<s>[INST] Second order IVP differential equation So the given material and equations I have are
$$(1):x=C_1\cos(t)+C_2\sin(t)$$
$$x \left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}$$
$$x'\left(\frac{\pi}{3}\right)=0$$
First I sub in the $\frac{\pi}{3}$ and the $\frac{\sqrt{3}}{2}$ into my first equation and get 
$$\frac{\sqrt{3}}{2}=\frac{1}{2}C_1+\frac{\sqrt{3}}{2}C_2$$
Then finding $x'$and subbing in the same I get
$$0=-\frac{\sqrt{3}}{2}C_1+\frac{1}{2}C_2$$
Now I am supposed to add the two equations together to find either $C_1$ or $C_2$ like so 
$$\frac{\sqrt{3}}{2}=\frac{1}{2}C_1+\frac{\sqrt{3}}{2}C_2$$
$$0=-\frac{\sqrt{3}}{2}C_1+\frac{1}{2}C_2$$
But neither of the terms cancel out and I'm not sure if i need to find $x''$ to proceed 
 [/INST] $$0=-\frac{\sqrt{3}}{2}C_1+\frac{1}{2}C_2$$
$$  \implies \sqrt{3}C_1=C_2$$
Plug that in first equation.
$$\frac{\sqrt{3}}{2}=\frac{1}{2}C_1+\frac{\sqrt{3}}{2}C_2$$
$$\frac{\sqrt{3}}{2}=\frac{1}{2}C_1+\frac{{3}}{2}C_1$$
$$C_1=\frac{\sqrt{3}}{4}$$
$$  \implies C_2=\sqrt{3}C_1=\frac {3} 4$$
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