
#1
Apr112, 02:43 AM

P: 2,453

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. 



#2
Apr112, 04:45 AM

Sci Advisor
HW Helper
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P: 26,167

hi cragar!
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! ) 



#3
Apr112, 05:04 AM

P: 2,453

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




#4
Apr112, 05:10 AM

Sci Advisor
HW Helper
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P: 26,167

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) 



#5
Apr112, 05:16 AM

P: 2,453

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



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