Question

solve the 2nd order non-homogeneous linear recurrence relation with constant coefficients: f(n+2)+f(n+1)-6f(n)=-8n+22 subject to the start-up conditions f(0)=3; f(1)=2

Answer 1

we will let you solve this one

Answer 2

i wish i knew how to solve it ..i wouldn't be asking for help if i knew

Answer 3

look at previous one ;solved by others for you

Answer 4

i know i am looking at it and trying to make sense of it

Answer 5

hi mathmate

Answer 6

ok so the first step is \[r ^{2}+r-6=8n+22 \]?

Answer 7

r= -3,2 ?

Answer 8

solve for homogenous first

Answer 9

then plug in to solve for particular case

Answer 10

hold on let me try to figure out

Answer 11

I'm here, but not finished looking at the numbers.

Answer 12

okay no problem :) ..im also trying to figure out how to do this by looking at the previous problems

Answer 13

OK, I've got the solution.
It takes a little while because we can convert the given inhomogeneous relation to a 4th order homogeneous, which we know how to solve easily.

Answer 14

It's a little more complicated than that because of the right-hand side.

Answer 15

My aim is to eliminate the rhs by elimination, and it turns out that it takes two steps.

Answer 16

right-hand-side.

Answer 17

What we do, is to increase n by 1, and write the relation as
f(n+3)+f(n+2)-6f(n+1)=-8(n+1)+22

Answer 18

Subtract the original relation from it to get:
f(n+3)-7f(n+1)+6f(n)=-8

Answer 19

We still have 8 on the rhs, so we do the same thing again to get
f(n+4)-f(n+3)-7f(n+2)+13f(n+1)-6f(n)=0
You're OK with that so far?

Answer 20

hold on im still reading the second part u wrote

Answer 21

where did u get the 7 from in 7f(n+1)?

Answer 22

and what happened to the 22?

Answer 23

When you subtract the first equation from the second, the 22 disappears, and the -7f(n+1) comes from -6f(n+1)-f(n+1).

Answer 24

If it helps, give me 4-5 minutes, I will write it up properly so it will be easier to follow. Is that OK?

Answer 25

oh sure it's no problem at all
thank yyou

Answer 26

Will be right back.

Answer 27

please take your time :)

Answer 28

Question: to solve
f(n+2)+f(n+1)-6f(n)=-8n+22 ...(1)
given f(0)=3, f(1)=2.

Since the equation is non-homogeneous, we will try to eliminate the right-hand-side by elimination, so that the characteristic equation method will work.

The elimination is done by increasing n, and subtracting.

Substitute n+1 for n in (1) gives

f(n+3) + f(n+2) - 6f(n+1) = -8(n+1) + 22 ....(2)
Subtract (1):
f(n+2) + f(n+1) - 6f(n) = -8(n) + 22 ....(1)
to get

f(n+3) - 7f(n+1) + 6f(n) = 8 .........(3)

Now, repeat the same process to eliminate 8:

f(n+4) - 7f(n+2) + 6f(n+1) = 8 ...(3A)
subtract:
f(n+3) - 7f(n+1) + 6f(n) = 8 ....(3)
to get
f(n+4)-f(n+3)-7f(n+2)+13f(n+1)-6f(n) = 0 ..,.(4)

Equation (4) is a homogeneous equation which can be solved by the characteristic equation method.

Solving 4 gives the roots as x=1,1,2,3 (double root at 1).

So the general solution is:
f(n)=(c1 + c2 n) + c3.2^n + c4.(-3)^n

To solve for c1-c4, we need 4 values, of which we were given two:
f(0)=3
f(1)=2

so calculate from equation (1)
f(2)=22
f(3)=-12
f(4)=134 (for checking later on).

Set up a linear system by substituting
f(0)=3=c1+c2(0)+c3+c4
f(1)=2=c1+c2+2C3-3C4
f(2)=22=c1+2c2+4c3+9c4
f(3)=-12=c1+3c2+8c3-27c4

Solve the system for c1-c4 as
[0,2,9/5,6/5]
which means that the solution is:

f(n)=2n+(9/5)2^n+(6/5)(-3)^n

Check that f(n) works for at least n=0-4.