Question

(10) 7. Use the Annihilator method to find a particular solution of the equation y" + y - 2y = cos 32. [15] 8. (a) Check if the matrix A is defective or not. (b) Use the results of (a) to find the general solution to the system x' = Ax if 1-(2)

Answer 1

7 We have Y"' + y' - 2 y COS 3 het Yp = a cos 3x + osin 3x Yp' - za sin3x + 3 bcos3 X Yp" 5-9 acos 3x 9b singx Putting these values in eq (1) because y a solution of (-9acos 32 - 9 bsin3x) + (-3 asin3x +3 bcos 3x) 2 (acos 3x + b sin 3x) - cos 3 x => [-9a + 3b - 22 )COS 3X + (-96-32-2b)sings = cos 3x comparing coefficient of cos 3x + sin 3 x -92 +3b-29 -11a + 3b = 1 -9b-39-2b = 0 -34-11b = - 11/26 from 1Q +3b = 1 (putting a= LIM I b 121b+ 3b = 1 → b 130 Thus a = 14 130 Thus particular solution Up -11 Cos 3x + 3 sinax. 130 130

(a) A = (1-2) De fective matrix: A square matrix of onder men In 2x2 is called defective if have complete basis of eigen vectors or m - linearly eignvectors. A (2) both vectors (1) and (-2) independent because any vector is mot multiple of other vector: Thus A = ) is not defective. are linearly 2 4 (b) We have to find solution of X'= AX We know that solution of this differential equation X = $(tc where ølt eis fundamental matrix. To find fundamental matrix, first we will calculate eigen values. X be eigen value of A JA-111=0 het This I- - 2 0 + (1-2) (4-2) + 2 =0 4-À 12-58+4 +2=0 T

12_51 +6=0 (1-2)(1-3)=0 = 2, d=3 het eigen vector corresponding to d = 2 in Vet (m, xo)" They Avq =2v1 (A-21V=0 1-2 -2 2:1:1:1:13 4 - 2x2=0 - x = -222 het x2=1 then nu=-2 Thus Ve= (-2,1) Again het eigen vector corresponding to 4= 3 is V, - (91,4)* Then (?-?)Cl]-[0], -14, -- 4, = -42 het "= 1 then y,=-1 Thus Thus V2 =(-1, 1)T fundamental matrix tal matrix pct = [vere vorent] $(t) = -zeit -103 182t 37 ze This solution of g'AX ús x = 0lt) C. - 2021 - p3t C, X = TS where ca [: et 3t C2