Solving an ODE with eulers formula.

[cragar]

If I have $y''+y'+2y=sin(x)+cos(x)$
can I just say $y=Ae^{ix}$
and then find y' and y'' and then plug them in and solve for A.
so I get that $A= \frac{1}{1+i}$
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!

If I have $y''+y'+2y=sin(x)+cos(x)$
can I just say $y=Ae^{ix}$

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.