Differential Equation - Determining frequency of beats/rapid oscillations

[cse63146]

Homework Statement

y'' + 6y = 2cos3t

a)Determine frequency of the beats
b)Determine frequency of the rapid oscillations
c)Use the information from parts a) and b) to give a rough sketch of a typical solution

Homework Equations

The Attempt at a Solution

Not sure how to do the first 2 parts. I know the natural period is $$\frac{2 \pi}{B}$$ and natural frequency is $$\frac{B}{2 \pi}$$

 

[Mark44]

The characteristic equation of the homogeneous equation is r² + 6 = 0, so its solutions are r = +/- i$\sqrt{6}$. This means that
$$y_h = Ae^{i\sqrt{6}t} + Be^{-i\sqrt{6}t}$$
It's more convenient to write a different linear combination of these to get
$$y_h = c_1cos(\sqrt{6}t) + c_2sin(\sqrt{6}t)$$

For the nonhomogeneous equation, look for a solution y_p = Acos(3t) + Bsin(3t).

The general solution will be y_h + y_p.

For a, you'll have to figure out how the functions in the homogeneous solution interact with what I'm pretty sure will be the one particular solution function, and how often the waves of the two periodic parts reinforce each other (the beats).

For b, you need to find the rapid oscillation frequency. You have in essence a long wave (low frequency) imposed on a short wave (high frequency).

Hope that helps. It's getting late where I am, so I'm going to turn in.

 

[HallsofIvy]

Homework Statement

y'' + 6y = 2cos3t

a)Determine frequency of the beats
b)Determine frequency of the rapid oscillations
c)Use the information from parts a) and b) to give a rough sketch of a typical solution

Homework Equations

The Attempt at a Solution

Not sure how to do the first 2 parts. I know the natural period is $$\frac{2 \pi}{B}$$ and natural frequency is $$\frac{B}{2 \pi}$$

Since there is no "B" in the statement of your problem, that makes no sense! Mark44 has already pointed out that the general solution to the associated homogeneous equation is $A cos(\sqrt{6}t)+ B sin(\sqrt{6}t)$. That tells you that the natural period is $2\pi/\sqrt{6}$.

Trig identity: cos(s+ t)= cos(s)cos(t)- sin(s)sin(t) so if you have a formula of the form Acos(2\pi t/\sqrt{6})+ Bcos(3t) you can write that, by careful manipulation of A and B, as $cos((2\pi/\sqrt{6}+ 3)t)$.

 

[cse63146]

It's not a B. It's supposed to be the greek uppercase Beta.

as in - $A cos(Bt)+ B sin(Bt)$. It's how my textbook/prof write it.

In this case B (Beta) = $\sqrt{6}$