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D.E. Simple annihilator method question

  1. Mar 14, 2013 #1
    Does the sin^2() matter when finding the annihilator of 5xe^(4x)sin^2(pi*x)? The 5 is ignored but I have no notes or references on what to do if sin or cos has a power. Do I ignore it?
     
  2. jcsd
  3. Mar 14, 2013 #2

    LCKurtz

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    Ignore it at your peril. Express your ##sin^2(\pi x)## in terms of a double angle.
     
  4. Mar 15, 2013 #3
    ok. I did that and now I have another oddity. 5/2*x*(e^x)^4-5/2*x*(e^x)^4*cos(2pi*x) I don't know what to do with the 4th power that e^x is raised to..
     
  5. Mar 15, 2013 #4

    HallsofIvy

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    You seem to be trying to do differential equations without the prerequisites. You can "ignore" the "5" because the derivative of a constant is 0. That is NOT true for a function of x. You should have learned that in your first Calculus class. Further [itex](e^x)^2= e^{2x}[/itex]. You should have learned that in algebra or pre-Calculus.
     
  6. Mar 15, 2013 #5
    I've taken the pre-reqs. I just promptly forgot most of it after each test :)
     
  7. Mar 15, 2013 #6

    LCKurtz

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    So when when you graduate in Engineering or whatever your major is, are you going to explain to your future employers that you don't remember anything you learned?
     
  8. Mar 15, 2013 #7
    you don't use 95% of what you learned anyway. School is mostly a waste of time. The most valuable stuff isn't taught in school and the intelligence and creativity you need to succeed is primarily genetic. I've met and worked with so many people without degrees that are far more intelligent and competent that others I've met with E.E. degrees.
     
  9. Mar 15, 2013 #8
    To get back on track, I got 5/2*x*e^(4x)-5/2*x*e^(4x)*cos(2pi*x) for which the annihilator seems to be: (D-4)^2*[D^2-8D+16+pi^2]

    The weird part is the pi^2. It doesn't seem right.
     
  10. Mar 15, 2013 #9

    LCKurtz

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    It's close but not quite right. Remember ##(D-a)^2+b^2## annihilates ##e^{ax}\cos(bx)## and ##e^{ax}\sin(bx))## so you do get a constant in there, but your ##\pi^2## isn't quite right. Also you need to account for the fact that that exponential times a cosine is multiplied by ##x##.
     
  11. Mar 15, 2013 #10

    Mark44

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    Really? On what do you base this opinion?
    Then why are you bothering to spend the money and time to be in school?
    Such as?
    I agree with you here, but at some point the intelligence and creativity that most people have isn't enough, and they have to put in some effort.
    I don't doubt that there are a few people with EE degrees who aren't the sharpest tools in the drawer, and some people without those degrees might be more competent, but with everything being equal, if I had to hire someone for an EE-type slot, I'd be looking at folks with the degree and with some experience in the real world.

    I feel somewhat sorry for you, that you have such a negative attitude about your education. If you were in some major other than engineering, your approach of studying for the exam and nothing more probably wouldn't be so troublesome, but you've picked a major with a ton of math in it. Forgetting most of what you learned right after a test isn't a recipe for success, that I can see.
     
  12. Mar 15, 2013 #11
    From the notes I have:
    To annihilate: x^n*e^(ax)*cos or sin(b*x)
    Use: [D^2-2aD+a^2+b^2]

    When using that, I get [D^2-8D+16+pi^2]. Is there something wrong with the formula?
     
  13. Mar 15, 2013 #12

    LCKurtz

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    Please quote from the post to which you are replying for context. Remember the argument of your cosine is ##2\pi x##. And factors of ##x## multiplying it affect the annihilator.
     
  14. Mar 16, 2013 #13
    ok. So it would be [D^2-8D+16+(2pi)^2] ?
     
  15. Mar 16, 2013 #14

    LCKurtz

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    If you can't be bothered to quote the post to which you are replying I will quit responding. And why did you ignore what I have highlighted in red?
     
  16. Mar 16, 2013 #15
    Ok. Any factors of x multiplying it affect the annihilator. That doesn't tell me anything. That's just like walking by and saying, "that's wrong" .

    Our instructor only gave us 3 scenarios and the only one that matches is this:
    To annihilate: x^n*e^(ax)*cos or sin(b*x)
    You use: [D^2-2aD+a^2+b^2]

    If that's wrong, what should it be?
     
  17. Mar 16, 2013 #16

    LCKurtz

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    Well, it's obviously wrong isn't it, because if you change ##n## on the function you would expect it to make a difference in the annihilator. Look here:

    http://www.utdallas.edu/dept/abp/PDF_Files/DE_Folder/Annihilator_Method.pdf [Broken]
     
    Last edited by a moderator: May 6, 2017
  18. Mar 16, 2013 #17
    I previously left out n+1. So:
    To annihilate: x^n*e^(ax)*cos or sin(b*x)
    The instructor's notes say to use: [D^2-2aD+a^2+b^2]^(n+1)

    The pdf has something different:
    To annihilate: x^n*e^(ax)*cos or sin(b*x)
    Use: [(D–α)^2+b^2]^(n+1)

    Which one should I use? Or are they both equivalent?
     
    Last edited by a moderator: May 6, 2017
  19. Mar 16, 2013 #18

    vela

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    Really?
     
  20. Mar 18, 2013 #19
    So to confirm:
    If you're given:
    5/2*x*e^(4x)-5/2*x*e^(4x)*cos(2pi*x)
    The annihilator is:
    (D-4)^2*[D^2-8D+16+4pi^2]^2
     
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