Solving an ODE with eulers formula.

cragar

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.

tiny-tim

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! )

cragar

if i substitute what you said. Do I take the real or imaginary part in the end. Or do I just take it all.

tiny-tim

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)

cragar

thanks for your answers. I like to push the limits of Eulers formula. And see what it can do.