[{"_1":2,"_207":-5,"_208":-5},"loaderData",{"_3":4,"_9":10},"root",{"_5":-5,"_6":7,"_8":-5},"registerModalConfig","url","https://prouniversitario.com/exercicios/b671dd2b-5169-42b6-aa00-1d0d377fefbb","user","pages/questions/show/index",{"_11":12,"_13":14,"_15":14,"_16":17,"_86":87,"_88":89,"_29":170,"_192":193,"_203":204,"_205":206},"comments",[],"hasVideoHash",false,"isAiGenerated","question",{"_18":19,"_20":19,"_21":22,"_23":24,"_56":57,"_80":19,"_81":82,"_83":84,"_85":14},"correctOptionId","","difficulty","id","b671dd2b-5169-42b6-aa00-1d0d377fefbb","metadata",{"_25":26,"_52":53,"_54":55},"pd",{"_27":28,"_29":30,"_31":32,"_39":40,"_41":42,"_43":44,"_45":46,"_47":44,"_48":49,"_50":51},"area","Outros","tags",[],"subjects",[33],{"_21":34,"_35":36,"_37":38},678643,"name","Análise Estrutural","alias","analise-estrutural","materialId","119876319","answerCount",1,"isIndexable",true,"mainSubject",{"_21":34,"_35":36,"_37":38},"isIndexableSEO","materialFormat","Text","studentDegreeLevel","Graduate","source","PD","externalId","165949295","options",[58,64,68,72,76],{"_21":59,"_60":61,"_62":63},"f5dec689-9946-47a9-b6e0-cd278f3679d5","label","A","text","0 kN.",{"_21":65,"_60":66,"_62":67},"d9a14be6-260c-4deb-9da3-ea1cf10bfa00","B","10 kN.",{"_21":69,"_60":70,"_62":71},"0cc2d12b-beb9-4f7e-b58a-1789d1182b04","C","40 kN.",{"_21":73,"_60":74,"_62":75},"5e81e261-9c2a-4d05-a749-f3b0f64c9449","D","50 kN.",{"_21":77,"_60":78,"_62":79},"047b936e-c56a-4169-b2a9-34061287d784","E","60 kN.","solution_explanation","statement","

O valor, em módulo, da soma das forças de reação horizontais devido aos apoios A e B, é igual a:

","status","pending_classification","hasRelatedVideo","relatedVideos",[],"similarQuestions",[90,131,149],{"_18":91,"_20":19,"_21":92,"_56":93,"_80":108,"_81":109,"_29":110},"112831da-be9c-458e-94cb-e8292d0dba12","3c0b9441-b878-4896-8bac-fa4ca61e3f13",[94,97,100,103,106],{"_21":95,"_60":61,"_62":96},"265bd71a-6749-41f5-b291-67efd646ca64","y'' + 2y' - 3y = 0",{"_21":98,"_60":66,"_62":99},"6d3c3a1d-8e80-4ff3-b97a-d5b5bca91940","6y'' - y' - y = 0",{"_21":101,"_60":70,"_62":102},"4168ac67-64fd-46ed-a69c-a01dbfcde7ea","y'' - 2y' + y = 0",{"_21":104,"_60":74,"_62":105},"cc7bc8df-6995-4264-ac5a-32ea9b4ed7db","9y'' + 6y' + y = 0",{"_21":91,"_60":78,"_62":107},"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

",[111],{"_21":112,"_35":113,"_37":114,"_115":116,"_117":118,"_119":120,"_121":120,"_23":122,"_129":130},"b6e37f32-f8e5-4025-bf0f-69ffc00fb331","Engenharia Civil","engenharia-civil","tag_group_id","c1ce2cf2-b217-4cc7-8182-81c56bd3522d","parent_id","eb9315c5-bd04-4116-aa46-02c4fde320c0","created_at","2025-04-29T02:19:12.828432+00:00","updated_at",{"_123":124,"_125":126,"_127":128},"childrenCount",21,"content_count",0,"question_count",45865,"slug","engenharia-civil-3",{"_18":132,"_20":19,"_21":133,"_56":134,"_80":146,"_81":147,"_29":148},"6831a9b9-c9a4-4a76-bd7a-2ce98bfea875","93a32dd4-5fc6-4ccb-a76f-4724bf113742",[135,138,141,143],{"_21":136,"_60":61,"_62":137},"13e3f476-a0a5-421f-b47e-0969e8e9a7e0","

Verdadeiro, verdadeiro, falso

",{"_21":139,"_60":66,"_62":140},"c9a20855-d84b-4860-8b35-9784818282cf","

Falso, verdadeiro, verdadeiro

",{"_21":132,"_60":70,"_62":142},"

Verdadeiro, falso, verdadeiro

",{"_21":144,"_60":74,"_62":145},"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?

",[111],{"_18":150,"_20":19,"_21":151,"_56":152,"_80":167,"_81":168,"_29":169},"3ce109b1-6cdb-4b4d-8816-a6af4d74bed0","c478a46c-46a8-4942-ae14-2e79def3dda2",[153,156,159,161,164],{"_21":154,"_60":61,"_62":155},"a5a16a42-3e58-4c76-80f2-c291aa2d8e49","Sim",{"_21":157,"_60":66,"_62":158},"5e6d0254-696c-4da0-b572-82e3a19f433e","Não",{"_21":150,"_60":70,"_62":160},"Indeterminado",{"_21":162,"_60":74,"_62":163},"35403f68-a880-429d-97f4-db507c58de38","Todos",{"_21":165,"_60":78,"_62":166},"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????",[111],[111,171,180,185],{"_21":172,"_35":173,"_37":174,"_115":116,"_117":112,"_119":175,"_121":175,"_23":176,"_129":179},"c5bf57d4-c3d8-4365-b53e-085fa14a3a95","Estruturas","estruturas","2025-04-29T03:44:32.303369+00:00",{"_123":177,"_125":126,"_127":178},5,8852,"estruturas-1",{"_21":181,"_35":36,"_37":38,"_115":116,"_117":172,"_119":182,"_121":182,"_23":183,"_129":38},"e9d666f5-dcb8-4aa3-abc7-27cfb929e3b6","2025-04-29T04:00:06.9766+00:00",{"_123":126,"_125":126,"_127":184},4694,{"_21":118,"_35":186,"_37":187,"_115":116,"_117":-5,"_119":188,"_121":188,"_23":189,"_129":187},"Engenharia","engenharia","2025-04-29T01:55:38.635246+00:00",{"_123":190,"_125":126,"_127":191},29,351586,"tagsTree",[194],{"_21":118,"_35":186,"_37":187,"_115":116,"_117":-5,"_119":188,"_121":188,"_23":189,"_129":187,"_195":196},"children",[197],{"_21":112,"_35":113,"_37":114,"_115":116,"_117":118,"_119":120,"_121":120,"_23":122,"_129":130,"_195":198},[199],{"_21":172,"_35":173,"_37":174,"_115":116,"_117":112,"_119":175,"_121":175,"_23":176,"_129":179,"_195":200},[201],{"_21":181,"_35":36,"_37":38,"_115":116,"_117":172,"_119":182,"_121":182,"_23":183,"_129":38,"_195":202},[],"videoDislikes",{},"videoLikes",{},"actionData","errors"]

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