- #1

- 325

- 0

a) Solve the initial value problem: [tex] x'' + x = 2cos(\omega t) ; x(0) = 0 ; x'(0) = 0 ; \omega != 1 [/tex]

b) repeat but with \omega = 1

c) use the trigonometric identity: [tex] cos(\alpha) - cos(\beta) = -2sin(\frac{\alpha + \beta}{2})sin(\frac{\alpha - \beta}{2}) [/tex]

i.e. as a sinusoidal function oscillating inside an envelope A(t).

Determine the function A(t). Show that it is a slowly oscillating function if ! is close

to 1.

## The Attempt at a Solution

a) [tex] \lambda^2 + 1 = 0 [/tex] Solving for general inhomogeneous.

[tex] \lambda = \frac{+}{} i [/tex]

Therefore: [tex] x_{gi} = Ae^{it} + Be^{-it} [/tex] Where A and B are arbitrary constants.

Now solving for homogeneous solution, take form: [tex] x_p = asin(\omega t) + bcos(\omega t) [/tex]

where a and b are arbitrary constants, therefore, [tex] x'_p = a\omega cos(\omega t) - b\omega sin(\omega t) [/tex]

therefore [tex] x''_p = -a\omega^2 sin(\omega t) - b\omega^2 cos(\omega t) [/tex]

Subbing these values in for the IVP yields:

[tex] -a\omega^2 sin(\omega t) - b\omega^2 cos(\omega t) + x_p = asin(\omega t) + bcos(\omega t) = 2cos(\omega t) [/tex]

By equating co-efficients; [tex] b - b\omega^2 = 2 [/tex]

and [tex] a - a\omega^2 = 0 [/tex]

Therefore the homogeneous solution = [tex] \frac{2cos(\omega t)}{1-\omega^2} - \frac{2\omega^2 cos(\omega t)}{1-\omega^2} [/tex]

Common factor taken out [tex] \frac{2cos(\omega t)}{1-\omega^2}(1-\omega^2) = 2cos(\omega t) [/tex]

General homogeneous solution = [tex] Ae^{it} + Be^{-it} + 2cos(\omega t) [/tex] By using the initial conditions:

[tex] A + B + 2 = 0 [/tex]

and [tex] Ai - Bi = 0 [/tex]

This implies: [tex] A = B = -1 [/tex]

**Therefore particular homogeneous solution**= [tex] x(t) = -e^{it} - e^{-it} + 2cos(\omega t) [/tex]

b) This is the first part that confuses me, before I factored out [tex] 1 - \omega^2 [/tex] and cancelled, I could not have omega = 1 because it would be dividing by zero, but since I cancelled it allows me to. Did I make an error in cancelling? It would seem logically that I did due to the nature of the question. So I know my answer is going to either be undefined, or:

[tex] -e^{-it} - e^{it} [/tex]

c) Once again I'm not entirely sure how to approach this, I can see that my solution:

[tex] \frac{2cos(\omega t)}{1-\omega^2} - \frac{2\omega^2 cos(\omega t)}{1-\omega^2} [/tex]

loosely fits the trig identity given, but it has constant coefficients and I'm not sure what to do with them. Also the terms inside the cosine function are identical, so it's not really:

[tex] cos(\alpha) + cos(\beta) [/tex] unless alpha = beta which is possible, but I am doubting it due to the way the question is worded.

So I assume that either it's correct, and I just need someone to read over it so that I know if I'm on the right track, or waaay off, or it's full of mistakes.

Thanks in advance,