Complexification of rsin(at)+rcos(at)

[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 don't 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.

 

[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 648e^i9t

 

[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

 

[Ben Niehoff]

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

 

[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).

 

[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?

 

[marmoset]

R and a not unknown, they are determined by the right hand side of your equation. In general,

asin(x) + bcos(x) = Rcos(x+\alpha)

where R and alpha are determined by a and b. See this for how to find R and alpha:

http://www.mathcentre.ac.uk/resources/workbooks/mathcentre/web-rcostheta-alphaetc.pdf

In case you haven't got time to read that now, use

sin(t)+cos(t) = \sqrt{2} cos(t-\pi/4)