**Solve ar - 7 ar - 1 + 10 ar - 2 = 6 + 8r**

$a_o = 1$ $a_1 = 2$

**Solution:**

**Homogeneous equation.**

**ar - 7 ar - 1 + 10 ar - 2 = 0**

$\therefore$ **Characteristics equation.**

$a^2 - 7a + 10 = 0$

$\therefore$ Robt's of equation are a = 2, 5

$\therefore$ Homogeneous solution

$ar^{(n)} = A_1 \ (2)^r + A_2 \ (5)^r$

$\therefore$ RHS is a polynomial

$\therefore$ $ar^{(p)} = p_o + p_1^r$

ar = po + p1 r

$a_r-1 = p_o + p_1 (r-1)$

$a_r-2 = p_o + p_1 (r-2)$

Substituting these in given equation.

$( p_o + p_1^r) - 7 (po + p_1 (r-1) ) + 10 [po + p_1 (r-2)] = 6 + 8 r$

Solving $p_o$ = 8 and $p_1$ = 2

$\therefore$ $ar^{(p)} = 8 + 2 r$

$ar = ar^{(n)} + ar^{(p)}$

$= A1 (2)^r + A_2 (5)^r + 8 + 2r$

Using initial conditions $a_o = 1$ and $a_1 = 2$

we get, $A_1 = -9$ and $A_2 = 2$

$\therefore$ $ar = -9(2)^r + 2 (5)^r + 8 + 2r$