Second Order Differential Equation with constant complex coefficients and real solutions what is the condition on $r_1$ and $r_2$ the roots of a second order homogeneous differential equation with constant complex coefficients so that it accepts real solutions. Given:
$$ay'' + by'+cy=0 \space\space\space\space\space\space a,b,c\in \Bbb C$$
$$\exists\lambda,\mu \in \Bbb C \space\space\space  \forall x : \space\space Im(y)=0$$
I wrote the case in another way I don't know if it helps
$$\exists\lambda,\mu\in \Bbb C  \space\space\space  \forall x : \space\space$$
$$e^{Re(r_1)}Re(\lambda)\sin(Im(r_1)x) + e^{Re(r_1)}Im(\lambda)\cos(Im(r_1)x) +e^{Re(r_2)}Re(\mu)\sin(Im(r_2)x) +e^{Re(r_2)}Im(\mu)\cos(Im(r_2)x) = 0$$
Since
$$y=\lambda e^{r_1x}+\mu e^{r_2x} \space\space\space\space \lambda,\mu,r_1,r_2 \in \Bbb C$$
And again I'm looking for a condition on $r_1$ and $r_2$ so that if and only if it's true (satisfied) the solution $y\space$ will accept real solutions (Other than $y=0$).
 A: For $r_n \in \Bbb C$, to maintain $\Im[y(x)] =0$

*

*the addends of $y(x)$ will be jointly either spiraling in to the origin, spiraling out from the origin, or maintaining circular paths about the origin.


*the addends must maintain the same radii and angular rate around the origin. The exponential makes sure that any mismatch in initial radius and hence mismatched radial growth between the 2 addends will make it impossible to keep $\Im[y(x)] =0$ for all $x$, no matter what the angular rates.


*The addends can follow opposite angular directions, forcing the addends to be complex conjugates, or they can follow the same angular direction, forcing the addends to be the same radius, but separated by $\pi$ radians.
So there are 2 cases to consider for the addends
Case 1:
$$\mu e^{r_2 x} = \overline{\lambda e^{r_1 x}}$$
thus
$$\mu = \overline{\lambda}$$
$$r_2 = \overline{r_1}$$
Case 2:
$$\mu e^{r_2 x} = \lambda e^{i\pi} e^{r_1 x} =-\lambda e^{r_1 x}$$
thus
$$\mu = -\lambda$$
$$r_2 = r_1$$
Since $a, b \in \Bbb C$, it is possible to have a double root for which $\Im(r) \ne 0$.
Case 2 is probably not so interesting, as it results in $y(x) = 0$.
