Solução_Calculo_Stewart_6e

F.
SECTION 18.3 APPLICATIONS TX.10 OF SECOND-ORDER DIFFERENTIAL EQUATIONS ET SECTION 17.3 ¤ 315
9. The differential equation is mx 00 + kx = F 0 cos ω 0t and ω 0 6=ω = k/m. Here the auxiliary equation is mr 2 + k =0
with roots ± k/m i = ±ωi so x c (t) =c 1 cos ωt + c 2 sin ωt. Sinceω 0 6=ω,tryx p (t) =A cos ω 0 t + B sin ω 0 t.
Then we need (m) −ω 2 0 (A cos ω0 t + B sin ω 0 t)+k(A cos ω 0 t + B sin ω 0 t)=F 0 cos ω 0 t or A k − mω 2 0 = F0 and
B
k − mω 2 F 0
F 0
0 =0. Hence B =0and A = =
k − mω 2 0 m(ω 2 − ω 2 0) since ω2 = k . Thus the motion of the mass is given
m
by x(t) =c 1 cos ωt + c 2 sin ωt +
F 0
m(ω 2 − ω 2 0 ) cos ω 0t.
11. From Equation 6, x(t) =f(t)+g(t) where f(t) =c 1 cos ωt + c 2 sin ωt and g(t) =
F 0
m(ω 2 − ω 2 0 ) cos ω 0t. Thenf
is periodic, with period 2π ,andifω 6=ω ω 0, g is periodic with period 2π
ω 0
.If ω ω 0
is a rational number, then we can say
ω
ω 0
= a b
⇒ a = bω
ω 0
where a and b are non-zero integers. Then
x t + a · 2π ω
so x(t) is periodic.
= f
t + a · 2π
ω
+ g t + a · 2π
ω = f(t)+g
t + bω
ω 0
· 2π = f(t)+g t + b · 2π
ω
ω 0
= f(t)+g(t) =x(t)
13. Here the initial-value problem for the charge is Q 00 +20Q 0 +500Q =12, Q(0) = Q 0 (0) = 0. Then
Q c (t) =e −10t (c 1 cos 20t + c 2 sin 20t) and try Q p (t) =A ⇒ 500A =12or A = 3
125 .
The general solution is Q(t) =e −10t (c 1 cos 20t + c 2 sin 20t)+ 3
125 .But0=Q(0) = c 1 + 3
125 and
Q 0 (t) =I(t) =e −10t [(−10c 1 +20c 2) cos 20t +(−10c 2 − 20c 1)sin20t] but 0=Q 0 (0) = −10c 1 +20c 2. Thus the charge
is Q(t) =− 1
250 e−10t (6 cos 20t +3sin20t)+ 3
125 and the current is I(t) =e−10t 3
5
sin 20t.
15. As in Exercise 13, Q c (t) =e −10t (c 1 cos 20t + c 2 sin 20t) but E(t) =12sin10t so try
Q p(t) =A cos 10t + B sin 10t. Substituting into the differential equation gives
(−100A +200B +500A)cos10t +(−100B − 200A + 500B)sin10t =12sin10t
⇒
400A +200B =0and 400B − 200A =12. Thus A = − 3 , B = 3 and the general solution is
250 125
Q(t) =e −10t (c 1 cos 20t + c 2 sin 20t) − 3
3
cos 10t + sin 10t. But0=Q(0) = c 250 125 1 − 3 so c 250 1 = 3 . 250
Also Q 0 (t) = 3 25 sin 10t + 6 25 cos 10t + e−10t [(−10c 1 +20c 2) cos 20t +(−10c 2 − 20c 1)sin20t] and
0=Q 0 (0) = 6 − 10c 25 1 +20c 2 so c 2 = − 3 . Hence the charge is given by
500
Q(t) =e −10t 3
3
cos 20t − sin 20t − 3
3
cos 10t + sin 10t.
250 500 250 125
c1
17. x(t) =A cos(ωt + δ) ⇔ x(t) =A[cos ωt cos δ − sin ωt sin δ ] ⇔ x(t) =A
A cos ωt + c
2
A sin ωt where
cos δ = c 1/A and sin δ = −c 2/A ⇔ x(t) =c 1 cos ωt + c 2 sin ωt. [Notethatcos 2 δ +sin 2 δ =1 ⇒ c 2 1 + c 2 2 = A 2 .]