rmiranda99.github.io · Exerc cio 4 Exerc cio Encontre a solu˘c~ao do PVI _x = 3x −3y, _y = 4x −3y. Logo, exp(At)= Mexp(Bt)M−1 = ™ Œ ﬂ cos› √ 3t”− √ 3sin›

Exercıcios sobre sistemas lineares e

sistemas nao-homogeneos

Calculo III -Turma B, IMECC - UNICAMP

Mayara Duarte de Araujo Caldas

03/07/2020

Exercıcio 1Exercıcio

Encontre a solucao do PVI x = 6x − 7y , y = x − 2y .

Seja A = ( 6 −71 −2

). A solucao do PVI e dada por

x(t) = exp(At)x(0),

entao precisamos determinar exp(At). Temos que,

p(λ) = λ2 − tr(A)λ + det(A) = λ2 − 4λ − 5 = (λ − 5)(λ + 1),

assim, os autovalores de A sao λ1 = 5 e λ2 = −1.Para λ1 = 5,

(A − 5I )u = 0⇒ ( 1 −71 −7

)( ab

) = ( 00

)⇒ a − 7b = 0.

Sendo assim, u = (7,1).



Seja A = ( 6 −71 −2

).

A solucao do PVI e dada por

x(t) = exp(At)x(0),




(A − 5I )u = 0⇒ ( 1 −71 −7

)( ab

) = ( 00

)⇒ a − 7b = 0.




Seja A = ( 6 −71 −2


x(t) = exp(At)x(0),




(A − 5I )u = 0⇒ ( 1 −71 −7

)( ab

) = ( 00

)⇒ a − 7b = 0.




Seja A = ( 6 −71 −2


x(t) = exp(At)x(0),

entao precisamos determinar exp(At).

Temos que,



(A − 5I )u = 0⇒ ( 1 −71 −7

)( ab

) = ( 00

)⇒ a − 7b = 0.




Seja A = ( 6 −71 −2


x(t) = exp(At)x(0),


p(λ) = λ2 − tr(A)λ + det(A) =

λ2 − 4λ − 5 = (λ − 5)(λ + 1),


(A − 5I )u = 0⇒ ( 1 −71 −7

)( ab

) = ( 00

)⇒ a − 7b = 0.




Seja A = ( 6 −71 −2


x(t) = exp(At)x(0),


p(λ) = λ2 − tr(A)λ + det(A) = λ2 − 4λ − 5 =

(λ − 5)(λ + 1),


(A − 5I )u = 0⇒ ( 1 −71 −7

)( ab

) = ( 00

)⇒ a − 7b = 0.




Seja A = ( 6 −71 −2


x(t) = exp(At)x(0),




(A − 5I )u = 0⇒ ( 1 −71 −7

)( ab

) = ( 00

)⇒ a − 7b = 0.




Seja A = ( 6 −71 −2


x(t) = exp(At)x(0),



assim, os autovalores de A sao λ1 = 5 e λ2 = −1.

Para λ1 = 5,

(A − 5I )u = 0⇒ ( 1 −71 −7

)( ab

) = ( 00

)⇒ a − 7b = 0.




Seja A = ( 6 −71 −2


x(t) = exp(At)x(0),




(A − 5I )u = 0⇒

( 1 −71 −7

)( ab

) = ( 00

)⇒ a − 7b = 0.




Seja A = ( 6 −71 −2


x(t) = exp(At)x(0),




(A − 5I )u = 0⇒ ( 1 −71 −7

)( ab

) = ( 00

)⇒

a − 7b = 0.




Seja A = ( 6 −71 −2


x(t) = exp(At)x(0),




(A − 5I )u = 0⇒ ( 1 −71 −7

)( ab

) = ( 00

)⇒ a − 7b = 0.




Seja A = ( 6 −71 −2


x(t) = exp(At)x(0),




(A − 5I )u = 0⇒ ( 1 −71 −7

)( ab

) = ( 00

)⇒ a − 7b = 0.


Exercıcio 1

Exercıcio


Para λ2 = −1,

(A + I )v = 0⇒

( 7 −71 −1

)( ab

) = ( 00

)⇒ a − b = 0.

Sendo assim, v = (1,1).Entao,

B = ( 5 00 −1

) e M = ( 7 11 1

) .

Sabemos que, M−1 = 1det(M) (

1 −1−1 7

) =⎛⎜⎝

16 −1

6

−16

76

⎞⎟⎠

.

Exercıcio 1

Exercıcio


Para λ2 = −1,

(A + I )v = 0⇒ ( 7 −71 −1

)( ab

) = ( 00

)⇒

a − b = 0.


B = ( 5 00 −1

) e M = ( 7 11 1

) .


1 −1−1 7

) =⎛⎜⎝

16 −1

6

−16

76

⎞⎟⎠

.

Exercıcio 1

Exercıcio


Para λ2 = −1,

(A + I )v = 0⇒ ( 7 −71 −1

)( ab

) = ( 00

)⇒ a − b = 0.


B = ( 5 00 −1

) e M = ( 7 11 1

) .


1 −1−1 7

) =⎛⎜⎝

16 −1

6

−16

76

⎞⎟⎠

.

Exercıcio 1

Exercıcio


Para λ2 = −1,

(A + I )v = 0⇒ ( 7 −71 −1

)( ab

) = ( 00

)⇒ a − b = 0.

Sendo assim, v = (1,1).

Entao,

B = ( 5 00 −1

) e M = ( 7 11 1

) .


1 −1−1 7

) =⎛⎜⎝

16 −1

6

−16

76

⎞⎟⎠

.

Exercıcio 1

Exercıcio


Para λ2 = −1,

(A + I )v = 0⇒ ( 7 −71 −1

)( ab

) = ( 00

)⇒ a − b = 0.


B = ( 5 00 −1

) e M = ( 7 11 1

) .


1 −1−1 7

) =⎛⎜⎝

16 −1

6

−16

76

⎞⎟⎠

.

Exercıcio 1

Exercıcio


Para λ2 = −1,

(A + I )v = 0⇒ ( 7 −71 −1

)( ab

) = ( 00

)⇒ a − b = 0.


B = ( 5 00 −1

) e M = ( 7 11 1

) .


1 −1−1 7

) =

⎛⎜⎝

16 −1

6

−16

76

⎞⎟⎠

.

Exercıcio 1

Exercıcio


Para λ2 = −1,

(A + I )v = 0⇒ ( 7 −71 −1

)( ab

) = ( 00

)⇒ a − b = 0.


B = ( 5 00 −1

) e M = ( 7 11 1

) .


1 −1−1 7

) =⎛⎜⎝

16 −1

6

−16

76

⎞⎟⎠

.

Exercıcio 1

Exercıcio


Alem disso, exp(Bt) = ( e5t 00 e−t ).

Logo,

exp(At) =Mexp(Bt)M−1 =⎛⎜⎝

− e−t

6 +7e5t

67e−t

6 −7e5t

6

− e−t

6 +e5t

67e−t

6 −e5t

6

⎞⎟⎠.

Portanto, se x(0) = (α,β) e a condicao inicial, temos que asolucao e dada por

x(t) =⎛⎜⎝

− e−t

6 +7e5t

67e−t

6 −7e5t

6

− e−t

6 +e5t

67e−t

6 −e5t

6

⎞⎟⎠

⎛⎜⎝

α

β

⎞⎟⎠.

Exercıcio 1

Exercıcio


Alem disso, exp(Bt) = ( e5t 00 e−t ). Logo,

exp(At) =Mexp(Bt)M−1 =

⎛⎜⎝

− e−t

6 +7e5t

67e−t

6 −7e5t

6

− e−t

6 +e5t

67e−t

6 −e5t

6

⎞⎟⎠.


x(t) =⎛⎜⎝

− e−t

6 +7e5t

67e−t

6 −7e5t

6

− e−t

6 +e5t

67e−t

6 −e5t

6

⎞⎟⎠

⎛⎜⎝

α

β

⎞⎟⎠.

Exercıcio 1

Exercıcio




− e−t

6 +7e5t

67e−t

6 −7e5t

6

− e−t

6 +e5t

67e−t

6 −e5t

6

⎞⎟⎠.


x(t) =⎛⎜⎝

− e−t

6 +7e5t

67e−t

6 −7e5t

6

− e−t

6 +e5t

67e−t

6 −e5t

6

⎞⎟⎠

⎛⎜⎝

α

β

⎞⎟⎠.

Exercıcio 1

Exercıcio




− e−t

6 +7e5t

67e−t

6 −7e5t

6

− e−t

6 +e5t

67e−t

6 −e5t

6

⎞⎟⎠.

Portanto, se x(0) = (α,β) e a condicao inicial,

temos que asolucao e dada por

x(t) =⎛⎜⎝

− e−t

6 +7e5t

67e−t

6 −7e5t

6

− e−t

6 +e5t

67e−t

6 −e5t

6

⎞⎟⎠

⎛⎜⎝

α

β

⎞⎟⎠.

Exercıcio 1

Exercıcio




− e−t

6 +7e5t

67e−t

6 −7e5t

6

− e−t

6 +e5t

67e−t

6 −e5t

6

⎞⎟⎠.


x(t) =⎛⎜⎝

− e−t

6 +7e5t

67e−t

6 −7e5t

6

− e−t

6 +e5t

67e−t

6 −e5t

6

⎞⎟⎠

⎛⎜⎝

α

β

⎞⎟⎠.



-2 -1 0 1 2

-2

-1

0

1

2


Encontre a solucao do PVI x = −2x + y , y = x − 2y .

Seja A = ( −2 11 −2


x(t) = exp(At)x(0),


p(λ) = λ2 − tr(A)λ + det(A) = λ2 + 4λ + 3 = (λ + 1)(λ + 3),

assim, os autovalores de A sao λ1 = −1 e λ2 = −3.Para λ1 = −1,

(A + I )u = 0⇒ ( −1 11 −1

)( ab

) = ( 00

)⇒ a − b = 0.




Seja A = ( −2 11 −2

).


x(t) = exp(At)x(0),




(A + I )u = 0⇒ ( −1 11 −1

)( ab

) = ( 00

)⇒ a − b = 0.




Seja A = ( −2 11 −2


x(t) = exp(At)x(0),




(A + I )u = 0⇒ ( −1 11 −1

)( ab

) = ( 00

)⇒ a − b = 0.




Seja A = ( −2 11 −2


x(t) = exp(At)x(0),


Temos que,



(A + I )u = 0⇒ ( −1 11 −1

)( ab

) = ( 00

)⇒ a − b = 0.




Seja A = ( −2 11 −2


x(t) = exp(At)x(0),


p(λ) = λ2 − tr(A)λ + det(A) =

λ2 + 4λ + 3 = (λ + 1)(λ + 3),


(A + I )u = 0⇒ ( −1 11 −1

)( ab

) = ( 00

)⇒ a − b = 0.




Seja A = ( −2 11 −2


x(t) = exp(At)x(0),


p(λ) = λ2 − tr(A)λ + det(A) = λ2 + 4λ + 3 =

(λ + 1)(λ + 3),


(A + I )u = 0⇒ ( −1 11 −1

)( ab

) = ( 00

)⇒ a − b = 0.




Seja A = ( −2 11 −2


x(t) = exp(At)x(0),




(A + I )u = 0⇒ ( −1 11 −1

)( ab

) = ( 00

)⇒ a − b = 0.




Seja A = ( −2 11 −2


x(t) = exp(At)x(0),



assim, os autovalores de A sao λ1 = −1 e λ2 = −3.

Para λ1 = −1,

(A + I )u = 0⇒ ( −1 11 −1

)( ab

) = ( 00

)⇒ a − b = 0.




Seja A = ( −2 11 −2


x(t) = exp(At)x(0),




(A + I )u = 0⇒

( −1 11 −1

)( ab

) = ( 00

)⇒ a − b = 0.




Seja A = ( −2 11 −2


x(t) = exp(At)x(0),




(A + I )u = 0⇒ ( −1 11 −1

)( ab

) = ( 00

)⇒

a − b = 0.




Seja A = ( −2 11 −2


x(t) = exp(At)x(0),




(A + I )u = 0⇒ ( −1 11 −1

)( ab

) = ( 00

)⇒ a − b = 0.




Seja A = ( −2 11 −2


x(t) = exp(At)x(0),




(A + I )u = 0⇒ ( −1 11 −1

)( ab

) = ( 00

)⇒ a − b = 0.


Exercıcio 2

Exercıcio


Para λ2 = −3,

(A + 3I )v = 0⇒

( 1 11 1

)( ab

) = ( 00

)⇒ a + b = 0.

Sendo assim, v = (1,−1).Entao,

B = ( −1 00 −3

) e M = ( 1 11 −1

) .


−1 −1−1 1

) =⎛⎜⎝

12

12

12 −1

2

⎞⎟⎠

.

Exercıcio 2

Exercıcio


Para λ2 = −3,

(A + 3I )v = 0⇒ ( 1 11 1

)( ab

) = ( 00

)⇒

a + b = 0.


B = ( −1 00 −3

) e M = ( 1 11 −1

) .


−1 −1−1 1

) =⎛⎜⎝

12

12

12 −1

2

⎞⎟⎠

.

Exercıcio 2

Exercıcio


Para λ2 = −3,

(A + 3I )v = 0⇒ ( 1 11 1

)( ab

) = ( 00

)⇒ a + b = 0.


B = ( −1 00 −3

) e M = ( 1 11 −1

) .


−1 −1−1 1

) =⎛⎜⎝

12

12

12 −1

2

⎞⎟⎠

.

Exercıcio 2

Exercıcio


Para λ2 = −3,

(A + 3I )v = 0⇒ ( 1 11 1

)( ab

) = ( 00

)⇒ a + b = 0.

Sendo assim, v = (1,−1).

Entao,

B = ( −1 00 −3

) e M = ( 1 11 −1

) .


−1 −1−1 1

) =⎛⎜⎝

12

12

12 −1

2

⎞⎟⎠

.

Exercıcio 2

Exercıcio


Para λ2 = −3,

(A + 3I )v = 0⇒ ( 1 11 1

)( ab

) = ( 00

)⇒ a + b = 0.


B = ( −1 00 −3

) e M = ( 1 11 −1

) .


−1 −1−1 1

) =⎛⎜⎝

12

12

12 −1

2

⎞⎟⎠

.

Exercıcio 2

Exercıcio


Para λ2 = −3,

(A + 3I )v = 0⇒ ( 1 11 1

)( ab

) = ( 00

)⇒ a + b = 0.


B = ( −1 00 −3

) e M = ( 1 11 −1

) .


−1 −1−1 1

) =

⎛⎜⎝

12

12

12 −1

2

⎞⎟⎠

.

Exercıcio 2

Exercıcio


Para λ2 = −3,

(A + 3I )v = 0⇒ ( 1 11 1

)( ab

) = ( 00

)⇒ a + b = 0.


B = ( −1 00 −3

) e M = ( 1 11 −1

) .


−1 −1−1 1

) =⎛⎜⎝

12

12

12 −1

2

⎞⎟⎠

.

Exercıcio 2

Exercıcio


Alem disso, exp(Bt) = ( e−t 00 e−3t ).

Logo,


e−3t

2 +e−t

2 − e−3t

2 +e−t

2

− e−3t

2 +e−t

2e−3t

2 +e−t

2

⎞⎟⎠.


x(t) =⎛⎜⎝

e−3t

2 +e−t

2 − e−3t

2 +e−t

2

− e−3t

2 +e−t

2e−3t

2 +e−t

2

⎞⎟⎠

⎛⎜⎝

α

β

⎞⎟⎠.

Exercıcio 2

Exercıcio


Alem disso, exp(Bt) = ( e−t 00 e−3t ). Logo,

exp(At) =Mexp(Bt)M−1 =

⎛⎜⎝

e−3t

2 +e−t

2 − e−3t

2 +e−t

2

− e−3t

2 +e−t

2e−3t

2 +e−t

2

⎞⎟⎠.


x(t) =⎛⎜⎝

e−3t

2 +e−t

2 − e−3t

2 +e−t

2

− e−3t

2 +e−t

2e−3t

2 +e−t

2

⎞⎟⎠

⎛⎜⎝

α

β

⎞⎟⎠.

Exercıcio 2

Exercıcio




e−3t

2 +e−t

2 − e−3t

2 +e−t

2

− e−3t

2 +e−t

2e−3t

2 +e−t

2

⎞⎟⎠.


x(t) =⎛⎜⎝

e−3t

2 +e−t

2 − e−3t

2 +e−t

2

− e−3t

2 +e−t

2e−3t

2 +e−t

2

⎞⎟⎠

⎛⎜⎝

α

β

⎞⎟⎠.

Exercıcio 2

Exercıcio




e−3t

2 +e−t

2 − e−3t

2 +e−t

2

− e−3t

2 +e−t

2e−3t

2 +e−t

2

⎞⎟⎠.



x(t) =⎛⎜⎝

e−3t

2 +e−t

2 − e−3t

2 +e−t

2

− e−3t

2 +e−t

2e−3t

2 +e−t

2

⎞⎟⎠

⎛⎜⎝

α

β

⎞⎟⎠.

Exercıcio 2

Exercıcio




e−3t

2 +e−t

2 − e−3t

2 +e−t

2

− e−3t

2 +e−t

2e−3t

2 +e−t

2

⎞⎟⎠.


x(t) =⎛⎜⎝

e−3t

2 +e−t

2 − e−3t

2 +e−t

2

− e−3t

2 +e−t

2e−3t

2 +e−t

2

⎞⎟⎠

⎛⎜⎝

α

β

⎞⎟⎠.



-2 -1 0 1 2

-2

-1

0

1

2


Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .

Seja A = ( 1 −22 5


x(t) = exp(At)x(0),


p(λ) = λ2 − tr(A)λ + det(A) = λ2 − 6λ + 9 = (λ − 3)2,

assim, a matriz A possui apenas um autovalor que e λ = 3.Para λ = 3,

(A − 3I )u = 0⇒ ( −2 −22 2

)( ab

) = ( 00

)⇒ 2a + 2b = 0.

Sendo assim, u = (1,−1).



Seja A = ( 1 −22 5

).


x(t) = exp(At)x(0),


p(λ) = λ2 − tr(A)λ + det(A) = λ2 − 6λ + 9 = (λ − 3)2,


(A − 3I )u = 0⇒ ( −2 −22 2

)( ab

) = ( 00

)⇒ 2a + 2b = 0.




Seja A = ( 1 −22 5


x(t) = exp(At)x(0),


p(λ) = λ2 − tr(A)λ + det(A) = λ2 − 6λ + 9 = (λ − 3)2,


(A − 3I )u = 0⇒ ( −2 −22 2

)( ab

) = ( 00

)⇒ 2a + 2b = 0.




Seja A = ( 1 −22 5


x(t) = exp(At)x(0),


Temos que,

p(λ) = λ2 − tr(A)λ + det(A) = λ2 − 6λ + 9 = (λ − 3)2,


(A − 3I )u = 0⇒ ( −2 −22 2

)( ab

) = ( 00

)⇒ 2a + 2b = 0.




Seja A = ( 1 −22 5


x(t) = exp(At)x(0),


p(λ) = λ2 − tr(A)λ + det(A) =

λ2 − 6λ + 9 = (λ − 3)2,


(A − 3I )u = 0⇒ ( −2 −22 2

)( ab

) = ( 00

)⇒ 2a + 2b = 0.




Seja A = ( 1 −22 5


x(t) = exp(At)x(0),


p(λ) = λ2 − tr(A)λ + det(A) = λ2 − 6λ + 9 =

(λ − 3)2,


(A − 3I )u = 0⇒ ( −2 −22 2

)( ab

) = ( 00

)⇒ 2a + 2b = 0.




Seja A = ( 1 −22 5


x(t) = exp(At)x(0),


p(λ) = λ2 − tr(A)λ + det(A) = λ2 − 6λ + 9 = (λ − 3)2,


(A − 3I )u = 0⇒ ( −2 −22 2

)( ab

) = ( 00

)⇒ 2a + 2b = 0.




Seja A = ( 1 −22 5


x(t) = exp(At)x(0),


p(λ) = λ2 − tr(A)λ + det(A) = λ2 − 6λ + 9 = (λ − 3)2,

assim, a matriz A possui apenas um autovalor que e λ = 3.

Para λ = 3,

(A − 3I )u = 0⇒ ( −2 −22 2

)( ab

) = ( 00

)⇒ 2a + 2b = 0.




Seja A = ( 1 −22 5


x(t) = exp(At)x(0),


p(λ) = λ2 − tr(A)λ + det(A) = λ2 − 6λ + 9 = (λ − 3)2,


(A − 3I )u = 0⇒

( −2 −22 2

)( ab

) = ( 00

)⇒ 2a + 2b = 0.




Seja A = ( 1 −22 5


x(t) = exp(At)x(0),


p(λ) = λ2 − tr(A)λ + det(A) = λ2 − 6λ + 9 = (λ − 3)2,


(A − 3I )u = 0⇒ ( −2 −22 2

)( ab

) = ( 00

)⇒

2a + 2b = 0.




Seja A = ( 1 −22 5


x(t) = exp(At)x(0),


p(λ) = λ2 − tr(A)λ + det(A) = λ2 − 6λ + 9 = (λ − 3)2,


(A − 3I )u = 0⇒ ( −2 −22 2

)( ab

) = ( 00

)⇒ 2a + 2b = 0.




Seja A = ( 1 −22 5


x(t) = exp(At)x(0),


p(λ) = λ2 − tr(A)λ + det(A) = λ2 − 6λ + 9 = (λ − 3)2,


(A − 3I )u = 0⇒ ( −2 −22 2

)( ab

) = ( 00

)⇒ 2a + 2b = 0.


Exercıcio 3

Exercıcio


Vamos agora determinar v a partir de u,

(A − 3I )v = u⇒

( −2 −22 2

)( ab

) = ( 1−1

)⇒ 2a + 2b = −1.


Entao,

B = ( 3 10 3

) e M = ( 1 1−1 −3

2

) .


−32 −1

1 1) = ( 3 2

−2 −2).

Exercıcio 3

Exercıcio



(A − 3I )v = u⇒ ( −2 −22 2

)( ab

) = ( 1−1

)⇒

2a + 2b = −1.


Entao,

B = ( 3 10 3

) e M = ( 1 1−1 −3

2

) .


−32 −1

1 1) = ( 3 2

−2 −2).

Exercıcio 3

Exercıcio



(A − 3I )v = u⇒ ( −2 −22 2

)( ab

) = ( 1−1

)⇒ 2a + 2b = −1.


Entao,

B = ( 3 10 3

) e M = ( 1 1−1 −3

2

) .


−32 −1

1 1) = ( 3 2

−2 −2).

Exercıcio 3

Exercıcio



(A − 3I )v = u⇒ ( −2 −22 2

)( ab

) = ( 1−1

)⇒ 2a + 2b = −1.


Entao,

B = ( 3 10 3

) e M = ( 1 1−1 −3

2

) .


−32 −1

1 1) = ( 3 2

−2 −2).

Exercıcio 3

Exercıcio



(A − 3I )v = u⇒ ( −2 −22 2

)( ab

) = ( 1−1

)⇒ 2a + 2b = −1.


Entao,

B = ( 3 10 3

) e M = ( 1 1−1 −3

2

) .


−32 −1

1 1) = ( 3 2

−2 −2).

Exercıcio 3

Exercıcio



(A − 3I )v = u⇒ ( −2 −22 2

)( ab

) = ( 1−1

)⇒ 2a + 2b = −1.


Entao,

B = ( 3 10 3

) e M = ( 1 1−1 −3

2

) .


−32 −1

1 1) =

( 3 2−2 −2

).

Exercıcio 3

Exercıcio



(A − 3I )v = u⇒ ( −2 −22 2

)( ab

) = ( 1−1

)⇒ 2a + 2b = −1.


Entao,

B = ( 3 10 3

) e M = ( 1 1−1 −3

2

) .


−32 −1

1 1) = ( 3 2

−2 −2).

Exercıcio 3

Exercıcio


Alem disso, exp(Bt) = ( e3t te3t

0 e3t).

Logo,

exp(At) =Mexp(Bt)M−1 = ( e3t(1 − 2t) −2e3tt2e3tt e3t(2t + 1) ) .


x(t) = ( e3t(1 − 2t) −2e3tt2e3tt e3t(2t + 1) )( α

β) .

Exercıcio 3

Exercıcio



0 e3t). Logo,




β) .

Exercıcio 3

Exercıcio



0 e3t). Logo,





β) .

Exercıcio 3

Exercıcio



0 e3t). Logo,




β) .



-2 -1 0 1 2

-2

-1

0

1

2


Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .

Seja A = ( 3 −34 −3


x(t) = exp(At)x(0),entao precisamos determinar exp(At). Temos que,

p(λ) = λ2 − tr(A)λ + det(A) = λ2 + 3,

assim, os autovalores de A sao λ1 =√

3i e λ2 = −√

3i .Para λ1 =

√3i ,

[A − (√

3i)I ]u = 0⇒ ( 3 −√

3i −3

4 −3 −√

3i)( a

b) = ( 0

0)

⇒ 4a + (−3 −√

3i)b = 0⇒ a = (3 +√

3i

4)b⇒ u = (3 +

√3i ,4).



Seja A = ( 3 −34 −3

).



p(λ) = λ2 − tr(A)λ + det(A) = λ2 + 3,


3i e λ2 = −√

3i .Para λ1 =

√3i ,

[A − (√

3i)I ]u = 0⇒ ( 3 −√

3i −3

4 −3 −√

3i)( a

b) = ( 0

0)

⇒ 4a + (−3 −√

3i)b = 0⇒ a = (3 +√

3i

4)b⇒ u = (3 +

√3i ,4).



Seja A = ( 3 −34 −3


x(t) = exp(At)x(0),


p(λ) = λ2 − tr(A)λ + det(A) = λ2 + 3,


3i e λ2 = −√

3i .Para λ1 =

√3i ,

[A − (√

3i)I ]u = 0⇒ ( 3 −√

3i −3

4 −3 −√

3i)( a

b) = ( 0

0)

⇒ 4a + (−3 −√

3i)b = 0⇒ a = (3 +√

3i

4)b⇒ u = (3 +

√3i ,4).



Seja A = ( 3 −34 −3


x(t) = exp(At)x(0),entao precisamos determinar exp(At).

Temos que,

p(λ) = λ2 − tr(A)λ + det(A) = λ2 + 3,


3i e λ2 = −√

3i .Para λ1 =

√3i ,

[A − (√

3i)I ]u = 0⇒ ( 3 −√

3i −3

4 −3 −√

3i)( a

b) = ( 0

0)

⇒ 4a + (−3 −√

3i)b = 0⇒ a = (3 +√

3i

4)b⇒ u = (3 +

√3i ,4).



Seja A = ( 3 −34 −3



p(λ) = λ2 − tr(A)λ + det(A) =

λ2 + 3,


3i e λ2 = −√

3i .Para λ1 =

√3i ,

[A − (√

3i)I ]u = 0⇒ ( 3 −√

3i −3

4 −3 −√

3i)( a

b) = ( 0

0)

⇒ 4a + (−3 −√

3i)b = 0⇒ a = (3 +√

3i

4)b⇒ u = (3 +

√3i ,4).



Seja A = ( 3 −34 −3



p(λ) = λ2 − tr(A)λ + det(A) = λ2 + 3,


3i e λ2 = −√

3i .Para λ1 =

√3i ,

[A − (√

3i)I ]u = 0⇒ ( 3 −√

3i −3

4 −3 −√

3i)( a

b) = ( 0

0)

⇒ 4a + (−3 −√

3i)b = 0⇒ a = (3 +√

3i

4)b⇒ u = (3 +

√3i ,4).



Seja A = ( 3 −34 −3



p(λ) = λ2 − tr(A)λ + det(A) = λ2 + 3,


3i e λ2 = −√

3i .

Para λ1 =√

3i ,

[A − (√

3i)I ]u = 0⇒ ( 3 −√

3i −3

4 −3 −√

3i)( a

b) = ( 0

0)

⇒ 4a + (−3 −√

3i)b = 0⇒ a = (3 +√

3i

4)b⇒ u = (3 +

√3i ,4).



Seja A = ( 3 −34 −3



p(λ) = λ2 − tr(A)λ + det(A) = λ2 + 3,


3i e λ2 = −√

3i .Para λ1 =

√3i ,

[A − (√

3i)I ]u = 0⇒

( 3 −√

3i −3

4 −3 −√

3i)( a

b) = ( 0

0)

⇒ 4a + (−3 −√

3i)b = 0⇒ a = (3 +√

3i

4)b⇒ u = (3 +

√3i ,4).



Seja A = ( 3 −34 −3



p(λ) = λ2 − tr(A)λ + det(A) = λ2 + 3,


3i e λ2 = −√

3i .Para λ1 =

√3i ,

[A − (√

3i)I ]u = 0⇒ ( 3 −√

3i −3

4 −3 −√

3i)( a

b) = ( 0

0)

⇒

4a + (−3 −√

3i)b = 0⇒ a = (3 +√

3i

4)b⇒ u = (3 +

√3i ,4).



Seja A = ( 3 −34 −3



p(λ) = λ2 − tr(A)λ + det(A) = λ2 + 3,


3i e λ2 = −√

3i .Para λ1 =

√3i ,

[A − (√

3i)I ]u = 0⇒ ( 3 −√

3i −3

4 −3 −√

3i)( a

b) = ( 0

0)

⇒ 4a + (−3 −√

3i)b = 0⇒

a = (3 +√

3i

4)b⇒ u = (3 +

√3i ,4).



Seja A = ( 3 −34 −3



p(λ) = λ2 − tr(A)λ + det(A) = λ2 + 3,


3i e λ2 = −√

3i .Para λ1 =

√3i ,

[A − (√

3i)I ]u = 0⇒ ( 3 −√

3i −3

4 −3 −√

3i)( a

b) = ( 0

0)

⇒ 4a + (−3 −√

3i)b = 0⇒ a = (3 +√

3i

4)b⇒

u = (3 +√

3i ,4).



Seja A = ( 3 −34 −3



p(λ) = λ2 − tr(A)λ + det(A) = λ2 + 3,


3i e λ2 = −√

3i .Para λ1 =

√3i ,

[A − (√

3i)I ]u = 0⇒ ( 3 −√

3i −3

4 −3 −√

3i)( a

b) = ( 0

0)

⇒ 4a + (−3 −√

3i)b = 0⇒ a = (3 +√

3i

4)b⇒ u = (3 +

√3i ,4).

Exercıcio 4

Exercıcio


Sendo assim, u = (3,4) + i(√

3,0).

Entao,

B = ( 0 −√

3√3 0

) e M = ( 3√

34 0

) .


0 −√

3−4 3

) =⎛⎜⎝

0 14

1√3−√34

⎞⎟⎠

.

Alem disso,

exp(Bt) = ( cos(√

3t) − sen(√

3t)sen(

√3t) cos(

√3t) ) .

Exercıcio 4

Exercıcio



3,0).Entao,

B = ( 0 −√

3√3 0

) e M = ( 3√

34 0

) .


0 −√

3−4 3

) =⎛⎜⎝

0 14

1√3−√34

⎞⎟⎠

.

Alem disso,

exp(Bt) = ( cos(√

3t) − sen(√

3t)sen(

√3t) cos(

√3t) ) .

Exercıcio 4

Exercıcio



3,0).Entao,

B = ( 0 −√

3√3 0

) e M = ( 3√

34 0

) .


0 −√

3−4 3

) =

⎛⎜⎝

0 14

1√3−√34

⎞⎟⎠

.

Alem disso,

exp(Bt) = ( cos(√

3t) − sen(√

3t)sen(

√3t) cos(

√3t) ) .

Exercıcio 4

Exercıcio



3,0).Entao,

B = ( 0 −√

3√3 0

) e M = ( 3√

34 0

) .


0 −√

3−4 3

) =⎛⎜⎝

0 14

1√3−√34

⎞⎟⎠

.

Alem disso,

exp(Bt) = ( cos(√

3t) − sen(√

3t)sen(

√3t) cos(

√3t) ) .

Exercıcio 4

Exercıcio



3,0).Entao,

B = ( 0 −√

3√3 0

) e M = ( 3√

34 0

) .


0 −√

3−4 3

) =⎛⎜⎝

0 14

1√3−√34

⎞⎟⎠

.

Alem disso,

exp(Bt) = ( cos(√

3t) − sen(√

3t)sen(

√3t) cos(

√3t) ) .

Exercıcio 4

Exercıcio


Logo, exp(At) =Mexp(Bt)M−1

=⎛⎜⎝

cos (√

3t) −√

3 sin (√

3t)√

3 sin (√

3t)

−4 sin(√3t)√3

cos (√

3t) + sin (√

3t)√

3

⎞⎟⎠.


x(t) =⎛⎜⎝

cos (√

3t) −√

3 sin (√

3t)√

3 sin (√

3t)

− 4 sin(√

3t)√

3cos (

√3t) + sin (

√3t)

√3

⎞⎟⎠

⎛⎜⎝α

β

⎞⎟⎠.

Exercıcio 4

Exercıcio



=⎛⎜⎝

cos (√

3t) −√

3 sin (√

3t)√

3 sin (√

3t)

−4 sin(√3t)√3

cos (√

3t) + sin (√

3t)√

3

⎞⎟⎠.


x(t) =⎛⎜⎝

cos (√

3t) −√

3 sin (√

3t)√

3 sin (√

3t)

− 4 sin(√

3t)√

3cos (

√3t) + sin (

√3t)

√3

⎞⎟⎠

⎛⎜⎝α

β

⎞⎟⎠.

Exercıcio 4

Exercıcio



=⎛⎜⎝

cos (√

3t) −√

3 sin (√

3t)√

3 sin (√

3t)

−4 sin(√3t)√3

cos (√

3t) + sin (√

3t)√

3

⎞⎟⎠.



x(t) =⎛⎜⎝

cos (√

3t) −√

3 sin (√

3t)√

3 sin (√

3t)

− 4 sin(√

3t)√

3cos (

√3t) + sin (

√3t)

√3

⎞⎟⎠

⎛⎜⎝α

β

⎞⎟⎠.

Exercıcio 4

Exercıcio



=⎛⎜⎝

cos (√

3t) −√

3 sin (√

3t)√

3 sin (√

3t)

−4 sin(√3t)√3

cos (√

3t) + sin (√

3t)√

3

⎞⎟⎠.


x(t) =⎛⎜⎝

cos (√

3t) −√

3 sin (√

3t)√

3 sin (√

3t)

− 4 sin(√

3t)√

3cos (

√3t) + sin (

√3t)

√3

⎞⎟⎠

⎛⎜⎝α

β

⎞⎟⎠.



-2 -1 0 1 2

-2

-1

0

1

2


Encontre a solucao do PVI x = 2x − y + 2, y = x + 2y + 1.

Sejam A = ( 2 −11 2

) e b = ( 21

), entao

x = Ax + b⇒ x −Ax = b⇒ e−At x − e−AtAx = e−Atb

⇒ d

dt(e−Atx(t)) = e−Atb⇒ e−Atx(t) = ∫ e−Atb dt.

Temos que

−A = ( −2 1−1 −2

) ,

assim,

e−At =⎛⎜⎝

e−2t cos(−t) −e−2t sen(−t)e−2t sen(−t) e−2t cos(−t)

⎞⎟⎠=

⎛⎜⎝

e−2t cos(t) e−2t sen(t)−e−2t sen(t) e−2t cos(t)

⎞⎟⎠



Sejam A = ( 2 −11 2

) e b = ( 21

),

entao


⇒ d


Temos que

−A = ( −2 1−1 −2

) ,

assim,

e−At =⎛⎜⎝


⎞⎟⎠=

⎛⎜⎝


⎞⎟⎠



Sejam A = ( 2 −11 2

) e b = ( 21

), entao

x = Ax + b⇒

x −Ax = b⇒ e−At x − e−AtAx = e−Atb

⇒ d


Temos que

−A = ( −2 1−1 −2

) ,

assim,

e−At =⎛⎜⎝


⎞⎟⎠=

⎛⎜⎝


⎞⎟⎠



Sejam A = ( 2 −11 2

) e b = ( 21

), entao

x = Ax + b⇒ x −Ax = b⇒

e−At x − e−AtAx = e−Atb

⇒ d


Temos que

−A = ( −2 1−1 −2

) ,

assim,

e−At =⎛⎜⎝


⎞⎟⎠=

⎛⎜⎝


⎞⎟⎠



Sejam A = ( 2 −11 2

) e b = ( 21

), entao


⇒

d


Temos que

−A = ( −2 1−1 −2

) ,

assim,

e−At =⎛⎜⎝


⎞⎟⎠=

⎛⎜⎝


⎞⎟⎠



Sejam A = ( 2 −11 2

) e b = ( 21

), entao


⇒ d

dt(e−Atx(t)) = e−Atb⇒

e−Atx(t) = ∫ e−Atb dt.

Temos que

−A = ( −2 1−1 −2

) ,

assim,

e−At =⎛⎜⎝


⎞⎟⎠=

⎛⎜⎝


⎞⎟⎠



Sejam A = ( 2 −11 2

) e b = ( 21

), entao


⇒ d


Temos que

−A = ( −2 1−1 −2

) ,

assim,

e−At =⎛⎜⎝


⎞⎟⎠=

⎛⎜⎝


⎞⎟⎠



Sejam A = ( 2 −11 2

) e b = ( 21

), entao


⇒ d


Temos que

−A = ( −2 1−1 −2

) ,

assim,

e−At =⎛⎜⎝


⎞⎟⎠=

⎛⎜⎝


⎞⎟⎠



Sejam A = ( 2 −11 2

) e b = ( 21

), entao


⇒ d


Temos que

−A = ( −2 1−1 −2

) ,

assim,

e−At =⎛⎜⎝


⎞⎟⎠=

⎛⎜⎝


⎞⎟⎠



Sejam A = ( 2 −11 2

) e b = ( 21

), entao


⇒ d


Temos que

−A = ( −2 1−1 −2

) ,

assim,

e−At =⎛⎜⎝


⎞⎟⎠=

⎛⎜⎝


⎞⎟⎠



Desta forma,

e−Atb =⎛⎜⎝

2e−2t cos(t) + e−2t sen(t)

e−2t cos(t) − 2e−2t sen(t)

⎞⎟⎠.

Entao,

∫ e−Atb dt =⎛⎜⎝

−e−2t cos(t) + c1e−2t sen(t) + c2

⎞⎟⎠.

Note que

eAt =⎛⎜⎝

e2t cos(t) −e2t sen(t)

e2t sen(2t) e2t cos(t)

⎞⎟⎠.



Desta forma,

e−Atb =⎛⎜⎝



⎞⎟⎠.

Entao,



⎞⎟⎠.

Note que

eAt =⎛⎜⎝



⎞⎟⎠.



Desta forma,

e−Atb =⎛⎜⎝



⎞⎟⎠.

Entao,



⎞⎟⎠.

Note que

eAt =⎛⎜⎝



⎞⎟⎠.

Exercıcio 5

Exercıcio


Portanto,

x(t) =⎛⎜⎝



⎞⎟⎠

⎛⎜⎝


⎞⎟⎠

=⎛⎜⎝

e2t cos(t) (c1 − e−2t cos(t)) − e2t sen(t) (c2 + e−2t sen(t))

e2t (c1 − e−2t cos(t)) sen(t) + e2t cos(t) (c2 + e−2t sen(t))

⎞⎟⎠.

Exercıcio 5

Exercıcio


Portanto,

x(t) =⎛⎜⎝



⎞⎟⎠

⎛⎜⎝


⎞⎟⎠

=⎛⎜⎝

e2t cos(t) (c1 − e−2t cos(t)) − e2t sen(t) (c2 + e−2t sen(t))

e2t (c1 − e−2t cos(t)) sen(t) + e2t cos(t) (c2 + e−2t sen(t))

⎞⎟⎠.



-2 -1 0 1 2

-2

-1

0

1

2


Encontre a solucao do PVI x = 3x + 4y , y = 3x + 2y + t2.

Sejam A = ( 3 43 2

) e b(t) = ( 0t2

), entao

x = Ax + b(t)⇒ x −Ax = b(t)⇒ e−At x − e−AtAx = e−Atb(t)

⇒ d

dt(e−Atx(t)) = e−Atb(t)⇒ e−Atx(t) = ∫ e−Atb(t) dt.

Temos que

−A = ( −3 −4−3 −2

) ,

assim,

e−At =⎛⎜⎝

4e−6t

7 + 3et

74e−6t

7 − 4et

7

3e−6t

7 − 3et

73e−6t

7 + 4et

7

⎞⎟⎠.



Sejam A = ( 3 43 2

) e b(t) = ( 0t2

),

entao


⇒ d


Temos que

−A = ( −3 −4−3 −2

) ,

assim,

e−At =⎛⎜⎝

4e−6t

7 + 3et

74e−6t

7 − 4et

7

3e−6t

7 − 3et

73e−6t

7 + 4et

7

⎞⎟⎠.



Sejam A = ( 3 43 2

) e b(t) = ( 0t2

), entao

x = Ax + b(t)⇒

x −Ax = b(t)⇒ e−At x − e−AtAx = e−Atb(t)

⇒ d


Temos que

−A = ( −3 −4−3 −2

) ,

assim,

e−At =⎛⎜⎝

4e−6t

7 + 3et

74e−6t

7 − 4et

7

3e−6t

7 − 3et

73e−6t

7 + 4et

7

⎞⎟⎠.



Sejam A = ( 3 43 2

) e b(t) = ( 0t2

), entao

x = Ax + b(t)⇒ x −Ax = b(t)⇒

e−At x − e−AtAx = e−Atb(t)

⇒ d


Temos que

−A = ( −3 −4−3 −2

) ,

assim,

e−At =⎛⎜⎝

4e−6t

7 + 3et

74e−6t

7 − 4et

7

3e−6t

7 − 3et

73e−6t

7 + 4et

7

⎞⎟⎠.



Sejam A = ( 3 43 2

) e b(t) = ( 0t2

), entao


⇒

d


Temos que

−A = ( −3 −4−3 −2

) ,

assim,

e−At =⎛⎜⎝

4e−6t

7 + 3et

74e−6t

7 − 4et

7

3e−6t

7 − 3et

73e−6t

7 + 4et

7

⎞⎟⎠.



Sejam A = ( 3 43 2

) e b(t) = ( 0t2

), entao


⇒ d

dt(e−Atx(t)) = e−Atb(t)⇒

e−Atx(t) = ∫ e−Atb(t) dt.

Temos que

−A = ( −3 −4−3 −2

) ,

assim,

e−At =⎛⎜⎝

4e−6t

7 + 3et

74e−6t

7 − 4et

7

3e−6t

7 − 3et

73e−6t

7 + 4et

7

⎞⎟⎠.



Sejam A = ( 3 43 2

) e b(t) = ( 0t2

), entao


⇒ d


Temos que

−A = ( −3 −4−3 −2

) ,

assim,

e−At =⎛⎜⎝

4e−6t

7 + 3et

74e−6t

7 − 4et

7

3e−6t

7 − 3et

73e−6t

7 + 4et

7

⎞⎟⎠.



Sejam A = ( 3 43 2

) e b(t) = ( 0t2

), entao


⇒ d


Temos que

−A = ( −3 −4−3 −2

) ,

assim,

e−At =⎛⎜⎝

4e−6t

7 + 3et

74e−6t

7 − 4et

7

3e−6t

7 − 3et

73e−6t

7 + 4et

7

⎞⎟⎠.



Sejam A = ( 3 43 2

) e b(t) = ( 0t2

), entao


⇒ d


Temos que

−A = ( −3 −4−3 −2

) ,

assim,

e−At =⎛⎜⎝

4e−6t

7 + 3et

74e−6t

7 − 4et

7

3e−6t

7 − 3et

73e−6t

7 + 4et

7

⎞⎟⎠.

Exercıcio 6

Exercıcio


Desta forma,

e−Atb(t) =⎛⎜⎝

4t2e−6t

7 − 4t2et

7

3t2e−6t

7 + 4t2et

7

⎞⎟⎠.

Entao,

∫ e−Atb(t) dt =⎛⎜⎝− 4

7(et (t2 − 2t + 2) + 1

108e−6t (18t2 + 6t + 1)) + c1

47et (t2 − 2t + 2) − 1

252e−6t (18t2 + 6t + 1) + c2

⎞⎟⎠.

Note que

eAt =⎛⎜⎝

3e−t

7+ 4e6t

7− 4e−t

7+ 4e6t

7

− 3e−t

7+ 3e6t

74e−t

7+ 3e6t

7

⎞⎟⎠.

Exercıcio 6

Exercıcio


Desta forma,

e−Atb(t) =⎛⎜⎝

4t2e−6t

7 − 4t2et

7

3t2e−6t

7 + 4t2et

7

⎞⎟⎠.

Entao,

∫ e−Atb(t) dt =⎛⎜⎝− 4

7(et (t2 − 2t + 2) + 1

108e−6t (18t2 + 6t + 1)) + c1

47et (t2 − 2t + 2) − 1

252e−6t (18t2 + 6t + 1) + c2

⎞⎟⎠.

Note que

eAt =⎛⎜⎝

3e−t

7+ 4e6t

7− 4e−t

7+ 4e6t

7

− 3e−t

7+ 3e6t

74e−t

7+ 3e6t

7

⎞⎟⎠.

Exercıcio 6

Exercıcio


Desta forma,

e−Atb(t) =⎛⎜⎝

4t2e−6t

7 − 4t2et

7

3t2e−6t

7 + 4t2et

7

⎞⎟⎠.

Entao,

∫ e−Atb(t) dt =⎛⎜⎝− 4

7(et (t2 − 2t + 2) + 1

108e−6t (18t2 + 6t + 1)) + c1

47et (t2 − 2t + 2) − 1

252e−6t (18t2 + 6t + 1) + c2

⎞⎟⎠.

Note que

eAt =⎛⎜⎝

3e−t

7+ 4e6t

7− 4e−t

7+ 4e6t

7

− 3e−t

7+ 3e6t

74e−t

7+ 3e6t

7

⎞⎟⎠.

Exercıcio 6

Exercıcio


Portanto,

x(t) =

⎛

⎜⎜

⎝

3e−t

7+

4e6t

7−

4e−t

7+

4e6t

7

−3e−t

7+

3e6t

74e−t

7+

3e6t

7

⎞

⎟⎟

⎠

⎛

⎜⎜

⎝

−47(et (t2 − 2t + 2) + 1

108e−6t (18t2 + 6t + 1)) + c1

47et (t2 − 2t + 2) − 1

252e−6t (18t2 + 6t + 1) + c2

⎞

⎟⎟

⎠

=

⎛

⎜⎜

⎝

1189

e−t (108c2 (−1 + e7t) + 27c1 (3 + 4e7t) − 7et (18t2 − 30t + 31))

1252

e−t (108c1 (−1 + e7t) + 36c2 (4 + 3e7t) + 7et (18t2 − 42t + 41))

⎞

⎟⎟

⎠

.

Exercıcio 6

Exercıcio


Portanto,

x(t) =

⎛

⎜⎜

⎝

3e−t

7+

4e6t

7−

4e−t

7+

4e6t

7

−3e−t

7+

3e6t

74e−t

7+

3e6t

7

⎞

⎟⎟

⎠

⎛

⎜⎜

⎝

−47(et (t2 − 2t + 2) + 1

108e−6t (18t2 + 6t + 1)) + c1

47et (t2 − 2t + 2) − 1

252e−6t (18t2 + 6t + 1) + c2

⎞

⎟⎟

⎠

=

⎛

⎜⎜

⎝

1189

e−t (108c2 (−1 + e7t) + 27c1 (3 + 4e7t) − 7et (18t2 − 30t + 31))

1252

e−t (108c1 (−1 + e7t) + 36c2 (4 + 3e7t) + 7et (18t2 − 42t + 41))

⎞

⎟⎟

⎠

.

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rmiranda99.github.io · Exerc cio 4 Exerc cio Encontre a solu˘c~ao do PVI _x = 3x −3y, _y = 4x −3y. Logo, exp(At)= Mexp(Bt)M−1 = ™ Œ ﬂ cos› √ 3t”− √ 3sin›