How do I resolve a recurrence relation when the characteristic equation has fewer roots than terms? I know how to solve "simple" recurrence relations. For instance, say you have:
$$c_0 = 20$$
$$c_1 = 30$$
$$c_n = 3 c_{n-1} - 2 c_{n-2}$$
We can write the characteristic equation as:
$$3x^{n-1} - 2x^{n-2} = x^n$$
Solving this with $n=2$, we get $x = 1$ or $x = 2$. This lets us write the relation $c_n = \alpha_1 1^n + \alpha_2 2^n$, and we can solve for $\alpha_1$ and $\alpha_2$ with the initial states $c_0$ and $c_1$.
However, this depends on the fact that $3x^{n-1} - 2x^{n-2} = x^n$ has two roots.
Now, I'm stuck on another problem where the characteristic equation has fewer roots than terms.
Say I have this recurrence relation instead:
$$a_0 = 0$$
$$a_1 = 2$$
$$a_2 = −1$$
$$a_n = 9a_{n-1} - 15a_{n-2} - 25a_{n-3}$$
The characteristic equation would be:
$$9x^{n-1} - 15x^{n-2} - 25x^{n-3} = x^n$$
However, solving with $n=3$, we only get two roots: $x=-1$ or $x=5$. There are not enough roots to write a relation in the form of $a_n = \alpha_1 r_1^n + \alpha_2r_2^n + \alpha_3r_3^n$. How do I proceed?
 A: The characteristic equation is actually $x^3-9x^2+15x+25 = 0$; it doesn’t depend on $n$. After factoring this becomes $(x+1)(x-5)^2 = 0$, with a double root at $x=5$. In this case the general solution has the form $a_n = \alpha_1(-1)^n + \alpha_2 \cdot 5^n + \alpha_3n \cdot 5^n$, and you can use the known values of $a_0,a_1,a_2$ to solve for $\alpha_1,\alpha_2,\alpha_3$. 
More generally, if $r$ is a root of the characteristic equation of multiplicity $m$, it gives rise to these $m$ terms in the general solution:$$\alpha_1r^n + \alpha_2nr^n + \alpha_3n^2 r^n + \dots + \alpha_m n^{m-1}r^n.$$Thus, you will always have as many terms as the degree of the characteristic equation.
A: Factor the characteristic polynomial to get
$$
x^3-9x^2+15x+25=(x+1)(x-5)^2
$$
The $x+1$ factor requires a term of the form $a(-1)^k$, but the $(x-5)^2$ term requires $(b+ck)5^k$.  This is because both $5^k$ and $k\:5^k$ are annihilated by the difference operator $(S-5)^2$ (where $S$ is the shift operator: $Sa_n=a_{n+1}$).  Now, just find $a$, $b$, and $c$ to fit your initial data.
For factors of $(x-a)^n$, use $(b_0+b_1 k+b_2 k^2+...+b_{n-1}k^{n-1})a^k$ since this is annihilated by $(S-a)^n$.
A: Use generating function directly: define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence with no subtractions in indices:
$$
a_{n + 3} = 9 a_{n + 2} - 15 a_{n + 1} - 25 a_n
$$
Multiply by $z^n$ and sum over $n \ge 0$, recognize:
$$
\sum_{n \ge 0} a_{n + k} z^n 
  = \frac{A(z) - a_0 - a_1 z - \ldots - a_{k - 1} z^{k + 1}}{z^k}
$$
to get:
$$
\frac{A(z) - 2 z + z^2}{z^3}
  = 9 \frac{A(z) - 2 z}{z^2}
      - 15 \frac{A(z)}{z}
      - 25 A(z)
$$
Written as partial fractions:
$$
A(z) = \frac{53}{60 (1 - 5 z)} - \frac{3}{10 (1 - 5 z)^2} - \frac{7}{12 (1 + z)}
$$
The generalized binomial theorem lets you read off coefficients:
\begin{align}
a_n &= \frac{53}{60} \cdot 5^n 
        - \frac{3}{10} \binom{-2}{n} (-5)^n
        - \frac{7}{12} \cdot (-1)^n \\
    &= \frac{53}{60} \cdot 5^n
        - \frac{3}{10} \binom{n + 2 - 1}{2 - 1} \cdot 5^n
        - \frac{7}{12} \cdot (-1)^n \\
    &= \frac{(35 - 18 n) \cdot 5^n - 35 \cdot (-1)^n}{60}
\end{align}
Repeat roots give terms including $\binom{-m}{n} = \binom{n + m - 1}{m - 1}$, which is just a polynomial of degree $m - 1$ in $n$.
A: $\newcommand{\+}{^{\dagger}}
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You can always insert a parameter $\ds{\epsilon}$ to 'remove' the equality of the roots. At the end, take a proper limit to restore the original recurrence. In order to illustrate the method I'll solve the simple example
$$
a_{n + 1} - 2a_{n} + a_{n - 1} = 0\qquad\mbox{with}\qquad a_{0} = 0\,,\quad a_{1} = 1
$$
The solution is, of course, $\color{#66f}{\LARGE \ds{a_{n} = n}}$ and the 'characteristic equation' has the double root equal to one.

Instead, I'll solve $\ds{a_{n + 1} - 2a_{n} + \epsilon a_{n - 1} = 0}$. The 'characteristic equation' has roots
  $\ds{r_{\pm} = 1 \pm \underbrace{\root{1 - \epsilon^{2}}}_{\ds{\equiv \delta}}}$ 
  \begin{align}
a_{n}&=
Ar_{-}^{n} + Br_{+}^{n}\quad\imp\quad A + B = 0\,,\quad Ar_{-} + Br_{+} = 1
\quad\imp\quad
\left\lbrace\begin{array}{rcl}
A & = & {1 \over r_{-} - r_{+}}
\\
B & = & -A
\end{array}\right.
\\[3mm]
a_{n} &={r_{-}^{n} - r_{+}^{n} \over r_{-} - r_{+}}
\end{align}

The limit $\ds{\epsilon \to 1}$ yields:
$$
\lim_{\epsilon \to 1}{r_{-}^{n} - r_{+}^{n} \over r_{-} - r_{+}}
=\lim_{\delta \to 0}{\pars{1 - \delta}^{n} - \pars{1 + \delta}^{n} \over -2\delta}
=\lim_{\delta \to 0}
{n\pars{1 - \delta}^{n - 1}\pars{-1} - n\pars{1 + \delta}^{n - 1} \over -2}
=\color{#66f}{\LARGE n}
$$
