Consider the following ODE, defined on the interval I = $\mathbb{R}$, $$\frac{d^3 y}{dx^3}+\frac{dy}{dx}=x$$ and its homogeneous differential equation, $$\frac{d^3 y}{dx^3}+\frac{dy}{dx}=0$$ whose characteristic polynomial yields the roots $\lambda_{1} = 0$, $\lambda_{2} = i$ and $\lambda_{3} = -i$, each with multiplicity 1. At it this point, there is a fundamental system of complex solutions, $(\phi_{1}, \phi_{2}, \phi_{3}) = (1, e^{ix}, e^{-ix})$ and a fundamental system of real solutions, $(\psi_{1}, \psi_{2}, \psi_{3}) = (1, cos(x), sin(x))$. Upon applying the undetermined coefficients method, I've come up with a particular solution for the original ODE, $\eta_{p}(x) = \frac{x^2}{2}$. Since the coefficients of the given ODE are all real numbers, what is the form of the general solution? Should it be $\psi (x) = \eta_{p} + c_{1} + c_{2}cos(x) +c_{3}sin(x)$ or rather $\phi (x) = \eta_{p} + c_{1} + c_{2}e^{ix} +c_{3}e^{-ix}$ ? And, in each case, are $c_{1},c_{2},c_{3} \in \mathbb{R}$ or $\mathbb{C}$? Thank you kindly.

  • $\begingroup$ @RobertThingum They both satisfy the original ODE, so I'm guessing they both work. $\endgroup$ – minplanck Nov 11 '18 at 0:41

Your solution is correct

The real solution is preferred because most of the time you are solving a real life problem and you are looking for a real solution.

The coefficients are real as well.

  • $\begingroup$ In this case, the constants $c_{1}, c_{2}, c_{3}$ (which are determined by the initial conditions), should, too, be real, am I right? $\endgroup$ – minplanck Nov 11 '18 at 0:43
  • $\begingroup$ Yes they are real as well $\endgroup$ – Mohammad Riazi-Kermani Nov 11 '18 at 0:48

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