 #1
 160
 14
Homework Statement:

Find the particular solution
y''y'+9y=3sin3t
My question is related to this and the homogeneous sol'n and the general sol'n
Relevant Equations:

possibly
[r=+a]>cosh(ax),sinh(ax)
or
[r=+bi]>cos(bx),sin(bx)
So to answer my question I need to reference another problem, I hope i won't get flagged for this... it is only to make a point about the way I am trying to approach this current problem.
the previous problem stated:
y''+2y'y=10
so first I am finding Y_(homogeneous) and going straight to the auxiliary equation:
after completing the square→(r+1)^2=2→r=1±√2 [since when r=±a, we get cosh(ax) and sinh(ax), the solution becomes...]
Y_(H)=e^((1±√2)t) →e^(t)⋅e^((±√2)t)→e^t[C1cosh((√2)t)+C2sinh((√2)t)] etc. etc.
My question: with the current problem y''y'+9y=3sin3t, when finding the homogeneous equation(solution?) after finding my r values which I will show now:
Aux: r^2r+9=0 and solving for r I get: r=(1/2)±(√35)/2 using this to find the homogeneous equation (shown 3 lines up with the other problem) am I using the same property of r, which is when r=±a we get cosh(ax) and sinh(ax)? or am I using the property that says when r=±√# we get ±i out of it?
Sorry if this question is confusing, I tried to explain my logic as much as possible.
I see similarities in both r equations:
r=1±√2 I know you use cosh and sinh with this one?
r=(1/2)±(√35)/2 If i don't use cosh and sinh with this one why not?
I've been looking at my textbook, the internet, and my notes and cannot why you would get "i" for some r's and cosh and sinh for others.
Thanks!
the previous problem stated:
y''+2y'y=10
so first I am finding Y_(homogeneous) and going straight to the auxiliary equation:
after completing the square→(r+1)^2=2→r=1±√2 [since when r=±a, we get cosh(ax) and sinh(ax), the solution becomes...]
Y_(H)=e^((1±√2)t) →e^(t)⋅e^((±√2)t)→e^t[C1cosh((√2)t)+C2sinh((√2)t)] etc. etc.
My question: with the current problem y''y'+9y=3sin3t, when finding the homogeneous equation(solution?) after finding my r values which I will show now:
Aux: r^2r+9=0 and solving for r I get: r=(1/2)±(√35)/2 using this to find the homogeneous equation (shown 3 lines up with the other problem) am I using the same property of r, which is when r=±a we get cosh(ax) and sinh(ax)? or am I using the property that says when r=±√# we get ±i out of it?
Sorry if this question is confusing, I tried to explain my logic as much as possible.
I see similarities in both r equations:
r=1±√2 I know you use cosh and sinh with this one?
r=(1/2)±(√35)/2 If i don't use cosh and sinh with this one why not?
I've been looking at my textbook, the internet, and my notes and cannot why you would get "i" for some r's and cosh and sinh for others.
Thanks!