Originally Posted by
HallsofIvy So two solutions are of the form a+ bi and a- bi and the other two are ak+ bki and ak- bki for some constant k?
If two solutions to the characteristic equation are a+ bi and a- bi then two independent solutions are $\displaystyle e^{ax}cos(bx)$ and $\displaystyle ^{ax}sin(bx)$. The other solutions to the characteristic, ak+ bki and ak- bki, give $\displaystyle e^{akx}cos(bkx)$ and $\displaystyle e^{akx}sin(bkx)$.
(The fact that the second pair of solutions to the characteristic equation are proportional to the first pair is not really of importance. If one pair of solutions is $\displaystyle a_1+b_1i$, $\displaystyle a_1- b_1i$ and the other is l$\displaystyle a_2+ b_2i$ and $\displaystyle a_2- b_2i$ then the four independent solutions to the differential equation are $\displaystyle e^{a_1x}cos(b_1x)$, $\displaystyle e^{a_1x}sin(b_1x)$, $\displaystyle e^{a_2x}cos(b_2x)$, and $\displaystyle e^{a_2x}sin(b_2x)$.