Q:

Probability of ruin

The game is heads or tails. If you guess right you win £$1$ and if you guess wrong you lose £$1$. Initial capital is £$x$. Play until capital is £$m$ or £$0$.
The probability of ruin given you win the first toss is $p(x+1)$ and if you lose the first toss it is $p(x-1)$.
This gives $p(x) = \dfrac{1}{2} [p(x+1) + p(x-1)]$  for  $1 \le x \le (m-1)$, 
where $p(0) = 1$ and $p(m) = 0$.
So far I am clear about all of this.
The text book next says:

The solution is  the linear function:
  $p(x) = C_1 + C_2x$

My question is how do we know it is of the form $p(x) = C_1 + C_2x$ ?

A:

Rearrange the terms to get
$$p(x+1)-p(x)=p(x)-p(x-1)$$
Since this is true for every $x$ in the domain, it must be some constant $d$.
Therefore
$$p(x+1)-p(x)=d$$
for every $x$ in the domain. This is nothing but an arithmetic progression with common difference $d$. And the form of the terms in a AP is 
$$C_1+C_2x$$

