The solution of the differential equation 

$\frac{{\mathrm{d}}^{2}x}{\mathrm{d}{t}^2}+x=0;x\left(0\right)=1,x^{\mathrm{\prime }}\left(0\right)=0$

Consider the equation $m^2+1=0,m=\pm i$

Hence $x=C_1\cos t+C_2\sin t$ (see theory)

Now $x\left(0\right)=1\Rightarrow C_1=1$ and $x^{\mathrm{\prime }}=-C_1\sin t+C_2\cos t$ so

$C_2=0.$Hence $x=\cos t$ is the required solution which is a 

periodic function.

Alternate Solution:

We have $2\frac{\mathrm{d}x}{\mathrm{d}t}\frac{{\mathrm{d}}^{2}x}{\mathrm{d}{t}^2}=-2x\frac{\mathrm{d}x}{\mathrm{d}t}\Rightarrow {\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)}^2=-x^2+C$

Since $x^{\mathrm{\prime }}\left(0\right)=0$ and $x\left(0\right)=1$so $C=1$Hence $\frac{\mathrm{d}x}{\mathrm{d}t}=\pm \sqrt{1-x^2}$

$\begin{array}{l}\Rightarrow \frac{dx}{\sqrt{1-x^2}}=\pm t\\ \Rightarrow {\sin }^{-1} x=t+C_1\text{ or }{\cos }^{-1}x=t+C_2\\ x=\sin \left(t+C_1\right)\text{ or }x=\cos \left(t+C_2\right)\end{array}$

When $t=0,x=1$so $C_1=\pi /2$ and $C_2=0$

Hence $x=\cos t$.