Second order differential equation.(Damped oscillation)

[mr pizzle]

Hi could do with a little help with this question please!

The question
A damped oscillation with no external forces can be modeled by the equation:

$\frac{d^2x}{dt^2}$+2$\frac{dx}{dt}$+2x=0

Where x mm is amplitude of the oscillation at time seconds. The initial amplitude of the oscillation is 3mm (i.e. when t=0) and the intial velocity is 5mm/s.
Solve the equation for x.

Ok! so far I have;

$\frac{d^2x}{dt^2}$+2$\frac{dx}{dt}$+2x=0

$m^2+2m+2=0$

$\frac{-b±\sqrt{b^2-4ac}}{2a}$

$\frac{-2±\sqrt{2^2-4x1x2}}{2x1}$

$\frac{-2±\sqrt{-4}}{2}$

$\frac{-2}{2}$±$\frac{\sqrt{-4}}{2}$

m=-1±j$\frac{\sqrt{4}}{2}$

Equating this with m=α±jβ

Gives α=-1 β=1

Substituting into the general solution (complex roots)

x=e^-1t[Acos(t)+Bsin(t)]

x=3 t=0

3=e^-1x0[Acos(0)+Bsin(0)]

3=A

This is the part we are unsure of!

$\frac{dx}{dt}$=e^-1t[-Asin(t)+Bcos(t)]-1e^-1t[Acos(t)+Bsin(t)]

By the product rule;

=e^-1t[B-A]cos(t)-[A-B]sin(t)

x=0 $\frac{dx}{dt}$=5

5=e⁰[B-A]cos(0)-[A-B]sin(0)

5=[B-A]

We know A=3 so therefore B=8

x=e^-1t[3cos(t)+8sin(t)]

Is this correct? If not any help would be greatly appreciated.

 

[HallsofIvy]

Yes, that is correct. I'm not sure why you had any question about it.

 

[mr pizzle]

Just wasn't sure. Followed it from a book from a similar question and don't fully understand it! The question was different enough to have doubts!

 

[bigfooted]

when you have the solution, you can substitute it into your differential equation to see if you end up with 0.