Quiz — equadiff.pdf

Question 1 sur 5 10:00

[{"id":49745,"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,y)","option_d":"y = f(x,y')","option_e":"","option_f":"","bonne_reponse":"a","explication":"La forme générale d'une équation différentielle linéaire du premier ordre est y' + a(x)y = b(x), où a(x) et b(x) sont des fonctions données.","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":49746,"question":"Quel est le facteur intégrant pour résoudre y' + 2y = e^x ?","option_a":"e^(2x)","option_b":"e^x","option_c":"e^(x\/2)","option_d":"e^(-2x)","option_e":"","option_f":"","bonne_reponse":"d","explication":"Le facteur intégrant est e^(∫2 dx) = e^(2x). Cependant, pour cette équation, le facteur intégrant est e^(∫2 dx) = e^(2x), mais la résolution nécessite de multiplier par e^(2x) pour simplifier l'équation.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"d\", \"options\": {\"a\": \"e^(2x)\", \"b\": \"e^x\", \"c\": \"e^(x\/2)\", \"d\": \"e^(-2x)\"}}","_debug_options_count":4},{"id":49747,"question":"Quelle est la solution générale de y'' - 4y' + 4y = 0 ?","option_a":"y = C1e^(2x) + C2e^(-2x)","option_b":"y = (C1 + C2x)e^(2x)","option_c":"y = C1e^(x) + C2e^(-x)","option_d":"y = C1cos(2x) + C2sin(2x)","option_e":"","option_f":"","bonne_reponse":"b","explication":"L'équation caractéristique r² - 4r + 4 = 0 a une racine double r = 2, donc la solution générale est y = (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 = C1e^(2x) + C2e^(-2x)\", \"b\": \"y = (C1 + C2x)e^(2x)\", \"c\": \"y =","_debug_options_count":4},{"id":49748,"question":"Pour l'équation y' = 3y, quelle est la solution vérifiant y(0) = 2 ?","option_a":"y = 2e^(3x)","option_b":"y = 2e^(-3x)","option_c":"y = 6x + 2","option_d":"y = 3x + 2","option_e":"","option_f":"","bonne_reponse":"a","explication":"La solution générale est y = Ce^(3x). Avec y(0) = 2, on trouve C = 2, donc y = 2e^(3x).","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^(3x)\", \"b\": \"y = 2e^(-3x)\", \"c\": \"y = 6x + 2\", \"d\": \"y = 3","_debug_options_count":4},{"id":49749,"question":"Quel est le type d'une équation différentielle de la forme y' = xy² ?","option_a":"Linéaire du premier ordre","option_b":"Séparable","option_c":"Du second ordre","option_d":"Non linéaire","option_e":"","option_f":"","bonne_reponse":"b","explication":"Cette équation est séparable car elle peut s'écrire sous la forme dy\/y² = x dx, ce qui permet une intégration directe.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0,"_debug_answer_data_type":"string","_debug_answer_data_preview":"{\"correct\": \"b\", \"options\": {\"a\": \"Linéaire du premier ordre\", \"b\": \"Séparable\", \"c\": \"Du second o","_debug_options_count":4}]