Thursday, April 1, 2010

y''(t) + 4y'(t) + 5y(t) = 6sin(t)

The roots of the auxiliary/characteristic equation of
y''(t) + 4y'(t) + 5y(t) = 6sin(t)
are complex because 42 <>´1´
5.

The roots are t = -2 + i and -2 – i.
Then the complementary solution/function must be e-2t{Acos(t) + Bsin(t)}

which can be rewritten in the form of Ce-2tsin(ω0t+φ),
where ω0+φ are constants corresponding to the natural frequency and phase of the system (because they have been found by experiment to so correspond).

The inhomogeneous term 6sin(t) corresponds to a forcing oscillation, from which in this case the particular integral/solution can be shown to eventually (t ®¥) prevail over the more transient complementary solution/function because the 4y'(t) term is not a sufficient damping. Therefore the amplitude will eventually be the coefficient of the forcing function; which in your case is 6.

It is more customary to write the dimensionless ωt instead of the t that you have, where ω corresponds to the angular frequency in rad s-1. In your case ω = 1 rad s-1.

To explain more will require more algebra than I am prepared to write in an email, but I am sure you can find it in your textbook, now that you know what to look for. Otherwise make an appointment.

AB