Which function to kill: Sine or Cos? I got an equation which was a solution to some familiar Differential Equation I am solving, the solution takes the form of:
$$V=Ce^{-ix}$$
but
$$Ce^{-ix}=A\cos(x)+B\sin(x)$$
so
$$V=A\cos(x)+B\sin(x)$$
Typically, evaluating the equation above involves killing either the $\cos$ term or $\sin$ term, or keeping both.
My question is, how do we know which function to kill or to keep?
 A: If the differential equation itself is unchanged by complex conjugation (i.e. all coefficients are real) then if $e^{-ix}$ is a solution then so is its complex conjugate $e^{ix}$.  (Here I'm assuming $x$ is real; if it's not then the foregoing sentence needs emendation.)
If it's a linear differential equation then $$Ae^{ix}+Be^{-ix} \tag 1$$ is a solution, for any two complex numbers $A$ and $B$.
Now let's suppose you want real solutions.  Then you have
\begin{align}
Ae^{ix}+Be^{-ix} & = A(\cos x+i\sin x)+B(\cos x-i\sin x) \\[6pt]
& = (A+B)\cos x + i(A-B)\sin x \\[6pt]
& = C\cos x + D\sin x. \tag 2
\end{align}
This is real if and only if
\begin{align}
A+B & = C \text{ is real, and} \\[6pt]
i(A-B) & = D \text{ is real.}
\end{align}
In order that $A+B$ be real, it is necessary and sufficient that the complex parts of $A$ and $B$ cancel out when added.
In order that $i(A-B)$ be real, it is necessary and sufficient that $A-B$ be a pure imaginary, so the real parts of $A$ and $B$ cancel out when subtracted.
This $A$ and $B$ must be complex conjugates of each other.  That is equivalent to $C$ and $D$ in $(2)$ above being real.
Either $(1)$ or $(2)$ gives the same set of solutions if $A,B,C,D$ are allowed to be arbitrary complex numbers.
If you want to know all solutions, then that's what you need.  If you want a particular solution satisfying specified initial conditions, then you need to find the values of $A$ and $B$, or of $C$ and $D$, that satisfy those conditions.
For example if $y(0)=1$ and $y'(0)=0$, then you need $C=1$ and $D=0$.
