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Exercícios resolvidos Equações diferenciais elementares 9 ed Boyce e DiPrima

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t
u ′ − (1− 1
4t2
)u .
Set x1 = u and x2 = u ′. It follows that x ′1 = x2 and
x ′2 = u
′′ = −1
t
u ′ − (1− 1
4t2
)u .
We obtain the system of equations
x ′1 = x2
x ′2 = −(1−
1
4t2
)x1 − 1
t
x2 .
294 Chapter 7. Systems of First Order Linear Equations
5. Let x1 = u and x2 = u ′; then u ′′ = x′2 . In terms of the new variables, we have
x′2 + 0.25x2 + 4x1 = 2 cos 3t
with the initial conditions x1(0) = 1 and x2(0) = −2 . The equivalent first order
system is
x′1 = x2
x′2 = −4x1 − 0.25x2 + 2 cos 3t
with the above initial conditions.
7.(a) Solving the first equation for x2 , we have x2 = x ′1 + 2x1 . Substitution into
the second equation results in
(x ′1 + 2x1)
′ = x1 − 2(x ′1 + 2x1) .
That is, x ′′1 + 4x
′
1 + 3x1 = 0 . The resulting equation is a second order differential
equation with constant coefficients. The general solution is
x1(t) = c1e−t + c2e−3t .
With x2 given in terms of x1 , it follows that
x2(t) = c1e−t − c2e−3t .
(b) Imposing the specified initial conditions, we obtain
c1 + c2 = 2
c1 − c2 = 3 ,
with solution c1 = 5/2 and c2 = −1/2 . Hence
x1(t) =
5
2
e−t − 1
2
e−3t and x2(t) =
5
2
e−t +
1
2
e−3t .
(c)
7.1 295
10.(a) Solving the first equation for x2 , we obtain x2 = (x1 − x ′1)/2 . Substitution
into the second equation results in
(x1 − x ′1)′/2 = 3x1 − 2(x1 − x ′1) .
Rearranging the terms, the single differential equation for x1 is
x ′′1 + 3x
′
1 + 2x1 = 0 .
(b) The general solution is
x1(t) = c1e−t + c2e−2t .
With x2 given in terms of x1 , it follows that
x2(t) = c1e−t +
3
2
c2e
−2t .
Invoking the specified initial conditions, c1 = −7 and c2 = 6 . Hence
x1(t) = −7e−t + 6 e−2t and x2(t) = −7e−t + 9 e−3t .
(c)
11.(a) Solving the first equation for x2 , we have x2 = x ′1/2 . Substitution into the
second equation results in
x ′′1 /2 = −2x1 .
The resulting equation is x ′′1 + 4x1 = 0 .
(b) The general solution is
x1(t) = c1 cos 2t+ c2 sin 2t .
With x2 given in terms of x1 , it follows that
x2(t) = −c1 sin 2t+ c2 cos 2t .
Imposing the specified initial conditions, we obtain c1 = 3 and c2 = 4 . Hence
x1(t) = 3 cos 2t+ 4 sin 2t and x2(t) = −3 sin 2t+ 4 cos 2t .
296 Chapter 7. Systems of First Order Linear Equations
(c)
12.(a) Solving the first equation for x2 , we obtain x2 = x ′1/2 + x1/4 . Substitution
into the second equation results in
x ′′1 /2 + x
′
1/4 = −2x1 − (x ′1/2 + x1/4)/2 .
Rearranging the terms, the single differential equation for x1 is
x ′′1 + x
′
1 +
17
4
x1 = 0 .
(b) The general solution is
x1(t) = e−t/2 [c1 cos 2t+ c2 sin 2t] .
With x2 given in terms of x1 , it follows that
x2(t) = e−t/2 [−c1 cos 2t+ c2 sin 2t] .
Imposing the specified initial conditions, we obtain c1 = −2 and c2 = 2 . Hence
x1(t) = e−t/2 [−2 cos 2t+ 2 sin 2t] and x2(t) = e−t/2 [2 cos 2t+ 2 sin 2t] .
(c)
7.1 297
13. Solving the first equation for V , we obtain V = L · I ′. Substitution into the
second equation results in
L · I ′′ = − I
C
− L
RC
I ′ .
Rearranging the terms, the single differential equation for I is
LRC · I ′′ + L · I ′ +R · I = 0 .
15. Direct substitution results in
(c1x1(t) + c2x2(t))′ = p11(t) [c1x1(t) + c2x2(t)] + p12(t) [c1y1(t) + c2y2(t)]
(c1y1(t) + c2y2(t))′ = p21(t) [c1x1(t) + c2x2(t)] + p22(t) [c1y1(t) + c2y2(t)] .
Expanding the left-hand-side of the first equation,
c1x
′
1(t) + c2x
′
2(t) = c1 [p11(t)x1(t) + p12(t)y1(t)]+
+ c2 [p11(t)x2(t) + p12(t)y2(t)] .
Repeat with the second equation to show that the system of ODEs is identically
satisfied.
16. Based on the hypothesis,
x ′1(t) = p11(t)x1(t) + p12(t)y1(t) + g1(t)
x ′2(t) = p11(t)x2(t) + p12(t)y2(t) + g1(t) .
Subtracting the two equations,
x ′1(t)− x ′2(t) = p11(t) [x ′1(t)− x ′2(t)] + p12(t) [y ′1(t)− y ′2(t)] .
Similarly,
y ′1(t)− y ′2(t) = p21(t) [x ′1(t)− x ′2(t)] + p22(t) [y ′1(t)− y ′2(t)] .
Hence the difference of the two solutions satisfies the homogeneous ODE.
17. For rectilinear motion in one dimension, Newton’s second law can be stated as∑
F = mx ′′.
The resisting force exerted by a linear spring is given by Fs = k δ , in which δ is
the displacement of the end of a spring from its equilibrium configuration. Hence,
with 0 < x1 < x2 , the first two springs are in tension, and the last spring is in
compression. The sum of the spring forces on m1 is
F 1s = −k1x1 − k2(x2 − x1) .
The total force on m1 is∑
F 1 = −k1x1 + k2(x2 − x1) + F1(t) .
298 Chapter 7. Systems of First Order Linear Equations
Similarly, the total force on m2 is∑
F 2 = −k2(x2 − x1)− k3x2 + F2(t) .
18. One of the ways to transform the system is to assign the variables
y1 = x1, y2 = x ′1, y3 = x2, y4 = x
′
2.
Before proceeding, note that
x ′′1 =
1
m1
[−(k1 + k2)x1 + k2x2 + F1(t)]
x ′′2 =
1
m2
[k2x1 − (k2 + k3)x2 + F2(t)] .
Differentiating the new variables, we obtain the system of four first order equations
y ′1 = y2
y ′2 =
1
m1
[−(k1 + k2)y1 + k2y3 + F1(t)]
y ′3 = y4
y ′4 =
1
m2
[k2y1 − (k2 + k3)y3 + F2(t)] .
19.(a) Taking a clockwise loop around each of the paths, it is easy to see that
voltage drops are given by V1 − V2 = 0 , and V2 − V3 = 0 .
(b) Consider the right node. The current in is given by I1 + I2 . The current leaving
the node is −I3 . Hence the current passing through the node is (I1 + I2)− (−I3).
Based on Kirchhoff’s first law, I1 + I2 + I3 = 0 .
(c) In the capacitor,
C V ′1 = I1.
In the resistor,
V2 = RI2 .
In the inductor,
LI ′3 = V3 .
(d) Based on part (a), V3 = V2 = V1. Based on part (b),
C V ′1 +
1
R
V2 + I3 = 0 .
It follows that
C V ′1 = −
1
R
V1 − I3 and LI ′3 = V1.
7.1 299
21. Let I1, I2, I3,and I4 be the current through the resistors, inductor, and capac-
itor, respectively. Assign V1, V2, V3,and V4 as the respective voltage drops. Based
on Kirchhoff’s second law, the net voltage drops, around each loop, satisfy
V1 + V3 + V4 = 0, V1 + V3 + V2 = 0 and V4 − V2 = 0 .
Applying Kirchhoff’s first law to the upper-right node,
I3 − (I2 + I4) = 0 .
Likewise, in the remaining nodes,
I1 − I3 = 0 and I2 + I4 − I1 = 0 .
That is,
V4 − V2 = 0, V1 + V3 + V4 = 0 and I2 + I4 − I3 = 0 .
Using the current-voltage relations,
V1 = R1I1, V2 = R2I2, L I ′3 = V3, C V
′
4 = I4.
Combining these equations,
R1I3 + LI ′3 + V4 = 0 and C V
′
4 = I3 −
V4
R2
.
Now set I3 = I and V4 = V , to obtain the system of equations
LI ′ = −R1I − V and C V ′ = I − V
R2
.
23.(a)
Let Q1(t) and Q2(t) be the amount of salt in the respective tanks at time t . Note
that the volume of each tank remains constant. Based on conservation of mass, the
rate of increase of salt, in any given tank, is given by
rate of increase = rate in − rate out.
For Tank 1, the rate of salt flowing into Tank 1 is
rin =
[
q1
oz
gal
] [
3
gal
min
]
+
[
Q2
100
oz
gal
] [
1
gal
min
]
= 3 q1 +
Q2
100
oz
min
.
300 Chapter 7. Systems of First Order Linear Equations
The rate at which salt flow out of Tank 1 is
rout =
[
Q1
60
oz
gal
] [
4
gal
min
]
=
Q1
15
oz
min
.
Hence
dQ1
dt
= 3 q1 +
Q2
100
− Q1
15
.
Similarly, for Tank 2,
dQ2
dt
= q2 +
Q1
30
− 3Q2
100
.
The process is modeled by the system of equations
Q ′1 = −
Q1
15
+
Q2
100
+ 3 q1
Q ′2 =
Q1
30
− 3Q2
100
+ q2 .
The initial conditions are Q1(0) = Q01 and Q2(0) = Q
0
2 .
(b) The equilibrium values are obtained by solving the system
−Q1
15
+
Q2
100
+ 3 q1 = 0
Q1
30
− 3Q2
100
+ q2 = 0 .
Its solution leads to QE1 = 54 q1 + 6 q2 and Q
E
2 = 60 q1 + 40 q2 .
(c) The question refers to a possible solution of the system
54 q1 + 6 q2 = 60
60 q1 + 40 q2 = 50 .
It is possible for formally solve the system