For instance consider one arbitrary solution of a Differential equation:

$$x_1 = {C_1}'\,e^{-\gamma\,t + i\,\omega\,t} + {C_2}'\,e^{-\gamma\,t-i\,\omega\,t}$$

$\left[\textsf{this is the Solution of $\ddot{x}_1(t)+ 2\,\gamma\,\dot{x}_1(t)+{w_0}^2\,x_1(t) = 0$ where $\omega = \sqrt{{w_0}^2-\gamma^2}$}\right]$

I took it for granted you can only rewrite it like:

$$ \begin{align}&x_1 = ({C_1}'+{C_2}')\,\cos(\omega\,t) + i\,({C_1}'-{C_2}')\,\sin(\omega\,t) \\[12pt] &x_1= C_1\,\cos(\omega\,t)+C_2\sin(\omega\,t)\end{align}$$

Apparently it is possible to write instead $$x_1 = B_1\,\cos(\omega\,t+\varphi)$$

Is this due to some kind of addition theorems? I'm really bad at those, so don't right see the connection. The reason being I've seen this often recently I need some clarification.


Actually I found an attempt to the problem in my tattered documents:

Write $C_1 = c\,e^{i\,\varphi}$ and $C_2 = c\,e^{-i\,\varphi}$ (complex constant in polar form)

Hence the solution becomes: $$\begin{align} &x_1 = c\,e^{i\,\varphi}\,e^{-\gamma\,t + i\,\omega\,t} + c\,e^{-i\,\varphi}\,e^{-\gamma\,t-i\,\omega\,t} \\\\ &x_1 = c\,e^{-\gamma\,t}\,\left(\,e^{i\,(\varphi+\omega\,t)}+\,e^{-i\,(\varphi+\omega\,t)}\right) \\\\ &\text{importantly: $\cos(x) = \dfrac{e^{i\,x}+e^{-i\,x}}{ 2}$} \\\\ & \quad\Rightarrow \quad x_1 = 2\,c\,e^{-\gamma\,t}\,\cos(\varphi+\omega\,t) \\\\\\ &\text{Also $2\,c$ can be renamed to "$B_1$" e.g.} \end{align}$$

My only problem remaining is how $C_1$ and $C_2$ both have same magnitude $c$. So it seems to work only if $C_1$ and $C_2$ are complex conjugate.

  • 1
    $\begingroup$ Do you remember the formula for $\cos(A+B)?$ $\endgroup$ Jun 1 at 0:54
  • $\begingroup$ $\cos(A+B) = \cos(A)\,\cos(B)-\sin(A)\,\sin(B)$. There is no multiplication however? $\endgroup$
    – Leon
    Jun 1 at 10:05

Hint: $\varphi$ is such that $$\cos\varphi =\frac{C_1}{\sqrt{C_1^2+C_2^2}},\\\sin\varphi=\frac{C_2}{\sqrt{C_1^2+C_2^2}}$$

When $C_1,C_2$ are real, there is a real $\varphi.$

For complex $C_i,$ there is a complex $\varphi$ unless $C_1^2+C_2^2=0.$


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