Solução_Calculo_Stewart_6e

F.
398 ¤ CHAPTER 9 DIFFERENTIAL EQUATIONS
TX.10
23. Setting u = y 1−n , du
dx =(1− n) dy dy
y−n or
dx dx =
becomes un/(1−n)
1 − n
yn du
1 − n dx = un/(1−n)
1 − n
du
. Then the Bernoulli differential equation
dx
du
dx + P (x)u1/(1−n) = Q(x)u n/(1−n) or du +(1− n)P (x)u = Q(x)(1 − n).
dx
25. Here y 0 + 2 x y = y3
x ,son =3, P (x) = 2 2 x and Q(x) = 1 x .Settingu = 2 y−2 , u satisfies u 0 − 4u x = − 2 x . 2
Then I(x) =e (−4/x) dx = x −4 and u = x 4 − 2 2
x dx + C = x 4 6 5x + C = Cx 4 + 2
5 5x .
Thus, y = ± Cx 4 + 2 −1/2
.
5x
27. (a) 2 dI
dI
+10I =40or
dt dt +5I =20. Then the integrating factor is e 5 dt = e 5t . Multiplying the differential equation
by the integrating factor gives e 5t dI
dt +5Ie5t =20e 5t ⇒ (e 5t I) 0 =20e 5t ⇒
I(t) =e −5t 20e 5t dt + C =4+Ce −5t .But0=I(0) = 4 + C,soI(t) =4− 4e −5t .
(b) I(0.1) = 4 − 4e −0.5 ≈ 1.57 A
29. 5 dQ
dt +20Q =60with Q(0) = 0 C. Then the integrating factor is e 4 dt = e 4t , and multiplying the differential
31.
equation by the integrating factor gives e 4t dQ
dt +4e4t Q =12e 4t ⇒ (e 4t Q) 0 =12e 4t ⇒
Q(t) =e −4t 12e 4t dt + C =3+Ce −4t .But0=Q(0) = 3 + C so Q(t) =3(1− e −4t ) isthechargeattimet
and I = dQ/dt =12e −4t is the current at time t.
dP
dt + kP = kM,soI(t) =e kdt = e kt . Multiplying the differential equation
by I(t) gives e kt dP
dt + kPekt = kMe kt ⇒ (e kt P ) 0 = kMe kt ⇒
P (t) =e −kt kMe kt dt + C = M + Ce −kt , k>0. Furthermore,itis
reasonable to assume that 0 ≤ P (0) ≤ M,so−M ≤ C ≤ 0.
33. y(0) = 0 kg. Salt is added at a rate of 0.4 kg
5 L
=2 kg . Since solution is drained from the tank at a rate of
L min min
3 L/min,butsaltsolutionisaddedatarateof5 L/min, the tank, which starts out with 100 L of water, contains (100 + 2t) L
of liquid after t min. Thus, the salt concentration at time t is
y(t)
100 + 2t
kg
3 L
=
L min
3y
100 + 2t
y(t)
100 + 2t
kg
. Salt therefore leaves the tank at a rate of
L
kg
. Combining the rates at which salt enters and leaves the tank, we get
min