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Exercícios resolvidos Equações diferenciais elementares 9 ed Boyce e DiPrima

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point. Note that p0 = q0 = 0 .
(b) The indicial equation is given by
r(r − 1) = 0 ,
that is, r2 − r = 0 , with roots r1 = 1 and r2 = 0 .
(c) In order to find the solution corresponding to r1 = 1 , set y = x
∑∞
n=0 anx
n.
Upon substitution into the ODE, we have
∞∑
n=0
(n+ 2)(n+ 1)an+1 xn+1 + 2
∞∑
n=0
(n+ 1)anxn+1 + 6 ex
∞∑
n=0
anx
n+1 = 0 .
After adjusting the indices in the first two series, and expanding the exponential
function,
∞∑
n=1
n(n+ 1)an xn + 2
∞∑
n=1
nan−1xn + 6 a0x+ (6a0 + 6a1)x2+
+ (6a2 + 6a1 + 3a0)x3 + (6a3 + 6a2 + 3a1 + a0)x4 + . . . = 0 .
Equating the coefficients, we obtain the system of equations
2a1 + 2a0 + 6a0 = 0
6a2 + 4a1 + 6a0 + 6a1 = 0
12a3 + 6a2 + 6a2 + 6a1 + 3a0 = 0
20a4 + 8a3 + 6a3 + 6a2 + 3a1 + a0 = 0
...
Setting a0 = 1 , solution of the system results in a1 = −4, a2 = 17/3, a3 = −47/12,
a4 = 191/120 , . . . . Therefore one solution is
y1(x) = x− 4x2 + 173 x
3 − 47
12
x4 + . . . .
5.6 213
The exponents differ by an integer. So for a second solution, set
y2(x) = a y1(x) ln x + 1 + c1x+ c2x2 + . . .+ cnxn + . . . .
Substituting into the ODE, we obtain
aL [y1(x)] · ln x+ 2a y ′1(x) + 2a y1(x)− a
y1(x)
x
+ L
[
1 +
∞∑
n=1
cn x
n
]
= 0 .
Since L [y1(x)] = 0 , it follows that
L
[
1 +
∞∑
n=1
cn x
n
]
= −2a y ′1(x)− 2a y1(x) + a
y1(x)
x
.
More specifically,
∞∑
n=1
n(n+ 1)cn+1xn + 2
∞∑
n=1
n cnx
n + 6 + (6 + 6c1)x+
+ (6c2 + 6c1 + 3)x2 + . . . = −a+ 10ax− 613 ax
2 +
193
12
ax3 + . . . .
Equating the coefficients, we obtain the system of equations
6 = −a
2c2 + 8c1 + 6 = 10a
6c3 + 10c2 + 6c1 + 3 = −613 a
12c4 + 12c3 + 6c2 + 3c1 + 1 =
193
12
a
...
Solving these equations for the coefficients, a = −6 . In order to solve the remaining
equations, set c1 = 0 . Then c2 = −33, c3 = 449/6 , c4 = −1595/24 , . . . . Therefore
a second solution is
y2(x) = −6 y1(x) ln x+
[
1− 33x2 + 449
6
x3 − 1595
24
x4 + . . .
]
.
15.(a) Note the p(x) = 6x/(x− 1) and q(x) = 3x−1(x− 1)−1 . Furthermore, x p(x) =
6x2/(x− 1) and x2q(x) = 3x/(x− 1) . It follows that
p0 = lim
x→0
6x2
x− 1 = 0
q0 = lim
x→0
3x
x− 1 = 0
and therefore x = 0 is a regular singular point.
(b) The indicial equation is given by
r(r − 1) = 0 ,
that is, r2 − r = 0 , with roots r1 = 1 and r2 = 0 .
214 Chapter 5. Series Solutions of Second Order Linear Equations
(c) In order to find the solution corresponding to r1 = 1 , set y = x
∑∞
n=0 anx
n.
Upon substitution into the ODE, we have
∞∑
n=1
n(n+ 1)an xn+1 −
∞∑
n=1
n(n+ 1)an xn+
+ 6
∞∑
n=0
(n+ 1)anxn+2 + 3
∞∑
n=0
anx
n+1 = 0 .
After adjusting the indices, it follows that
∞∑
n=2
n(n− 1)an−1 xn −
∞∑
n=1
n(n+ 1)an xn+
+ 6
∞∑
n=2
(n− 1)an−2xn + 3
∞∑
n=1
an−1xn = 0 .
That is,
−2a1 + 3a0 +
∞∑
n=2
[−n(n+ 1)an + (n2 − n+ 3)an−1 + 6(n− 1)an−2]xn = 0.
Setting the coefficients equal to zero, we have a1 = 3a0/2 , and for n ≥ 2 ,
n(n+ 1)an = (n2 − n+ 3)an−1 + 6(n− 1)an−2 .
If we assign a0 = 1 , then we obtain a1 = 3/2 , a2 = 9/4 , a3 = 51/16 , . . . . Hence
one solution is
y1(x) = x+
3
2
x2 +
9
4
x3 +
51
16
x4 +
111
40
x5 + . . . .
The exponents differ by an integer. So for a second solution, set
y2(x) = a y1(x) ln x + 1 + c1x+ c2x2 + . . .+ cnxn + . . . .
Substituting into the ODE, we obtain
2ax y ′1(x)− 2a y ′1(x) + 6ax y1(x)− a y1(x) + a
y1(x)
x
+ L
[
1 +
∞∑
n=1
cn x
n
]
= 0,
since L [y1(x)] = 0 . It follows that
L
[
1 +
∞∑
n=1
cn x
n
]
= 2a y ′1(x)− 2ax y ′1(x) + a y1(x)− 6ax y1(x)− a
y1(x)
x
.
Now
L
[
1 +
∞∑
n=1
cn x
n
]
= 3 + (−2c2 + 3c1)x+ (−6c3 + 5c2 + 6c1)x2+
+ (−12c4 + 9c3 + 12c2)x3 + (−20c5 + 15c4 + 18c3)x4 + . . . .
Substituting for y1(x) , the right hand side of the ODE is
a+
7
2
ax+
3
4
ax2 +
33
16
ax3 − 867
80
ax4 − 441
10
ax5 + . . . .
5.6 215
Equating the coefficients, we obtain the system of equations
3 = a
−2c2 + 3c1 = 72a
−6c3 + 5c2 + 6c1 = 34a
−12c4 + 9c3 + 12c2 = 3316a
...
We find that a = 3. In order to solve the second equation, set c1 = 0. Solution
of the remaining equations results in c2 = −21/4 , c3 = −19/4 , c4 = −597/64 , . . ..
Hence a second solution is
y2(x) = 3 y1(x) ln x+
[
1− 21
4
x2 − 19
4
x3 − 597
64
x4 + . . .
]
.
16.(a) After multiplying both sides of the ODE by x , we find that x p(x) = 0 and
x2q(x) = x . Both of these functions are analytic at x = 0 , hence x = 0 is a regular
singular point.
(b) Furthermore, p0 = q0 = 0 . So the indicial equation is r(r − 1) = 0 , with roots
r1 = 1 and r2 = 0 .
(c) In order to find the solution corresponding to r1 = 1 , set y = x
∑∞
n=0 anx
n.
Upon substitution into the ODE, we have
∞∑
n=1
n(n+ 1)an xn +
∞∑
n=0
anx
n+1 = 0 .
That is,
∞∑
n=1
[n(n+ 1)an + an−1] xn = 0 .
Setting the coefficients equal to zero, we find that for n ≥ 1 ,
an =
−an−1
n(n+ 1)
.
It follows that
an =
−an−1
n(n+ 1)
=
an−2
(n− 1)n2(n+ 1) = . . . =
(−1)na0
(n!)2(n+ 1)
.
Hence one solution is
y1(x) = x− 12x
2 +
1
12
x3 − 1
144
x4 +
1
2880
x5 + . . . .
The exponents differ by an integer. So for a second solution, set
y2(x) = a y1(x) ln x + 1 + c1x+ c2x2 + . . .+ cnxn + . . . .
216 Chapter 5. Series Solutions of Second Order Linear Equations
Substituting into the ODE, we obtain
aL [y1(x)] · ln x+ 2a y ′1(x)− a
y1(x)
x
+ L
[
1 +
∞∑
n=1
cn x
n
]
= 0 .
Since L [y1(x)] = 0 , it follows that
L
[
1 +
∞∑
n=1
cn x
n
]
= −2a y ′1(x) + a
y1(x)
x
.
Now
L
[
1 +
∞∑
n=1
cn x
n
]
= 1 + (2c2 + c1)x+ (6c3 + c2)x2 + (12c4 + c3)x3+
+ (20c5 + c4)x4 + (30c6 + c5)x5 + . . . .
Substituting for y1(x) , the right hand side of the ODE is
−a+ 3
2
ax− 5
12
ax2 +
7
144
ax3 − 1
320
ax4 + . . . .
Equating the coefficients, we obtain the system of equations
1 = −a
2c2 + c1 =
3
2
a
6c3 + c2 = − 512a
12c4 + c3 =
7
144
a
...
Evidently, a = −1 . In order to solve the second equation, set c1 = 0 . We then find
that c2 = −3/4 , c3 = 7/36 , c4 = −35/1728 , . . . . Therefore a second solution is
y2(x) = −y1(x) ln x+
[
1− 3
4
x2 +
7
36
x3 − 35
1728
x4 + . . .
]
.
19.(a) After dividing by the leading coefficient, we find that
p0 = lim
x→0
x p(x) = lim
x→0
γ − (1 + α+ β)x
1− x = γ .
q0 = lim
x→0
x2q(x) = lim
x→0
−αβ x
1− x = 0 .
Hence x = 0 is a regular singular point. The indicial equation is r(r − 1) + γ r = 0 ,
with roots r1 = 1− γ and r2 = 0 .
(b) For x = 1,
p0 = lim
x→1
(x− 1)p(x) = lim
x→1
−γ + (1 + α+ β)x
x
= 1− γ + α+ β .
5.6 217
q0 = lim
x→1
(x− 1)2q(x) = lim
x→1
αβ(x− 1)
x
= 0 .
Hence x = 1 is a regular singular point. The indicial equation is
r2 − (γ − α− β) r = 0 ,
with roots r1 = γ − α− β and r2 = 0 .
(c) Given that r1 − r2 is not a positive integer, we can set y =
∑∞
n=0 anx
n. Sub-
stitution into the ODE results in
x(1− x)
∞∑
n=2
n(n− 1)anxn−2 + [γ − (1 + α+ β)x]
∞∑
n=1
nanx
n−1 − αβ
∞∑
n=0
anx
n = 0.
That is,
∞∑
n=1
n(n+ 1)an+1xn −
∞∑
n=2
n(n− 1)anxn + γ
∞∑
n=0
(n+ 1)an+1xn−
− (1 + α+ β)
∞∑
n=1
nanx
n − αβ
∞∑
n=0
anx
n = 0.
Combining the series, we obtain
γ a1 − αβ a0 + [(2 + 2γ)a2 − (1 + α+ β + αβ)a1]x+
∞∑
n=2
An x
n = 0 ,
in which
An = (n+ 1)(n+ γ)an+1 − [n(n− 1) + (1 + α+ β)n+ αβ] an .
Note that n(n− 1) + (1 + α+ β)n+ αβ = (n+ α)(n+ β) . Setting the coefficients
equal to zero, we have γ a1 − αβ a0 = 0 , and
an+1 =
(n+ α)(n+ β)
(n+ 1)(n+ γ)
an
for n ≥ 1 . Hence one solution is
y1(x) = 1 +
αβ
γ ·

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