# Solving linear homogeneous difference equation

Let $a_{n+2} + a_{n+1} - 6a_n = 0$ and $a_1=1, a_0 = 0$ be given. How do we solve such a difference equation?

I set $a_n =Aw^n$. Then, we get $Aw^{n+2}-Aw^{n+1}-6Aw^n =0$. After that we have $(r^2+r-6)=0$ and $r=w^n = -3,2$.

I plug $-3$ in the first equation and it is satisfied but $2$ does not satisyfing it. What can I do next?

• What do you mean when you say you 'plugged it in to the first equation'? – Sean Henderson Jan 23 '18 at 17:23

There is a nice way to solve this type of recurrence with operators.

1- Define the operator $E^k$ which does $E^ka_n=a_{n+k}.$

2- So this recurrence can be writen in the form $$(E^2+E-6)a_n=0$$

this kind of operator have an factorization (like polynomials) $$(E+3)(E-2)a_n=0$$

3- An operator in the form (E-r) annihilates $Cr^n$ $$(E-r)Cr^n=Cr^{n+1}-rCr^n=0.$$

4- It can be shown that, in this case $$a_n=c_1(-3)^n+c_2(2)^n.$$

5- But we don't need to stop there. We can find the contants $c_1$ and $c_2$, in a easy way with those operators.

Apply $E+3$

$$a_{n+1}+3a_n=c_2(2)^{n+1}+3c_2(2)^n,$$ take $n=0$ and use the initial conditions $a_0=0$, $a_1=1$ $$a_1+3a_0=2c_2+3c_2= 1$$ so $c_2=1/5$.

Now apply $E-2$

$$a_{n+1}-2a_n=c_1(-3)^{n+1} -2c_1(-3)^n,$$

take $n=0$ again $$1=-c_1.3-2c_1,$$

so $$c_1=-1/5.$$

The solution is $$a_n=((2)^n-(-3)^n)/5 .$$

The characteristic equation is $$r^2+r-6=0\iff(r+3)(r-2)=0\iff r=-3,2.$$

Thus, the solution is $$a_n=c_{1}(-3)^{n}+c_{2}2^n$$ for some $c_{1},c_{2}\in\mathbb{R}$ (as $(-3)^n$ and $2^n$ are linearly independent in a two dimensional solution space). Use the fact that $a_{0}=0$ and $a_{1}=1$ to determine the $c_{i}$.

• Is this the way we solve difference equations in general? – Ninja Jan 23 '18 at 17:22