1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    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


    User Avatar
    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]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook