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Universidade Federal do Rio de Janeiro
INSTITUTO DE MATEMÁTICA
Departamento de Matemática
Exerćıcios de equações diferenciais lineares de segunda ordem - Mônica Merkle
1. Resolva:
(a) y′′ + 2y′ − 3y = 0
(b) 4y′′ + 4y′ + y = 0
(c) 6y′′ − y′ − y = 0
(d) 2y′′ − 3y′ + y = 0
(e) y′′ − y = 0
(f) y′′ − 2y′ + y = 0
(g) y′′ − 2y′ + 2y = 0
(h) y′′ − 2y′ + 6y = 0
(i) y′′ + 2y′ − 8y = 0
(j) y′′ + 2y′ + 2y = 0
(k) 9y′′ − 6y′ + y = 0
(l) y′′ + 6y′ + 13y = 0
(m) y′′ + 5y′ = 0
(n) y′′ − 9y′ + 9y = 0
(o) y′′ − 2y′ − 2y = 0
(p) y′′ + y′ − 2y = 0
2. Resolva:
(a)

y′′ + y′ − 2y = 0
y(0) = 1
y′(0) = 1
(b)

y′′ − 6y′ + 9y = 0
y(0) = 0
y′(0) = 2
(c)

y′′ + 4y = 0
y(0) = 0
y′(0) = 1
(d)

y′′ + 4y′ + 5y = 0
y(0) = 1
y′(0) = 0
(e)

y′′ + 8y′ − 9y = 0
y(1) = 1
y′(1) = 0
3. Resolva:
(a) y′′ − 5y′ + 6y = 2ex
(b) y′′ − y′ − 2y = 2e−x
(c) y′′ + 2y′ + y = 3e−x
(d) 2y′′ + 3y′ + y = x2 + 3 senx
(e) y′′ + y′ + 4y = 2 senh x
(f) y′′ − 4y′ = 2x2
(g) y′′ + y′ = sen 2x
(h) y′′ − 8y′ + 7y = 14
(i) y′′ + y = cosx
(j) y′′ − 2y′ + 10y = ex + sen (3x)
4. Resolva:
(a)

y′′ + 4y = sen x
y(0) = 1
y′(0) = 1
(b)

y′′ − 2y′ = e2x + x2 − 1
y(0) = 1/8
y′(0) = 1
5. Resolver a equação
d2x
dt2
+ w2x = A sen (pt), examinando os casos (a) p 6= w e (b) p = w.
Respostas
1. (a) y = C1e
x + C2e
−3x
(b) y = C1e
−x/2 + C2xe
−x/2
(c) y = C1e
x/2 + C2e
−x/3
(d) y = C1e
x/2 + C2e
x
(e) y = C1e
x + C2e
−x
(f) y = C1e
x + C2xe
x
(g) y = C1e
x cosx + C2e
x senx
(h) y = C1e
x cos (
√
5x) + C2e
x sen (
√
5x)
(i) y = C1e
2x + C2e
−4x
(j) y = C1e
−x cosx + C2e
−x senx
(k) y = C1e
x/3 + C2xe
x/3
(l) y = e−3x(C1 cos (2x) + C2 sen (2x))
(m) y = C1 + C2e
−5x
(n) y = C1e
(9+3
√
5)x/2 + C2e
(9−3
√
5)x/2
(o) y = C1e
(1+
√
3)x + C2e
(1−
√
3)x
(p) y = C1e
−2x + C2e
x
2. (a) y = ex
(b) y = 2xe3x
(c) y = 1
2
sen (2x)
(d) y = e−2x cosx + 2e−2x senx
(e) y = 1
10
e−9(x−1) + 9
10
ex−1
3. (a) y = C1e
2x + C2e
3x + ex
(b) y = C1e
2x + C2e
−x − 2
3
xe−x
(c) y = C1e
−x + C2xe
−x + 3
2
x2e−x
(d) y = C1e
−x + C2e
−x/2 + x2 − 6x + 14− 3
10
senx− 9
10
cosx
(e) y =
[
C1 cos
(√
15x
2
)
+ C2 sen
(√
15x
2
)]
e−x/2 + e
x
6
− e−x
4
(f) y = C1 + C2e
4x − x3
6
− x2
8
− x
16
(g) y = C1 + C2e
−x + x
2
+ 1
20
(2 cos (2x)− sen (2x))
(h) y = C1e
7x + C2e
x + 2
(i) y = C1 cosx + C2 senx +
x
2
senx
(j) y = ex(C1 cos(3x) + C2 sen (3x)) +
1
9
ex + 1
37
( sen (3x) + 6 cos(3x))
4. (a) y = sen (2x)
3
+ cos (2x) + senx
3
(b) y = 1
8
e2x(4x + 1)− x3
6
− x2
4
+ x
4
5. (a) x = C1 cos (wt) + C2 sen (wt) +
A
w2−p2 sen (pt)
(b) x = C1 cos (wt) + C2 sen (wt)− A2w t cos (wt)
2