Lista 11 Sistemas de EDO's

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UFBA - Ca´lculo C - 2017/1
Lista de Exerc´ıcios 11 - Sistema de EDO’s
(1) Escreva o sistema de EDO na forma matricial.
(a)
dx
dt
= 3x− 5y
dy
dt
= 4x + 8y.
(b)
dx
dt
= −3x + 4y − 9z
dy
dt
= 6x− y
dz
dt
= 10x + 4y + 3z.
(c)
dx
dt
= x− y + z + t− 1
dy
dt
= 2x + y − z − 3t2
dz
dt
= x + y + z + t2 − t + 2.
(2) Escreva o sistema de EDO sem usar matrizes.
(a) X ′ =
(
4 2
−1 3
)
X +
(
1
−1
)
et.
(b) X ′ =
 1 −1 23 −4 1
−2 5 6
X +
 12
2
 e−t −
 3−1
1
 t.
(3) Verifique que o vetor dado satisfaz o sistema de EDO’s.
(a) X ′ =
(
3 −2
2 −2
)
X, X1 =
(
4
2
)
e2t.
(b) X ′ =
(
2 −1
3 −2
)
X +
(
1
−1
)
et, X1 =
(
1
0
)
et + 2
(
1
1
)
tet.
(c) X ′ =
 1 1 12 1 −1
0 −1 1
X, X1 =
 6−8
−4
 e−t + 2
 01
−1
 e2t.
(4) Dado o sistema de EDO’s
X ′ =
 0 6 01 0 1
1 1 0
X,
mostre que os vetores
X1 =
 6−1
−5
 e−t X2 =
 −31
1
 e−2t X3 =
 21
1
 e3t
formam um conjunto fundamental de soluc¸o˜es.
(5) Encontre a soluc¸a˜o geral do sistema de EDO’s.
(a) X ′ =
(
3 −2
2 −2
)
X.
(b) X ′ =
(
2 −1
3 −2
)
X.
(c) X ′ =
( −2 1
1 −2
)
X.
1
2
(d)
dx
dt
= 4x− 3y
dx
dt
= 8x− 6y.
(6) Encontre a soluc¸a˜o geral do sistema de EDO’s.
(a) X ′ =
 1 1 21 2 1
2 1 1
X.
(b) X ′ =
 3 2 42 0 2
4 2 3
X.
(c)
dx
dt
= x + y + z
dy
dt
= 2x + y − z
dz
dt
= −8x− 5y − 3z.
(7) Encontre a soluc¸a˜o do sistema de EDO’s com condic¸o˜es iniciais.
(a) X ′ =
(
5 −1
3 1
)
X, X(0) =
(
2
−1
)
(b) X ′ =
 1 1 20 2 2
−1 1 3
X, X(0) =
 20
−1

(8) Encontre a soluc¸a˜o do sistema de EDO’s.
(a) X ′ =
(
3 −4
1 −1
)
X.
(b) X ′ =
(
4 −2
8 −4
)
X.
(c) X ′ =
( −32 1
−14 −12
)
X.
(d)
dx
dt
= −3x + 5
2
y
dy
dt
= −5
2
x + 2y.
(9) Encontre a soluc¸a˜o do sistema de EDO’s com condic¸o˜es iniciais.
(a)
dx
dt
= x− 4y, x(0) = 3
dy
dt
= 4x− 7y, y(0) = 2.
(b) X ′ =
(
3 9
−1 −3
)
X, X(0) =
(
2
4
)
.
(10) Encontre a soluc¸a˜o do sistema de EDO’s.
(a) X ′ =
(
3 −2
4 −1
)
X.
(b) X ′ =
( −1 −4
1 −1
)
X.
(c) X ′ =
(
2 −5
1 −2
)
X.
(d)
dx
dt
= 2x− 5
2
y
dy
dt
=
9
5
x− y.
(11) Encontre a soluc¸a˜o do sistema de EDO’s com condic¸o˜es iniciais.
(a)
dx
dt
= x− 5y, x(0) = 1
3
dy
dt
= x− 3y, y(0) = 1.
(b) X ′ =
( −3 2
−1 −1
)
X, X(0) =
(
1
−2
)
.
Respostas
(1) (a) X ′ =
(
3 −5
4 8
)
X
(b) X ′ =
 −3 4 96 −1 0
10 4 3
X
(c) X ′ =
 1 −1 12 1 −1
1 1 1
X +
 t− 1−3t2
t2 − t + 2

(2) (a)
dx
dt
= 4x + 2y + et
dy
dt
= −x + 3y − et.
(b)
dx
dt
= x− y + 2z + e−t − 3t
dy
dt
= 3x− 4y + z + 2e−t + t
dz
dt
= −2x + 5y + 6z + 2e−t − t.
(3)
(4)
(5) (a) X = c1
(
1
2
)
e−t + c2
(
2
1
)
e2t
(b) X = c1
(
1
1
)
et + c2
(
1
3
)
e−t
(c) X = c1
(
1
−1
)
e−3t + c2
(
1
1
)
e−t
(d) x = 3c1 + c2e
−2t, y = 4c1 + 2c2e−2t
(6) (a) X = c1
 11
1
 e4t + c2
 1−2
1
 et + c3
 10
−1
 e−t
(b) X = c1
 1−4
1
 e−t + c2
 10
−1
 e−t + c3
 21
2
 e8t
(c) x = 4c1e
−2t + 3c2e−t
y = −5c1e−2t − 4c2e−t + c3e2t
z = −7c1e−2t − 2c2e−t − c3e2t
(7) (a) X = −3
2
(
1
3
)
e2t +
7
2
(
1
1
)
e4t
(b) X =
 0−2
1
 et + 2
 11
0
 e2t
(8) (a) X = c1
(
2
1
)
et + c2
[(
2
1
)
tet +
(
1
0
)
et
]
(b) X = c1
(
1
2
)
+ c2
[(
1
2
)
t−
(
0
1
2
)]
(c) X = c1
(
2
1
)
e−t + c2
[(
2
1
)
te−t +
(
0
2
)
e−t
]
(d) x = c1e
−t/2 + c2te−t/2
4
y = c1e
−t/2 + c2
(
te−t/2 + 25e
−t/2)
(9) (a) x = (3 + 4t)e−3t
y = (2 + 4t)e−3t
(b) X = 2
(
1
2
)
+ 14
(
3
−1
)
t
(10) (a) X = c1
(
cos(2t)
cos(2t) + sen(2t)
)
et + c2
(
sen(2t)
− cos(2t) + sen(2t)
)
et
(b) X = c1
(
2 cos(2t)
sen(2t)
)
e−t + c2
( −2 sen(2t)
cos(2t)
)
e−t
(c) X = c1
(
5 cos(t)
2 cos(t) + sen(t)
)
+ c2
(
sen(t)
− cos(t) + 2 sen(t)
)
(d) x = 5c1e
t/2 cos
(
3
2 t
)
+ 5c2e
t/2 sen
(
3
2 t
)
y = 3c1e
t/2
(
cos
(
3
2 t
)
+ sen
(
3
2 t
))
+ 3c2e
t/2
(− cos (32 t)+ sen (32 t))
(11) (a) x = (cos(t)− 3 sen(t))e−t
y = (cos(t)− sen(t))e−t
(b) x = (cos(t)− 5 sen(t))e−2t
y = (−2 cos(t)− 3 sen(t))e−2t