# Complexification of rsin(at)+rcos(at)

1. Mar 1, 2009

### marmot

Sorry for this dumb questions! its that i am just trying to find multiple ways of solving problems: Basically i have

y'' + 9y' = 648sin(9 t) + 648cos(9 t)

i dont know how can you complexify this, would it be p(D)y=648exp(9t)?

if so, what part do i extract, the imaginary or real? because it has both....

i can solve this with uknown coefficients but i am trying to look for efficient ways of doing this.

2. Mar 1, 2009

### tiny-tim

complexify?

Hi marmot!

Do you mean y'' + 9y' = i648sin(9 t) + 648cos(9 t)?

(and do you mean 3t?)

if so, yes, that's 648ei9t

3. Mar 1, 2009

### marmot

no, it lacks the imaginary number, its just like that. i am trying to complexify it so that the differential equation becomes relatively straightforward. the problem i encounter is that generally when i want to complexify something, its either sin(x) or cos(x) that is in the right of the equation, not both. so i am wondering how to complexify it and then, whether if i have to extract from the solution the real or imaginary or both parts.

thanks

4. Mar 1, 2009

### Ben Niehoff

You need both $e^{i\omega t}$ and its complex conjugate.

5. Mar 1, 2009

### marmoset

I would rewrite the right hand side in the form R*cos(9t+a), where R and a are two new constants, because then it is the real part of R*e^i(9t+a).

6. Mar 1, 2009

### marmot

thank you!

however marmoset

i am trying to do this

p(D)y=P(D)exp(at)

where p(D)=D^2+D

thus

the solution would be

y=exp(at)/P(a)

however, how can i do this when i have two unknown variables, which in your case, are R and a?

7. Mar 1, 2009

### marmoset

Last edited: Mar 2, 2009