As Equações Diferenciais e Séries
Por: Leonardo Oliveira • 12/6/2019 • Abstract • 1.537 Palavras (7 Páginas) • 136 Visualizações
F(x)= e2x[pic 1]
F’(x)= eu . u’ = e2x . (2x)’
= e2x . (2)
= 2e2x
[pic 2]
F(x)= ln (3x)
F’(x)= [ ln . u]’ . u’
F’(x)= . u’ = . (3x)’[pic 3][pic 4]
F(x)= . (3x)’= . 3 = [pic 5][pic 6][pic 7]
[pic 8]
F(x)= [ln (x)] . [sem (x)]
F’(x)= u’.v + u.v’
F’(x)= . sem (x) + ln (x) . cos (x) [pic 9]
F(x)= + ln (x) . cos(x)[pic 10]
[pic 11]
F(x)= sen [pic 12]
F’(x)= cos (u).u’
= cos .’[pic 13][pic 14]
= cos . = /3[pic 15][pic 16][pic 17]
[pic 18]
F(x)= [cos(x)]13
F’(x)= 13. [cos(x)]13-1. [cos (x)]’
= 13. [cos(x)]12. [-sem(x)]
= -13. sen (x). [cos(x)] 12
[pic 19]
dx = = [pic 20][pic 21][pic 22][pic 23][pic 24]
= . [ln(u)] + k[pic 25]
= . [ln(3x)] + k[pic 26]
[pic 27]
= = 2 + k[pic 28][pic 29][pic 30]
= 2 + k[pic 31]
[pic 32]
[pic 33][pic 34]
= [pic 35]
= [pic 36]
[pic 37]
= 4[pic 38][pic 39][pic 40]
= 4 . sem (u) + k
= 4 sem [pic 41]
→ λ2 + 3λ + 2 = 0 [pic 42]
∆= b2 – 4 .a.c [pic 43]
∆= (3)2 – 4.(1).(2)
∆= 9 – 8 = 1 → >0
λ = → λ= → λ= [pic 44][pic 45][pic 46]
λ1= = 1 λ2= = -2[pic 49][pic 47][pic 48]
x= Ae-1.t + Be-2.t
x=Ae-t + Be-2t
[pic 50]
→ λ2 - 8 λ +16= 0[pic 51]
∆= b2 – 4 .a.c
∆= (-8)2 – 4.(1).(16)[pic 52]
∆= 64 – 64 = 0 → λ1 = λ2
λ1 = λ2 = = 4[pic 53]
x= Ae4t + Bte4t
[pic 54]
→ λ2 – 9 = 0[pic 55]
λ2 = 9
λ = [pic 56]
λ1 = 3 λ2 = - 3[pic 57]
x = Ae3t + Be-3t
[pic 58]
→ λ2 - 2 λ -3 = 0[pic 59]
∆= b2 – 4 .a.c
∆= (-2)2 – 4.(1).(-3)
∆= 4+12 = 16
λ = → λ = → λ = [pic 60][pic 61][pic 62]
λ1 = 3 λ2 = -1[pic 63]
λ1 ≠ λ2
x=Ae3t + Be-t
[pic 64]
→ λ2 - 4 = 0[pic 65]
λ2 = 4
λ = [pic 66]
λ1 = 2 λ2 = -2[pic 67]
x = Ae2t + Be-2t
[pic 68]
→ λ2 - 2λ = 0[pic 69]
λ (λ-2) = 0
λ1 = 0 / λ2 = 2
x= Ae0.t + Be2t
x= A + Be2t
→ λ2 - 3λ = 0[pic 70]
∆= b2 – 4 .a.c[pic 71]
∆= (3)2 – 4.(1).(0)
∆= 9
λ = → λ = → λ1 = 0[pic 72][pic 73]
λ2 = -3
x= Ae0.t + Be-3.t
x= A + Be-3t
[pic 74]
→ λ2 + 6λ + 9 = 0[pic 75]
∆= b2 – 4 .a.c
∆= (6)2 – 4.(1).(9)
∆= 0
λ = → λ = → λ1 = -3[pic 76][pic 77]
λ2 = -3
λ1 = λ2
x= Ae-3t + Bte-3t
[pic 78]
[pic 79]
x= ke-2t + e-2t.[pic 80]
x= ke-2t + e2t.[pic 81]
x= ke-2t + [pic 82]
=[pic 83]
=[pic 84]
= [pic 85]
= [pic 88][pic 86][pic 87]
=[pic 89]
[pic 90]
2t=u
2dt=du
dt=[pic 91]
[pic 92]
→ λ2 – [pic 93][pic 94]
λ2 = → λ = → λ = → λ1 = [pic 95][pic 96][pic 97][pic 98]
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