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3y = 0",{"_43":206,"_81":87,"_83":207},"6d3c3a1d-8e80-4ff3-b97a-d5b5bca91940","6y'' - y' - y = 0",{"_43":209,"_81":91,"_83":210},"4168ac67-64fd-46ed-a69c-a01dbfcde7ea","y'' - 2y' + y = 0",{"_43":212,"_81":95,"_83":213},"cc7bc8df-6995-4264-ac5a-32ea9b4ed7db","9y'' + 6y' + y = 0",{"_43":199,"_81":99,"_83":215},"y'' - 2y' + 6y = 0","Somente em y''−2y'+6y=0 o polinômio característico r²−2r+6=0 tem Δ<0, gerando raízes complexas conjugadas (1±i√5) e, portanto, solução em termos de exponencial modulando seno e cosseno. Nos demais itens (a–d) o discriminante é ≥0, produzindo raízes reais (distintas ou repetidas).","

Resolva as seguintes EDOs determinando a solução geral e, quando houver condições iniciais do PVI, a solução particular.

(a) y'' + 2y' - 3y = 0

(b) 6y'' - y' - y = 0

(d) 9y'' + 6y' + y = 0

(e) y'' - 2y' + 6y = 0

(f) 4y'' + 9y = 0

(g) 6y'' - 5y' + y = 0; \\, y(0) = 4; \\, y'(0) = 0

(h) y'' + 4y' + 4y = 0; \\, y(-1) = 2; \\, y'(-1) = 1

(i) y'' + y = 0; \\, y\bigg(\\frac{\\pi}{3}\\bigg) = 2; \\, y'\\bigg(\\frac{\\pi}{3}\\bigg) = -4

(j) y'' - 2y' - 3y = 3e^{2t}

(k) y'' + 9y = t^2 e^{3t} + 6

(l) 2y'' + 3y' + y = t^2 + 3 \\sin(t)

(m) y'' + y' - 2y = 2t; \\, y(0) = 0; \\, y'(0) = 1

(n) y'' - 2y' + y = te^t + 4; \\, y(0) = 1; \\, y'(0) = 1

(o) y'' + 4y = 3 \\sin(2t); \\, y(0) = 2; \\, y'(0) = -1

",[219],{"_43":220,"_57":221,"_59":222,"_150":151,"_152":223,"_123":165,"_125":165,"_45":224,"_127":227},"861ad829-750d-4a44-9569-2f18b430b4df","Engenharia Mecatrônica","engenharia-mecatronica","eb9315c5-bd04-4116-aa46-02c4fde320c0",{"_155":225,"_157":158,"_159":226},5,30614,"engenharia-mecatronica-1",{"_40":229,"_42":41,"_43":230,"_77":231,"_101":243,"_102":244,"_51":245},"6831a9b9-c9a4-4a76-bd7a-2ce98bfea875","93a32dd4-5fc6-4ccb-a76f-4724bf113742",[232,235,238,240],{"_43":233,"_81":82,"_83":234},"13e3f476-a0a5-421f-b47e-0969e8e9a7e0","

Verdadeiro, verdadeiro, falso

",{"_43":236,"_81":87,"_83":237},"c9a20855-d84b-4860-8b35-9784818282cf","

Falso, verdadeiro, verdadeiro

",{"_43":229,"_81":91,"_83":239},"

Verdadeiro, falso, verdadeiro

",{"_43":241,"_81":95,"_83":242},"3c98539b-1af9-4d44-aecd-c266754a2075","

Falso, falso, falso

","85,5 % decomposto → 14,5 % restante.\nN/N0 = 0,145 ⇒ t = ln(1/0,145)/(ln2/5730) ≈ 1,6×10⁴ anos.\nItens prováveis: (1) resta 14,5 % do C-14 (V); (2) idade ≈ 1,2×10⁴ anos ou 3 meias-vidas (F); (3) t½ do C-14 = 5730 anos (V). Sequência: V-F-V, correspondendo à alternativa c.","

Qual é a idade da madeira na qual 85,5\\% do C-14 se decompos?

",[246],{"_43":220,"_57":221,"_59":222,"_150":151,"_152":223,"_123":165,"_125":165,"_45":247,"_127":227},{"_155":225,"_157":158,"_159":226},{"_40":249,"_42":41,"_43":250,"_77":251,"_101":266,"_102":267,"_51":268},"3ce109b1-6cdb-4b4d-8816-a6af4d74bed0","c478a46c-46a8-4942-ae14-2e79def3dda2",[252,255,258,260,263],{"_43":253,"_81":82,"_83":254},"a5a16a42-3e58-4c76-80f2-c291aa2d8e49","Sim",{"_43":256,"_81":87,"_83":257},"5e6d0254-696c-4da0-b572-82e3a19f433e","Não",{"_43":249,"_81":91,"_83":259},"Indeterminado",{"_43":261,"_81":95,"_83":262},"35403f68-a880-429d-97f4-db507c58de38","Todos",{"_43":264,"_81":99,"_83":265},"2e06d948-d37e-4350-afc7-ced04941bcaf","Nenhum","Os polinômios e números fornecidos estão corrompidos (símbolos “????”), impossibilitando aplicar Ruffini e concluir se são raízes. Sem dados completos, o problema é indeterminado.","Emplea la regla de Ruffini para dictaminar si los siguientes números son o no raíces de los polinomios citados:
a) ????3 de ????4????5
b) ????2 de ????2????????2
c) ????1 de 2????????1
d) ????1 de 2????2????",[269],{"_43":220,"_57":221,"_59":222,"_150":151,"_152":223,"_123":165,"_125":165,"_45":270,"_127":227},{"_155":225,"_157":158,"_159":226},[272,274,280,282,284],{"_43":220,"_57":221,"_59":222,"_150":151,"_152":223,"_123":165,"_125":165,"_45":273,"_127":227},{"_155":225,"_157":158,"_159":226},{"_43":275,"_57":58,"_59":60,"_150":151,"_152":220,"_123":276,"_125":276,"_45":277,"_127":279},"88fe96b3-2a17-4198-90d5-ddda3d6f5d45","2025-04-29T03:44:35.328997+00:00",{"_155":225,"_157":158,"_159":278},25741,"mecanica-3",{"_43":147,"_57":148,"_59":149,"_150":151,"_152":-5,"_123":153,"_125":153,"_45":281,"_127":149},{"_155":156,"_157":158,"_159":160},{"_43":164,"_57":58,"_59":60,"_150":151,"_152":147,"_123":165,"_125":165,"_45":283,"_127":169},{"_155":167,"_157":158,"_159":168},{"_43":223,"_57":285,"_59":286,"_150":151,"_152":-5,"_123":287,"_125":287,"_45":288,"_127":286},"Engenharia","engenharia","2025-04-29T01:55:38.635246+00:00",{"_155":289,"_157":158,"_159":290},29,351586,"tagsTree",[293,299],{"_43":147,"_57":148,"_59":149,"_150":151,"_152":-5,"_123":153,"_125":153,"_45":294,"_127":149,"_161":295},{"_155":156,"_157":158,"_159":160},[296],{"_43":164,"_57":58,"_59":60,"_150":151,"_152":147,"_123":165,"_125":165,"_45":297,"_127":169,"_161":298},{"_155":167,"_157":158,"_159":168},[],{"_43":223,"_57":285,"_59":286,"_150":151,"_152":-5,"_123":287,"_125":287,"_45":300,"_127":286,"_161":301},{"_155":289,"_157":158,"_159":290},[302],{"_43":220,"_57":221,"_59":222,"_150":151,"_152":223,"_123":165,"_125":165,"_45":303,"_127":227,"_161":304},{"_155":225,"_157":158,"_159":226},[305],{"_43":275,"_57":58,"_59":60,"_150":151,"_152":220,"_123":276,"_125":276,"_45":306,"_127":279,"_161":307},{"_155":225,"_157":158,"_159":278},[],"videoDislikes",{},"videoLikes",{},"actionData","errors"]

Comentários