Problem: Show that any function of the form:

$\displaystyle x = C_{1}\text{cosh}(\omega*t) + C_{2}\text{sinh}(\omega*t)$

satisfies the differential equation:

$\displaystyle x'' - \omega^2x = 0$

For x'' I have:

$\displaystyle x'' = C_{1}\text{cosh}(\omega*t) + C_{2}\text{sinh}(\omega*t)$

so...

$\displaystyle C_{1}\text{cosh}(\omega*t) + C_{2}\text{sinh}(\omega*t) - \omega^2[C_{1}\text{cosh}(\omega*t) + C_{2}\text{sinh}(\omega*t)] = 0$

where do I go from here?