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The question

A damped oscillation with no external forces can be modelled by the equation:

[itex]\frac{d^2x}{dt^2}[/itex]+2[itex]\frac{dx}{dt}[/itex]+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;

[itex]\frac{d^2x}{dt^2}[/itex]+2[itex]\frac{dx}{dt}[/itex]+2x=0

[itex] m^2+2m+2=0 [/itex]

[itex]\frac{-b±\sqrt{b^2-4ac}}{2a}[/itex]

[itex]\frac{-2±\sqrt{2^2-4x1x2}}{2x1}[/itex]

[itex]\frac{-2±\sqrt{-4}}{2}[/itex]

[itex]\frac{-2}{2}[/itex]±[itex]\frac{\sqrt{-4}}{2}[/itex]

m=-1±j[itex]\frac{\sqrt{4}}{2}[/itex]

Equating this with m=α±jβ

Gives α=-1 β=1

Substituting in to 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!

[itex]\frac{dx}{dt}[/itex]=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 [itex]\frac{dx}{dt}[/itex]=5

5=e

^{0}[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.