# difference equations

• Jul 9th 2007, 12:24 PM
charlotte_usa
difference equations
hi!

ive got three difference equations but could do with some help becuase i didnt think that they would be too long but they are my homework for this week and am slightly confused why i would get such a short homework!

a) Xn+2 - 5Xn+1 + 6Xn =0

b) Xn+2 - 5Xn+1 + 6Xn = 4

c) Xn+2 - 6Xn+1 + 9Xn =0

http://www.mathhelpforum.com/math-he...cons/icon7.gif
• Jul 9th 2007, 01:39 PM
ThePerfectHacker
Quote:

Originally Posted by charlotte_usa
a) Xn+2 - 5Xn+1 + 6X[SIZE=1][SIZE=2][SIZE=1]n [SIZE=2]=0

$x_{n+2}-5x_{n+1}+6x_n=0$

The charachteristic equation,
$k^2 - 5k + 6 = (x-2)(x-3)=0$

So,
$x_n = C_1 2^n + C_2 3^n$
But we cannot find,
$C_1,C_2$
Without initial conditions.
• Jul 9th 2007, 01:43 PM
ThePerfectHacker
Quote:

Xn+2 - 5Xn+1 + 6Xn = 4
Here we need a particular solution,
Look for a solution of the form $x_n = A$
Then,
$x_n=A \ \ x_{n+1} = A \ \ x_{n+2} = A$
Thus,
$A-5A+6A = 4$ implies $A=2$

Thus, the general solution is,
$x_n = C_1 2^n + C_2 3^n + 2$
• Jul 9th 2007, 01:44 PM
ThePerfectHacker
Quote:

Xn+2 - 6Xn+1 + 9Xn =0
The charachteristic equation is,
$k^2 - 6k + 9 = (k-3)(k-3)=0$

Since there is only one solution for $k=3$

It means the general solution is,
$x_n = C_13^n + C_2n3^n$
• Jul 10th 2007, 06:25 AM
charlotte_usa
Quote:

Originally Posted by ThePerfectHacker
$x_{n+2}-5x_{n+1}+6x_n=0$

The charachteristic equation,
$k^2 - 5k + 6 = (x-2)(x-3)=0$

So,
$x_n = C_1 2^n + C_2 3^n$
But we cannot find,
$C_1,C_2$
Without initial conditions.

but to get the complementary solution

$Yn= A2^n + B3^n$

right? and since Dn = 0 then there is no particular solution?
• Jul 10th 2007, 06:46 AM
charlotte_usa
Quote:

Originally Posted by ThePerfectHacker
Here we need a particular solution,
Look for a solution of the form $x_n = A$
Then,
$x_n=A \ \ x_{n+1} = A \ \ x_{n+2} = A$
Thus,
$A-5A+6A = 4$ implies $A=2$

Thus, the general solution is,
$x_n = C_1 2^n + C_2 3^n + 2$

here the c1 = 2 and c2 = 3 so ive been taught that for a particular solution, since c1 doesnt equal c2 then $Zn = DnC1^n
$

so then to try Zn = Cn

so from Xn+2 - 5Xn+1 + 6Xn = 4
it will become C(n+2) - 5(c(n+1)) + 6(Cn) = 4

and from setting n=-1 i get c-6c=4 so c doesnt equal 2 :S

where have i gone wrong?
• Jul 10th 2007, 07:18 AM
ThePerfectHacker
Quote:

Originally Posted by charlotte_usa
but to get the complementary solution

$Yn= A2^n + B3^n$

right? and since Dn = 0 then there is no particular solution?

Yes that is the full solution because it is a homogenous difference equation. Thus, there is no particular solution.

Quote:

here the c1 = 2 and c2 = 3 so ive been taught that for a particular solution, since c1 doesnt equal c2 then

so then to try Zn = Cn

so from Xn+2 - 5Xn+1 + 6Xn = 4
it will become C(n+2) - 5(c(n+1)) + 6(Cn) = 4

and from setting n=-1 i get c-6c=4 so c doesnt equal 2 :S

where have i gone wrong?
I have no idea what you did.
Why C_1=2 and C_2=3?
• Jul 10th 2007, 08:00 AM
Soroban
Hello, charlotte_usa!

I believe the second problem requires special handling . . .

Quote:

$b)\;X_{n+2} - 5X_{n+1} + 6X_n\;=\; 4$
$\begin{array}{ccccc}\text{We have:} & X_{n+2} - 5X_{n+1} + 6X_n & = & 4 & (1)\\
\text{and also:} & X_{n+3} - 5X_{n+2} + 6X_{n+1} & = & 4 & (2)\end{array}$

Subtract (1) from (2): . $X_{n+3} - 6X_{n+2} + 11X_{n+1} - 6X_n \;=\;0$

The characteristic equation is: . $m^3 - 6m^2 + 11m - 6 \:=\:0$

. . which factors: . $(m-1)(m-2)(m-3) \;=\;0$

. . and has roots: . $m \:=\:1,\:2,\:3$

Can you finish it now?

• Jul 10th 2007, 08:37 AM
charlotte_usa
Quote:

Originally Posted by ThePerfectHacker
Yes that is the full solution because it is a homogenous difference equation. Thus, there is no particular solution.

I have no idea what you did.
Why C_1=2 and C_2=3?

The charachteristic equation,
http://www.mathhelpforum.com/math-he...90262f46-1.gif

i mean that say r=2 and t=3 (the solutions to the characheristic equation) and since r doesnt equal t, then ive been taught to try Zn = Cn for the particular solution. then i put cn into the original equation

so from Xn+2 - 5Xn+1 + 6Xn = 4
it will become C(n+2) - 5(c(n+1)) + 6(Cn) = 4

and from setting n=0 then 2c-5c=4 so c=-3/4 and Zn (the particular solution) is 3/4n

so the general solution is $yn = A2^n + B3^n +3/4n$
• Jul 11th 2007, 08:42 AM
charlotte_usa