# Why do we seek real-valued solutions to second-order linear homogeneous ODEs when the characteristic equation has complex roots?

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?

• We live in the real world? – André Nicolas Sep 15 '14 at 4:52
• 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 – ClassicStyle Sep 15 '14 at 4:54
• Thanks for the insight, TylerHG. Besides ease of computation, is there any real-world interpretation of a non-real solution? – Jenq Sep 15 '14 at 14:36
• 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. – Jenq Sep 15 '14 at 14:40
• Here's an example of real-world uses of complex exponentials instead of trigonometric functions for analysis of real systems. – Ruslan Oct 7 '14 at 9:19

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.