If I have [itex] y''+y'+2y=sin(x)+cos(x) [/itex] can I just say [itex] y=Ae^{ix} [/itex] and then find y' and y'' and then plug them in and solve for A. so I get that [itex] A= \frac{1}{1+i} [/itex] then i multiply and divided by the complex conjugate. then I back substitute in Eulers formula. now since I have my original equation has both a real and an imaginary part. when I multiply out A times Eulers formula, I will take both real and imaginary parts. and I get that y=sin(x) and tested this and it works. But is what i did with Eulers formula ok.
hi cragar! not really for a particular solution for an RHS of sinx + cosx, you should use Ae^{ix} + Be^{-ix} (or Ccosx + Dsinx) in this case, you're lucky that B = 0 (but don't do it again! )
if i substitute what you said. Do I take the real or imaginary part in the end. Or do I just take it all.
you take it all but if you're expecting a real solution anyway, you might as well use Ccosx + Dsinx in the first place! (technically, C and D are both complex, but they'll come out real)