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Thread: Help with imaginary roots

  1. #1
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    Help with imaginary roots

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

    Now, as you know, $\displaystyle e^{ix}= cos(x)+ isin(x)$ and $\displaystyle e^{-ix}= cos(x)- i sin(x)$ (because cos(x) is an even function and sin(x) is an odd function). That gives $\displaystyle Ae^{ix}+ Be^{-ix}= A(cos(x)+ i sin(x))+ B(cos(x)- i sin(x))= (A+ B)cos(x)+ i(A- B)sin(x)$ 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 $\displaystyle e^{ix}$ and $\displaystyle e^{-ix}$ together that give both trig functions. It is NOT the case that $\displaystyle e^{ix}$ alone gives "cos(x)" and $\displaystyle e^{-ix}$ 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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