rmiranda99.github.io · exerc cio 4 exerc cio encontre a solu˘c~ao do pvi _x = 3x −3y, _y = 4x...
TRANSCRIPT
![Page 1: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/1.jpg)
Exercıcios sobre sistemas lineares e
sistemas nao-homogeneos
Calculo III -Turma B, IMECC - UNICAMP
Mayara Duarte de Araujo Caldas
03/07/2020
![Page 2: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/2.jpg)
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).
![Page 3: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/3.jpg)
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).
![Page 4: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/4.jpg)
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).
![Page 5: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/5.jpg)
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).
![Page 6: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/6.jpg)
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).
![Page 7: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/7.jpg)
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).
![Page 8: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/8.jpg)
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).
![Page 9: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/9.jpg)
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).
![Page 10: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/10.jpg)
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).
![Page 11: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/11.jpg)
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).
![Page 12: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/12.jpg)
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).
![Page 13: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/13.jpg)
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).
![Page 14: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/14.jpg)
Exercıcio 1
Exercıcio
Encontre a solucao do PVI x = 6x − 7y , y = x − 2y .
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
⎞⎟⎠
.
![Page 15: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/15.jpg)
Exercıcio 1
Exercıcio
Encontre a solucao do PVI x = 6x − 7y , y = x − 2y .
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
⎞⎟⎠
.
![Page 16: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/16.jpg)
Exercıcio 1
Exercıcio
Encontre a solucao do PVI x = 6x − 7y , y = x − 2y .
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
⎞⎟⎠
.
![Page 17: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/17.jpg)
Exercıcio 1
Exercıcio
Encontre a solucao do PVI x = 6x − 7y , y = x − 2y .
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
⎞⎟⎠
.
![Page 18: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/18.jpg)
Exercıcio 1
Exercıcio
Encontre a solucao do PVI x = 6x − 7y , y = x − 2y .
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
⎞⎟⎠
.
![Page 19: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/19.jpg)
Exercıcio 1
Exercıcio
Encontre a solucao do PVI x = 6x − 7y , y = x − 2y .
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
⎞⎟⎠
.
![Page 20: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/20.jpg)
Exercıcio 1
Exercıcio
Encontre a solucao do PVI x = 6x − 7y , y = x − 2y .
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
⎞⎟⎠
.
![Page 21: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/21.jpg)
Exercıcio 1
Exercıcio
Encontre a solucao do PVI x = 6x − 7y , y = x − 2y .
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
⎞⎟⎠.
![Page 22: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/22.jpg)
Exercıcio 1
Exercıcio
Encontre a solucao do PVI x = 6x − 7y , y = x − 2y .
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
⎞⎟⎠.
![Page 23: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/23.jpg)
Exercıcio 1
Exercıcio
Encontre a solucao do PVI x = 6x − 7y , y = x − 2y .
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
⎞⎟⎠.
![Page 24: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/24.jpg)
Exercıcio 1
Exercıcio
Encontre a solucao do PVI x = 6x − 7y , y = x − 2y .
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
⎞⎟⎠.
![Page 25: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/25.jpg)
Exercıcio 1
Exercıcio
Encontre a solucao do PVI x = 6x − 7y , y = x − 2y .
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
⎞⎟⎠.
![Page 26: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/26.jpg)
Exercıcio 1Exercıcio
Encontre a solucao do PVI x = 6x − 7y , y = x − 2y .
-2 -1 0 1 2
-2
-1
0
1
2
![Page 27: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/27.jpg)
Exercıcio 2Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
Seja A = ( −2 11 −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 + 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.
Sendo assim, u = (1,1).
![Page 28: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/28.jpg)
Exercıcio 2Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
Seja A = ( −2 11 −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 + 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.
Sendo assim, u = (1,1).
![Page 29: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/29.jpg)
Exercıcio 2Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
Seja A = ( −2 11 −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 + 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.
Sendo assim, u = (1,1).
![Page 30: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/30.jpg)
Exercıcio 2Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
Seja A = ( −2 11 −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 + 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.
Sendo assim, u = (1,1).
![Page 31: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/31.jpg)
Exercıcio 2Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
Seja A = ( −2 11 −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 + 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.
Sendo assim, u = (1,1).
![Page 32: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/32.jpg)
Exercıcio 2Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
Seja A = ( −2 11 −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 + 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.
Sendo assim, u = (1,1).
![Page 33: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/33.jpg)
Exercıcio 2Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
Seja A = ( −2 11 −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 + 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.
Sendo assim, u = (1,1).
![Page 34: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/34.jpg)
Exercıcio 2Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
Seja A = ( −2 11 −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 + 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.
Sendo assim, u = (1,1).
![Page 35: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/35.jpg)
Exercıcio 2Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
Seja A = ( −2 11 −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 + 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.
Sendo assim, u = (1,1).
![Page 36: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/36.jpg)
Exercıcio 2Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
Seja A = ( −2 11 −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 + 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.
Sendo assim, u = (1,1).
![Page 37: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/37.jpg)
Exercıcio 2Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
Seja A = ( −2 11 −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 + 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.
Sendo assim, u = (1,1).
![Page 38: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/38.jpg)
Exercıcio 2Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
Seja A = ( −2 11 −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 + 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.
Sendo assim, u = (1,1).
![Page 39: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/39.jpg)
Exercıcio 2
Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
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
) .
Sabemos que, M−1 = 1det(M) (
−1 −1−1 1
) =⎛⎜⎝
12
12
12 −1
2
⎞⎟⎠
.
![Page 40: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/40.jpg)
Exercıcio 2
Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
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
) .
Sabemos que, M−1 = 1det(M) (
−1 −1−1 1
) =⎛⎜⎝
12
12
12 −1
2
⎞⎟⎠
.
![Page 41: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/41.jpg)
Exercıcio 2
Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
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
) .
Sabemos que, M−1 = 1det(M) (
−1 −1−1 1
) =⎛⎜⎝
12
12
12 −1
2
⎞⎟⎠
.
![Page 42: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/42.jpg)
Exercıcio 2
Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
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
) .
Sabemos que, M−1 = 1det(M) (
−1 −1−1 1
) =⎛⎜⎝
12
12
12 −1
2
⎞⎟⎠
.
![Page 43: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/43.jpg)
Exercıcio 2
Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
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
) .
Sabemos que, M−1 = 1det(M) (
−1 −1−1 1
) =⎛⎜⎝
12
12
12 −1
2
⎞⎟⎠
.
![Page 44: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/44.jpg)
Exercıcio 2
Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
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
) .
Sabemos que, M−1 = 1det(M) (
−1 −1−1 1
) =
⎛⎜⎝
12
12
12 −1
2
⎞⎟⎠
.
![Page 45: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/45.jpg)
Exercıcio 2
Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
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
) .
Sabemos que, M−1 = 1det(M) (
−1 −1−1 1
) =⎛⎜⎝
12
12
12 −1
2
⎞⎟⎠
.
![Page 46: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/46.jpg)
Exercıcio 2
Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
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
⎞⎟⎠.
Portanto, se x(0) = (α,β) e a condicao inicial, temos que asolucao e dada por
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
⎞⎟⎠.
![Page 47: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/47.jpg)
Exercıcio 2
Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
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
⎞⎟⎠.
Portanto, se x(0) = (α,β) e a condicao inicial, temos que asolucao e dada por
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
⎞⎟⎠.
![Page 48: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/48.jpg)
Exercıcio 2
Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
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
⎞⎟⎠.
Portanto, se x(0) = (α,β) e a condicao inicial, temos que asolucao e dada por
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
⎞⎟⎠.
![Page 49: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/49.jpg)
Exercıcio 2
Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
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
⎞⎟⎠.
Portanto, se x(0) = (α,β) e a condicao inicial,
temos que asolucao e dada por
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
⎞⎟⎠.
![Page 50: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/50.jpg)
Exercıcio 2
Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
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
⎞⎟⎠.
Portanto, se x(0) = (α,β) e a condicao inicial, temos que asolucao e dada por
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
⎞⎟⎠.
![Page 51: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/51.jpg)
Exercıcio 2Exercıcio
Encontre a solucao do PVI x = −2x + y , y = x − 2y .
-2 -1 0 1 2
-2
-1
0
1
2
![Page 52: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/52.jpg)
Exercıcio 3Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Seja A = ( 1 −22 5
). 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 − 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).
![Page 53: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/53.jpg)
Exercıcio 3Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Seja A = ( 1 −22 5
).
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 − 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).
![Page 54: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/54.jpg)
Exercıcio 3Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Seja A = ( 1 −22 5
). 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 − 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).
![Page 55: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/55.jpg)
Exercıcio 3Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Seja A = ( 1 −22 5
). 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 − 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).
![Page 56: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/56.jpg)
Exercıcio 3Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Seja A = ( 1 −22 5
). 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 − 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).
![Page 57: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/57.jpg)
Exercıcio 3Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Seja A = ( 1 −22 5
). 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 − 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).
![Page 58: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/58.jpg)
Exercıcio 3Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Seja A = ( 1 −22 5
). 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 − 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).
![Page 59: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/59.jpg)
Exercıcio 3Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Seja A = ( 1 −22 5
). 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 − 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).
![Page 60: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/60.jpg)
Exercıcio 3Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Seja A = ( 1 −22 5
). 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 − 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).
![Page 61: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/61.jpg)
Exercıcio 3Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Seja A = ( 1 −22 5
). 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 − 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).
![Page 62: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/62.jpg)
Exercıcio 3Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Seja A = ( 1 −22 5
). 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 − 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).
![Page 63: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/63.jpg)
Exercıcio 3Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Seja A = ( 1 −22 5
). 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 − 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).
![Page 64: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/64.jpg)
Exercıcio 3
Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Vamos agora determinar v a partir de u,
(A − 3I )v = u⇒
( −2 −22 2
)( ab
) = ( 1−1
)⇒ 2a + 2b = −1.
Sendo assim, v = (1,−32).
Entao,
B = ( 3 10 3
) e M = ( 1 1−1 −3
2
) .
Sabemos que, M−1 = 1det(M) (
−32 −1
1 1) = ( 3 2
−2 −2).
![Page 65: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/65.jpg)
Exercıcio 3
Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Vamos agora determinar v a partir de u,
(A − 3I )v = u⇒ ( −2 −22 2
)( ab
) = ( 1−1
)⇒
2a + 2b = −1.
Sendo assim, v = (1,−32).
Entao,
B = ( 3 10 3
) e M = ( 1 1−1 −3
2
) .
Sabemos que, M−1 = 1det(M) (
−32 −1
1 1) = ( 3 2
−2 −2).
![Page 66: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/66.jpg)
Exercıcio 3
Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Vamos agora determinar v a partir de u,
(A − 3I )v = u⇒ ( −2 −22 2
)( ab
) = ( 1−1
)⇒ 2a + 2b = −1.
Sendo assim, v = (1,−32).
Entao,
B = ( 3 10 3
) e M = ( 1 1−1 −3
2
) .
Sabemos que, M−1 = 1det(M) (
−32 −1
1 1) = ( 3 2
−2 −2).
![Page 67: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/67.jpg)
Exercıcio 3
Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Vamos agora determinar v a partir de u,
(A − 3I )v = u⇒ ( −2 −22 2
)( ab
) = ( 1−1
)⇒ 2a + 2b = −1.
Sendo assim, v = (1,−32).
Entao,
B = ( 3 10 3
) e M = ( 1 1−1 −3
2
) .
Sabemos que, M−1 = 1det(M) (
−32 −1
1 1) = ( 3 2
−2 −2).
![Page 68: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/68.jpg)
Exercıcio 3
Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Vamos agora determinar v a partir de u,
(A − 3I )v = u⇒ ( −2 −22 2
)( ab
) = ( 1−1
)⇒ 2a + 2b = −1.
Sendo assim, v = (1,−32).
Entao,
B = ( 3 10 3
) e M = ( 1 1−1 −3
2
) .
Sabemos que, M−1 = 1det(M) (
−32 −1
1 1) = ( 3 2
−2 −2).
![Page 69: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/69.jpg)
Exercıcio 3
Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Vamos agora determinar v a partir de u,
(A − 3I )v = u⇒ ( −2 −22 2
)( ab
) = ( 1−1
)⇒ 2a + 2b = −1.
Sendo assim, v = (1,−32).
Entao,
B = ( 3 10 3
) e M = ( 1 1−1 −3
2
) .
Sabemos que, M−1 = 1det(M) (
−32 −1
1 1) =
( 3 2−2 −2
).
![Page 70: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/70.jpg)
Exercıcio 3
Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Vamos agora determinar v a partir de u,
(A − 3I )v = u⇒ ( −2 −22 2
)( ab
) = ( 1−1
)⇒ 2a + 2b = −1.
Sendo assim, v = (1,−32).
Entao,
B = ( 3 10 3
) e M = ( 1 1−1 −3
2
) .
Sabemos que, M−1 = 1det(M) (
−32 −1
1 1) = ( 3 2
−2 −2).
![Page 71: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/71.jpg)
Exercıcio 3
Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Alem disso, exp(Bt) = ( e3t te3t
0 e3t).
Logo,
exp(At) =Mexp(Bt)M−1 = ( e3t(1 − 2t) −2e3tt2e3tt e3t(2t + 1) ) .
Portanto, se x(0) = (α,β) e a condicao inicial, temos que asolucao e dada por
x(t) = ( e3t(1 − 2t) −2e3tt2e3tt e3t(2t + 1) )( α
β) .
![Page 72: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/72.jpg)
Exercıcio 3
Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Alem disso, exp(Bt) = ( e3t te3t
0 e3t). Logo,
exp(At) =Mexp(Bt)M−1 = ( e3t(1 − 2t) −2e3tt2e3tt e3t(2t + 1) ) .
Portanto, se x(0) = (α,β) e a condicao inicial, temos que asolucao e dada por
x(t) = ( e3t(1 − 2t) −2e3tt2e3tt e3t(2t + 1) )( α
β) .
![Page 73: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/73.jpg)
Exercıcio 3
Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Alem disso, exp(Bt) = ( e3t te3t
0 e3t). Logo,
exp(At) =Mexp(Bt)M−1 = ( e3t(1 − 2t) −2e3tt2e3tt e3t(2t + 1) ) .
Portanto, se x(0) = (α,β) e a condicao inicial,
temos que asolucao e dada por
x(t) = ( e3t(1 − 2t) −2e3tt2e3tt e3t(2t + 1) )( α
β) .
![Page 74: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/74.jpg)
Exercıcio 3
Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
Alem disso, exp(Bt) = ( e3t te3t
0 e3t). Logo,
exp(At) =Mexp(Bt)M−1 = ( e3t(1 − 2t) −2e3tt2e3tt e3t(2t + 1) ) .
Portanto, se x(0) = (α,β) e a condicao inicial, temos que asolucao e dada por
x(t) = ( e3t(1 − 2t) −2e3tt2e3tt e3t(2t + 1) )( α
β) .
![Page 75: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/75.jpg)
Exercıcio 3Exercıcio
Encontre a solucao do PVI x = x − 2y , y = 2x + 5y .
-2 -1 0 1 2
-2
-1
0
1
2
![Page 76: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/76.jpg)
Exercıcio 4Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Seja A = ( 3 −34 −3
). 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 + 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).
![Page 77: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/77.jpg)
Exercıcio 4Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Seja A = ( 3 −34 −3
).
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 + 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).
![Page 78: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/78.jpg)
Exercıcio 4Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Seja A = ( 3 −34 −3
). 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 + 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).
![Page 79: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/79.jpg)
Exercıcio 4Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Seja A = ( 3 −34 −3
). 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 + 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).
![Page 80: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/80.jpg)
Exercıcio 4Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Seja A = ( 3 −34 −3
). 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 + 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).
![Page 81: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/81.jpg)
Exercıcio 4Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Seja A = ( 3 −34 −3
). 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 + 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).
![Page 82: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/82.jpg)
Exercıcio 4Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Seja A = ( 3 −34 −3
). 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 + 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).
![Page 83: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/83.jpg)
Exercıcio 4Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Seja A = ( 3 −34 −3
). 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 + 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).
![Page 84: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/84.jpg)
Exercıcio 4Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Seja A = ( 3 −34 −3
). 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 + 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).
![Page 85: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/85.jpg)
Exercıcio 4Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Seja A = ( 3 −34 −3
). 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 + 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).
![Page 86: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/86.jpg)
Exercıcio 4Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Seja A = ( 3 −34 −3
). 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 + 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).
![Page 87: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/87.jpg)
Exercıcio 4Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Seja A = ( 3 −34 −3
). 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 + 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).
![Page 88: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/88.jpg)
Exercıcio 4
Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Sendo assim, u = (3,4) + i(√
3,0).
Entao,
B = ( 0 −√
3√3 0
) e M = ( 3√
34 0
) .
Sabemos que, M−1 = 1det(M) (
0 −√
3−4 3
) =⎛⎜⎝
0 14
1√3−√34
⎞⎟⎠
.
Alem disso,
exp(Bt) = ( cos(√
3t) − sen(√
3t)sen(
√3t) cos(
√3t) ) .
![Page 89: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/89.jpg)
Exercıcio 4
Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Sendo assim, u = (3,4) + i(√
3,0).Entao,
B = ( 0 −√
3√3 0
) e M = ( 3√
34 0
) .
Sabemos que, M−1 = 1det(M) (
0 −√
3−4 3
) =⎛⎜⎝
0 14
1√3−√34
⎞⎟⎠
.
Alem disso,
exp(Bt) = ( cos(√
3t) − sen(√
3t)sen(
√3t) cos(
√3t) ) .
![Page 90: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/90.jpg)
Exercıcio 4
Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Sendo assim, u = (3,4) + i(√
3,0).Entao,
B = ( 0 −√
3√3 0
) e M = ( 3√
34 0
) .
Sabemos que, M−1 = 1det(M) (
0 −√
3−4 3
) =
⎛⎜⎝
0 14
1√3−√34
⎞⎟⎠
.
Alem disso,
exp(Bt) = ( cos(√
3t) − sen(√
3t)sen(
√3t) cos(
√3t) ) .
![Page 91: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/91.jpg)
Exercıcio 4
Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Sendo assim, u = (3,4) + i(√
3,0).Entao,
B = ( 0 −√
3√3 0
) e M = ( 3√
34 0
) .
Sabemos que, M−1 = 1det(M) (
0 −√
3−4 3
) =⎛⎜⎝
0 14
1√3−√34
⎞⎟⎠
.
Alem disso,
exp(Bt) = ( cos(√
3t) − sen(√
3t)sen(
√3t) cos(
√3t) ) .
![Page 92: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/92.jpg)
Exercıcio 4
Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Sendo assim, u = (3,4) + i(√
3,0).Entao,
B = ( 0 −√
3√3 0
) e M = ( 3√
34 0
) .
Sabemos que, M−1 = 1det(M) (
0 −√
3−4 3
) =⎛⎜⎝
0 14
1√3−√34
⎞⎟⎠
.
Alem disso,
exp(Bt) = ( cos(√
3t) − sen(√
3t)sen(
√3t) cos(
√3t) ) .
![Page 93: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/93.jpg)
Exercıcio 4
Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Logo, exp(At) =Mexp(Bt)M−1
=⎛⎜⎝
cos (√
3t) −√
3 sin (√
3t)√
3 sin (√
3t)
−4 sin(√3t)√3
cos (√
3t) + sin (√
3t)√
3
⎞⎟⎠.
Portanto, se x(0) = (α,β) e a condicao inicial, temos que asolucao e dada por
x(t) =⎛⎜⎝
cos (√
3t) −√
3 sin (√
3t)√
3 sin (√
3t)
− 4 sin(√
3t)√
3cos (
√3t) + sin (
√3t)
√3
⎞⎟⎠
⎛⎜⎝α
β
⎞⎟⎠.
![Page 94: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/94.jpg)
Exercıcio 4
Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Logo, exp(At) =Mexp(Bt)M−1
=⎛⎜⎝
cos (√
3t) −√
3 sin (√
3t)√
3 sin (√
3t)
−4 sin(√3t)√3
cos (√
3t) + sin (√
3t)√
3
⎞⎟⎠.
Portanto, se x(0) = (α,β) e a condicao inicial, temos que asolucao e dada por
x(t) =⎛⎜⎝
cos (√
3t) −√
3 sin (√
3t)√
3 sin (√
3t)
− 4 sin(√
3t)√
3cos (
√3t) + sin (
√3t)
√3
⎞⎟⎠
⎛⎜⎝α
β
⎞⎟⎠.
![Page 95: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/95.jpg)
Exercıcio 4
Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Logo, exp(At) =Mexp(Bt)M−1
=⎛⎜⎝
cos (√
3t) −√
3 sin (√
3t)√
3 sin (√
3t)
−4 sin(√3t)√3
cos (√
3t) + sin (√
3t)√
3
⎞⎟⎠.
Portanto, se x(0) = (α,β) e a condicao inicial,
temos que asolucao e dada por
x(t) =⎛⎜⎝
cos (√
3t) −√
3 sin (√
3t)√
3 sin (√
3t)
− 4 sin(√
3t)√
3cos (
√3t) + sin (
√3t)
√3
⎞⎟⎠
⎛⎜⎝α
β
⎞⎟⎠.
![Page 96: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/96.jpg)
Exercıcio 4
Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
Logo, exp(At) =Mexp(Bt)M−1
=⎛⎜⎝
cos (√
3t) −√
3 sin (√
3t)√
3 sin (√
3t)
−4 sin(√3t)√3
cos (√
3t) + sin (√
3t)√
3
⎞⎟⎠.
Portanto, se x(0) = (α,β) e a condicao inicial, temos que asolucao e dada por
x(t) =⎛⎜⎝
cos (√
3t) −√
3 sin (√
3t)√
3 sin (√
3t)
− 4 sin(√
3t)√
3cos (
√3t) + sin (
√3t)
√3
⎞⎟⎠
⎛⎜⎝α
β
⎞⎟⎠.
![Page 97: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/97.jpg)
Exercıcio 4Exercıcio
Encontre a solucao do PVI x = 3x − 3y , y = 4x − 3y .
-2 -1 0 1 2
-2
-1
0
1
2
![Page 98: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/98.jpg)
Exercıcio 5Exercıcio
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)
⎞⎟⎠
![Page 99: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/99.jpg)
Exercıcio 5Exercıcio
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)
⎞⎟⎠
![Page 100: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/100.jpg)
Exercıcio 5Exercıcio
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)
⎞⎟⎠
![Page 101: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/101.jpg)
Exercıcio 5Exercıcio
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)
⎞⎟⎠
![Page 102: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/102.jpg)
Exercıcio 5Exercıcio
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)
⎞⎟⎠
![Page 103: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/103.jpg)
Exercıcio 5Exercıcio
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)
⎞⎟⎠
![Page 104: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/104.jpg)
Exercıcio 5Exercıcio
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)
⎞⎟⎠
![Page 105: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/105.jpg)
Exercıcio 5Exercıcio
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)
⎞⎟⎠
![Page 106: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/106.jpg)
Exercıcio 5Exercıcio
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)
⎞⎟⎠
![Page 107: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/107.jpg)
Exercıcio 5Exercıcio
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)
⎞⎟⎠
![Page 108: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/108.jpg)
Exercıcio 5Exercıcio
Encontre a solucao do PVI x = 2x − y + 2, y = x + 2y + 1.
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)
⎞⎟⎠.
![Page 109: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/109.jpg)
Exercıcio 5Exercıcio
Encontre a solucao do PVI x = 2x − y + 2, y = x + 2y + 1.
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)
⎞⎟⎠.
![Page 110: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/110.jpg)
Exercıcio 5Exercıcio
Encontre a solucao do PVI x = 2x − y + 2, y = x + 2y + 1.
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)
⎞⎟⎠.
![Page 111: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/111.jpg)
Exercıcio 5
Exercıcio
Encontre a solucao do PVI x = 2x − y + 2, y = x + 2y + 1.
Portanto,
x(t) =⎛⎜⎝
e2t cos(t) −e2t sen(t)
e2t sen(2t) e2t cos(t)
⎞⎟⎠
⎛⎜⎝
−e−2t cos(t) + c1e−2t sen(t) + c2
⎞⎟⎠
=⎛⎜⎝
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))
⎞⎟⎠.
![Page 112: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/112.jpg)
Exercıcio 5
Exercıcio
Encontre a solucao do PVI x = 2x − y + 2, y = x + 2y + 1.
Portanto,
x(t) =⎛⎜⎝
e2t cos(t) −e2t sen(t)
e2t sen(2t) e2t cos(t)
⎞⎟⎠
⎛⎜⎝
−e−2t cos(t) + c1e−2t sen(t) + c2
⎞⎟⎠
=⎛⎜⎝
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))
⎞⎟⎠.
![Page 113: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/113.jpg)
Exercıcio 5Exercıcio
Encontre a solucao do PVI x = 2x − y + 2, y = x + 2y + 1.
-2 -1 0 1 2
-2
-1
0
1
2
![Page 114: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/114.jpg)
Exercıcio 6Exercıcio
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
⎞⎟⎠.
![Page 115: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/115.jpg)
Exercıcio 6Exercıcio
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
⎞⎟⎠.
![Page 116: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/116.jpg)
Exercıcio 6Exercıcio
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
⎞⎟⎠.
![Page 117: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/117.jpg)
Exercıcio 6Exercıcio
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
⎞⎟⎠.
![Page 118: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/118.jpg)
Exercıcio 6Exercıcio
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
⎞⎟⎠.
![Page 119: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/119.jpg)
Exercıcio 6Exercıcio
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
⎞⎟⎠.
![Page 120: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/120.jpg)
Exercıcio 6Exercıcio
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
⎞⎟⎠.
![Page 121: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/121.jpg)
Exercıcio 6Exercıcio
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
⎞⎟⎠.
![Page 122: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/122.jpg)
Exercıcio 6Exercıcio
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
⎞⎟⎠.
![Page 123: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/123.jpg)
Exercıcio 6
Exercıcio
Encontre a solucao do PVI x = 3x + 4y , y = 3x + 2y + t2.
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
⎞⎟⎠.
![Page 124: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/124.jpg)
Exercıcio 6
Exercıcio
Encontre a solucao do PVI x = 3x + 4y , y = 3x + 2y + t2.
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
⎞⎟⎠.
![Page 125: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/125.jpg)
Exercıcio 6
Exercıcio
Encontre a solucao do PVI x = 3x + 4y , y = 3x + 2y + t2.
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
⎞⎟⎠.
![Page 126: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/126.jpg)
Exercıcio 6
Exercıcio
Encontre a solucao do PVI x = 3x + 4y , y = 3x + 2y + t2.
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))
⎞
⎟⎟
⎠
.
![Page 127: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/127.jpg)
Exercıcio 6
Exercıcio
Encontre a solucao do PVI x = 3x + 4y , y = 3x + 2y + t2.
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))
⎞
⎟⎟
⎠
.
![Page 128: 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›](https://reader034.vdocuments.site/reader034/viewer/2022042413/5f2d34592e086277dc61ac45/html5/thumbnails/128.jpg)
Muito Obrigada ©