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Math Help - Help with imaginary roots

  1. #1
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    Help with imaginary roots

    y''''+y=0
    The characteristic equation yields r^4=1
    The four roots of unity are 1,-1, i, and -i
    I get that y_{1}=e^x and y_{2}=e^{-x}, but my book says that y_{3}=e^{ix} reduces to y_{3}=\cos{x} because of Euler's equation. I can't figure out where the +i\sin{x} goes. The same with y_{4}=\sin{x}
    Any help?
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  2. #2
    A Plied Mathematician
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    Isn't the characteristic equation r^{4}+1=0, or r^{4}=-1? Then you'd be trying to find the fourth roots of -1.
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  3. #3
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    Yes, I typed it wrong. It should have read y''''-y=0. Thanks for looking.
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  4. #4
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    Quote Originally Posted by polarbear73 View Post
    y''''+y=0
    The characteristic equation yields r^4=1
    The four roots of unity are 1,-1, i, and -i
    I get that y_{1}=e^x and y_{2}=e^{-x}, but my book says that y_{3}=e^{ix} reduces to y_{3}=\cos{x} because of Euler's equation. I can't figure out where the +i\sin{x} goes. The same with y_{4}=\sin{x}
    Any help?
    I suspect you are misreading your text book. If i and -i are roots of the characteristic equation, then, yes, e^{ix} and e^{-ix} are solutions and the general solution will be of the form Ae^{ix}+ Be{-ix} (I am dropping the e^x and e^{-x} since they are not relevant to this point).

    Now, as you know, e^{ix}= cos(x)+ isin(x) and e^{-ix}= cos(x)- i sin(x) (because cos(x) is an even function and sin(x) is an odd function). That gives Ae^{ix}+ Be^{-ix}= A(cos(x)+ i sin(x))+ B(cos(x)- i sin(x))= (A+ B)cos(x)+ i(A- B)sin(x) and, defining C to be A+ B and D to be (A- B)i, this is Ccos(x)+D sin(x).

    That is, it is the e^{ix} and e^{-ix} together that give both trig functions. It is NOT the case that e^{ix} alone gives "cos(x)" and e^{-ix} alone gives "sin(x)".
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  5. #5
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    Hey- Thanks for your time. Unfortunately, I'm not misreading. And it's actually not a text book, it has the word "Dummies" in the title, and that might be part of the problem. I'm going to forge on and hope I can learn something. I'm going to see if I can arrive at the correct general solution to that DE.
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