lista para ap2 edo (2)

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Disciplina: equações diferenciais 
Professora: Vanessa Monteiro 
Dupla: Rahely Victoria Santos de Santana 151030192 
Camila Batista 152030444 
 
 
Orientações 
1. Data de entrega dia 18/05 ate as 18:35 para o email: 
vanessa.monteiro@area1.edu.br, colocar no assunto a disciplina, 
senão não será corrigido 
2. Fazer em dupla 
 
Lista para Ap2 
 
1. Resolva as Edo abaixo: 
a) 𝑦4 − 16𝑦 = 3𝑠𝑒𝑛(2𝑥) 
b) 𝑦" − 2𝑦′ + 𝑦 = 0 
c) 𝑦4 − 4𝑦" − 2𝑦′ = 3𝑥2 − 2𝑥 + 1 
d) 𝑦′′′ + 3𝑦" − 4𝑦 = 0 
 
 
 
 
 
 
 
 
 
 
 
mailto:vanessa.monteiro@area1.edu.br
 RESPOSTAS 
 
𝒂) 𝒚𝟒 − 𝟏𝟔𝒚 = 𝟑𝒔𝒆𝒏(𝟐𝒙) 
𝑚4 − 16 = 0 
𝑚4 = 𝑢 
𝑢 − 16 = 0 −→ 𝑢 = 16 
𝑚4 = 16 
𝑚 = √16
4
= ±𝟐  Raízes reais e diferentes 
Solução: 𝒀𝒉 = 𝒄𝟏𝒆𝟐𝒙 + 𝒄𝟐𝒆−𝟐𝒙 
Passo 2 : Achar Yp 
Q(x)= 3sen(2x) 
𝑦 = 𝐴𝑥𝑠𝑒𝑛(2𝑥) + 𝐵𝑥𝑐𝑜𝑠(2𝑥) 
𝒚′ = 𝒔𝒆𝒏𝑨(𝟐𝒙) + 𝟐𝑨𝒙𝒄𝒐𝒔(𝟐𝒙) + 𝒄𝒐𝒔𝑩(𝟐𝒙) − 𝟐𝑩𝒙𝒔𝒆𝒏(𝟐𝒙) 
𝑦′′ = 2𝐴𝑐𝑜𝑠(2𝑥) + 2𝐴𝑐𝑜𝑠(2𝑥) − 4𝐴𝑥𝑠𝑒𝑛(2𝑥) − 2𝐵𝑠𝑒𝑛(2𝑥) − 2𝐵𝑠𝑒𝑛(2𝑥) − 4𝐵𝑥𝑐𝑜𝑠(2𝑥) 
𝒚′′ = 𝟒𝑨𝒄𝒐𝒔(𝟐𝒙) − 𝟒𝑨𝒙𝒔𝒆𝒏(𝟐𝒙) − 𝟒𝑩𝒔𝒆𝒏(𝟐𝒙) − 𝟒𝑩𝒙𝒄𝒐𝒔(𝟐𝒙) 
𝑦′′′ = −8𝐴𝑠𝑒𝑛(2𝑥) − 4𝐴𝑠𝑒𝑛(2𝑥) − 8𝐴𝑥𝑐𝑜𝑠(2𝑥) − 8𝐵𝑐𝑜𝑠(2𝑥) − 4𝐵𝑐𝑜𝑠(2𝑥) + 8𝐵𝑥𝑠𝑒𝑛(2𝑥) 
𝒚′′′ = −𝟏𝟐𝑨𝒔𝒆𝒏(𝟐𝒙) − 𝟖𝑨𝒙𝒄𝒐𝒔(𝟐𝒙) − 𝟏𝟐𝑩𝒄𝒐𝒔(𝟐𝒙) + 𝟖𝑩𝒙𝒔𝒆𝒏(𝟐𝒙) 
𝑦′′′′ = −24𝐴𝑐𝑜𝑠(2𝑥) − 8𝐴𝑐𝑜𝑠(2𝑥) + 16𝐴𝑥𝑠𝑒𝑛(2𝑥) + 24𝐵𝑠𝑒𝑛(2𝑥) + 8𝐵𝑠𝑒𝑛(2𝑥)
+ 16𝐵𝑥𝑐𝑜𝑠(2𝑥) 
𝒚𝟒 = −𝟑𝟐𝑨𝒄𝒐𝒔(𝟐𝒙) + 𝟏𝟔𝑨𝑿𝒔𝒆𝒏(𝟐𝒙) + 𝟑𝟐𝑩𝒔𝒆𝒏(𝟐𝒙) + 𝟏𝟔𝑩𝒙𝒄𝒐𝒔(𝟐𝒙) 
 
Passo 3: Substituir na Função 
𝒚𝟒 − 𝟏𝟔𝒚 = 𝟑𝒔𝒆𝒏(𝟐𝒙) 
−32𝐴𝑐𝑜𝑠(2𝑥) + 16𝐴𝑋𝑠𝑒𝑛(2𝑥) + 32𝐵𝑠𝑒𝑛(2𝑥) + 16𝐵𝑥𝑐𝑜𝑠(2𝑥)
− 16(𝐴𝑥𝑠𝑒𝑛(2𝑥) + 𝐵𝑥𝑐𝑜𝑠(2𝑥)) = 3𝑠𝑒𝑛(2𝑥) 
−32𝐴𝑐𝑜𝑠(2𝑥) + 16𝐴𝑋𝑠𝑒𝑛(2𝑥) + 32𝐵𝑠𝑒𝑛(2𝑥) + 16𝐵𝑥𝑐𝑜𝑠(2𝑥) − 16𝐴𝑥𝑠𝑒𝑛(2𝑥)
− 16𝐵𝑥𝑐𝑜𝑠(2𝑥) = 3𝑠𝑒𝑛(2𝑥) 
−32𝐴𝑐𝑜𝑠(2𝑥) + 32𝐵𝑠𝑒𝑛(2𝑥) = 3𝑠𝑒𝑛(2𝑥) 
32𝐵 = 3 
𝑩 =
𝟑
𝟑𝟐
 
−32𝐴 = 0 
𝐴 =
0
−32
 
𝑨 = 𝟎 
 
𝑦𝑝 = 𝐴𝑥𝑠𝑒𝑛(2𝑥) + 𝐵𝑥𝑐𝑜𝑠(2𝑥) 
𝒚𝒑 =
𝟑
𝟑𝟐
𝒙𝒄𝒐𝒔(𝟐𝒙) 
 
𝑺𝒐𝒍𝒖çã𝒐 𝑭𝒊𝒏𝒂𝒍: 𝒀 = 𝒄𝟏𝒆𝟐𝒙 + 𝒄𝟐𝒆−𝟐𝒙 +
𝟑
𝟑𝟐
𝒙𝒄𝒐𝒔(𝟐𝒙) 
 
 
 
 
 
 
 
 
 
 
 
 
 
𝒃)𝒚" − 𝟐𝒚′ + 𝒚 = 𝟎 
𝑚2 − 2𝑚 + 1 = 0 
∆= 𝑏2 − 4𝑎𝑐 
∆= (−2)2 − 4(1) ∗ (1) = 0 
∆= 0 
 
𝑋 = 
−𝑏±√∆
2𝑎
 
𝑋 = 
2 ± 0
2
 
𝑿𝟏 = 
2 + 0
2
= 𝟏 
𝑿𝟐 = 
2 − 0
2
= 𝟏 
 
Solução: 𝒀 = 𝒄𝟏𝒆𝒙 + 𝒄𝟐𝒙𝒆𝒙 
 
 
 
 
 
 
 
 
 
 
 
 
 
Reais e Iguais 
𝒄) 𝒚𝟒 − 𝟒𝒚" − 𝟐𝒚′ = 𝟑𝒙𝟐 − 𝟐𝒙 + 𝟏 
𝑚4 − 4𝑚2 − 2𝑚 
𝑚(𝑚3 − 4𝑚 − 2) 
𝒎𝟏 = 𝟎 
𝑚(𝑚3 − 4𝑚 − 2) 
𝒎𝟐 = −𝟎, 𝟓𝟒 
(−0,54)3 − 4(−0,54) − 2 = 0 
−0,16 + 2,16 − 2 = 0 
0 = 0 
-0,54 1 0 -4 -2 
 1 -0,54 -3,7084 0,002536 
 
𝑚2 − 0,54𝑚 − 3,7084 = 0,002536 
𝑚2 − 0,54𝑚 − 3,71 = 0 
∆= 𝑏2 − 4𝑎𝑐 
∆= (−0,54)2 − 4(1) ∗ (−3,71) 
∆= 0,2916 + 14,84 
∆= 15,1316 
√∆= 𝟑, 𝟖𝟗 
𝑥 =
−𝑏 ± √∆
2𝑎
 
𝒙𝟏 =
0,54 + 3,89
2
=
4,43
2
 = 𝟐, 𝟐𝟏 
𝒙𝟐 =
0,54 − 3,89
2
= −
3,35
2
= −𝟏, 𝟔𝟕 
 
𝑺𝒐𝒍𝒖çã𝒐 𝟏 
𝑦ℎ = 𝑐1𝑒0𝑥 + 𝑐2𝑒−0,54𝑥 + 𝑐3𝑒2,21𝑥 + 𝑐4𝑒−1,67𝑥 
𝒚𝒉 = 𝒄𝟏 + 𝒄𝟐𝒆−𝟎,𝟓𝟒𝒙 + 𝒄𝟑𝒆𝟐,𝟐𝟏𝒙 + 𝒄𝟒𝒆−𝟏,𝟔𝟕𝒙 
 
Achar Yp: 
𝑄(𝑥) = 3𝑥2 − 2𝑥 + 1 
𝒚 = 𝑨𝒙𝟑 − 𝑩𝒙𝟐 + 𝑪𝒙 
𝑦′ = 3𝐴𝑥2 − 2𝐵𝑥 + 𝐶 
𝑦′′ = 6𝐴𝑥 − 2𝐵 
𝑦′′′ = 6𝐴 
𝑦4 = 0 
𝒚𝟒 − 𝟒𝒚" − 𝟐𝒚′ = 𝟑𝒙𝟐 − 𝟐𝒙 + 𝟏 
0 − 4(6𝐴𝑥 − 2𝐵) − 2(3𝐴𝑥2 − 2𝐵𝑥 + 𝐶) = 3𝑥2 − 2𝑥 + 1 
−24𝐴𝑥 + 8𝐵 − 6𝐴𝑥2 + 4𝐵𝑥 − 2𝐶 = 3𝑥2 − 2𝑥 + 1 
−6𝐴𝑥2 − 24𝐴𝑥 + 8𝐵 + 4𝐵𝑥 − 2𝐶 = 3𝑥2 − 2𝑥 + 1 
−6𝐴 = 3 
𝑨 = −
𝟑
𝟔
= −
𝟏
𝟐
 
−24𝐴 + 4𝐵 = −2 
−24 (−
1
2
) + 4𝐵 = −2 
24
2
+ 4𝐵 = −2 
𝐵 = −
14
2
 
𝑩 = −
𝟕
𝟐
 
8𝐵 − 2𝐶 = 1 
8(−
7
2
) − 2𝐶 = 1 
−28 − 2𝐶 = 1 
−2𝐶 = 29 
𝑪 = −
𝟐𝟗
𝟐
 
 
𝒚𝒑 = 𝑨𝒙𝟑 − 𝑩𝒙𝟐 + 𝑪𝒙 
𝒚𝒑 = −
𝟏
𝟐
𝒙𝟑 +
𝟕
𝟐
𝒙𝟐 −
𝟐𝟗
𝟐
𝒙 
 
𝑺𝒐𝒍𝒖çã𝒐 𝑭𝒊𝒏𝒂𝒍: 𝒀 = 𝒄𝟏 + 𝒄𝟐𝒆−𝟎,𝟓𝟒𝒙 + 𝒄𝟑𝒆𝟐,𝟐𝟏𝒙 + 𝒄𝟒𝒆−𝟏,𝟔𝟕𝒙 −
𝟏
𝟐
𝒙𝟑 +
𝟕
𝟐
𝒙𝟐 −
𝟐𝟗
𝟐
𝒙 
 
 
 
 
 
 
 
 
 
 
 
 
𝒅) 𝒚′′′ + 𝟑𝒚" − 𝟒𝒚 = 𝟎 
𝑚3 + 3𝑚2 − 4 = 0 
(1)3 + 3(1)2 − 4 = 0 
0 = 0; 𝑚 = 1 é 𝑎 𝑝𝑟𝑖𝑚𝑒𝑖𝑟𝑎 𝑟𝑎𝑖𝑧 
Utilizando Briot Ruffini 
1 1 3 0 -4 
 1 4 4 0 
 
𝑚2 + 4𝑚 + 4 = 0 
∆= 𝑏2 − 4𝑎𝑐 
∆= 42 − 4(1) ∗ (4) 
∆= 0 
 
𝑋 = 
−𝑏 ± √∆
2𝑎
 
𝑋 = 
−4 ± 0
2
 
𝑿𝟏 = 
−4 + 0
2
= −𝟐 
𝑿𝟐 = 
−4 − 0
2
= −𝟐 
 
Solução: 𝒀 = 𝒄𝟏𝒆−𝟐𝒙 + 𝒄𝟐𝒙𝒆−𝟐𝒙 + 𝒄𝟑𝒙𝟐𝒆𝒙 
 
 
Reais e Iguais