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Trial Solutions of Particular Integrals?

  1. Apr 12, 2008 #1
    Hiya, I'm having trouble understanding particular integrals at the moment. From what I understand, you have to try trial solutions, unfortunately I've got no idea how to start? Google doesn't turn up anything .. I think I need to examine the equation, and somehow determine its form?

    For example this is what i'm currently trying to solve :

    Y" + 2y` +2y = cos x


    From my notes, if f(x) = A cos wx + A1 sin wx then y = a0 cos wx + a1 sin wx
    From this, I'm guessing A1=0? so y=a1 sin wx
    and y' is a1 w cos wx
    and y'' is -a1 w^2 wx

    And then I'm stuck, how do I now find the PI?

    ps. Does anyone have any good notes on the PI and relavent methods to determine what form its in?

    Sorry for the messy post, the working came to me as I was typing out the post!
    Thanks in advance!
  2. jcsd
  3. Apr 12, 2008 #2


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    Hi Davio! :smile:

    Don't guess!

    Don't take short-cuts ("more haste, less speed")! :frown:

    Learn the method in your notes, and follow it!

    Use a0 and a1, and you should get a solution! :smile:
  4. Apr 12, 2008 #3
    After a bit of thought ... and more deciphering of my notes (missing just one lecture makes everything so much more hard!!!), I've come up with:
    PI= 1/5 cos x + 2/5 sin x

    from me putting in the equations, and equating coefficients with A0=1 and A1=0 (on the premise that 0 sin x = 0).
    Putting that back into the original equation.. doesn't get me the right answer, I'll muse on it, thanks for the reply :D
  5. Apr 12, 2008 #4
    Hey Davio
    Have you guys learnt about D operators?
    I found that method to help immensely when dealing with linear ODE's.
    Good luck!

    Edit: They're not going to help us here with the particular integral since [tex]\cos x[/tex] is infinitely differentiable. D operators are still worth a look though :)
    Last edited: Apr 12, 2008
  6. Apr 12, 2008 #5
    D operators? Nope never heard of them :-D, I'll learn it after my exams, since I think they would be quite strict if I used something they didn't teach me to solve their questions :-p

    I now have PI= 1/10 cos x + 1/5 sin x, which equals to 1/2 cos x = 1 cos x ..
    Still not sure why? I've checked my working a few times, but I don't see any obvious errors.

    a0 (c-aw^2)+a1 . b . w =A0

    -a0 . b .w + a1 (c-aw^2) =A1

    With A1 =0 and A0 = 1
    and a=1 b=2 c=2

    a0=? a1=?
  7. Apr 12, 2008 #6
    Oh God, I was right the first time, thanks guys! :-p
  8. Apr 12, 2008 #7


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    Hi Davio! :smile:

    Your original solution, "PI= 1/5 cos x + 2/5 sin x", was correct.

    Just put it back into the equation, and it gives cosx. Try it again! :smile:

    Then I used your equations (with w = 1), and I got the same result, 1/5 cos x + 2/5 sin x.

    I don't understand how you keep making these mistakes. :confused:

    If you type out your whole working, I'll try to spot where you went wrong.

    ok, now you've got a PI, what about the general solution?

    (btw, I always use the D method … what method have you been taught?)
  9. Apr 13, 2008 #8
    Hey, yeah, I made a mistake the second time round, probabily due to me staring at the question too long, thanks for help though!
    I worked out the gen solution to be e^x (C cos x + D cos x) + 1/5 cos x + 2/5 sin x. Which in comparson to my friends working, is correct! :-D
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