# Using complex exponentials as solution of ODE

I'm having trouble wrapping my head around the following issue. My book solves a problem without using complex exponential solution like $C_1 e^{it}$ and using either $A \cos(t) + B \sin(t)$ or $A \cos(t+\phi)$ to a second-order "undamped" linear ODE. I understand that the solution can be purely real or complex. The trouble arises when I use the solution to this ODE inside another function which cubes this solution. Cubing $C_1 e^{it}$ results in another complex sinusoid with 3 times the frequency, which is a very different effect from cubing something like $A \cos(t+\phi)$

Here is my work. I've highlighted where I think things have gone awry.

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Your book suggestion is not to replace $A\cos t + B\sin t$ by $Ce^{it}$, but rather with $Ce^{it}+De^{-it}$. –  AD. Oct 17 '10 at 4:55
I am not sure what you mean. The real part of e^it is (e^it - e^-it)/2. Is that what you are mentioning? The book does not solve these problems with complex exponentials. –  Gus Oct 17 '10 at 6:07

If you cube $cos(t)$ you get $(3cos(t) + cos(3t))/4$ so the triple frequency shows up here, too.

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Right, so there are two frequency components. One at 1 and one at 3, whereas $e^{3it}$ contains only a single component so I'm still confused. –  Gus Oct 17 '10 at 5:40
But if you use $Ce^{it}+De^{−it}$ and cube it, you get terms in frequency 1 as well as 3. –  Ross Millikan Oct 17 '10 at 15:30
Yes, well that is because $Ce^{it}+De^{-it}$ is a real value. I was confused why things broke if I assumed that complex solutions existed. I think I have partially answered my own question. –  Gus Nov 8 '10 at 20:28
Why do you presume that $Ce^{it}+De^{-it}$ is a real value? It's true in certain special cases (e.g., C real, D=C), but not at all so in general (e.g., $e^{it}+2e^{-it}$ certainly isn't always real). –  Steven Stadnicki Nov 8 '10 at 21:21
@Steven Thanks for pointing that out. I assumed there was a a constraint on C and D since I am not sure why it would make sense to have a 4-dimensional family of solutions to a 2nd order ODE. At the end it looks like I get two ODEs in two complex variables: $2i\cdot\partial_TC+3C^2D = 0$ $2i\cdot\partial_TD+3CD^2 = 0$ –  Gus Nov 12 '10 at 6:39
The complex exponential solutions make sense for linear ODEs because you can get a characteristic polynomial since the exponential is the eigenfunction of the derivative operator. The reason why a complex solution times any constant is still a complex solution is due to the linearity in the linear ODE. If $x(t)$ is a solution and $i*y(t)$ is a solution then any linear combination is also a solution. Additivity and homogeneity are not properties of the solutions of a non-linear ODE and so it makes no sense to assume that the solution is of the form $e^{st}$ .