# Thread: hyperbolic sin/cos differential equation

1. ## hyperbolic sin/cos differential equation

Problem: Show that any function of the form:
$x = C_{1}\text{cosh}(\omega*t) + C_{2}\text{sinh}(\omega*t)$
satisfies the differential equation:
$x'' - \omega^2x = 0$

For x'' I have:
$x'' = C_{1}\text{cosh}(\omega*t) + C_{2}\text{sinh}(\omega*t)$

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

where do I go from here?

2. Originally Posted by sgcb
Problem: Show that any function of the form:

For x'' I have:
$x'' = C_{1}\text{cosh}(\omega*t) + C_{2}\text{sinh}(\omega*t)$
where do I go from here?
This is wrong. By chain rule, each time you take derivative with respect to t, you'll get one more factor of $\omega$ and so

$x'' = \omega^2 x$