# Third Order DE Using Complex Exponential

1. Sep 23, 2010

### Phyzeeks

1. The problem statement, all variables and given/known data

find three independent solutions using complex exponentials, but express answer in real form.
d^3(f(t))/dt^3 - f(t) = 0

2. Relevant equations

3. The attempt at a solution
after taking the derivative of z = Ce^(rt) three times
I put it in the following form:

Ce^(rt)(r^3 -1) = 0

and solved for r:

r^3 = 1
1 = e^(i*0)
and so
r = e^(i*0) = 1

1 is one of the solutions right? but I don't understand how it's possible to find 3 independent solutions to that differential equation.

2. Sep 23, 2010

### Thaakisfox

What other complex numbers are there which satisfy r^3=1?

3. Sep 23, 2010

### Phyzeeks

is it e^(i*0) , e^(i*2pi/3) and e^(i*4pi/3)?

4. Sep 23, 2010

### vela

Staff Emeritus
Yes.

You could also find the other roots by factoring: r3-1 = (r-1)(r2+r+1)

5. Sep 23, 2010

### Phyzeeks

oh right, thank you very much for your help and also this is still in the complex exponential form right? I need to put them into real form, do i do this by e^ix = Re{cosx + isinx}?

6. Sep 23, 2010

### collinsmark

Gah! Don't do that.

If there are multiple solutions to a linear differential equation, then any linear combination (i.e. sum, subtraction, constant scaling) of those solutions is also a solution to the differential equation. For example if the differential equation is

$$\dddot f(t) = f(t)$$

and two of those solutions are g(t) and h(t), then (C1g(t) + C2h(t)) is also a solution.

(Hint: you're on the right track with sines and cosines. But just don't take the Real part all willy nilly. Sin(θ) can be expressed as a linear combination of complex exponentials [and nowhere do you explicitly need to take the "Re{}" of anything]. Cos(θ) can be expressed by a different linear combination of complex exponentials. Find these relationships, and you should be finish the problem without too much trouble.)

Last edited: Sep 23, 2010
7. Sep 23, 2010

### Phyzeeks

Oh so since the solutions are e^(i*0), e^(i*(3pi/2) and e^(i*(3pi/4) then i just take the linear combinations of the cos and sin by adding and subtracting the equations in the form e^ix = cosx + isinx,
then i would end up with three solutions in the form of cos and sines right?

8. Sep 24, 2010

### collinsmark

Go back to Vela's last post. That post about the roots of r are important.

With that, keep my last post in mind, and memorize the following:

$$\cos \theta = \frac{e^{i \theta} + e^{- i \theta} }{2}$$

$$\sin \theta = \frac{ e^{i \theta} - e^{-i \theta} }{2i}$$

These are are relationships worth memorizing, or at least recognizing, and at least keeping in your back pocket such that you can recognize when they come up. If you continue advanced physics, these relationships will come up quite a bit and it wouldn't be a bad idea to keep these relationships handy.

For this particular problem, these relationships will help with putting two of the three relationships in real form. For the third, go back to Vela's post involving solving for r.

[Edit you can check the above relationships noting that $e^{i \phi} = \cos \phi + i \sin \phi [/tex], along with noting that [itex] \cos (- \phi) = \cos \phi$ and $\sin (-\phi) = -\sin \phi$.]

Last edited: Sep 24, 2010