Diff Eq Solution for E&M Example: Solving y'' and z'' with Author's Method

[hitmeoff]

Homework Statement

This is part of a solution to a E&M example from a book. I am in the middle of following the solution and I come across this system of diff eq's:$y''=\omega z'$
$z''=\omega (\frac{E}{B}-y')$

Homework Equations

Solution according to author is:

$y(t)=C_{1}\cos (\omega t)+C_{2}\sin (\omega t)+(\frac{E}{B})t+C_{3}$
$z(t)=C_{2}\cos (\omega t)-C_{1}\sin (\omega t)+C_{4}$

The Attempt at a Solution

So the author just says that this is easily solvable by differentiating the first and using the second to eliminate z''. Now I tried that, but correct me if I am wrong, this works out to :

$z''=\frac{y'''}{\omega}$
$y'''+\omega^{2}y''-\omega^{2}\frac{E}{B}=0$

So now do I solve this diff eq, to get y(t) and then use substitution to solve for z(t)?

 

[Dick]

Sure, that looks like the right strategy.