IVP Forced Mechanical vibration

[jrsweet]

Homework Statement

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.

Homework Equations

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?

 

[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.

 

[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!

 

[LCKurtz]

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!

Your solution for x(t) satisfies both the equation and the initial conditions, so your solution is correct. I'm not sure what x_tr 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.

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

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