# IVP Forced Mechanical vibration

1. Oct 29, 2009

### jrsweet

1. The problem statement, all variables and given/known data

The solution to the Initial value problem, x''+2x'+5x=0, is the sum of the steady periodic solution x_sp and x_tr. Find both.

2. Relevant equations

3. The attempt at a solution
I already found x_sp ( the particular solution). It is (-44/533)cos(7t)+(14/533)sin(7t).

r^2+2r+5=0
-1=+- 2i
x_tr = e^(-t)(Acos(2t)+Bcos(2t))

x(t) = e^(-t)(Acos(2t)+Bsin(2t)) + (-44/533)cos(7t)+(14/533)sin(7t).
x'(t) = e^(-t)(-2Asin(2t)+2Bcos(2t)) - e^(-t)(Acos(2t)+Bsin(2t)) +
(308/533)sin(7t)+(98/533)cos(7t).

x(0)=1(A+0)-(44/553) ==> A = 44/533
x'(0)= 1(2B)-1(A)+(98/533) ==> B= -27/533

So, x_tr=(44/533)cos(2t)-(27/533)sin(2t)
this is not right however. Can someone see where I might have messed up?

Last edited: Oct 29, 2009
2. Oct 29, 2009

### LCKurtz

Where did the cos(7t) and sin(7t) come from? Did you fail to state the whole problem? What are the initial conditions? Is there supposed to be a non-homogeneous term? No wonder there are no replies.

3. Oct 30, 2009

### jrsweet

Yeah, sorry about that!! I was pretty tired when I wrote it I think.

x''+2x'+5x=4cos(7t), x(0)=x'(0)=0

Haha. I feel stupid now. Sorry again!

4. Oct 30, 2009

### LCKurtz

Your solution for x(t) satisfies both the equation and the initial conditions, so your solution is correct. I'm not sure what xtr means, but I'm guessing it means the transient part of the solution. That would be the part that fades away as t approaches infinity. It would be the terms with the e-t factors. The pure sine and cosine terms are the periodic part.

 On rechecking it I find A = 44/533 and B = -27/533. This gives the solution:

$$x(t)=e^{-t}(\frac {44}{533}\cos{2t}-\frac{27}{533}\cos{2t})-\frac{44}{533}\cos{7t}+\frac{14}{533}\sin{7t}$$

Last edited: Oct 30, 2009