1. ## difference equation

Am I on the right track with this problem
I am trying to find the particular solution
$\displaystyle a_{n+2} - 2a _{n+1} -63a_{n} = 64n$, n>= 0 with $\displaystyle a_{0} = 0 a_{1} = 7$

(x-9)(x+7) = 0
x=9, x=-7

$\displaystyle u_{n} = A*9^n + B * -7^n$ for arbitrary constants A and B

RHS
64n =
$\displaystyle V_{n} = (1^n) * (C_{n} + D) = (C_{n} + D)$

substitute $\displaystyle a_{n} = V_{n}$
$\displaystyle a_{n+2} - 2a_{n+1} - 63an$
= (C(n+2)+D) - 2(C(n+1)+D -63 (Cn + D) = 64n
then
(-64 + -C) = 0
-64D = 0

implying C = -64
D = 0

thus Vn = n+ -64

Therefore the general solution is
$\displaystyle A9^n + B(-7)^n + n - 64$ for arbitrary constants A and B

$\displaystyle a_{0} =0 = A * 9^0 + B*-7^0 + 0 -64$
$\displaystyle a_{1} = 7 = A * 9^1 + B*-7^1 + 1 -64$ <-- I will work on these if I am on the right track

how am I doing?
(no good I think)

2. Originally Posted by dunsta
Am I on the right track with this problem
I am trying to find the particular solution
$\displaystyle a_{n+2} - 2a _{n+1} -63a_{n} = 64n$, n>= 0 with $\displaystyle a_{0} = 0 a_{1} = 7$

(x-9)(x+7) = 0
x=9, x=-7

$\displaystyle u_{n} = A*9^n + B * -7^n$ for arbitrary constants A and B

RHS
64n =
$\displaystyle V_{n} = (1^n) * (C_{n} + D) = (C_{n} + D)$

substitute $\displaystyle a_{n} = V_{n}$
$\displaystyle a_{n+2} - 2a_{n+1} - 63an$
= (C(n+2)+D) - 2(C(n+1)+D) -63 (Cn + D) = 64n Mr F says: The red bracket will make a difference.

then
(-64 + -C) = 0
-64D = 0

Mr F says: The above two equations make no sense. Please show clearly where they come from.

implying C = -64
D = 0

thus Vn = n+ -64

Therefore the general solution is
$\displaystyle A9^n + B(-7)^n + n - 64$ for arbitrary constants A and B

$\displaystyle a_{0} =0 = A * 9^0 + B*-7^0 + 0 -64$
$\displaystyle a_{1} = 7 = A * 9^1 + B*-7^1 + 1 -64$ <-- I will work on these if I am on the right track

how am I doing?
(no good I think)
Your particular solution is wrong. I get $\displaystyle a_n = -n$ as a particular solution.

3. = (C(n+2)+D) - 2(C(n+1)+D) -63 (Cn + D) = 64n Mr F says: The red bracket will make a difference.

LHS
=(Cn + 2C + D) - 2(Cn + C + D) - 63Cn - 63CD
= Cn + 2C + D -2Cn -2C -2D -63Cn -63CD
= Cn - 2Cn -63Cn +2C -2C +D -2D -63D
= -64Cn -64D

RHS = 64n

-64Cn -64D = 64n
-64Cn -64n -64D = 0
n(-64C -64) -64D = 0

(after this I split the equation into two, as below, and dropped the n (I am just going off the most similar example in my notes)

then
(-64 + -C) = 0
-64D = 0

Mr F says: The above two equations make no sense. Please show clearly where they come from.

4. Originally Posted by dunsta
= (C(n+2)+D) - 2(C(n+1)+D) -63 (Cn + D) = 64n Mr F says: The red bracket will make a difference.

LHS
=(Cn + 2C + D) - 2(Cn + C + D) - 63Cn - 63CD Mr F says: ?
= Cn + 2C + D -2Cn -2C -2D -63Cn -63CD Mr F says: ?
= Cn - 2Cn -63Cn +2C -2C +D -2D -63D
= -64Cn -64D

RHS = 64n

-64Cn -64D = 64n
-64Cn -64n -64D = 0
n(-64C -64) -64D = 0

(after this I split the equation into two, as below, and dropped the n (I am just going off the most similar example in my notes)

then
(-64 + -C) = 0 Mr F says: Shouldn't this be -64C - 64 = 0?
-64D = 0 Mr F says: I got -63D, but I can't be bothered re-checking. D = 0 is the correct result in both cases.

[snip]
Did you check the particular solution you found (in both of your posts) to see that it worked? If you had done so, you would have realised it was wrong.

5. Can anybody point me towards a similar problem with a solution. So far in Discrete maths I have had the best results from looking at a problem with a soution and then working out the hows and whys, I have read my notes on this topic over and over but they are long winded explanations in words, with no examples.
So if any one can point me to a similar problem/exercise I would be appreciative. At the moment I seem to be lost with this problem.

$\displaystyle a_{n+2} - 2a _{n+1} -63a_{n} = 64n$, n>= 0 with $\displaystyle a_{0} = 0 a_{1} = 7$

(x-9)(x+7) = 0
x=9, x=-7 <---is this part correct?

$\displaystyle u_{n} = A*9^n + B * -7^n$ for arbitrary constants A and B

RHS
64n =
$\displaystyle V_{n} = (1^n) * (C_{n} + D) = (C_{n} + D)$ <-- is this correct?? Mr Fantastic is this the particular solution you refer to that should be $\displaystyle a_{n} = -n$

$\displaystyle A9^n + B(-7)^n + n - 64$ for arbitrary constants A and B

$\displaystyle a_{0} =0 = A * 9^0 + B*-7^0 + 0 -64$
$\displaystyle a_{1} = 7 = A * 9^1 + B*-7^1 + 1 -64$ <-- I think my goal is to try and find A and B for these expressions

6. Originally Posted by dunsta
Can anybody point me towards a similar problem with a solution. So far in Discrete maths I have had the best results from looking at a problem with a soution and then working out the hows and whys, I have read my notes on this topic over and over but they are long winded explanations in words, with no examples.
So if any one can point me to a similar problem/exercise I would be appreciative. At the moment I seem to be lost with this problem.

$\displaystyle a_{n+2} - 2a _{n+1} -63a_{n} = 64n$, n>= 0 with $\displaystyle a_{0} = 0 a_{1} = 7$

(x-9)(x+7) = 0
x=9, x=-7 <---is this part correct?

$\displaystyle u_{n} = A*9^n + B * -7^n$ for arbitrary constants A and B Mr F says: To check if it's correct, substitute it into the homogenous difference equation. Does it work? (Do you understand the thinking behind what you have done above?)

RHS
64n =
$\displaystyle V_{n} = (1^n) * (C_{n} + D) = (C_{n} + D)$ <-- is this correct?? Mr Fantastic is this the particular solution you refer to that should be $\displaystyle a_{n} = -n$ Mr F says: Yes.

$\displaystyle A9^n + B(-7)^n + n - 64$ for arbitrary constants A and B

$\displaystyle a_{0} =0 = A * 9^0 + B*-7^0 + 0 -64$
$\displaystyle a_{1} = 7 = A * 9^1 + B*-7^1 + 1 -64$ <-- I think my goal is to try and find A and B for these expressions
..