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

Mathematics
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  • hba
Is it \[y^{n+1} \ or \ y_{n+1}\]
The second one
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, \(y^(p)=c_1(4)^n\)

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Other answers:

*therefore, \(y^{(p)}=c_1(4)^n\)
@experimentX Can you give a hand here?
any info given on y_0 ?
yn=a4^n+b1^n
do you have initial cond.
no it is not given
it is done then..
it is not hom. diff. eq, but we make hom. diff. eq..
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
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is it clear..
this is the solution ..\[\Large y_{n}=a4^{n}+b1^{n}\]
that looks pretty nice.
i dnt see that screenshot properly
http://www.wolframalpha.com/input/?i=y_%28n%2B1%29+-+4+y_%28n%29+%3D+8 Wolf says only one variable.
\[ y_{n+2} = 5 y_{n+1} -4y_n \\ y_{n+1} = y_{n+1} \\ \] \[u_{n+1} = \begin{bmatrix} 5 & -4\\ 1 & 0 \end{bmatrix} u_n\]
http://imageshack.us/photo/my-images/836/fsedfe.png/
I dnt get the second step?
multiply top eq. by -1 and add them
I mean y do u make it y_n+2-4y_n+1=8
in order to get rid of 8
to make it hom. diff. eq..
but I dont know how to find a and b if initial condition is not given..
Really sorry. I still dnt get the conversion part
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.
How do u get y_n+2-5y_n+1+4y_n=0
\[ y_0 = c_1 + c_2 \\ y_1 = 8 + 4y_0\\ y_1 = 4c_1 +c_2 \\ \] --------------------- 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.
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\[ u_{n+1} = \begin{bmatrix} 5 & -4\\ 1 & 0 \end{bmatrix} u_n \] 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.
so there will not be any particular solutions right?
no ... unless you have initial value i.e y_0
did you check the final answer? http://www.wolframalpha.com/input/?i=y_%28n%2B1%29+-+4+y_%28n%29+%3D+8
Ya. I checked it out. Bt got confused seeing 32 and all that:(
just put c_1 = k + 8/3 and c_2 = -8/3 .. rest is just manipulation of numbers by WA
oh k. Thanxx a lot.
no probs at all. using this method, you can solve for n-th Fibonacci number.

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