LC circuit poor man theory
The solution to a differential equation involves arbitrary constants (\$c_1\$ and \$c_2\$
in this instance) whose value cannot be determined unless the initial conditions are
known.  It is not necessary that one of \$c_1\$ and \$c_2\$ be zero. Both could be
nonzero, and you will still observe a (single)
sinusoid of the same frequency. As @pjc50 cogently
observed, it is a question of phase (that is, the initial phase value). 
Note that if both \$a\$ and \$b\$ are nonzero, then the
sum \$a\cos(\omega t) + b\sin(\omega t)\$ is not the sum
of two sinusoids of different frequencies, but rather 
a single sinusoid of frequency \$\omega\$ rad/s 
of a different initial phase:

$$\begin{align*}
a\cos(\omega t) + b\sin(\omega t) 
&amp;= \sqrt{a^2+b^2}\left[\frac{a}{\sqrt{a^2+b^2}} \cos(\omega t)
+ \frac{b}{\sqrt{a^2+b^2}}\sin(\omega t)\right]\\
&amp;=  \left.\left.\sqrt{a^2+b^2}\right[\sin(\theta) \cos(\omega t)
+ \cos(\theta)\sin(\omega t)\right]\\
&amp;= \sqrt{a^2+b^2}\sin(\omega t + \theta)\\
&amp;= \sqrt{a^2+b^2}\cos(\omega t + (\theta - \pi/2))
\end{align*}$$
where \$\theta = \arctan\left(\frac{a}{b}\right)\$ and we have used
the identity \$\sin(x) = \cos(x-\pi/2)\$ in getting to that last line.