Quiz — Equations differentielles lineaires.pdf

Question 1 sur 10 20:00

[{"id":68186,"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":"y² + a(x)y = b(x)","option_d":"y' = f(x,y)","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\": \"y²","_debug_options_count":4},{"id":68187,"question":"La solution générale d'une équation différentielle linéaire homogène est toujours de la forme y = Ce^(∫a(x)dx).","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"C'est exact : la solution générale d'une équation homogène y' + a(x)y = 0 est y = Ce^(∫a(x)dx).","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":68188,"question":"Quelle méthode utilise-t-on pour trouver une solution particulière d'une équation différentielle linéaire non homogène ?","option_a":"Méthode de variation de la constante","option_b":"Méthode des coefficients indéterminés","option_c":"Méthode de substitution","option_d":"Méthode graphique","option_e":"","option_f":"","bonne_reponse":"b","explication":"La méthode des coefficients indéterminés est couramment utilisée pour les seconds membres simples (polynômes, exponentielles, etc.).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"Méthode de variation de la constante\", \"b\": \"Méthode des coeffi","_debug_options_count":4},{"id":68189,"question":"Si y1 et y2 sont deux solutions d'une équation différentielle linéaire homogène, alors y = y1 + y2 est aussi une solution.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"C'est une propriété fondamentale des équations différentielles linéaires : la somme de deux solutions est aussi une solution.","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":68190,"question":"Résoudre l'équation différentielle y' - 2y = 0.","option_a":"y = Ce^(2x)","option_b":"y = Ce^(-2x)","option_c":"y = 2Ce^x","option_d":"y = Ce^x","option_e":"","option_f":"","bonne_reponse":"a","explication":"L'équation y' - 2y = 0 a pour solution y = Ce^(2x), car la solution de l'équation homogène est y = Ce^(∫2dx) = Ce^(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 = Ce^(2x)\", \"b\": \"y = Ce^(-2x)\", \"c\": \"y = 2Ce^x\", \"d\": \"y = Ce","_debug_options_count":4},{"id":68191,"question":"Quelle est la solution générale de l'équation y' + y = e^(-x) ?","option_a":"y = Ce^(-x) + x","option_b":"y = Ce^(-x) + xe^(-x)","option_c":"y = Ce^(-x) + 1","option_d":"y = Ce^(-x) + x²e^(-x)","option_e":"","option_f":"","bonne_reponse":"b","explication":"La solution générale est y = Ce^(-x) + xe^(-x), où Ce^(-x) est la solution homogène et xe^(-x) la solution particulière.","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 = Ce^(-x) + x\", \"b\": \"y = Ce^(-x) + xe^(-x)\", \"c\": \"y = Ce^(-x)","_debug_options_count":4},{"id":68192,"question":"Une équation différentielle linéaire du second ordre peut s'écrire sous la forme y'' + a(x)y' + b(x)y = c(x).","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"C'est la forme générale d'une équation différentielle linéaire du second ordre.","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":68193,"question":"Quelle est la solution générale de l'équation y'' + 4y = 0 ?","option_a":"y = C1cos(2x) + C2sin(2x)","option_b":"y = C1e^(2x) + C2e^(-2x)","option_c":"y = C1cos(x) + C2sin(x)","option_d":"y = C1x + C2","option_e":"","option_f":"","bonne_reponse":"a","explication":"L'équation caractéristique est r² + 4 = 0, dont les racines sont r = ±2i. La solution générale est donc y = C1cos(2x) + C2sin(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 = C1cos(2x) + C2sin(2x)\", \"b\": \"y = C1e^(2x) + C2e^(-2x)\", \"c\":","_debug_options_count":4},{"id":68194,"question":"Pour résoudre une équation différentielle linéaire, il faut toujours commencer par résoudre l'équation homogène associée.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"a","explication":"Oui, la résolution de l'équation homogène est une étape essentielle avant de chercher une solution particulière.","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":68195,"question":"Quelle est la solution du problème de Cauchy y' + y = 0 avec y(0) = 2 ?","option_a":"y = 2e^(-x)","option_b":"y = 2e^x","option_c":"y = e^(-x) + 1","option_d":"y = 2cos(x)","option_e":"","option_f":"","bonne_reponse":"a","explication":"La solution générale est y = Ce^(-x). Avec y(0) = 2, on trouve C = 2, donc y = 2e^(-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\": \"y = 2e^(-x)\", \"b\": \"y = 2e^x\", \"c\": \"y = e^(-x) + 1\", \"d\": \"y = 2","_debug_options_count":4}]