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Math Help - Explanation requested

  1. #1
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    Explanation requested

    Given this equation

    \varphi''+\lambda\varphi=0

    \varphi_1(x)=\cos(x\sqrt{\lambda})

    \displaystyle\varphi_2(x)=\frac{\sin(x\sqrt{\lambd  a})}{\sqrt{\lambda}}

    The book goes from equation to solution without explanation.

    When I do the problem, I get:

    m^2+\lambda=0\Rightarrow m=\pm\sqrt{-\lambda}

    \displaystyle \exp(ix\pm\sqrt{\lambda})

    \exp(ix\sqrt{\lambda})=\cos(x\sqrt{\lambda})+\sin(  x\sqrt{\lambda})

    Why is sine divided by \sqrt{\lambda} in the 2nd solution, and why is the solution broken up?
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  2. #2
    A Plied Mathematician
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    Generally, it's six of one, a half-dozen of the other. If you get imaginary roots to the characteristic equation, you can either go with complex exponentials or with trig functions. If the quantities in the original DE have to be real (for example, if they correspond to measurements in a lab), then often you'd go with trig functions, or at least end up with them after, possibly, using exponentials to ease the computations. On the other hand, if you get real roots to the characteristic equation, you can either go with real exponentials, or with hyperbolic trig functions. They're entirely equivalent. I can't say I know why they chose one over the other.

    As to why they divided by the square root of lambda, I can't say I know why there, either. However, it is a perfectly valid thing to do: you're talking about the eigenfunctions of a linear operator. A scalar multiple of an eigenfunction is an eigenfunction. My guess is that that form of the solution makes something later on look simpler, but I could be way off on that.

    Incidentally, I think you should have

    e^{ix\sqrt{\lambda}}=\cos(x\sqrt{\lambda})+i\sin(x  \sqrt{\lambda}).

    Your i got poked out.
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    Generally, it's six of one, a half-dozen of the other. If you get imaginary roots to the characteristic equation, you can either go with complex exponentials or with trig functions. If the quantities in the original DE have to be real (for example, if they correspond to measurements in a lab), then often you'd go with trig functions, or at least end up with them after, possibly, using exponentials to ease the computations. On the other hand, if you get real roots to the characteristic equation, you can either go with real exponentials, or with hyperbolic trig functions. They're entirely equivalent. I can't say I know why they chose one over the other.

    As to why they divided by the square root of lambda, I can't say I know why there, either. However, it is a perfectly valid thing to do: you're talking about the eigenfunctions of a linear operator. A scalar multiple of an eigenfunction is an eigenfunction. My guess is that that form of the solution makes something later on look simpler, but I could be way off on that.

    Incidentally, I think you should have

    e^{ix\sqrt{\lambda}}=\cos(x\sqrt{\lambda})+i\sin(x  \sqrt{\lambda}).

    Your i got poked out.
    Here are the pages from the book:

    The problem starts at "Let us consider...."



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