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This is a random question that has been bugging me. In first-year calculus we learned that the second-order linear homogeneous ODE with complex roots $a\pm ib$ to its characteristic equation , has a real-valued general solution of the form: $y(x)=e^{ax}\left(c_1cos(bx)+c_2sin(bx)\right)$.

To get to this real-valued general solution, some intermediate steps were performed on the original complex-valued general solution, $y(x)=Pe^{(a+ib)x}+Qe^{(a-ib)x}$.

I'm interested in learning why the real-valued case is of particular interest only. Is there a particular reason for doing so?

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    $\begingroup$ We live in the real world? $\endgroup$ – André Nicolas Sep 15 '14 at 4:52
  • $\begingroup$ Actually, in some cases, the complex case is preferred to the trig case. It's really just a matter of what you are doing. The trig case can be more accessible to first year students and usually makes more physical sense. The complex case can make computations a little easier $\endgroup$ – ClassicStyle Sep 15 '14 at 4:54
  • $\begingroup$ Thanks for the insight, TylerHG. Besides ease of computation, is there any real-world interpretation of a non-real solution? $\endgroup$ – Jenq Sep 15 '14 at 14:36
  • $\begingroup$ André: Thanks for your comment. Yes, it is obvious that real-valued solutions would make more sense in the real world that we live in. However, i was just curious: might there be an interpretation for the imaginary solutions? I was hoping someone on stackexchange might be able to offer some insights on that. $\endgroup$ – Jenq Sep 15 '14 at 14:40
  • $\begingroup$ Here's an example of real-world uses of complex exponentials instead of trigonometric functions for analysis of real systems. $\endgroup$ – Ruslan Oct 7 '14 at 9:19
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If the initial conditions $y(0)$ and $y'(0)$ are real numbers, then the solution $y(x)$ must be real-valued. And this situation is what you have in most applications.

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In quantum mechanics, a running plane wave is an example of a solution of Schrödinger's equation, which is necessarily complex. It still remains the solution when you solve time-independent Schrödinger's equation, i.e. when the ODE is real in itself. If you chose the solution as real, you'd instead get a standing wave.

So it's not always true that only real-valued solutions are of interest. Sometimes they are not the answer one looks for.

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