[{"_1":2,"_269":-5,"_270":-5},"loaderData",{"_3":4,"_9":10},"root",{"_5":-5,"_6":7,"_8":-5},"registerModalConfig","url","https://prouniversitario.com/exercicios/f4ffc447-470f-4410-9359-d4e78a94aa64?video=true","user","pages/questions/show/index",{"_11":12,"_13":14,"_15":14,"_16":17,"_86":87,"_170":171,"_29":241,"_257":258,"_265":266,"_267":268},"comments",[],"hasVideoHash",false,"isAiGenerated","question",{"_18":19,"_20":19,"_21":22,"_23":24,"_56":57,"_80":19,"_81":82,"_83":84,"_85":44},"correctOptionId","","difficulty","id","f4ffc447-470f-4410-9359-d4e78a94aa64","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},672071,"name","Mecânica","alias","mecanica","materialId","135938266","answerCount",1,"isIndexable",true,"mainSubject",{"_21":34,"_35":36,"_37":38},"isIndexableSEO","materialFormat","Text","studentDegreeLevel","Graduate","source","PD","externalId","135938289","options",[58,64,68,72,76],{"_21":59,"_60":61,"_62":63},"bda805d7-2382-44e0-afe7-357623ae497e","label","A","text","-x \\, \text{cos}(x) + \text{sen}(x) + C.",{"_21":65,"_60":66,"_62":67},"d35c5055-3716-4919-98dc-f4bc83a239af","B","-x \\, \text{cos}(x) + \text{cos}(x) + C.",{"_21":69,"_60":70,"_62":71},"6dea11e5-893f-4e39-b886-8a7c6f45a18f","C","-x \\, \text{cos}(x) - \text{sen}(x) + C.",{"_21":73,"_60":74,"_62":75},"46de62ca-5b93-4fe7-b8d8-0c9e16d7960c","D","-x \\, \text{sen}(x) - \text{cos}(x) + C.",{"_21":77,"_60":78,"_62":79},"2a6c6d11-76f9-4220-ba3b-a41ea893e13d","E","x \\, \text{cos}(x) - \text{sen}(x) + C.","solution_explanation","statement","

A integração por partes é uma técnica utilizada para integrar o produto de duas funções. Calcule a integral indefinida de x \\, \text{sen}(x)

","status","pending_classification","hasRelatedVideo","relatedVideos",[88],{"_89":90,"_108":109,"_29":124,"_150":151,"_169":125},"content",{"_21":91,"_35":92,"_93":-5,"_94":95,"_96":97,"_98":-5,"_23":99,"_102":103,"_104":103,"_105":-5,"_106":107},"468e3e62-35e0-47be-a15f-68dd43bb32bb","Óptica Geométrica - Parte 1","description","content_type_id","1051eee7-c62e-4b2c-9900-77a29fcec080","resource_id","fa6aaf42-4cc4-4d76-8915-7a67081f7e9a","user_id",{"_100":101},"external_id","4963","created_at","2025-08-20T12:27:41.608438+00:00","updated_at","content_group_id","slug","optica-geometrica-parte-1","prometheusData",{"_110":111,"_117":118,"_119":120,"_121":122,"_6":123},"provider",{"_21":112,"_113":114,"_115":116},"06f120be-2194-4548-8336-7ba44ffbae14","key","grupo_q_proteus","kind","proteus","reference","d80487a0-0f98-42c8-99bb-551165ba3068","state","completed","thumbnail_url","https://proteus.grupoq.io/videos/d80487a0-0f98-42c8-99bb-551165ba3068/thumbnailsthumb.0000000.jpg","https://proteus.grupoq.io/videos/d80487a0-0f98-42c8-99bb-551165ba3068/levels.m3u8",[125],{"_21":126,"_35":127,"_37":128,"_129":130,"_131":-5,"_102":132,"_104":132,"_23":133,"_106":128,"_140":141},"9772e958-72c6-4499-ad47-877cf94d670d","Exatas","exatas","tag_group_id","c1ce2cf2-b217-4cc7-8182-81c56bd3522d","parent_id","2025-04-29T01:56:45.284172+00:00",{"_134":135,"_136":137,"_138":139},"childrenCount",18,"content_count",0,"question_count",135489,"children",[142],{"_21":143,"_35":36,"_37":38,"_129":130,"_131":126,"_102":144,"_104":144,"_23":145,"_106":148,"_140":149},"ca6fee0e-79c6-4f1c-b500-ee081a071949","2025-04-29T02:19:12.828432+00:00",{"_134":146,"_136":137,"_138":147},10,19948,"mecanica-2",[],"video",{"_21":97,"_152":92,"_93":-5,"_153":154,"_155":156,"_6":-5,"_157":-5,"_110":156,"_158":159,"_121":-5,"_160":-5,"_161":-5,"_162":-5,"_163":-5,"_23":164,"_165":137,"_166":-5,"_83":167,"_102":168,"_104":168},"title","duration",20,"source_type","prometheus","iframe_url","provider_video_id","1cd3d633-d690-4f4b-9995-9b0e44994e35","format","quality","file_path","file_size",{"_100":101},"views_count","last_played_at","draft","2025-08-19T11:56:20.758718+00:00","tag","similarQuestions",[172,202,220],{"_18":173,"_20":19,"_21":174,"_56":175,"_80":190,"_81":191,"_29":192},"112831da-be9c-458e-94cb-e8292d0dba12","3c0b9441-b878-4896-8bac-fa4ca61e3f13",[176,179,182,185,188],{"_21":177,"_60":61,"_62":178},"265bd71a-6749-41f5-b291-67efd646ca64","y'' + 2y' - 3y = 0",{"_21":180,"_60":66,"_62":181},"6d3c3a1d-8e80-4ff3-b97a-d5b5bca91940","6y'' - y' - y = 0",{"_21":183,"_60":70,"_62":184},"4168ac67-64fd-46ed-a69c-a01dbfcde7ea","y'' - 2y' + y = 0",{"_21":186,"_60":74,"_62":187},"cc7bc8df-6995-4264-ac5a-32ea9b4ed7db","9y'' + 6y' + y = 0",{"_21":173,"_60":78,"_62":189},"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

",[193],{"_21":194,"_35":195,"_37":196,"_129":130,"_131":197,"_102":144,"_104":144,"_23":198,"_106":201},"861ad829-750d-4a44-9569-2f18b430b4df","Engenharia Mecatrônica","engenharia-mecatronica","eb9315c5-bd04-4116-aa46-02c4fde320c0",{"_134":199,"_136":137,"_138":200},5,30614,"engenharia-mecatronica-1",{"_18":203,"_20":19,"_21":204,"_56":205,"_80":217,"_81":218,"_29":219},"6831a9b9-c9a4-4a76-bd7a-2ce98bfea875","93a32dd4-5fc6-4ccb-a76f-4724bf113742",[206,209,212,214],{"_21":207,"_60":61,"_62":208},"13e3f476-a0a5-421f-b47e-0969e8e9a7e0","

Verdadeiro, verdadeiro, falso

",{"_21":210,"_60":66,"_62":211},"c9a20855-d84b-4860-8b35-9784818282cf","

Falso, verdadeiro, verdadeiro

",{"_21":203,"_60":70,"_62":213},"

Verdadeiro, falso, verdadeiro

",{"_21":215,"_60":74,"_62":216},"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?

",[193],{"_18":221,"_20":19,"_21":222,"_56":223,"_80":238,"_81":239,"_29":240},"3ce109b1-6cdb-4b4d-8816-a6af4d74bed0","c478a46c-46a8-4942-ae14-2e79def3dda2",[224,227,230,232,235],{"_21":225,"_60":61,"_62":226},"a5a16a42-3e58-4c76-80f2-c291aa2d8e49","Sim",{"_21":228,"_60":66,"_62":229},"5e6d0254-696c-4da0-b572-82e3a19f433e","Não",{"_21":221,"_60":70,"_62":231},"Indeterminado",{"_21":233,"_60":74,"_62":234},"35403f68-a880-429d-97f4-db507c58de38","Todos",{"_21":236,"_60":78,"_62":237},"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????",[193],[193,242,248,249,250],{"_21":243,"_35":36,"_37":38,"_129":130,"_131":194,"_102":244,"_104":244,"_23":245,"_106":247},"88fe96b3-2a17-4198-90d5-ddda3d6f5d45","2025-04-29T03:44:35.328997+00:00",{"_134":199,"_136":137,"_138":246},25741,"mecanica-3",{"_21":126,"_35":127,"_37":128,"_129":130,"_131":-5,"_102":132,"_104":132,"_23":133,"_106":128},{"_21":143,"_35":36,"_37":38,"_129":130,"_131":126,"_102":144,"_104":144,"_23":145,"_106":148},{"_21":197,"_35":251,"_37":252,"_129":130,"_131":-5,"_102":253,"_104":253,"_23":254,"_106":252},"Engenharia","engenharia","2025-04-29T01:55:38.635246+00:00",{"_134":255,"_136":137,"_138":256},29,351586,"tagsTree",[125,259],{"_21":197,"_35":251,"_37":252,"_129":130,"_131":-5,"_102":253,"_104":253,"_23":254,"_106":252,"_140":260},[261],{"_21":194,"_35":195,"_37":196,"_129":130,"_131":197,"_102":144,"_104":144,"_23":198,"_106":201,"_140":262},[263],{"_21":243,"_35":36,"_37":38,"_129":130,"_131":194,"_102":244,"_104":244,"_23":245,"_106":247,"_140":264},[],"videoDislikes",{},"videoLikes",{},"actionData","errors"]

Comentários