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The solution to oscillatory equations of motion can be solved using complex number terms, so it would look something like:

$y(t)=c_1e^{iat}+c_2e^{-iat} \tag{1}$

which is then rearranged applying Euler's identity:

$y(t)=(c_1+c_2)\cos(at)+i(c_1-c_2)\sin(at) \tag{2}$

where next, you can get rid of the complex part by redefining:

$c_2'=i(c_1-c_2) \tag{3}$

I've done this many times now, but I find the last step is somewhat strange. I know that the integration constants are supposed to be arbitrary constants, and I know that you can solve these ODE's without even using complex numbers. But doesn't equation (3) assume, that the newly defined $c_2'\in \mathbb{C}$ ?

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    $\begingroup$ Hint: $\begin{pmatrix}c_1'\\c_2'\end{pmatrix}=\begin{pmatrix}1&1\\i&-i\end{pmatrix}\begin{pmatrix}c_1\\c_2\end{pmatrix}$ and $\mathrm{det}\begin{pmatrix}1&1\\i&-i\end{pmatrix}=-2i\neq0$. $\endgroup$ – AccidentalFourierTransform Dec 9 '17 at 13:09
  • $\begingroup$ I'm supposed to understand that, but I skipped linear algebra for now. Thanks, I will try to figure that out! $\endgroup$ – lthz Dec 9 '17 at 13:14
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    $\begingroup$ Isn't it just that, for $y(t)$ to be real, the second term on the right-hand side of equation (1) must be the complex conjugate of the first? $\endgroup$ – Alfred Centauri Dec 9 '17 at 13:33
  • $\begingroup$ Sorry, I don't get it. For $y$ to be real, doesn't still $Im(z)$ have to be zero? In other words, wouldn't $c_2'$ always be zero then? $\endgroup$ – lthz Dec 9 '17 at 13:42
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    $\begingroup$ Might Mathematics be better suited for this math question? $\endgroup$ – Kyle Kanos Dec 9 '17 at 15:03
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Sorry, I don't get it. For y to be real, doesn't still Im(z) have to be zero? In other words, wouldn't c′2 always be zero then?

Write $c_1 = a_1 + ib_1$ and $c_2 = a_2 + ib_2$ where the a's and b's are real numbers. Then your equation (2) becomes

$$y(t) = (a_1 + ib_1 + a_2 + ib_2)\cos(at) + i(a_1 + ib_1 - a_2 - ib_2)\sin(at)$$

Now, see that $y(t)$ is real only if $a_2 = a_1$ and $b_2 = -b_1$. That is, $y(t)$ is real only if $c_2 = c_1^*$ and then your equation (1) becomes

$$y(t) = c_1e^{iat} + c_1^*e^{-iat} = c_1e^{iat} + \left(c_1e^{iat}\right)^* = 2\mathfrak{Re}\{c_1e^{iat}\} = 2a_1\cos(at) - 2b_1\sin(at)$$

Where I have used the fact that $\mathfrak{Re}\{z\} = \frac{1}{2}(z + z^*)$

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  • $\begingroup$ Oh hey, perfect, thank you! Going through it, I also noticed that it corresponds to the fact that you get complex conjugates as solutions to the characteristic equation. $\endgroup$ – lthz Dec 9 '17 at 16:52
  • $\begingroup$ Might I add that your Equation (1) seems to also be a way more meaningful way to provide the solution to these ODE. It may seem obvious, but people really only write these things down for beginners anyway. $\endgroup$ – lthz Dec 9 '17 at 16:57

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