J_{n}$, the Jacobsthal numbers, see Oeis A001045, which leads to $$ P_{n} = \frac{3 \, J_{n}}{2^n} = 1 + \left(- \frac{1}{2}\right)^{n}. $$ The component $Q_{n}$ is then given by $$ Q_{n} = 4 - \frac{3 \, J_{n}}{2^{n-1}} = 2 \, \left( 1 - \left(- \frac{1}{2}\right)^n \right). $$
As the plotting the graphs of $P_{n}$ and $Q_{n}$ it should be noted that if $n < 0$ then the plots should grow large because a leading factor will be $2^n$, and will oscillate due to the $(-1)^n$ component. For $n > 0$ the graphs are dominated by $2^{-n}$ and will give increasingly small oscillations until they are nearly $1$ and $2$, respectively. This part is seen by: \begin{align} \lim_{n \to \infty} P_{n} &= 1 \\ \lim_{n \to \infty} Q_{n} &= 2. \end{align}