I'm trying to justify, why the graph of the following system:
$$\begin{aligned} v(n) = 0.6 \cdot v(n-1) \\ p(n) = 0.13 \cdot v(n) + 0.87 \cdot p(n-1) + 25 \end{aligned}$$
with initial values $v(0) \approx 1441.67$, $p(0) = 3000$ and $p(1) = 3500$,
seems to show exponential decay for both $v(n)$ and $p(n)$ (after some initial growth).
Justifying this for $v(n)$ is easy:
$$v(n)=0.6^nv(0) \implies v(n) \text{ decays exponentially}$$
but what about $p(n)$?
I can write for example $p(3)$:
$$\begin{aligned}p(3)=a*b^3*v(0)+c*\bigg[a*b^2*v(0)+c*[a*b*v(0)+c*p(0)+25]\bigg]\\=a*b^3*v(0)+c*\bigg[a*b^2*v(0)+c*a*b*v(0)+c^2*p(0)+c*25\bigg] \\=a*b^3*v(0)+c*a*b^2*v(0)+c^2*a*b^2*v(0)+c^3*p(0)+c^2*25\end{aligned}$$
where $a = 13/100,
b = 60/100, c = 87/100$
and one can see exponential terms, but the whole expression is too complicated in order to infer whether the decaying is "clean" exponential decay or whether it would exhibit some sort of other more complex curves.
Use z-transform as $$V(z)=0.6z^{-1}V(z)+v(0)\\P(z)=0.13V(z)+0.87z^{-1}P(z)+p(0)+25$$ to get a linear set of equations. Solving this linear equation yields: $$\begin{align} V(z)&=\frac{v(0)}{1-0.6z^{-1}}\\ P(z)&=\frac{p(0)+25}{1-0.87z^{-1}}+\frac{0.13v(0)}{(1-0.87z^{-1})(1-0.6z^{-1})} \end{align}$$ Now use the inverse z-transform to get: $$\begin{align} v(n)&=0.6^n\,v(0)\\ p(n)&=0.87^n\,\left(25+ p(0)\right)+v(0)\left(0.419\times 0.87^n-0.289\times 0.6^n\right) \end{align}$$
Addendum:
You can also use a discrete-time dynamical system model for this set of equations to better understand its behavior. Here I would define: $$x_n=\pmatrix{v(n)\\p(n)}$$ and obtain a model in the form of $x_n=Ax_{n-1}+Bu_n$, where: $$A=\pmatrix{0&0.6\\0.078&0.87},\qquad B=\pmatrix{0\\25}$$ now it is sufficient to find the (unit)step-response of this system.