Question

(3e-4 -8t +9 Consider the vector-valued functions xi(t) = | (-2+2 + 3t) and 22(t) = 3e-4t a. Compute the Wronskian of these two vectors. Wx(t) = (67 – 33t+27)e-4t), b. On which intervals are the vectors linearly independent? If there is more than one interval, enter a comma-separated list of intervals. The vectors are linearly independent on the interval(s): (-infinity,1),(1,4.5),(4.5, infinity), help (intervals). c. Find a matrix P(t) = (Pu(t) P12(t)) so that 21 and 22 are fundamental solutions to the system homogeneous differential equations a' = P(t)x. P21(t) P22(t)) Pul(t) = (4 + 86-+11) )*'. • pale) = 2 P21(t) = en p22(t) =

Answer 1

24 (+) = 1 +8++ 9 P+P+3+), 39 (t) = ( 34-45 1 3 e − 41 ) WX (t) = det (ki(t), (+)) = det 1-849 1-2 +2+36 3e-4t 3e-ut ) = 31-8t+3)e-45.- 3(-3+2+3+)eut = 3(-8"+9+3+-37) e-ut. = 3124?_j1t+g)e-ut. = (6t?-330 +27) e-4t . Two Solutions are linearly independent if Wx (t)=o. = (6+2-33+ +27) e 4to =>(6+2-334 +37) +0 ( e-ut to) => 3 lata 11t+8) +0. = (at-9)(+-) €0. => t + 1 and t The interval where | ( - 4, 4)/(1 3 1 (3,$). the solutions are linearly inde Since Wxlt) #0 le PHI 1-8t+9 3e-ut. 1-2t²t24 3e4t fore tE (-00, 1) U (1, 2) u (22,0).