Solve a non-homogeneous recurrence relation

Question

I need to solve this equation. I know how to solve the homogeneous side, but when I get to the non-homogeneous it becomes a problem.

$$ a_n = 7a_{n-1}- 10a_{n-2}+5^n-6$$

So far I have this

$$ a_n = 7a_{n-1}- 10a_{n-2}$$

$$ \frac{r^n}{r^{n-2}} = \frac{7r^{n-1}}{r^{n-2}}- \frac{10r^{n-2}}{r^{n-2}}$$

Then the characteristic equation for the homogeneous part is $$ r^2 = 7r - 10 $$ where $ r_1=2, r_2=5$.

So the homogeneous part is

$$ a_n^h=\alpha_1r_1^n + \alpha_2r_2^n = \alpha_1(2)^n+\alpha_2(5)^n$$

and I think the equation of the non-homogeneous part should be

$$ a_n^p = A(5^n)n+B$$

$$An(5^n)+B = 7(A(5^{n-1})(n-1)+B) -10(A(5^{n-2})(n-2)+B)$$

$$An(5^n)+B = 7(A(5^{n-1})(n-1))+7B-10(A(5^{n-2})(n-2))-10B$$

And I know there's a way to find $A$ and $B$ easily with like terms, but I don't know if I can use it here or how to use it.

Answer 1

The original recurrence relation is $$a_n = 7a_{n-1} - 10a_{n-2} + 5^n - 6 \tag1$$ Assume particular solution $$a_n^p = An5^n + B \tag2$$ Substitute $(2)$ into $(1)$: \begin{align} An5^n + B &= 7(A(n-1)5^{n-1} + B) - 10(A(n-2)5^{n-2} + B) + 5^n - 6 \\ &= 7An5^{n-1} - 7A5^{n-1} + 7B - 10An5^{n-2} + 20A5^{n-2} -10B + 5^n - 6 \\ &= \left(\frac{7}{5}-\frac{10}{25}\right)An5^n + \left(-\frac{7}{5}+\frac{20}{25}\right)A5^n -3B + 5^n - 6 \\ &= An5^n + \left(1-\frac{3A}{5}\right)5^n - 3B - 6 \end{align} Comparing like terms yields: \begin{align} 0 &= 1-\frac{3A}{5} \\ B &= -3B - 6 \end{align} So $A=5/3$ and $B=-3/2$, which implies particular solution $$a_n^p = \frac{5}{3}n5^n - \frac{3}{2} = \frac{n5^{n+1}}{3} - \frac{3}{2}$$ and general solution $$a_n = a_n^h + a_n^p = \alpha_1 2^n+\alpha_2 5^n + \frac{n5^{n+1}}{3} - \frac{3}{2}$$ Use any two provided initial conditions (which you did not specify) to solve for $\alpha_1$ and $\alpha_2$.