Baixe o app para aproveitar ainda mais
Prévia do material em texto
UTFPR - Universidade Tecnolo´gica Federal do Parana´ Pato Branco Engenharias Lista de Exerc´ıcios Sistemas de equac¸o˜es diferenciais lineares-VARIAC¸A˜O DOS PARAˆMETROS 1- Aplique a variac¸a˜o dos paraˆmetros para resolver o sistema dado. a) dx dt = 3x− 3y + 4 dy dt = 2x− 2y − 1 b) X ′ = ( 3 −5 3 4 −1 ) X + ( 1 −1 ) e t 2 c) X ′ = ( 0 2 −1 3 ) X + ( 1 −1 ) et d) X ′ = ( 1 8 1 −1 ) X + ( 12 12 ) t e) X ′ = ( 3 2 −2 −1 ) X + ( 2e−t e−t ) f) X ′ = ( 0 −1 1 0 ) X + ( sec t 0 ) g) X ′ = ( 1 −1 1 1 ) X + ( cos t sin t ) et h) X ′ = ( 0 1 −1 0 ) X + ( 0 sec t tan t ) Gilson Tumelero 2 i) X ′ = ( 1 2 −1 2 1 ) X + ( csc t sec t ) et j) X ′ = 1 1 01 1 0 0 0 3 X + ete2t te3t k) X ′ = ( 2 −1 3 −2 ) X + ( et t ) l) X ′ = ( 2 −5 1 −2 ) X + ( − cos t sin t ) m) X ′ = ( 1 1 4 −2 ) X + ( e−2t −2et ) n) X ′ = ( 4 −2 8 −4 ) X + ( t−3 −t−2 ) , t > 0 o) X ′ = ( −4 2 2 −1 ) X + ( t−1 2t−1 + 4 ) , t > 0 p) X ′ = ( 1 1 4 1 ) X + ( 2 −1 ) et q) X ′ = ( 2 −1 3 −2 ) X + ( 1 −1 ) et r) X ′ = ( −3 √2√ 2 −2 ) X + ( e−t −e−t ) s) X ′ = ( 2 −5 1 −2 ) X + ( 0 cos t ) , 0 < t < Π t) X ′ = ( 2 −5 1 −2 ) X + ( csc t sec t ) , Π 2 < t < Π Gilson Tumelero 3 2- Resolva o sistema dado sujeito a` condic¸a˜o inicial indicada. a) X ′ = ( 3 −1 −1 3 ) X + ( 4e2t 4e4t ) , X(0) = ( 1 1 ) b) X ′ = ( 4 1 6 5 ) X + ( 50e7t 0 ) , X(0) = ( 5 −5 )
Compartilhar