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ajprincess

  • 2 years ago

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

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  1. hba
    • 2 years ago
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    Is it \[y^{n+1} \ or \ y_{n+1}\]

  2. ajprincess
    • 2 years ago
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    The second one

  3. ajprincess
    • 2 years ago
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    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
    • 2 years ago
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    @Callisto

  5. ajprincess
    • 2 years ago
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    *therefore, \(y^{(p)}=c_1(4)^n\)

  6. Callisto
    • 2 years ago
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    @experimentX Can you give a hand here?

  7. experimentX
    • 2 years ago
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    any info given on y_0 ?

  8. cinar
    • 2 years ago
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    yn=a4^n+b1^n

  9. cinar
    • 2 years ago
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    do you have initial cond.

  10. ajprincess
    • 2 years ago
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    no it is not given

  11. cinar
    • 2 years ago
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    it is done then..

  12. cinar
    • 2 years ago
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    it is not hom. diff. eq, but we make hom. diff. eq..

  13. ajprincess
    • 2 years ago
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    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
    • 2 years ago
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  15. cinar
    • 2 years ago
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    is it clear..

  16. cinar
    • 2 years ago
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    this is the solution ..\[\Large y_{n}=a4^{n}+b1^{n}\]

  17. experimentX
    • 2 years ago
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    that looks pretty nice.

  18. ajprincess
    • 2 years ago
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    i dnt see that screenshot properly

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

  20. experimentX
    • 2 years ago
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    \[ y_{n+2} = 5 y_{n+1} -4y_n \\ y_{n+1} = y_{n+1} \\ \] \[u_{n+1} = \begin{bmatrix} 5 & -4\\ 1 & 0 \end{bmatrix} u_n\]

  21. cinar
    • 2 years ago
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    http://imageshack.us/photo/my-images/836/fsedfe.png/

  22. ajprincess
    • 2 years ago
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    I dnt get the second step?

  23. cinar
    • 2 years ago
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    multiply top eq. by -1 and add them

  24. ajprincess
    • 2 years ago
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    I mean y do u make it y_n+2-4y_n+1=8

  25. cinar
    • 2 years ago
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    in order to get rid of 8

  26. cinar
    • 2 years ago
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    to make it hom. diff. eq..

  27. cinar
    • 2 years ago
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    but I dont know how to find a and b if initial condition is not given..

  28. ajprincess
    • 2 years ago
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    Really sorry. I still dnt get the conversion part

  29. experimentX
    • 2 years ago
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    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
    • 2 years ago
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    How do u get y_n+2-5y_n+1+4y_n=0

  31. experimentX
    • 2 years ago
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    \[ y_0 = c_1 + c_2 \\ y_1 = 8 + 4y_0\\ y_1 = 4c_1 +c_2 \\ \] --------------------- 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
    • 2 years ago
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  33. experimentX
    • 2 years ago
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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.

  34. ajprincess
    • 2 years ago
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    so there will not be any particular solutions right?

  35. experimentX
    • 2 years ago
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    no ... unless you have initial value i.e y_0

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

  37. ajprincess
    • 2 years ago
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    Ya. I checked it out. Bt got confused seeing 32 and all that:(

  38. experimentX
    • 2 years ago
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    just put c_1 = k + 8/3 and c_2 = -8/3 .. rest is just manipulation of numbers by WA

  39. ajprincess
    • 2 years ago
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    oh k. Thanxx a lot.

  40. experimentX
    • 2 years ago
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    no probs at all. using this method, you can solve for n-th Fibonacci number.

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