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The Helmholtz wave equation is given as :

$$\nabla^2 \vec E =\mu\epsilon \frac{\partial^2 \vec E}{\partial t^2}$$

Considering $\vec E=E_x(z) e^{j \omega t}$ the Helmholtz wave equation now takes the form

$$\frac{\partial^2 E_x}{\partial z^2} + \omega^2 \mu \epsilon E_x=0$$

I know the solution to this equation can take many forms.

But which solution is generally used?

I mean some books use $$E_x=(E_x^+ e^{-jkz}+E_x^- e^{jkz})e^{j \omega t} = E_x^+ e^{j(wt-kz)}+E_x^- e^{j(wt+kz)}$$

while some other books use $$E_x=E_{0x}\cos(wt-kz)$$ where $E_{0x}$ is some value obtained from initial conditions.

Am I at liberty to use any of these solutions considering what is applicable in the problem in front of me?

I see that for formulation equations of polarization the cosine solution is used and not the exponential solution.

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Your first solution refers to the two waves propagating in the opposite directions ($+z$ and $-z$), while the second one has a wave propagating in one direction only. If propagation direction is specifically stated in your problem, or you don't need two-directional propagation to illustrate a concept, you can use the 2nd one of course.

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  • $\begingroup$ The information as to which of the many solutions you should use is provided by boundary or initial condition data such as how are the wave being created and so on. $\endgroup$
    – mike stone
    Jun 6, 2021 at 13:54

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