Question

Please help:)
Solve the difference equation
\(\Large y_{n+1}-4y_{n}=8\)

Answer 1

Is it

\[y^{n+1} \ or \ y_{n+1}\]

Answer 2

The second one

Answer 3

I found homogeneous solution in this way
\(y_{n+1}-4y_n=0\)
Substituting \(y_n=A\zeta^n\) where \(A\ne0\), a constant we get the characteristic equation as
\(A\zeta^{n+1}-4A\zeta^n=0\)
\(\zeta-4=0\)
whose root is \(\zeta=4\)
Therefore, \(y^(p)=c_1(4)^n\)

Answer 4

@Callisto

Answer 5

*therefore, \(y^{(p)}=c_1(4)^n\)

Answer 6

@experimentX Can you give a hand here?

Answer 7

any info given on y_0 ?

Answer 8

yn=a4^n+b1^n

Answer 9

do you have initial cond.

Answer 10

no it is not given

Answer 11

it is done then..

Answer 12

it is not hom. diff. eq, but we make hom. diff. eq..

Answer 13

we need to find the particular solution too. right? I assumed it to be of the form an+b.
then found a=0 and b=-8/3

Answer 14

is it clear..

Answer 15

this is the solution ..\[\Large y_{n}=a4^{n}+b1^{n}\]

Answer 16

that looks pretty nice.

Answer 17

i dnt see that screenshot properly

Answer 18

http://www.wolframalpha.com/input/?i=y_%28n%2B1%29+-+4+y_%28n%29+%3D+8
Wolf says only one variable.

Answer 19

\[ y_{n+2} = 5 y_{n+1} -4y_n \\
y_{n+1} = y_{n+1} \\
\]
\[u_{n+1} = \begin{bmatrix}
5 & -4\\
1 & 0
\end{bmatrix} u_n\]

Answer 20

http://imageshack.us/photo/my-images/836/fsedfe.png/

Answer 21

I dnt get the second step?

Answer 22

multiply top eq. by -1 and add them

Answer 23

I mean y do u make it
y_n+2-4y_n+1=8

Answer 24

in order to get rid of 8

Answer 25

to make it hom. diff. eq..

Answer 26

but I dont know how to find a and b if initial condition is not given..

Answer 27

Really sorry. I still dnt get the conversion part

Answer 28

yep ... the answer is
\[ y_n = c_1 4^n + c_2 1^n\]
I think we should be able to eliminate one constant. since the initial condition is not given, there will be one constant.

Answer 29

How do u get
y_n+2-5y_n+1+4y_n=0

Answer 30

\[ y_0 = c_1 + c_2 \\
y_1 = 8 + 4y_0\\
y_1 = 4c_1 +c_2 \\
\]
---------------------
solving this you get, http://www.wolframalpha.com/input/?i=a+%3D+x%2By+%2C+b+%3D+8+%2B+4a%2C+b+%3D+4x+%2By+
where x=c1 and y=c2 and a =y_0 and b=y1

you get the same answer as Wolf. there is one constant since you do not have initial condition.

Answer 31

\[ u_{n+1} = \begin{bmatrix}
5 & -4\\
1 & 0
\end{bmatrix} u_n \]
where
\[ u_n = \begin{bmatrix}{
y_{n+1}\\
y_n
}\end{bmatrix}\]
the solution will be \( u_n = c_1 \lambda _1^n x_1 + c_2 \lambda_2^2 x_2 \)
where \( \lambda \) and x are the eigen value and eigen vector of given matrix.

Answer 32

so there will not be any particular solutions right?

Answer 33

no ... unless you have initial value i.e y_0

Answer 34

did you check the final answer?
http://www.wolframalpha.com/input/?i=y_%28n%2B1%29+-+4+y_%28n%29+%3D+8

Answer 35

Ya. I checked it out. Bt got confused seeing 32 and all that:(

Answer 36

just put c_1 = k + 8/3 and c_2 = -8/3 .. rest is just manipulation of numbers by WA

Answer 37

oh k. Thanxx a lot.

Answer 38

no probs at all.
using this method, you can solve for n-th Fibonacci number.