Lista EDO - MTM125

Prévia do material em texto

x x0 = 0
y′′ − xy′ − y = 0
y(x) = a0
∞
n=0
x2n
(2n)!!
+ a1
∞
n=0
x2n+1
(2n+ 1)!!
,
(2n)!! = 2 · 4 · 6 . . . (2n) (2n+ 1)!! = 3 · 5 · 7 . . . (2n+ 1) 0!! = 1
x x0 = 0
y(x) = a0y1(x) + a1y2(x)
y1(x) y2(x) x
y′′ − y = 0
y′′ + k2x2y = 0 k
y′′ + xy′ + 2y = 0
y(0) = 4, y′(0) = 1
(1 + x2)y′′ − 4xy′ + 6y = 0
(4− x2)y′′ + 2y = 0
(3− x2)y′′ − 3xy′ − y = 0

(1− x)y′′ + xy′ − y = 0
y(0) = −3, y′(0) = 2
2y′′ + xy′ + 3y = 0
(1− x2)y′′ − 2xy′ + 2y = 0
an+2 =
an
n+ 2
, n ≥ 0
an+2 =
an
(n+ 2)(n+ 1)
, n ≥ 0,
y1(x) =
∞
n=0
x2n
(2n)! = coshx, y2(x) =
∞
n=0
x2n+1
(2n+1)! = senhx.
a2 = a3 = 0, an+4 = − k2an
(n+ 4)(n+ 3)
, n ≥ 2
y1(x) = 1+
∞
n=1
(−1)nk2nx4n
3 · 4 · ·7 · 8 . . . (4n− 1)(4n)
, y2(x) = x

1 +
∞
n=1
(−1)nk2nx4n+1
4 · 5 · 8 · 9 . . . (4n)(4n+ 1)

an+2 = − an
n+ 1
, n ≥ 0, y(x) = 4

1 +
∞
n=1
(−1)nx2n
(2n− 1)!!

+
∞
n=0
(−1)nx2n+1
(2n)!!
a2 = −3a0, a3 = −a1
3
, an+2 = − (n− 2)(n− 3)an
(n+ 2)(n+ 1)
, n ≥ 2
y1(x) = 1− 3x2, y2(x) = x− x3
3
an+2 =
(n− 2)an
4(n+ 2)
, n ≥ 2, y1(x) = 1− x2
4
, y2(x) = x−
∞
n=1
x2n+1
4n(2n− 1)(2n+ 1)
an+2 =
(n+ 1)an
3(n+ 2)
, n ≥ 0, y1(x) = 1 +
∞
n=1
(2n− 1)!!x2n
3n(2n)!!
, y2(x) =
∞
n=0
(2n)!!x2n+1
3n(2n+ 1)!!
an+2 =
(n+ 1)nan+1 − (n− 1)an
(n+ 2)(n+ 1)
, n ≥ 0, y(x) = −3

1 +
∞
n=2
xn
n!

+ 2x
an+2 = − (n+ 3)an
2(n+ 1)(n+ 2)
, n ≥ 0
y1(x) =
∞
n=0
(−1)n
(2n+ 1)!!
2n(2n)!
x2n, y2(x) =
∞
n=0
(−1)n
(2n+ 2)!!
2n+1(2n+ 1)!x2n+1
an+2 =
(n− 1)an
(n+ 1)
, n ≥ 0, y(x) = a0
 ∞
n=0
x2n
2n− 1

+ a1x
DECOMPOSIÇÃO POR FRAÇÕES PARCIAIS
c F (s) = L{f(t)}(s) s > a > 0
s > ac
L{f(ct)}(s) = 1
c
F
s
c

.
u = ct
6e−3t − t2 + 2t− 8
5− e2t + 6t2
t3 − tet + e4t cos t
t2 − 3t− 2e−tsen 3t
e3tsen 6t− t3 + et
t4 − t2 − t+ sen
√
2t
t4e5t − et cos
√
7t
e−2t cos
√
3t− t2e−2t
t2 + etsen 2t
3t2 − e2t
e−t cos 3t+ e6t − 1
3t4 − 2t2 + 1
2t2e−t − t+ cos 4t
e−2tsen 2t+ e3tt2
(t− 1)4
(1 + e−t)2
sen 3t cos 3t
sen 2t
e7tsen 2t
3
s2 + 4
4
(s− 1)3
2
s2 + 3s− 4
3s
s2 − s− 6
2s+ 2
s2 + 2s+ 5
2s− 3
s2 − 4
2s+ 1
s2 − 2s+ 2
8s2 − 4s+ 12
s(s2 + 4)
1− 2s
s2 + 4s+ 5
2s− 3
s2 + 2s+ 10

y′′ − y′ − 6y = 0,
y(0) = 1, y′(0) = −1

y′′ + 3y′ + 2y = 0,
y(0) = 1, y′(0) = 0

y′′ − 2y′ + 2y = 0,
y(0) = 0, y′(0) = 1
y′′ − 4y′ + 4y = 0,
y(0) = 1, y′(0) = 1
y′′ − 2y′ + 4y = 0,
y(0) = 2, y′(0) = 0
y′′ + 2y′ + 5y = 0,
y(0) = 2, y′(0) = −1


y(4) − 4y′′′ + 6y′′ − 4y′ + y = 0,
y(0) = 0 = y′′(0),
y′(0) = 1 = y′′′(0)


y(4) − y = 0
y(0) = 1 = y′′(0),
y′(0) = 0 = y′′′(0)


y(4) − 4y = 0,
y(0) = 1, y′(0) = 0,
y′′′(0) = 0, y′′(0) = −2

y′′ + ω2y = cos 2t,
y(0) = 1, y′(0) = 0

y′′ − 2y′ + 2y = cos t,
y(0) = 1, y′(0) = 0

y′′ − 2y′ + 2y = e−t,
y(0) = 0, y′(0) = 1

y′′ + 2y′ + y = 4e−t,
y(0) = 2, y′(0) = −1

y′′′ − y′′ + y′ − y = 0,
y(0) = 1 = y′(0), y′′(0) = 3

y′′′ + 4y′′ + y′ − 6y = −12,
y(0) = 1, y′(0) = 4, y′′(0) = −2

y′′′ + 3y′′ + 3y′ + y = 0,
y(0) = −4, y′(0) = 4, y′′(0) = −2
[0,∞)
u1(t) + 2u3(t)− 6u4(t)
u2(t)(t− 3)− u3(t)(t− 2)
u3(t)f(t− 3), f(t) = sen t
f(t) =

0, t 0
5
s − 1
s−2 + 12
s3 , s > 2
6
s4 − 1
(s−1)2 + s−4
(s−4)2+1 , s > 4
2
s3 − 3
s2 − 6
(s+1)2+9 , s > 0
6
(s−3)2+36 − 6
s4 + 1
s−1 , s > 3
s+2
(s+2)2+3 − 2
(s+2)3 , s > −2
24
(s−5)2 − s−1
(s−1)2+7 , s > 5
6
s3 − 1
s−2 , s > 2
2
s3 + 2
(s−1)2+4 , s > 1
6
s3 − 1
s−2 , s > 2
s+1
(s+1)2+9 + 1
s−6 − 1
s , s > 6
96
s5 − 4
s3 + 1
s , s > 0
4
(s+1)3 − 1
s2 + s
s2+16 s > 0
2
(s+2)2+4 + 2
(s−3)3 , s > 3
24
s5 − 24
s4 + 12
s3 − 4
s2 + 1
s , s > 0
1
s + 2
s+1 + 1
s+2 , s > 0
3
s2+36 , s > 0
1
2s − s
2(s2+4) , s > 0
1
2(s−7) − s−7
2(s−7)2+8 , s > 7
3
2 sen 2t
2t2et
2
5

et − e−4t

9
5e
3t + 6
5e
−2t
2e−t cos 2t
e2t
4 + 7
4e
2t
2et cos t+ 3etsen t
3− 2sen 2t+ 5 cos 2t
−2e−2t cos t+ 5e−2tsen t
2e−t cos 3t− 5
3e
−tsen 3t
y = 1
5 (e
3t + 4e−2t)
y = 2e−t − e−2t
y = etsen t
y = e2t − te2t
y = 1+
√
3
3 e(1+
√
3)t + 1−
√
3
3 e(1−
√
3)t
y = 2e−t cos 2t+ 1
2e
−tsen 2t
y = tet − t2et + 2
3 t
3et
y = et+e−t
2
y = cos
√
2t
y = (ω2 − 4)−1

(ω2 − 5) cosωt+ cos 2t

y = 1
5 (cos t− 2sen t+ 4et cos t− 2sen t)
y = 1
5 (e
−t − et cos t+ 7etsen t)
y = (2 + t+ 2t2)e−t
y = 2et − cos t− sen t
y = 1
5

et + 13e−2t − 9e−3t − 60

y = (t2 − 4)e−t
e−4s(2es+e3s−6)
s
− e−3s((s−1)es+s+1)
s2
e−3s
s2+1
2e−2s
s3
e−s(s2+2)
s3
e−2πs(−πs+eπs−1)
s2
2e−2s(s+1)
s2
e−3s(es−1)2
s2
y = 1− cos t+ sen t− uπ/2(1− sen t)
y = e−tsen t+ 1
2uπ(t)

1 + e−(t−π) cos t+ e−(t−π)sen t

− 1
2u2π(t)

1− e−(t−2π) cos t− e−(t−2π)sen t

y = 1
6 (1− u2π(t)(2sen t− sen 2t))
y = 1
6 (2sen t− sen 2t)− 1
6uπ(t)(2sen t+ sen 2t)
y = 1
2 + 1
2e
−2t − e−t − u10(t)

1
2 + 1
2e
−2(t−10) − e−(t−10)

y = e−t − e−2t + u2(t)

1
2 − e−(t−2) + 1
2e
−2(t−2)

y = cos t+ u3π(t) (1− cos(t− 3π))
y = h(t)− uπ(t)h

t− π
2

h(t) = 4
25

−4 + 5t+ 4e−t/2 cos t− 3e−t/2sen t

1
2 sen t+ 1
2 t− 1
2u6(t) (t− 6− sen (t− 6))
y = h(t) + uπ(t)h(t− π) h(t) = 4
17

−4 cos t+ sen t+ 4e−t/2 cos t+ e−t/2sen t

y = uπ(t)

1
4 − 1
4 cos (2t− 2π)

− u3π(t)

1
4 − 1
4 cos(2t− 6π)

y = u1(t)h(t− 1)− u2(t)h(t− 2) h(t) = −1 + 1
2 (cos t+ et − e−t)
y = h(t)− uπ(t)h(t− π) h(t) = 1
12 (3− 4 cos t+ cos2t)
e−s − e−3s e−s
e−2
y = e−t cos t+ e−tsen t+ uπ(t)e
−(t−π)sen (t− π)
y = 1
2uπ(t)sen 2(t− π)− 1
2u2π(t)sen 2(t− 2π)
y = − 1
2e
−2t + 1
2e
−t + u5(t)

−e−2(t−5) + e−(t−5)

+ 1
2u10(t)

1 + e−2(t−10) − e−(t−10)

y = 1
4

sen t− cos t+ e−t cos
√
2t+ (1/
√
2)u3π(t)e
−(t−3π) sen
√
2(t− 3π)

y = 1
2 [cos 2t+ u4π(t)sen 2(t− 4π)]
y = sen t+ u2π(t)sen (t− 2π)
y = uπ/4(t)sen 2(t− π/4)
y = uπ/2(t) [1− cos(t− π/2)] + 3u3π/2(t)sen (t− 3π/2)− u2π(t) [1− cos(t− 2π)]
y = 1√
31
uπ/6(t)e
− 1
4 (t−π/6)sen
√
31
4 (t− π/6)

y = 1
5

cos t+ 2sen t− e−t cos t− 3e−tsen t+ uπ/2(t)e
−(t−π/2)sen (t− π/2)

y = 1
4u1(t)

et−1 − e−(t−1) − 2sen (t− 1)

y = 4√
15
u1(t)e
−(t−1)/4sen
√
15
4 (t− 1)

t1 ≈ 2, 3613, y1 ≈ 0, 71153
y = 8
√
7
21 u1(t)e
−(t−1)/8 sen

3
√
7
8 (t− 1)

, t1 ≈ 2, 4569, y ≈ 0, 83351
t1 = 1 + π
2 , y1 = 1
k1 ≈ 2, 8108 k1 ≈ 2, 3995 k1 = 2
F (s) = 2
s2(s2+4)
F (s) = 1
(s+1)(s2+1)
F (s) = 1
s2(s−1)
F (s) = s
(s2+1)2
h(t) = 1
6
 t
0
(t− τ)3sen τ dτ
h(t) =
 t
0
eτ−t) cos 2τ dτ
h(t) = 1
2
 t
0
(t− τ)e−(t−τ)sen 2τ dτ
h(t) =
 t
0
sen (t− τ)g(τ ) dτ
y = 1
ω sen (ωt) +
1
ω
 t
0
sen (ω(t− τ))g(τ ) dτ
y =
 t
0
e−(t−τ)sen (t− τ)sen ατ dτ
y = 1
8
 t
0
e−(t−τ)/2sen (2(t− τ))g(τ ) dτ
y = e−t/2 cos t− 1
2e
−t/2sen t+
 t
0
e−(t−τ)/2sen (t− τ) (1− uπ(τ )) dτ
y(t) = (2− 9t)e−2t +
t
0
(t− τ)e−2(t−τ)g(τ )dτ
y = 2e−t − e−2t +
 t
0

e−(t−τ) − e−2(t−τ)

cosατ dτ
y = 1
4
 t
0

e(t−τ) − e(τ−t) − 2sen (t− τ)

g(τ ) dτ
y = 4
3 cos t− 1
3 cos 2t+
1
6
 t
0
(2sen (t− τ)− sen 2(t− τ)) g(τ ) dτ
f(t) = L−1{F (s)} F (s) = L{f(t)} =
∞
0
e−stf(t) dt
1
s
, s > 0
tn, n ∈ N
n!
sn+1
, s > 0
eat
1
s− a
, s > a
sen at
a
s2 + a2
, s > 0
cos at
s
s2 + a2
, s > 0
eat sen bt
b
(s− a)2 + b2
, s > a
eat cos bt
s− a
(s− a)2 + b2
, s > a
tneat, n ∈ N
n!
(s− a)n+1
, s > a
ectf(t) F (s− c)
f(ct)
1
c
F
s
c

, c > 0
uc(t)
e−cs
s
, s > 0
uc(t)f(t− c) e−csF (s)
δ(t− c) e−cs
f(t)δ(t− t0) f(t0)e
−st0 , s > 0
f ∗ g(t) =
 t
0
f(t− τ)g(τ ) dτ F (s)G(s)
(−t)nf(t) F (n)(s)
 t
0
f(τ ) dτ
1
s
F (s)
f (n)(t) snF (s)− sn−1f(0)− sn−2f ′(0)− . . .− sf (n−2)(0)− f (n−1)(0)