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I don't understand why the following non-homogeneous equation has been solved as shown below:

$h(x) = 1 + ph(x-1) + (1-p)h(x+1)$

Let $h(x) = Rx$

Hence, $Rx = 1 + pR(x-1) + (1-r)R(x+1)$

Which gives, $R = \frac{1}{2p - 1}$ Hence $h = \frac{x}{2p - 1}$

I don't understand how $R$ was obtained to be as such and why $h(x) = Rx$ was substituted into the first equation.

Thank You

user131983
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    What this shows is that this equation has a solution of this form. There could be others---ruling this possibility out is a task unto itself---but this one certainly exists. – Semiclassical Sep 29 '14 at 19:37
  • @Semiclassical Thank You, but I don't understand how we actually know that $R = \frac{1}{2p - 1}$ from $h = \frac{x}{2p - 1}$ from $Rx = 1 + pR(x-1) + (1-r)R(x+1)$? How is $h = \frac{x}{2p - 1}$ from $Rx = 1 + pR(x-1) + (1-r)R(x+1)$ solved? – user131983 Sep 29 '14 at 19:43
  • Your little "r" should be "p". 2) since the eqn needs to be true for all $x$, you can identify the coefficients of $x$ on the RHS and LHS. That should give the result.
    • – Semiclassical Sep 29 '14 at 21:26
  • @Semiclassical Thanks again. I'm still confused. Would you mind explaining what the last statement you made means in simple terms, for both 1) and 2)? – user131983 Sep 30 '14 at 01:08