For the fundamental solution set S={e^x, e^(2x), e^(3x)}
can we construct a linear ODE with constant coefficients?

Question

For the fundamental solution set S={e^x, e^(2x), e^(3x)}
can we construct a linear ODE with constant coefficients?

anonymous · Answer

I have verified that the set if linearly independent. But after that I am not sure how to proceed.

turingtest · Answer

well I suppose the characteristic polynomial for such a linear ODE would have to be$$(r-1)(r-2)(r-3)=0$$multiply that out and that should tell you how to get your ODE

anonymous · Answer

I get the char. eqn as $$r ^{3}-6r ^{2}+11r-6$$
Do I replace the r with respective derivative of y? i.e. $$r ^{2}$$ with y" ?

turingtest · Answer

exactly

anonymous · Answer

in that case I get:$$y''' - 6y'' + 11y - 6 = 0$$
Now if I subtitute y=e^x in this equation, since e^x is a solution, the LHS doesn't come out to 0? Shouldn't it equate to 0?

turingtest · Answer

careful, where is your y' term?

anonymous · Answer

Aha right it should be y′′′−6y′′+11y'−6y=0

turingtest · Answer

right, I hope that works
haven't checked myself

anonymous · Answer

Yeah that works fine. Well thanks a lot!

turingtest · Answer

welcome :)