Solução_Calculo_Stewart_6e

F.
314 ¤ CHAPTER 18 SECOND-ORDER DIFFERENTIAL EQUATIONS ETX.10
CHAPTER 17
18.3 Applications of Second-Order Differential Equations ET 17.3
1. By Hooke’s Law k(0.25) = 25 so k =100is the spring constant and the differential equation is 5x 00 + 100x =0.
The auxiliary equation is 5r 2 + 100 = 0 with roots r = ±2 √ 5 i, so the general solution to the differential equation is
x(t) =c 1 cos 2 √ 5 t + c 2 sin 2 √ 5 t . We are given that x(0) = 0.35 ⇒ c 1 =0.35 and x 0 (0) = 0 ⇒
2 √ 5 c 2 =0 ⇒ c 2 =0, so the position of the mass after t seconds is x(t) =0.35 cos 2 √ 5 t .
3. k(0.5) = 6 or k =12is the spring constant, so the initial-value problem is 2x 00 +14x 0 +12x =0, x(0) = 1, x 0 (0) = 0.
The general solution is x(t) =c 1e −6t + c 2e −t .But1=x(0) = c 1 + c 2 and 0=x 0 (0) = −6c 1 − c 2. Thus the position is
given by x(t) =− 1 5 e−6t + 6 5 e−t .
5. For critical damping we need c 2 − 4mk =0or m = c 2 /(4k) =14 2 /(4 · 12) = 49
12 kg.
7. We are given m =1, k = 100, x(0) = −0.1 and x 0 (0) = 0. From (3), the differential equation is d2 x
dt 2
with auxiliary equation r 2 + cr +100=0.
If c =10,wehavetwocomplexrootsr = −5 ± 5 √ 3 i, so the motion is underdamped and the solution is
+ c dx + 100x =0
dt
x = e −5t c 1 cos 5 √ 3 t + c 2 sin 5 √ 3 t .Then−0.1 =x(0) = c 1 and 0=x 0 (0) = 5 √ 3 c 2 − 5c 1 ⇒ c 2 = − 1
10 √ , 3
so x = e −5t −0.1cos 5 √ 3 t − 1
10 √ sin 5 √ 3 t .
3
If c =15, we again have underdamping since the auxiliary equation has roots r = − 15
x = e
c −15t/2 5 1 cos
√
7
t 5
2
+ c 2 sin
√
7
t 2
Thus x = e
−0.1cos
−15t/2 5 √
7
t 2
− 3
10 √ sin 5 √ 7
t 7 2
.
± 5 √ 7
2 2
,so−0.1 =x (0) = c 1 and 0=x 0 (0) = 5 √ 7
2
c2 −
15
2
For c =20, we have equal roots r 1 = r 2 = −10, so the oscillation is critically damped and the solution is
i. The general solution is
c1 ⇒ c2 = −
3
10 √ 7 .
x =(c 1 + c 2 t)e −10t .Then−0.1 =x(0) = c 1 and 0=x 0 (0) = −10c 1 + c 2 ⇒ c 2 = −1,sox =(−0.1 − t)e −10t .
If c =25the auxiliary equation has roots r 1 = −5, r 2 = −20, so we have overdamping and the solution is
x = c 1 e −5t + c 2 e −20t .Then−0.1 =x(0) = c 1 + c 2 and 0=x 0 (0) = −5c 1 − 20c 2 ⇒ c 1 = − 2 15 and c 2 = 1
30 ,
so x = − 2 15 e−5t + 1 30 e−20t .
If c =30we have roots r = −15 ± 5 √ 5, so the motion is
overdamped and the solution is x = c 1 e (−15 + 5 √ 5 )t + c 2 e (−15 − 5 √ 5 )t .
Then −0.1 =x(0) = c 1 + c 2 and
0=x 0 (0) = −15 + 5 √ 5 c 1 + −15 − 5 √ 5 c 2 ⇒
c 1 = −5 − 3 √ 5
and c
100 2 = −5+3√ 5
x =
e (−15 + 5 √
5)t +
−5 − 3 √ 5
100
100
,so
−5+3 √ 5
100
e (−15 − 5 √ 5)t .