# Underdamped harmonic motion

1. Jun 4, 2012

### giacomh

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

Find a particular solution to the differential equation using undetermined coefficients.

x$^{''}$+5x$^{'}$+4x=2sin2t
x(0)=1
x'(0)=0

I know that the equation is underdamped because c<W$_{0}$, and that W$_{0}$=2.

I know that the particular solution is x(t)=acos(2t)+bsin(2t)=Asin(2t)/(W$_{0}$$^{2}$-w$^{2}$)

Plugging the initial conditions in to x(t) and x$^{'}$(t) gives me a=1 and b=0.

x(t)=$\frac{8}{5}$e$^{-t}$-e(2/5)$^{-4t}$

How did he get this final answer? My book seems to set the solutions up differently, my professor hasn't been returning my e-mail, and my exam is tomorrow morning! Any help would be appreciated!

Last edited: Jun 4, 2012
2. Jun 4, 2012

### clamtrox

Do you remember how to solve differential equations? First you should solve the homogeneous equation,
$$x'' + 5x' + 4x = 0$$
Easiest way in this case is to try to find solutions of the form $x = e^{rt}$ where r is a constant.

3. Jun 4, 2012

### LCKurtz

Something is wrong here. The two sides of your professor's answer aren't equal to each other, and neither one could possibly give a sine function when you plug it into the equation. You should find the general solution of the homogeneous equation first, then look for a particular solution of the NH equation of the form $y_p=A\cos(2t)+B\sin(2t)$.

4. Jun 4, 2012

### vela

Staff Emeritus
What do you mean by c? The system is actually overdamped. This is clear because the roots to the characteristic equation are real.

5. Jun 4, 2012

### giacomh

Sorry in advanced for poor formatting. My internet is down and I have to use my phone. I agree that the equation is overdamped, I had under written in my notes. I found -4/3((e^-t)+(e^-4t)) for the homogeneous equation. I have no idea how my professor got his answer. Also, there was a 0 sin 2t at the end of his answer if that helps. I had to edit the original answer of his that I posted. There was an equal sign where there should have been a minus.

6. Jun 4, 2012

### giacomh

I just got -2/5(cos2t+sin2t) as my particular solution

7. Jun 4, 2012

### LCKurtz

That is incorrect. You should get $x_p=-\frac 1 5 \cos(2t)$. If you add that to the general solution of the homogeneous equation $x_h = Ae^{-4t} + Be^{-t}$ you will have the general solution to the DE. When you apply the initial conditions to figure out $A$ and $B$ you should find the unique solution to the initial value problem is$$x=-(\frac 2 5)e^{-4t}+(\frac 8 5)e^{-t}-\frac 2 5 \cos(2t)$$Your professor just left off the last term.