## ajprincess Group Title Please help:) Solve the difference equation $$\Large y_{n+1}-4y_{n}=8$$ one year ago one year ago

1. hba Group Title

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

2. ajprincess Group Title

The second one

3. ajprincess Group Title

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, $$y^(p)=c_1(4)^n$$

4. ajprincess Group Title

@Callisto

5. ajprincess Group Title

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

6. Callisto Group Title

@experimentX Can you give a hand here?

7. experimentX Group Title

any info given on y_0 ?

8. cinar Group Title

yn=a4^n+b1^n

9. cinar Group Title

do you have initial cond.

10. ajprincess Group Title

no it is not given

11. cinar Group Title

it is done then..

12. cinar Group Title

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

13. ajprincess Group Title

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

14. cinar Group Title

15. cinar Group Title

is it clear..

16. cinar Group Title

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

17. experimentX Group Title

that looks pretty nice.

18. ajprincess Group Title

i dnt see that screenshot properly

19. experimentX Group Title

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

20. experimentX Group Title

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

21. cinar Group Title
22. ajprincess Group Title

I dnt get the second step?

23. cinar Group Title

multiply top eq. by -1 and add them

24. ajprincess Group Title

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

25. cinar Group Title

in order to get rid of 8

26. cinar Group Title

to make it hom. diff. eq..

27. cinar Group Title

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

28. ajprincess Group Title

Really sorry. I still dnt get the conversion part

29. experimentX Group Title

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.

30. ajprincess Group Title

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

31. experimentX Group Title

$y_0 = c_1 + c_2 \\ y_1 = 8 + 4y_0\\ y_1 = 4c_1 +c_2 \\$ --------------------- 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.

32. cinar Group Title

33. experimentX Group Title

$u_{n+1} = \begin{bmatrix} 5 & -4\\ 1 & 0 \end{bmatrix} u_n$ 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.

34. ajprincess Group Title

so there will not be any particular solutions right?

35. experimentX Group Title

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

36. experimentX Group Title

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

37. ajprincess Group Title

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

38. experimentX Group Title

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

39. ajprincess Group Title

oh k. Thanxx a lot.

40. experimentX Group Title

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