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

The second one

*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..

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.

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

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

so there will not be any particular solutions right?

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

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.