0.2 Nonhomogeneous Linear Equations 19
b= 0 ,μ=ω:resonance. Now, sinceω=μ, the trial solution must be
revised to
up(t)=Atcos(μt)+Btsin(μt).Substitution into the differential equation and simple algebra giveA=0,B=
f 0 / 2 μ,or
up(t)= f^0
2 μtsin(μt).The general solution of the differential equation is
u(t)=f 0
2 μtsin(μt)+c^1 cos(μt)+c^2 sin(μt).(Remember thatb=0andω=μ.) The initial conditions givec 1 =c 2 =0, so
the solution of the initial value problem is
u(t)= f^0
2 μtsin(μt).The presence of the multipliertmeans that the amplitude of the oscillation is
increasing. This is the phenomenon of resonance.
b> 0 :damped motion. The ideas are straightforward applications of the
techniques developed earlier. The trial solution is a combination of cos(μt)
and sin(μt).Somewhatlesssimplealgebragives
up(t)=f^0(
(ω^2 −μ^2 )cos(μt)+μbsin(μt))
,
where =(ω^2 −μ^2 )^2 +μ^2 b^2 .Thegeneralsolutionofthedifferentialequation
may take different forms, depending on the relation betweenbandω.(See
Section 1.) Assuming the underdamped case holds, we have
u(t)=f^0(
(ω^2 −μ^2 )cos(μt)+μbsin(μt))
+e−bt/^2(
c 1 cos(γt)+c 2 sin(γt))
for the general solution of the differential equation. Here,γ=
√
ω^2 −(b/ 2 )^2
is real because we assumed underdamping.
Applying the initial conditions gives, after some nasty algebra,
c 1 =− f^0(
ω^2 −μ^2)
, c 2 =− f^0 γbω(^2) +μ 2
2.