I don't understand why X'(0)=0 and not X(0)=0 because in the previous problems, the initial conditions were X(0)=0.
Extracted text: Safari File Edit View History Bookmarks Window Help * 2 Mon 10:48 PM E A bartleby.com Updates Available Restart Do you want to restart to install these updates now or try tonight? Later = bartleby Search for textbooks, step-by-step explanations to homework . Ask an Expert Install SimUText Screen Shot 2021-03...8.09 PM E Bundle: Differential Equations with Bou... < chapter="" 12.3,="" problem="" 4e=""> Get live help whenever you need from online tutors! Try bartleby tutor today → kX" () Т () %—DX () T () X"(t) X(t) T'(t) kT(t) (4) = -1 Feb 18, 2021 at 11-18 P...pnglmage Lead to the two ordinary differential equations, X" (t) + 1X (t) = 0 (5) And, T' (t) + kAT (t) = 0 (6) Use boundary conditions (2) in u (x, t) = X (t) T (t) are, u (0, t) = X (0) T (t) =0 And u (L, t) = X (L) T (t) =0 Since T (t) + 0 for all t, so only condition which holds above when X GET 10 FREE QUESTIONS X' (0) =0 and X' (L) =0. Privacy - Terms The corresponding solutions of the differential equations, given by MAR 29 étv
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