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Why does $C_1e^{jx}+C_2e^{\bar jx}, C_1,C_2\in \mathbb{C}, j=\frac{1}{2}+i\frac{\sqrt 3}{2}$ has the following result:

$$S_{H_2}(\mathbb{R})=C_1e^{-1/2x}\cos(\frac{\sqrt 3x}{2})+C_2e^{-1/2x}\sin(\frac{\sqrt 3x}{2}), C_1,C_2\in \mathbb{R^2}$$

And not:

$$S_{H_2}(\mathbb{R})=C_1e^{-1/2x-\frac{i\sqrt 3x}{2}}+C_2e^{-1/2x+\frac{i\sqrt 3x}{2}}, C_1,C_2\in \mathbb{R^2}$$

$$S_{H_2}(\mathbb{R})=C_1(e^{-1/2x}\cos(\frac{\sqrt 3x}{2})-ie^{-1/2x}\sin(\frac{\sqrt 3x}{2}))+C_2(e^{-1/2x}\cos(\frac{\sqrt 3x}{2})+ie^{-1/2x}\sin(\frac{\sqrt 3x}{2})), C_1,C_2\in \mathbb{R^2}$$

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I don't know what you mean with $S_{H_2}(\mathbb{R})$, but if it is something like taking the real part of a complex function, then it's clear what you have to do. Your last expression has the form \begin{equation} (C_1+C_2) e^{-1/2x} \cos \frac{\sqrt{3} x}{2} + i (C_2 - C_1)e^{-1/2x} \sin \frac{\sqrt{3} x}{2}. \end{equation} Introduce $\alpha = C_1 + C_2$ and $\beta = i (C_1 - C_2)$, then the solution has the required form. Note that this implies that the solution is real valued if and only if we can write $C_{1,2}$ as \begin{equation} C_1 = \frac{\alpha - i \beta}{2},\quad C_2 = \frac{\alpha + i \beta}{2}, \end{equation} i.e. if and only if $C_1 = \overline{C_2}$.

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