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Exercícios resolvidos Equações diferenciais elementares 9 ed Boyce e DiPrima

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2n(2n+ 1)!
(x− 1)n)
y2(x) = 1 +
∞∑
n=1
(−1)n α(−α)(1− α
2)(4− α2) . . . ((n− 1)2 − α2)
n! · 3 · 5 . . . (2n− 1) (x− 1)
n .
13.(a) Here xp(x) = 1− x and x2q(x) = λx, which are both analytic at x = 0. In
fact,
p0 = lim
x→0
x p(x) = 1 and q0 = lim
x→0
x2q(x) = 0 .
(b) The indicial equation is r(r − 1) + r = 0 , with roots r1,2 = 0 .
(c) Set
y = a0 + a1x+ a2x2 + . . .+ anxn + . . . .
Substitution into the ODE results in
∞∑
n=2
n(n− 1)an xn−1 +
∞∑
n=1
nan x
n−1−
−
∞∑
n=0
nan x
n + λ
∞∑
n=0
anx
n = 0 .
5.6 207
That is,
∞∑
n=1
n(n+ 1)an+1 xn +
∞∑
n=0
(n+ 1)an+1 xn−
−
∞∑
n=1
nan x
n + λ
∞∑
n=0
anx
n = 0 .
It follows that
a1 + λa0 +
∞∑
n=1
[
(n+ 1)2an+1 − (n− λ)an
]
xn = 0 .
Setting the coefficients equal to zero, we find that a1 = −λa0 , and
an =
(n− 1− λ)
n2
an−1 , n = 2, 3, . . . .
That is, for n ≥ 2 ,
an =
(n− 1− λ)
n2
an−1 = . . . =
(−λ)(1− λ) . . . (n− 1− λ)
(n!)2
a0 .
Therefore one solution of the Laguerre equation is
y1(x) = 1 +
∞∑
n=1
(−λ)(1− λ) . . . (n− 1− λ)
(n!)2
xn.
Note that if λ = m , a positive integer, then an = 0 for n ≥ m+ 1 . In that case,
the solution is a polynomial
y1(x) = 1 +
m∑
n=1
(−λ)(1− λ) . . . (n− 1− λ)
(n!)2
xn .
5.6
2.(a) P (x) = 0 only for x = 0. Furthermore, xp(x) = −2− x and x2q(x) = 2 + x2.
It follows that
p0 = lim
x→0
(−2− x) = −2
q0 = lim
x→0
(2 + x2) = 2
and therefore x = 0 is a regular singular point.
(b) The indicial equation is given by
r(r − 1)− 2r + 2 = 0 ,
that is, r2 − 3r + 2 = 0 , with roots r1 = 2 and r2 = 1 .
4. The coefficients P (x) , Q(x) , and R(x) are analytic for all x ∈ R . Hence there
are no singular points.
208 Chapter 5. Series Solutions of Second Order Linear Equations
5.(a) P (x) = 0 only for x = 0. Furthermore, xp(x) = 3 sin xx and x
2q(x) = −2. It
follows that
p0 = lim
x→0
3
sin x
x
= 3
q0 = lim
x→0
−2 = −2
and therefore x = 0 is a regular singular point.
(b) The indicial equation is given by
r(r − 1) + 3r − 2 = 0 ,
that is, r2 + 2r − 2 = 0 , with roots r1 = −1 +
√
3 and r2 = −1−
√
3 .
6.(a) P (x) = 0 for x = 0 and x = −2. We note that p(x) = x−1(x+ 2)−1/2, and
q(x) = −(x+ 2)−1/2. For the singularity at x = 0,
p0 = lim
x→0
1
2(x+ 2)
=
1
4
q0 = lim
x→0
−x2
2(x+ 2)
= 0
and therefore x = 0 is a regular singular point.
For the singularity at x = −2,
p0 = lim
x→−2
(x+ 2)p(x) = lim
x→−2
1
2x
= −1
4
q0 = lim
x→−2
(x+ 2)2q(x) = lim
x→−2
−(x+ 2)
2
= 0
and therefore x = −2 is a regular singular point.
(b) For x = 0: the indicial equation is given by
r(r − 1) + 1
4
r = 0 ,
that is, r2 − 34r = 0 , with roots r1 = 34 and r2 = 0 .
For x = −2: the indicial equation is given by
r(r − 1)− 1
4
r = 0 ,
that is, r2 − 54r = 0 , with roots r1 = 54 and r2 = 0 .
7.(a) P (x) = 0 only for x = 0. Furthermore, xp(x) = 12 +
sin x
2x and x
2q(x) = 1. It
follows that
p0 = lim
x→0
xp(x) = 1
q0 = lim
x→0
x2q(x) = 1
and therefore x = 0 is a regular singular point.
5.6 209
(b) The indicial equation is given by
r(r − 1) + r + 1 = 0 ,
that is, r2 + 1 = 0 , with complex conjugate roots r = ± i .
8.(a) Note that P (x) = 0 only for x = −1. We find that p(x) = 3(x− 1)/(x+ 1),
and q(x) = 3/(x+ 1)2. It follows that
p0 = lim
x→−1
(x+ 1)p(x) = lim
x→−1
3(x− 1) = −6
q0 = lim
x→−1
(x+ 1)2q(x) = lim
x→−1
3 = 3
and therefore x = −1 is a regular singular point.
(b) The indicial equation is given by
r(r − 1)− 6r + 3 = 0,
that is, r2 − 7r + 3 = 0 , with roots r1 = (7 +
√
37 )/2 and r2 = (7−
√
37 )/2 .
10.(a) P (x) = 0 for x = 2 and x = −2. We note that p(x) = 2x(x− 2)−2(x+ 2)−1,
and q(x) = 3(x− 2)−1(x+ 2)−1. For the singularity at x = 2,
lim
x→2
(x− 2)p(x) = lim
x→2
2x
x2 − 4 ,
which is undefined. Therefore x = 0 is an irregular singular point. For the singu-
larity at x = −2,
p0 = lim
x→−2
(x+ 2)p(x) = lim
x→−2
2x
(x− 2)2 = −
1
4
q0 = lim
x→−2
(x+ 2)2q(x) = lim
x→−2
3(x+ 2)
x− 2 = 0
and therefore x = −2 is a regular singular point.
(b) The indicial equation is given by
r(r − 1)− 1
4
r = 0 ,
that is, r2 − 54r = 0 , with roots r1 = 54 and r2 = 0 .
11.(a) P (x) = 0 for x = 2 and x = −2. We note that p(x) = 2x/(4− x2), and
q(x) = 3/(4− x2). For the singularity at x = 2,
p0 = lim
x→2
(x− 2)p(x) = lim
x→2
−2x
x+ 2
= −1
q0 = lim
x→2
(x− 2)2q(x) = lim
x→2
3(2− x)
x+ 2
= 0
and therefore x = 2 is a regular singular point.
210 Chapter 5. Series Solutions of Second Order Linear Equations
For the singularity at x = −2 ,
p0 = lim
x→−2
(x+ 2)p(x) = lim
x→−2
2x
2− x = −1
q0 = lim
x→−2
(x+ 2)2q(x) = lim
x→−2
3(x+ 2)
2− x = 0
and therefore x = −2 is a regular singular point.
(b) For x = 2: the indicial equation is given by
r(r − 1)− r = 0 ,
that is, r2 − 2r = 0 , with roots r1 = 2 and r2 = 0 .
For x = −2: the indicial equation is given by
r(r − 1)− r = 0 ,
that is, r2 − 2r = 0 , with roots r1 = 2 and r2 = 0 .
12.(a) P (x) = 0 for x = 0 and x = −3. We note that p(x) = −2x−1(x+ 3)−1, and
q(x) = −1/(x+ 3)2. For the singularity at x = 0,
p0 = lim
x→0
x p(x) = lim
x→0
−2
x+ 3
= −2
3
q0 = lim
x→0
x2q(x) = lim
x→0
−x2
(x+ 3)2
= 0
and therefore x = 0 is a regular singular point.
For the singularity at x = −3,
p0 = lim
x→−3
(x+ 3)p(x) = lim
x→−3
−2
x
=
2
3
q0 = lim
x→−3
(x+ 3)2q(x) = lim
x→−3
(−1) = −1
and therefore x = −3 is a regular singular point.
(b) For x = 0: the indicial equation is given by
r(r − 1)− 2
3
r = 0 ,
that is, r2 − 53r = 0 , with roots r1 = 53 and r2 = 0 .
For x = −3: the indicial equation is given by
r(r − 1) + 2
3
r − 1 = 0 ,
that is, r2 − 13r − 1 = 0 , with roots r1 = (1 +
√
37 )/6 and r2 = (1−
√
37 )/6 .
13.(a) Note that p(x) = 1/x and q(x) = −1/x. Furthermore, xp(x) = 1 and
x2q(x) = −x. It follows that
p0 = lim
x→0
(1) = 1
q0 = lim
x→0
(−x) = 0
5.6 211
and therefore x = 0 is a regular singular point.
(b) The indicial equation is given by
r(r − 1) + r = 0 ,
that is, r2 = 0 , with roots r1 = r2 = 0 .
(c) Let y = a0 + a1x+ a2x2 + . . .+ anxn + . . . . Substitution into the ODE results
in
∞∑
n=0
(n+ 2)(n+ 1)an+2 xn+1 +
∞∑
n=0
(n+ 1)an+1xn −
∞∑
n=0
anx
n = 0 .
After adjusting the indices in the first series, we obtain
a1 − a0 +
∞∑
n=1
[n(n+ 1)an+1 + (n+ 1)an+1 − an]xn = 0 .
Setting the coefficients equal to zero, it follows that for n ≥ 0 ,
an+1 =
an
(n+ 1)2
.
So for n ≥ 1 ,
an =
an−1
n2
=
an−2
n2(n− 1)2 = . . . =
1
(n!)2
a0 .
With a0 = 1 , one solution is
y1(x) = 1 + x+
1
4
x2 +
1
36
x3 + . . .+
1
(n!)2
xn + . . . .
For a second solution, set y2(x) = y1(x) ln x + b1x+ b2x2 + . . .+ bnxn + . . . . Sub-
stituting into the ODE, we obtain
L [y1(x)] · ln x+ 2 y ′1(x) + L
[ ∞∑
n=1
bn x
n
]
= 0 .
Since L [y1(x)] = 0 , it follows that
L
[ ∞∑
n=1
bn x
n
]
= −2 y ′1(x) .
More specifically,
b1 +
∞∑
n=1
[n(n+ 1)bn+1 + (n+ 1)bn+1 − bn]xn =
= −2− x− 1
6
x2 − 1
72
x3 − 1
1440
x4 − . . . .
212 Chapter 5. Series Solutions of Second Order Linear Equations
Equating the coefficients, we obtain the system of equations
b1 = −2
4b2 − b1 = −1
9b3 − b2 = −1/6
16b4 − b3 = −1/72
...
Solving these equations for the coefficients, b1 = −2, b2 = −3/4, b3 = −11/108, b4 =
−25/3456 , . . . . Therefore a second solution is
y2(x) = y1(x) ln x+
[
−2x− 3
4
x2 − 11
108
x3 − 25
3456
x4 − . . .
]
.
14.(a) Here x p(x) = 2x and x2q(x) = 6xex . Both of these functions are analytic
at x = 0 , therefore x = 0 is a regular singular

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