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Solving an ODE with eulers formula.

by cragar
Tags: eulers, formula, solving
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cragar
#1
Apr1-12, 02:43 AM
P: 2,466
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.
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tiny-tim
#2
Apr1-12, 04:45 AM
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hi cragar!
Quote Quote by cragar View Post
If I have [itex] y''+y'+2y=sin(x)+cos(x) [/itex]
can I just say [itex] y=Ae^{ix} [/itex]
not really

for a particular solution for an RHS of sinx + cosx, you should use Aeix + Be-ix

(or Ccosx + Dsinx)
in this case, you're lucky that B = 0
(but don't do it again! )
cragar
#3
Apr1-12, 05:04 AM
P: 2,466
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
#4
Apr1-12, 05:10 AM
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Solving an ODE with eulers formula.

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
#5
Apr1-12, 05:16 AM
P: 2,466
thanks for your answers. I like to push the limits of Eulers formula. And see what it can do.


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