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Exercícios resolvidos Equações diferenciais elementares 9 ed Boyce e DiPrima

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given u2(x, t) = sin(x− at), we have
∂2u2
∂x2
= − sin(x− at)
∂2u2
∂t2
= −a2 sin(x− at)
Clearly, u2(x, t) is also a solution of the partial differential equation.
28. Given the function u(x, t) =
√
pi/t e−x
2/4α2t, the partial derivatives are
uxx = −
√
pi/t e−x
2/4α2t
2α2t
+
√
pi/t x2e−x
2/4α2t
4α4t2
ut = −
√
pit e−x
2/4α2t
2t2
+
√
pi x2e−x
2/4α2t
4α2t2
√
t
It follows that
α2 uxx = ut = −
√
pi (2α2t− x2)e−x2/4α2t
4α2t2
√
t
.
1.3 19
Hence u(x, t) is a solution of the partial differential equation.
29.(a)
W mg=
=
µ
µ
µ
T tension
L
(b) The path of the particle is a circle, therefore polar coordinates are intrinsic to
the problem. The variable r is radial distance and the angle θ is measured from
the vertical.
Newton’s Second Law states that
∑
F = ma. In the tangential direction, the
equation of motion may be expressed as
∑
Fθ = maθ, in which the tangential
acceleration, that is, the linear acceleration along the path is aθ = Ld2θ/dt2. ( aθ
is positive in the direction of increasing θ ). Since the only force acting in the
tangential direction is the component of weight, the equation of motion is
−mg sin θ = mLd
2θ
dt2
.
(c) Rearranging the terms results in the differential equation
d2θ
dt2
+
g
L
sin θ = 0 .
30.(a) The kinetic energy of a particle of mass m is given by T = 12 mv
2, in which
v is its speed. A particle in motion on a circle of radius L has speed L (dθ/dt),
where θ is its angular position and dθ/dt is its angular speed.
(b) Gravitational potential energy is given by V = mgh, where h is the height
above a certain datum. Choosing the lowest point of the swing as the datum
(V = 0), it follows from trigonometry that h = 1− cos θ.
(c) From parts (a) and (b),
E =
1
2
mL2(
dθ
dt
)2 +mgL(1− cos θ) .
Applying the Chain Rule for Differentiation,
dE
dt
= mL2
dθ
dt
d2θ
dt2
+mgL sin θ
dθ
dt
.
20 Chapter 1. Introduction
Setting dE/dt = 0 and dividing both sides of the equation by dθ/dt results in
mL2
d2θ
dt2
+mgL sin θ = 0,
which leads to equation (12).
31. Angular momentum is the moment (about a certain point) of linear momentum,
which is given by
mv = mL
dθ
dt
.
Taking a moment about the pivot point, the angular momentum is
Mp = mL2
dθ
dt
.
The moment of the gravitational force (about the same point) isMg = mg · L sin θ.
21
CH A P T E R
2
First Order Differential Equations
2.1
1.(a)
(b) Based on the direction field, all solutions seem to converge to a specific increas-
ing function.
(c) The integrating factor is µ(t) = e3t, and hence y(t) = t/3− 1/9 + e−2t + c e−3t.
It follows that all solutions converge to the function y1(t) = t/3− 1/9 .
22 Chapter 2. First Order Differential Equations
2.(a)
(b) All solutions eventually have positive slopes, and hence increase without bound.
(c) The integrating factor is µ(t) = e−2t, and hence y(t) = t3e2t/3 + c e2t. It is
evident that all solutions increase at an exponential rate.
3.(a)
(b) All solutions seem to converge to the function y0(t) = 1 .
(c) The integrating factor is µ(t) = et, and hence y(t) = t2e−t/2 + 1 + c e−t. It is
clear that all solutions converge to the specific solution y0(t) = 1 .
4.(a)
(b) Based on the direction field, the solutions eventually become oscillatory.
2.1 23
(c) The integrating factor is µ(t) = t , and hence the general solution is
y(t) =
3 cos 2t
4t
+
3
2
sin 2t+
c
t
in which c is an arbitrary constant. As t becomes large, all solutions converge to
the function y1(t) = 3(sin 2t)/2 .
5.(a)
(b) All solutions eventually have positive slopes, and hence increase without bound.
(c) The integrating factor is µ(t) = e−
∫
2dt = e−2t. The differential equation can
be written as e−2ty ′ − 2e−2ty = 3e−t, that is, (e−2ty)′ = 3e−t. Integration of both
sides of the equation results in the general solution y(t) = −3et + c e2t. It follows
that all solutions will increase exponentially.
6.(a)
(b) All solutions seem to converge to the function y0(t) = 0 .
(c) The integrating factor is µ(t) = t2 , and hence the general solution is
y(t) = −cos t
t
+
sin t
t2
+
c
t2
24 Chapter 2. First Order Differential Equations
in which c is an arbitrary constant (t > 0). As t becomes large, all solutions
converge to the function y0(t) = 0 .
7.(a)
(b) All solutions seem to converge to the function y0(t) = 0 .
(c) The integrating factor is µ(t) = et
2
, and hence y(t) = t2e−t
2
+ c e−t
2
. It is clear
that all solutions converge to the function y0(t) = 0 .
8.(a)
(b) All solutions seem to converge to the function y0(t) = 0 .
(c) Since µ(t) = (1 + t2)2, the general solution is
y(t) =
arctan t+ c
(1 + t2)2
.
It follows that all solutions converge to the function y0(t) = 0 .
2.1 25
9.(a)
(b) All solutions eventually have positive slopes, and hence increase without bound.
(c) The integrating factor is µ(t) = e
∫ 1
2dt = et/2. The differential equation can be
written as et/2y ′ + et/2y/2 = 3t et/2/2 , that is, (et/2 y/2)′ = 3t et/2/2. Integration
of both sides of the equation results in the general solution y(t) = 3t− 6 + c e−t/2.
All solutions approach the specific solution y0(t) = 3t− 6 .
10.(a)
(b) For y > 0 , the slopes are all positive, and hence the corresponding solutions
increase without bound. For y < 0 , almost all solutions have negative slopes, and
hence solutions tend to decrease without bound.
(c) First divide both sides of the equation by t (t > 0). From the resulting standard
form, the integrating factor is µ(t) = e−
∫ 1
t dt = 1/t . The differential equation can
be written as y ′/t− y/t2 = t e−t , that is, ( y/t)′ = t e−t. Integration leads to the
general solution y(t) = −te−t + c t . For c 6= 0 , solutions diverge, as implied by
the direction field. For the case c = 0 , the specific solution is y(t) = −te−t, which
evidently approaches zero as t → ∞ .
26 Chapter 2. First Order Differential Equations
11.(a)
(b) The solutions appear to be oscillatory.
(c) The integrating factor is µ(t) = et, and hence y(t) = sin 2t− 2 cos 2t+ c e−t. It
is evident that all solutions converge to the specific solution
y0(t) = sin 2t− 2 cos 2t.
12.(a)
(b) All solutions eventually have positive slopes, and hence increase without bound.
(c) The integrating factor is µ(t) = et/2. The differential equation can be written
as et/2y ′ + et/2y/2 = 3t2/2 , that is, (et/2 y/2)′ = 3t2/2. Integration of both sides
of the equation results in the general solution y(t) = 3t2 − 12t+ 24 + c e−t/2. It
follows that all solutions converge to the specific solution y0(t) = 3t2 − 12t+ 24 .
14. The integrating factor is µ(t) = e2t. After multiplying both sides by µ(t),
the equation can be written as (e2t y)′ = t . Integrating both sides of the equation
results in the general solution y(t) = t2e−2t/2 + c e−2t. Invoking the specified con-
dition, we require that e−2/2 + c e−2 = 0 . Hence c = −1/2 , and the solution to
the initial value problem is y(t) = (t2 − 1)e−2t/2 .
2.1 27
16. The integrating factor is µ(t) = e
∫ 2
t dt = t2 . Multiplying both sides by µ(t), the
equation can be written as (t2 y)′ = cos t . Integrating both sides of the equation
results in the general solution y(t) = sin t/t2 + c t−2. Substituting t = pi and setting
the value equal to zero gives c = 0 . Hence the specific solution is y(t) = sin t/t2.
17. The integrating factor is µ(t) = e−2t, and the differential equation can be
written as (e−2t y)′ = 1 . Integrating, we obtain e−2t y(t) = t+ c . Invoking the
specified initial condition results in the solution y(t) = (t+ 2)e2t.
19. After writing the equation in standard form, we find that the integrating factor
is µ(t) = e
∫ 4