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?
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 $D=0$.