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Consider-the-boundary-value-problem-y-2y-2y-0-y-a-c-y-b-d-1-If-this-problem-has-a-unique-solution-how-are-a-and-b-related-2-If-this-problem-has-no-solution-how-are-a-b-c-and-d-




Question Number 184306 by Mastermind last updated on 05/Jan/23
Consider the boundary value   problem y^(′′) −2y′+2y=0,       y(a)=c  ,y(b)=d.  1) If this problem has a unique  solution, how are a and b related?  2) If this problem has no solution,  how are a,b,c and d related?      Help!
Considertheboundaryvalueproblemy2y+2y=0,y(a)=c,y(b)=d.1)Ifthisproblemhasauniquesolution,howareaandbrelated?2)Ifthisproblemhasnosolution,howarea,b,canddrelated?Help!
Answered by FelipeLz last updated on 05/Jan/23
y′′−2y′+2y = 0  y = e^r  → r^2 e^r −2re^r +2e^r  = 0  r^2 −2r+2 = 0 → r = 1±i  y(x) = k_1 e^x cos(x)+k_2 e^x sin(x)     1. y(a) = c → k_1 e^a cos(a)+k_2 e^a sin(a) = c       y(b) = d → k_1 e^b cos(b)+k_2 e^b sin(b) = d        determinant (((e^a cos(a)),(e^a sin(a))),((e^b cos(b)),(e^b sin(b))))≠ 0       e^(a+b) cos(a)sin(b)−e^(a+b) sin(a)cos(b) ≠ 0       e^(a+b) sin(b−a) ≠ 0       sin(b−a) ≠ 0       b ≠ a+nπ  ∀n ∈ Z      2.  determinant (((e^a cos(a)),(e^a sin(a))),((e^b cos(b)),(e^b sin(b))))= 0       e^(a+b) cos(a)sin(b)−e^(a+b) sin(a)cos(b) = 0       e^(a+b) sin(b−a) = 0       sin(b−a) = 0       b = a+nπ  ∀n ∈ Z        determinant ((c,(e^a sin(a))),(d,(e^b sin(b))))≠ 0       ce^b sin(b)−de^a sin(a) ≠ 0       ce^(a+nπ) sin(a+nπ)−de^a sin(a) ≠ 0       ce^(a+nπ) sin(a)cos(nπ)−de^a sin(a) ≠ 0       e^a sin(a)[ce^(nπ) cos(nπ)−d] ≠ 0       d ≠ ce^(nπ) cos(nπ)        { ((b−a = nπ  ∀n ∈ Z)),((d ≠ ce^(b−a) cos(b−a))) :}
y2y+2y=0y=err2er2rer+2er=0r22r+2=0r=1±iy(x)=k1excos(x)+k2exsin(x)1.y(a)=ck1eacos(a)+k2easin(a)=cy(b)=dk1ebcos(b)+k2ebsin(b)=d|eacos(a)easin(a)ebcos(b)ebsin(b)|0ea+bcos(a)sin(b)ea+bsin(a)cos(b)0ea+bsin(ba)0sin(ba)0ba+nπnZ2.|eacos(a)easin(a)ebcos(b)ebsin(b)|=0ea+bcos(a)sin(b)ea+bsin(a)cos(b)=0ea+bsin(ba)=0sin(ba)=0b=a+nπnZ|ceasin(a)debsin(b)|0cebsin(b)deasin(a)0cea+nπsin(a+nπ)deasin(a)0cea+nπsin(a)cos(nπ)deasin(a)0easin(a)[cenπcos(nπ)d]0dcenπcos(nπ){ba=nπnZdcebacos(ba)

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