Ordinary Differential Equation, Beating and Oscillating $$y''+ 6y= \cos(2t)$$
[$\frac{d^2y}{dt^2} + 6y = \cos(2t)$]
How do I solve this equation, determine the frequency of the beats, determine the frequency of the rapid oscillations and use the information to draw a rough sketch?
I have already figured out that the eigenvalues are $\pm i\sqrt{6}$ and that the general solution, without damping, is
$$y(t) = k_1\cdot\cos(\sqrt{6}\,t) + k_2\cdot\sin(\sqrt{6}\,t),$$
but I don't know where to go from here. If anyone can help that would be much appreciated!
 A: When we are given an ordinary non-homogeneous linear differential equation we first solve the homogeneous equation $y'' + 6y = 0$ (as you have done) to find
$y = A \cos(\omega t) + B\sin(\omega t)$
where $\omega = \sqrt{6}$. 
To find the full solution a good method that almost always works is variation of the parameters (http://en.wikipedia.org/wiki/Variation_of_parameters), that is we seek a solution
$y = A(t) \cos(\omega t) + B(t) \sin(\omega t)$
We can follow then the standard recipe (see the wiki article for details) by calculate the determinant of the Wronskian $|W| = \omega$ and then find the two functions as
$A(t) = -\int \frac{\cos(2t)\cos(\omega t)dt}{\omega}$ 
$B(t) = \int \frac{\cos(2t)\sin(\omega t)dt}{\omega}$ 
For your problem however we can use a very simple method that often work for simple driving terms. We try a test-solution on the same form as the r.h.s. That is we put
$y = C\cos(a t)$
for some constant $C$ and $a$. Inserting this gives us
$C(6-a^2)\cos(a t) = \cos(2t)$
so for $a=2$ and $C = \frac{1}{2}$ we get the correct solution.
Thus the general solution is
$y = \frac{\cos(2t)}{2} + A \cos(\omega t) + B\sin(\omega t)$
where $A,B$ are determined by the initial conditions. Since you haven't given these, in the following I will assume that $B=0$ and $A=1/2$ just for the sake of simplicity. To find the beat frequency we can then use
$\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$
$\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)$
to get
$\cos(a+b) + \cos(a-b) = 2\cos(a)\cos(b)$
Now plug in $a+b = 2t$ and $a-b = \omega t$ i.e. $a = \frac{1}{2}(2 + \omega)t$ and $b = \frac{1}{2}(2 - \omega)t$ giving us
$y = \frac{\cos(2t) + \cos(\omega t)}{2} = \cos((1 + \omega/2)t)\cos((1 - \omega/2)t)$
The beat frequency (http://en.wikipedia.org/wiki/Beat_%28acoustics%29) can be read off as $f_{\rm beat} = f_1 - f_2 = \frac{\sqrt{6}-2}{2\pi}$ and the frequency of the rapid oscilations are $f_{\rm rapid} = f_1 + f_2 = \frac{\sqrt{6}+2}{2\pi}$
When the amplitudes of the two terms we added above aren't the same (which I assumed above by taking $A=1/2$) you can use the formula
$C\cos(2a) + D\cos(2b) = (C+D)\cos(a+b)\cos(a-b) - (C-D)\sin(a+b)\sin(a-b)$
In any case the beat frequency stays the same.
To draw the sketch: first draw the beating envelope i.e. $\pm \cos(t-\omega t/2)$. Now draw the rapid term $\cos(t+\omega t/2)$ but reduce the amplitude at each point so that this curve always stays inside the envelope you drew before. This will give you the solution. A plot can be found here https://www.wolframalpha.com/input/?i=plot+cos%28t+-+sqrt%286%29+t%2F2%29+cos%28t+%2B+sqrt%286%29+t%2F2%29%2C+cos%28t+%2B+sqrt%286%29+t%2F2%29%2Ccos%28t+-+sqrt%286%29+t%2F2%29%2C-cos%28t+-+sqrt%286%29+t%2F2%29
