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This comes from a problem sheet I found online, which since seems to have been removed. Either way, here's my attempt:

For the homogeneous case, we have $$w_{n+1}-5w_{n}+6w_{n-1}=2.$$ Its auxiliary equation $\lambda^2-5\lambda+6=0$ has solutions $\lambda=2$ and $\lambda=3$, so $w_n=A2^n+B3^n$, for constants $A,B$.

For the particular solution, I try guessing $v_n=C$, and get $$C-5C+6C=2,$$ which implies $C=1$.

So the general solution is $u_n=1+A2^n+B3^n$.

Substituting in the initial conditions, I get $$u_0=1+A+B=1$$ $$u_1=1+2A+3B=1,$$ which implies $A=B=0$.

So is the only solution is $u_n=1$ for all $n$, or is there a way to find other (less trivial) solutions?

GAVD
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man_in_green_shirt
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    Since the whole recurrence is determined by two successive values, there can only be one solution. To check your solution, why not just compute the next value or two from the initial values and the recurrence. You can prove the pattern persists by induction. – Mark Bennet Sep 26 '17 at 08:38
  • yes $$u_n=1$$ is the only solution – Dr. Sonnhard Graubner Sep 26 '17 at 08:38

5 Answers5

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Your method is correct. Rearranging the terms, we find that $u_{n+1} = 5u_n-6u_{n-1}+2$. When $u_0 = 1, u_1 = 1$ as in your problem, then $u_2$ is $5*1-6*1+2$, or $1$.

The key point is to note that this recurrence equation holds for any value of $n$. Since $u_{n+1} = u_n = 1$, letting $n = m+1$, then we have $u_{m+2} = u_{m+1}$. Continuing this pattern, we find out that every term will always equal $1$. This is the only solution, by definition of a recurrence equation.

Toby Mak
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You are right. There is only one solution $u_n = 1$.

Other way to see: Taking $w_n = u_{n+1}-2u_n + 1$, then $w_0=0$ and $w_n = 3w_{n-1} = 3^nw_0 = 0$. So $u_{n+1} -1 = 2(u_n-1) = 2^n(u_1-1) = 0$.

GAVD
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Easiest way

Let $X_n = (u_n,u_{n-1})$ Therefore since $u_{n+1}-5u_{n}+6u_{n-1}=2$ we have, $$X_{n+1}= AX_n+b$$

Where $$A=\begin{pmatrix}5&-6\\1&0 \end{pmatrix}\qquad and \qquad b=\begin{pmatrix}2\\0 \end{pmatrix} $$ Now let $$V_n = X_n-a$$

where the vector $a$ is chosen such that,

$$V_{n+1} = AV_n\Longleftrightarrow AX_n +b-a =AX_n -a\Longleftrightarrow a=(I-A)^{-1}b$$

Thus, it follows that,

$$ V_n =A^nV_0\qquad\text{with}\qquad V_0 = X_0-a. $$

that is $$X_n = V_n+a =A^n(X_0-a) +a$$

Find the eigenvectors and eigenvalues of $A$ which will help to find $A^n$. This holds by writing $A=PDP^{-1}$ with D diagonal. also compute the vector $ a=(I-A)^{-1}b$.

Guy Fsone
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Alternatively, you can reduce the difference equation of order 2 to order 1: $$u_{n+1}-5u_{n}+6u_{n-1}=2 \Rightarrow (\underbrace{u_{n+1}-3u_n}_{b_{n+1}})-2(\underbrace{u_n-3u_{n-1}}_{b_n})=2, b_0=-2.$$ whose solution is: $$b_{n+1}=-2.$$ Now solve the original equation: $$u_{n+1}-3u_n=-2,u_0=1 \Rightarrow u_n=1.$$

farruhota
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It would be more useful to have a solution to the general form

$$u_n=au_{n-1}+bu_{n-2}+c$$

To that end, let $u_n=f_n-p$, so that

$$f_n=af_{n-1}+bf_{n-2}+c-p(a+b-1)$$

Now let $p=\frac{c}{a+b-1}$ to get the canonical form of the generalized Fibonacci form,

$$f_n=af_{n-1}+bf_{n-2}$$

Using the methods that we have shown previously here, the general solution can be expressed as

$${{f}_{n}}=\frac{\left( {{f}_{1}}-{{f}_{0}}\beta \right)\,{{\alpha }^{n}}-\left( {{f}_{1}}-{{f}_{0}}\alpha\right)\,{{\beta }^{n}}}{\alpha -\beta}\\ \alpha,\beta=\frac{1\pm\sqrt{a^2+4b}}{2}$$

The rest is straightforward. The result for $u_n=f_n-p$. In the present case we have the rather trivial solutions $f_n=0$ and $u_n=1$. But the general solution is valid for any $a,b,c$. I have tested this numerically for $(a,b,c)\in[-5,5]$, integers and non-integers alike.

Cye Waldman
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