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Homework Help: Solving a recurrence equation

  1. Apr 8, 2008 #1
    Hi, I'm trying to solve this recurrence equation.

    1. The problem statement, all variables and given/known data
    Solve [tex]y[k+2]+y[k]=sin(k)[/tex].
    The answer is already given, it's
    [tex]y[k]=c_1sin(\frac{\pi}{2}k) + c_2cos(\frac{\pi}{2}k) + \frac{sin(k)+sin(k-2)}{2(1+cos(2))}[/tex]

    2. Relevant equations

    3. The attempt at a solution
    First I solve the homogeneous equation.
    The characteristic function is
    This has two complex roots [tex]r=i[/tex] and [tex]r=-i[/tex].
    Thus the general solution is
    If you change the notation for the complex numbers, then it gives
    [tex]y[k]=c_1(cos(\frac{\pi}{2}k)+isin(\frac{\pi}{2}k)) + c_2(cos(\frac{-\pi}{2}k)+isin(\frac{-\pi}{2}k))[/tex]
    After grouping together it gives
    [tex](c_1+c_2)cos(\frac{\pi}{2}k) + (c_1-c_2)isin(\frac{\pi}{2}k)[/tex]
    Apparently, the answer is without complex numbers so I guess they only keep the real part. Then it must be so that
    [tex]c_1-c_2=0\Leftrightarrow c_1=c_2[/tex]
    Then I get as solution
    which is not the same solution as the general solution in the answer. What did I do wrong?

    I also tried to find a particular solution for the non-homogeneous equation. I tried as a solution
    [tex]y[k]=\alpha sin(k)[/tex]
    with [tex]\alpha[/tex] the parameter to determine. I don't this is the right solution. Can anyone help me with this problem?


  2. jcsd
  3. Apr 8, 2008 #2
    They don't "only keep the real part". Plug the solution back into the homogeneous equation; the sine part falls out without imposing any constraint on c1 and c2.
  4. Apr 8, 2008 #3

    Thanks for your quick reply. I tried to do what you told me but when I plug it into the homogeneous equation, everything cancels out! I get 0=0. This is what I did:
    We have that
    I plug it into
    I get
    [tex]cos(\theta + \pi)=-cos(\theta)[/tex] and
    [tex]sin(\theta + \pi)=-sin(\theta)[/tex]
    I get that
    which makes
    How come?
  5. Apr 8, 2008 #4
    Actually, you are correct. That was a silly suggestion of me, because 0=0 simply confirms your solution satisfies the equation from which you derived the solution. I will edit this post if I come up with something.

    Edit: Wow I had been reading it wrong for a good 45 minutes, sorry! Consider complex coefficients and make the substitution d1=c1+c2 and d2=i(c1-c2).

    Can you find the particular solution from here?
    Last edited: Apr 8, 2008
  6. Apr 9, 2008 #5

    Yeah of course... just give different names as they are just parameters... Thanks a lot.
    I tried the substitution [tex]\alpha sin(k)[/tex] because there is [tex]sin(k)[/tex] on the right side of the equation. However, I can't find the right answer. This is what I have done:
    [tex]y[k]=\alpha sin(k)[/tex]
    Plug it into the equation:
    [tex]\alpha sin(k+2) + \alpha sin(k) = sin(k)[/tex]
    Then we get
    [tex]\alpha = \frac{sin (k)}{2sin(k+1)cos(1)}[/tex]
    by using the identity
    [tex]sin(\theta) + sin(\alpha) = 2sin(\frac{\theta + \alpha}{2})cos(\frac{\theta - \alpha}{2})[/tex]
    The answer for [tex]\alpha[/tex] is obviously not the right answer.
    Did I make the wrong substitution?
  7. Apr 10, 2008 #6
  8. Apr 10, 2008 #7


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    Science Advisor
    Homework Helper

    Hi yoran! :smile:

    Try [tex]y[k]=\alpha\,sin(k\,-\,1)[/tex] . :smile:
  9. Apr 13, 2008 #8
    Ok, I tried
    [tex]y[k]=\alpha \sin(k-1)[/tex].
    Then I get that
    [tex]\alpha = \frac{1}{2\cos 1}[/tex]
    So given the solution for the homogeneous equation and this solution, I have that
    [tex]y[k]=c_1\sin{\frac{\pi}{2}k} + c_2\cos{\frac{\pi}{2}k} + \frac{1}{2\cos{1}}\sin{k-1}[/tex]
    But the answer should be
    [tex]y[k]=c_1\sin{\frac{\pi}{2}k} + c_2\cos{\frac{\pi}{2}k} + \frac{\sin{k}+\sin(k-2)}{2(1+\cos{2})}[/tex]
    Where does it go wrong in my solution?
  10. Apr 13, 2008 #9


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    Science Advisor
    Homework Helper

    Hi yoran! :smile:
    :rofl: But they're the same! :rofl:​

    Standard trig equation: 1 + cos(2) = 2cos²(1).

    So [tex]\frac{\sin{k}+\sin(k-2)}{2(1+\cos{2})}[/tex] = … ? :smile:
  11. Apr 13, 2008 #10
    Hehe, of course... Thanks a lot!
    [tex]\frac{\sin{k}+\sin(k-2)}{2(1+\cos{2})}=\frac{2\sin(k-1)\cos(1)}{4\cos^2(1)}=\frac{\sin(k-1)}{2\cos 1}[/tex]
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