Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

SECTION 17.3 APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS 0 351
If c = 15, we again have underdamping since the auxi liary equation has roots r = - lf ± fL:{f-i. The general solution is
x = e- 15 t/'2 [c1 cos(fL:{f-t) + c2 sin( 2flt) ], so - 0.1 = x (O) = c1 and 0 = x'(O) = fL:{f-c2 - lf c1 => c2 = - 10
3
77
.
Thus x - e-lSt/ 2 [-o 1 cos( 5 .fit) - ~ sin(!b(J_t)]
- · 2 10 v7 2 ·
For c = 20, we have equal roots r 1 = r2 = -10, so the oscillation is critically damped and the solution is
x = (c1 + c2t)e- 10 t_ Then -0.1 = x(O) = c 1 and 0 = x'(O) = -10c1 + c2 => c2 = -1, sox= ( -0.1- t)e- 101 .
If c = 25 the auxiliary equation has roots r1 = --,-5, r2 = ...:.20, so we have overdamping and the solution is
X= c1e-st + C2e- 20 t. Then -0.1 = x(O) = Cl + C2 and 0 = x'(O) = -5cl- 20c2 => C1 =- { 5
and C2 = fo•
If c = 30 we have roots r = -15 ± 5 v's, so the motion is
overdamped and the solution is x = c1 e( -IS+ 5 v'5 )t + c2e( - 15 -
5
v'5 )t.
0.02
c = 10
c = 15
Then -0.1 = x(O) = C1 + c2 and
O=x'(O) = (-15+5v's)c1+(-15-5v's)c2 =>
-5 - 3 y'5 d -5 + 3 v'5
c1 = 100 an c2 = 100 , so
_ ( - s- 3 fi) (- 15+ s v'S)t + ( -s ± 3 v'5) e< -1s - & fi)t
x- wo e 100 · ·
-0.11
9. The differential equation is mx" + kx = Fo coswot and wo "1- w = .jkfm. Here the auxiliary equation is mr 2 + k = 0
with roots ±.jklmi = ±wi so Xc(t) = c1 coswt + C2 sinwt. Since wo "1- w, try x 1 ,(t) = Acoswot + B sinwot.
Then we need (m) ( -w5) (A coswot + B sinwot) + k(Acoswot + Bsinwot) = Fo coswot or A(k- mw~) = Fo and
B(k - ~w5) = 0. Hence B = 0 and A = k Fo 2 = ( ;"0 2 ) since w2 = k. Thus the motion of the mass is given
· - mw 0 m w - w 0 m
11. From Equation 6, x(t) = f(t) + g(t) where f(t) = c1 coswt + c2 sinwt and g(t) = ( ;o 2 ) cosw 0 t. Then f
mw - w 0
is periodic, with period :.:r, and if w "1- wo, g is periodic with period :,~.
~ = !..
wo 1 J wo
' => a = bw where a and b are non-zero integers. Then
If :' 0
is a rational number, then we can say
'x(t +a·:;-)= f(t +a· 2 ;) + g(t +a· :.:r) = f(t) + g(t + :~ · :.:r) = f(t) + g(t + b - :,~) = j(t) + g(t) = x(t)
so x(t) is periodic.
13. Here the initial-value problem for the charge is Q" + 20Q' + 500Q = 12, Q(O) = Q' (0) = 0. Then
Qc(t) = e- lOt(c1 cos 20t + c2 sin 20t) and try Qp (t) = A => 500A = 12 or A = 1 ~ 5 .
The general solution is Q(t) = e-IOt(c1 cos 20t + c2 sin 20t) + 1 ~ 5 • But 0 = Q(O) = c1 + 1 ~ 6 and
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