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3y = 0",{"_44":202,"_82":88,"_84":203},"6d3c3a1d-8e80-4ff3-b97a-d5b5bca91940","6y'' - y' - y = 0",{"_44":205,"_82":91,"_84":206},"4168ac67-64fd-46ed-a69c-a01dbfcde7ea","y'' - 2y' + y = 0",{"_44":208,"_82":95,"_84":209},"cc7bc8df-6995-4264-ac5a-32ea9b4ed7db","9y'' + 6y' + y = 0",{"_44":195,"_82":211,"_84":212},"E","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

",[216],{"_44":217,"_58":218,"_60":219,"_146":147,"_148":220,"_119":161,"_121":161,"_46":221,"_123":224},"861ad829-750d-4a44-9569-2f18b430b4df","Engenharia Mecatrônica","engenharia-mecatronica","eb9315c5-bd04-4116-aa46-02c4fde320c0",{"_151":222,"_153":154,"_155":223},5,30614,"engenharia-mecatronica-1",{"_40":226,"_42":43,"_44":227,"_78":228,"_97":240,"_98":241,"_52":242},"6831a9b9-c9a4-4a76-bd7a-2ce98bfea875","93a32dd4-5fc6-4ccb-a76f-4724bf113742",[229,232,235,237],{"_44":230,"_82":83,"_84":231},"13e3f476-a0a5-421f-b47e-0969e8e9a7e0","

Verdadeiro, verdadeiro, falso

",{"_44":233,"_82":88,"_84":234},"c9a20855-d84b-4860-8b35-9784818282cf","

Falso, verdadeiro, verdadeiro

",{"_44":226,"_82":91,"_84":236},"

Verdadeiro, falso, verdadeiro

",{"_44":238,"_82":95,"_84":239},"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?

",[243],{"_44":217,"_58":218,"_60":219,"_146":147,"_148":220,"_119":161,"_121":161,"_46":244,"_123":224},{"_151":222,"_153":154,"_155":223},{"_40":246,"_42":43,"_44":247,"_78":248,"_97":263,"_98":264,"_52":265},"3ce109b1-6cdb-4b4d-8816-a6af4d74bed0","c478a46c-46a8-4942-ae14-2e79def3dda2",[249,252,255,257,260],{"_44":250,"_82":83,"_84":251},"a5a16a42-3e58-4c76-80f2-c291aa2d8e49","Sim",{"_44":253,"_82":88,"_84":254},"5e6d0254-696c-4da0-b572-82e3a19f433e","Não",{"_44":246,"_82":91,"_84":256},"Indeterminado",{"_44":258,"_82":95,"_84":259},"35403f68-a880-429d-97f4-db507c58de38","Todos",{"_44":261,"_82":211,"_84":262},"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????",[266],{"_44":217,"_58":218,"_60":219,"_146":147,"_148":220,"_119":161,"_121":161,"_46":267,"_123":224},{"_151":222,"_153":154,"_155":223},[269,271,277,279,281],{"_44":217,"_58":218,"_60":219,"_146":147,"_148":220,"_119":161,"_121":161,"_46":270,"_123":224},{"_151":222,"_153":154,"_155":223},{"_44":272,"_58":59,"_60":61,"_146":147,"_148":217,"_119":273,"_121":273,"_46":274,"_123":276},"88fe96b3-2a17-4198-90d5-ddda3d6f5d45","2025-04-29T03:44:35.328997+00:00",{"_151":222,"_153":154,"_155":275},25741,"mecanica-3",{"_44":143,"_58":144,"_60":145,"_146":147,"_148":-5,"_119":149,"_121":149,"_46":278,"_123":145},{"_151":152,"_153":154,"_155":156},{"_44":160,"_58":59,"_60":61,"_146":147,"_148":143,"_119":161,"_121":161,"_46":280,"_123":165},{"_151":163,"_153":154,"_155":164},{"_44":220,"_58":282,"_60":283,"_146":147,"_148":-5,"_119":284,"_121":284,"_46":285,"_123":283},"Engenharia","engenharia","2025-04-29T01:55:38.635246+00:00",{"_151":286,"_153":154,"_155":287},29,351586,"tagsTree",[290,296],{"_44":143,"_58":144,"_60":145,"_146":147,"_148":-5,"_119":149,"_121":149,"_46":291,"_123":145,"_157":292},{"_151":152,"_153":154,"_155":156},[293],{"_44":160,"_58":59,"_60":61,"_146":147,"_148":143,"_119":161,"_121":161,"_46":294,"_123":165,"_157":295},{"_151":163,"_153":154,"_155":164},[],{"_44":220,"_58":282,"_60":283,"_146":147,"_148":-5,"_119":284,"_121":284,"_46":297,"_123":283,"_157":298},{"_151":286,"_153":154,"_155":287},[299],{"_44":217,"_58":218,"_60":219,"_146":147,"_148":220,"_119":161,"_121":161,"_46":300,"_123":224,"_157":301},{"_151":222,"_153":154,"_155":223},[302],{"_44":272,"_58":59,"_60":61,"_146":147,"_148":217,"_119":273,"_121":273,"_46":303,"_123":276,"_157":304},{"_151":222,"_153":154,"_155":275},[],"videoDislikes",{},"videoLikes",{},"actionData","errors"]

Comentários