Recursive sequence. Value is resetting to 0

Question

Solve $$\begin{cases} {a_n} = 5a_{n-1} - 6a_{n-2} + 6 \cdot 3^{n}\\ {a_0} = 1, a_1=50 \end{cases}$$

The problem I've got here is the C resetting. It happens to me sometimes and I don't know why. Hope someone can clear it up for me. Thanks.

Answer 1

You have found the general solution of the homogeneous part which is $$a_{hom}(n)=A2^n+B3^n.$$ As regards the particular solution $a_{*}(n)$, you should apply the method of undetermined coefficient in the correct way. Since $3$ is a solution of order $1$ of the characteristic polynomial, you have to try with $a_{*}(n)=Cn3^n$ (not simply $C3^n$) where $C$ is a constant to be determined: if we plug it into the recurrence we find $$Cn3^n = 5C(n-1)3^{n-1} - 6C(n-2)3^{n-2} + 6 \cdot 3^{n}$$ that is $$9Cn = 15C(n-1) - 6C(n-2) + 54$$ or $$0=-15C+12C+54$$ and we have that $C=54/3=18$.

Answer 2

$a_n-5a_{n-1}+6a_{n-2}=6\cdot 3^n$, with $a_0=1$, $a_1=50$.

First we find solution of homogenous recurrence relation

$$a_n-5a_{n-1}+6a_{n-2}=0.$$

$r^2-5r+6=0$, we have $r=2$ with multiplicity 1 and $r=3$ with multiplicity 1. So,

$$a_n^h=A\cdot 2^n+B\cdot 3^n.$$

Because in the right-hand side $$a_n-5a_{n-1}+6a_{n-2}=6\cdot 3^n,$$

Now we have $r=2$ with multiplicity 1 and $r=3$ with multiplicity 2. Let the particular solution

$$a_n^p=C\cdot 2^n+D\cdot 3^n + E\cdot n\cdot 3^n.$$

Now delete the terms which appear similar in the $a_n^h$,

$$a_n^p=E\cdot n\cdot 3^n.$$

Substituting new $a_n^p$ into $$a_n-5a_{n-1}+6a_{n-2}=6\cdot 3^n,$$

we have

\begin{eqnarray} a_n-5a_{n-1}+6a_{n-2}&=&6\cdot 3^n\\ E\cdot n\cdot 3^n-5E\cdot (n-1)\cdot 3^{n-1}+6E\cdot (n-2)\cdot 3^{n-2}&=&6\cdot 3^n\\ E\cdot n\cdot 3^n-\dfrac{5}{3}E\cdot (n-1)\cdot 3^{n}+\dfrac{2}{3}E\cdot (n-2)\cdot 3^{n}&=&6\cdot 3^n\\ E\cdot n\cdot 3^n-\dfrac{5}{3}E\cdot n\cdot 3^{n}+\dfrac{5}{3}E\cdot 3^{n}+\dfrac{2}{3}E\cdot n\cdot 3^{n}-\dfrac{4}{3}E\cdot 3^{n}&=&6\cdot 3^n\\ \dfrac{1}{3}E\cdot 3^{n}&=&6\cdot 3^n\\ \end{eqnarray}

Now, we have $E=18$. So, the particular solution will be $$a_n^p=18\cdot n\cdot 3^n.$$

Now,

$$a_n=a_n^h+a_n^p.$$

$$a_n=A\cdot 2^n+B\cdot 3^n+18\cdot n\cdot 3^n.$$

To find $A$ and $B$, we substitute initial value.

$$a_0=A\cdot 2^0+B\cdot 3^0+18\cdot 0\cdot 3^0=A+B=1.$$ $$a_1=A\cdot 2^1+B\cdot 3^1+18\cdot 1\cdot 3^1=2A+3B+54=50\iff 2A+3B=-4.$$

Solving last two equations, we get $A=7$ and $B=-6$. So we get the solution is

$$a_n=7\cdot 2^n-6\cdot 3^n+18\cdot n\cdot 3^n.$$