Quiz — eqdiff.pdf

Question 1 sur 10 20:00

[{"id":56142,"question":"Quelle est la forme générale d'une équation différentielle linéaire du premier ordre ?","option_a":"y' + a(x)y = b(x)","option_b":"y'' + a(x)y' + b(x)y = c(x)","option_c":"y = f(x)","option_d":"y' = f(x)","option_e":"","option_f":"","bonne_reponse":"a","explication":"La forme générale est y' + a(x)y = b(x), où a(x) et b(x) sont des fonctions continues.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"y' + a(x)y = b(x)\", \"b\": \"y'' + a(x)y' + b(x)y = c(x)\", \"c\": \"y =","_debug_options_count":4},{"id":56143,"question":"La solution générale d'une équation différentielle linéaire du second ordre à coefficients constants est toujours de la forme y(x) = C1e^(r1x) + C2e^(r2x).","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"b","explication":"Cette forme est valable uniquement si l'équation caractéristique admet deux racines réelles distinctes. Sinon, la solution diffère (racines doubles, complexes).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":56144,"question":"Quelle méthode permet de résoudre une équation différentielle de la forme y' = f(x)g(y) ?","option_a":"Intégration directe","option_b":"Séparation des variables","option_c":"Changement de variable","option_d":"Résolution par série","option_e":"","option_f":"","bonne_reponse":"b","explication":"La méthode de séparation des variables consiste à écrire dy\/g(y) = f(x)dx et à intégrer les deux membres.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"Intégration directe\", \"b\": \"Séparation des variables\", \"c\": \"Ch","_debug_options_count":4},{"id":56145,"question":"Si l'équation caractéristique d'une équation différentielle du second ordre a pour racines r1 = 2 et r2 = -3, quelle est la solution générale ?","option_a":"y(x) = C1e^(2x) + C2e^(-3x)","option_b":"y(x) = C1e^(2x) + C2e^(3x)","option_c":"y(x) = (C1 + C2x)e^(2x)","option_d":"y(x) = C1cos(2x) + C2sin(3x)","option_e":"","option_f":"","bonne_reponse":"a","explication":"La solution générale est y(x) = C1e^(r1x) + C2e^(r2x) avec r1 = 2 et r2 = -3.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"y(x) = C1e^(2x) + C2e^(-3x)\", \"b\": \"y(x) = C1e^(2x) + C2e^(3x)\", ","_debug_options_count":4},{"id":56146,"question":"Une équation différentielle du type y'' + y = 0 admet des solutions périodiques.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Les solutions sont de la forme y(x) = C1cos(x) + C2sin(x), qui sont périodiques de période 2π.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":56147,"question":"Quelle est la solution générale de l'équation différentielle y' - 2y = 0 ?","option_a":"y(x) = C1e^(2x)","option_b":"y(x) = C1e^(-2x)","option_c":"y(x) = C1 + C2x","option_d":"y(x) = C1x + C2","option_e":"","option_f":"","bonne_reponse":"a","explication":"L'équation caractéristique est r - 2 = 0, donc r = 2. La solution est y(x) = C1e^(2x).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"y(x) = C1e^(2x)\", \"b\": \"y(x) = C1e^(-2x)\", \"c\": \"y(x) = C1 + C2x\"","_debug_options_count":4},{"id":56148,"question":"Parmi ces équations, laquelle est une équation différentielle linéaire du second ordre ?","option_a":"y' = x²y + 1","option_b":"y'' + y'y = sin(x)","option_c":"y'' - 3y' + 2y = e^x","option_d":"y = x² + y'","option_e":"","option_f":"","bonne_reponse":"c","explication":"L'équation y'' - 3y' + 2y = e^x est linéaire du second ordre à coefficients constants.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"c\", \"options\": {\"a\": \"y' = x²y + 1\", \"b\": \"y'' + y'y = sin(x)\", \"c\": \"y'' - 3y' + 2y =","_debug_options_count":4},{"id":56149,"question":"La méthode du facteur intégrant s'applique uniquement aux équations différentielles du premier ordre.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Le facteur intégrant est une méthode spécifique aux équations linéaires du premier ordre de la forme y' + a(x)y = b(x).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4},{"id":56150,"question":"Quelle est la solution générale de l'équation différentielle y'' + 4y' + 4y = 0 ?","option_a":"y(x) = C1e^(-2x) + C2e^(-2x)","option_b":"y(x) = (C1 + C2x)e^(-2x)","option_c":"y(x) = C1cos(2x) + C2sin(2x)","option_d":"y(x) = C1e^(2x) + C2e^(-2x)","option_e":"","option_f":"","bonne_reponse":"b","explication":"L'équation caractéristique admet une racine double r = -2, donc la solution est y(x) = (C1 + C2x)e^(-2x).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"y(x) = C1e^(-2x) + C2e^(-2x)\", \"b\": \"y(x) = (C1 + C2x)e^(-2x)\", \"","_debug_options_count":4},{"id":56151,"question":"Si une équation différentielle admet une solution particulière yp(x), alors la solution générale est y(x) = yp(x) + yh(x), où yh(x) est la solution de l'équation homogène associée.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"C'est le principe de superposition pour les équations différentielles linéaires : la solution générale est la somme de la solution particulière et de la solution homogène.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"a\", \"options\": {\"a\": \"Vrai\", \"b\": \"Faux\", \"c\": \"\", \"d\": \"\"}}","_debug_options_count":4}]