General solution using Integration

Question

I have a DE like this,

$$X'' - 4a^2X = 0$$

and its solution is

$$X = c_1 \cosh 2ax + c_2 \sinh 2ax$$

Now, I want to know in detail how would I get this solution. I need all the steps to get the solution which I mentioned here.

Answer 1

Let $X=e^{mx},X'' =?$

$m=\pm2a$

So, $X=Ae^{2ax}+Be^{-2ax}$

Now $\cosh(y)\pm\sinh(y)=?$