Question

(1 point) 12 -9 2 Consider the initial value problem X, X(0) = 4 0 2 The eigenvalue is 11 = 12 = A corresponding eigenvector is K= 2 and a solution of (A - 111)P = K is given by P= :] The general solution can be written as ---( : )--( . )-(:). --[-DEL The solution of the Initial Value Problem is teht

Answer 1

Griven x = [12 :]* xro)=12] SA X = AX X10) = 20 To find the eigers values Ard Il = 0 we evaluate (2 . 위 4. 0 -9 -1 (12-0) 6-1) +36 = - 12x + x² + 36 X2_12x + 36 (106)² = so d= 6,6 6 ds we conclude To find the eigens vector (A-XI]* 12 19 {[% 11-18:]}}}1:1 W 6 *, -9% =0 4x, 6x₂ 지드를로 277 372 = 0 23, -330 so if "=2, n. Vis 2 To find w [A-XI ] = v 64,- gn Csscanned with Camscainer, - 6 M2 2 3

2%, "RE =) n = 14 314 22,- 372 = 1 Ir if a then x,- 2 So Vz - - X [":] (Cu,+ (tv, tv ₂)2] ett = [ ()+21+() +())]2** “[2] 2(13]te teattl [ ]ect] Gt e &c. k > (A-),I) P 3 S [(2-)-1: ;)]P-13] 19311) 2 t CS$canned with CamScanner

we get x = ( ) € (2) est 64 it 2(13) te (-)") x= 30, e8t + 3ętett teste w - x o + +2ątest 2c, est we have x,(0)=2 farol=2 2,(0) 30, + €2= 2 210) 2, = 2 C = 1 >) 3% 3ct (2=2 27 2- en C2 3-2 = C2 Ca - 2 2 t So sx = (2)=4-2[(*) te**+ (%)est] Anę CScanned with CamScanner
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