To solve this second order, linear, homogeneous differential equation the unknown function $x(t)$, is written as a power series:
\[
\begin{eqnarray}
x(t) &=& \sum_{n=0}^{\infty} a_n t^n \\
\Big( x(t) &=& a_0 + a_1 t + a_2 t^2 + a_3 t^3 + ... \Big)
\end{eqnarray}
\]
The first and second time derivatives of function $x(t)$ thus write:
\[
\begin{eqnarray}
\dot{x}(t) &=& \sum_{n=1}^{\infty} n a_n t^{n-1} \\
\Big( \dot{x}(t) &=& a_1 + 2a_2 t + 3a_3 t^2 + ... \Big) \\
\ddot{x}(t) &=& \sum_{n=2}^{\infty} n(n-1)a_n t^{n-2} \\
\Big( \ddot{x}(t) &=& 2a_2 + 6a_3 t + ... \Big)
\end{eqnarray}
\]
Using these expressions for $x(t)$ and $\ddot{x}(t)$ in the equation of motion we get:
\[
\sum_{n=2}^{\infty} n(n-1)a_n t^{n-2} + \omega_0^2 \sum_{n=0}^{\infty} a_n t^n = 0
\]
The next step is to keep only one sum ($\sum$)...
\[
\sum_{n=0}^{\infty}
\Big[ (n+2)(n+1)a_{n+2} + \omega_0^2 a_n \Big] t^n = 0
\]
Each coefficients should be equal to $0$ for the power series to equal $0$, so that:
\[
(n+2)(n+1)a_{n+2} + \omega_0^2 a_n = 0 \quad \forall n \in N \quad (n = 0,1,2,3,...)
\]
i.e.:
\[
a_{n+2} = - \frac{a_n \omega_0^2}{(n+2)(n+1)} \quad \forall n \in N \quad (n = 0,1,2,3,...)
\]
This last recurrence expression together with the initial conditions (chosen as the conditions at time $t=0$) given below allow to solve the equation of motion by finding the value for each coefficient $a_n$:
\[
\begin{eqnarray}
a_0 &=& x(0) \qquad \textrm{this is the position at the origin of time}\\
a_1 &=& \dot{x}(0) \qquad \textrm{this is the speed at the origin of time} \\
a_2 &=& - \frac{\omega_0^2}{2 \times 1} a_0 \\
a_3 &=& - \frac{\omega_0^2}{3 \times 2} a_1 \\
a_4 &=& - \frac{\omega_0^2}{4 \times 3} a_2 = \frac{\omega_0^4}{4!} a_0 \\
a_5 &=& - \frac{\omega_0^2}{5 \times 4} a_3 = \frac{\omega_0^4}{5!} a_1 \\
\end{eqnarray}
\]
and so on.
Finally, $x(t)$ is found to be equal to:
\[
\begin{eqnarray}
x(t) &=& \sum_{n=0}^{\infty} a_n t^n \\
x(t) &=& a_0 + a_1 t + a_2 t^2 + a_3 t^3 + a_4 t^4 + a_5 t^5 + ... \\
x(t) &=& a_0 + a_1 t
- \frac{\omega_0^2}{2!} a_0 t^2
- \frac{\omega_0^2}{3!} a_1 t^3
+ \frac{\omega_0^4}{4!} a_0 t^4
+ \frac{\omega_0^4}{5!} a_1 t^5
\end{eqnarray}
\]
Grouping the $a_0$ and $a_1$ terms, we get:
\[
\begin{eqnarray}
x(t) &=& a_0 \Big[ 1 - \frac{\omega_0^2 t^2}{2!} + \frac{\omega_0^4 t^4}{4!} - ... \Big]
+ a_1 \Big[ t - \frac{\omega_0^2 t^3}{3!} + \frac{\omega_0^4 t^5}{5!} - ... \Big] \\
x(t) &=& a_0 \Big[ 1 - \frac{(\omega_0 t)^2}{2!} + \frac{(\omega_0 t)^4}{4!} - ... \Big]
+ \frac{a_1}{\omega_0} \Big[ t - \frac{(\omega_0 t)^3}{3!} + \frac{(\omega_0 t)^5}{5!} - ... \Big] \\
x(t) &=& a_0 \cos (\omega_0 t) + \frac{a_1}{\omega_0} \sin (\omega_0 t)
\end{eqnarray}
\]
as the expressions between brackets corresponds to the definitions of the cosine and sine functions.
The solution of equation of motion is thus:
\[
x(t) = A\cos (\omega_0 t) + B \sin (\omega_0 t)
\]
where $A$ and $B$ are real quantities defined from the initial conditions.
The solution can also be written as:
\[
x(t) = M\cos(\omega_0 t - \alpha) \\
\]
with
\[
\begin{eqnarray}
M &=& \sqrt{A^2+B^2} \\
\tan \alpha &=& \frac{B}{A} \\
A &=& M \cos \alpha \\
B &=& M \sin \alpha
\end{eqnarray}
\]
In a complex form:
\[
x(t) = Ce^{i\omega_O t} + De^{-i\omega_0 t}
\]
where
\[
\begin{eqnarray}
C &=& A/2 + B/(2i) \\
D &=& A/2 - B/(2i)
\end{eqnarray}
\]