## ajprincess 2 years ago Please help:) Solve the difference equation $$\Large y_{n+1}-4y_{n}=8$$

1. hba

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

2. ajprincess

The second one

3. ajprincess

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

@Callisto

5. ajprincess

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

6. Callisto

@experimentX Can you give a hand here?

7. experimentX

any info given on y_0 ?

8. cinar

yn=a4^n+b1^n

9. cinar

do you have initial cond.

10. ajprincess

no it is not given

11. cinar

it is done then..

12. cinar

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

13. ajprincess

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

15. cinar

is it clear..

16. cinar

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

17. experimentX

that looks pretty nice.

18. ajprincess

i dnt see that screenshot properly

19. experimentX

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

20. experimentX

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

21. cinar
22. ajprincess

I dnt get the second step?

23. cinar

multiply top eq. by -1 and add them

24. ajprincess

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

25. cinar

in order to get rid of 8

26. cinar

to make it hom. diff. eq..

27. cinar

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

28. ajprincess

Really sorry. I still dnt get the conversion part

29. experimentX

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

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

31. experimentX

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

33. experimentX

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

so there will not be any particular solutions right?

35. experimentX

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

36. experimentX

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

37. ajprincess

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

38. experimentX

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

39. ajprincess

oh k. Thanxx a lot.

40. experimentX

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