y=C₁ˣ+C₂ˣ=e^(xln(C₁))+e^(xln(C₂)).
Compare this with y=Ae^(r₁x)+Be^(r₂x).
The DE resulting from eliminating A and B is:
y''-(r₁+r₂)y'+r₁r₂y=0. A, B, r₁, r₂ are constants.
So if A=B=1 and r₁=ln(C₁) and r₂=ln(C₂) then the DE is:
y''-ln(C₁C₂)y'+yln(C₁)ln(C₂)=0.
This DE eliminates A and B, arbitrary constants, and creates coefficient derived from C₁ and C₂.
Writing the De:
y''-ay'+by=0, C₁C₂=e^a and ln(C₁)ln(C₂)=b.
From this we get C₁=e^(a+√(a²-4b))/2, C₂=e^(a-√(a²-4b))/2.
Not sure if this is what the question was asking.