Question

2 4. Solve the initial value problem: X' 1 1 4 0 2 0 1 1 1 X X(0) = 6 0

Answer 1

4) The given initial value problem is x'a 1 1 O 0 2 X with X(0) - Nuevo No - Luxe have to soive the system: The vou can be solved by given inilia problem Eigenvalues - Eigenvecton method know that, if an initial problem is of the form x we valu АХ eue have to first and the . Then the determine the eigenvalues ti of a corresponding eigenvectens vi general solution of (ü) is given by X(+) = Ci viellit arbitrary füi) elit where are constants In the given system XER3 and can be written as X1 such that X2 X3 X1 X2 X3'

In the given problem, A = 1 1 4 a 0 2 — hoxe to corresponding First determine the rigen vawes and eigenwectors of A To find the eigen vawes Of A IA-AI) = 0 1-X 4 =0 2-1 0 T- (1-1) {(2-1)(1-1)-03-140-04 +90-12-1)}=0 GT a (2-1) (1-132-4(2-4)=0 my (2-1) *(1-1)-1} = 0 (2-x)=0 (1-1)-450 (-1)=4 *(1-1) = 12 (1-1)=2 and (1-1)=-2 OY = 2 MY = 1-2 Qy, 1=1+2 0 =-1 or X3 1331 Hence have, 1,= 2; =-1 and 13 = 3. Mow we have to determine the eigenvectors be the eigenver vector corresponding Let d c to 81 the eigenvalue x1 = 2 such that

AV, = 1,2 1 4 0 - 2 2 :)) B - o, +1 +471 221 2B1 2B1 di+ßitri 271 di+ß,+47, 20. - ditBitri=bri -0, +Bi +41,=0 ar dit ßi-8=0 Bi=d, -471 BIS-,+91 (u) from (1) and (4) di-4Y, = -0, +Y1 20,- 571 kiri (ui) From 6 : 4 - 4% | BE ri-ari r 2B1=-3r1 OY) (vii) since di, Biri are arbitrary satisfying (4=<; and BIE-Zri الا have to choose one variable arbitrarily (: three vanables and TUO equations) let us Choose Y = 2 such that di. = 5 and BI=-3

Hence U = - 3 2 Let "= to the ** (3) Com responding be the eigenvector BL Y eigenvalue 12:-) suih that, A = A₂ U2 - * C ;))-- 2 (1) B2 - - 1 -d2 of +82 +482 2B2 d₂ +82 + V2 -ß2 -82 į d₂ + 32 + 8 2 = -82 d₂ +82 +472 = -22 j 242 = -2 20₂ +82 +472=0 382=0 , 22, +472 = 0 [= B2 = 0] da + 272 = 0 or d₂ +82 +282=0 182=0 at d2+282 = 0 Thus eve have ß2 = 0 and dy Yz satisfying & + 2Y;= Let US choose 82=1 such that 2₂ = -2 V2 = 2 - Let "3= be the eigen vector corresponding to the eigenvalue 13 = 3 such that } Az = 73 43 ( ? :) B3

d) 2₂ + ß3 + 483 383 283 3 ß3 az + ß3 +²3 37 ar dz + B3 +4 %3 = 303 3 233 = 363 03+B3+ Y3-373 ~ 38, -233 = 0 - 203 +B3+4 Y3 - 0 -2ą +493-0 F: B3=0] 03+B3-293=0 OT 23-283 =0 01.23-2Yz = 0 [:B3-] Thus, we have ß3=0 and &z itz satisfying such that 83 23-28₂ = 0 Let us take 83 = 1 - 2 So, 235 2 Thus, the Solution of the system (D) is given by (From cü )) X (+) = civielt +C V2 (At +Cauzet where , and ca are constants er, X(+) EG 2 5 -3 2 e + C2 t 0 e + a € 3t wali) The initial condition given was ; *(0) = or a 5 -3 2 2x0 + - 0 e 30 + 3 () 6 c 54-262 +263 -34 + 0 + 0 20, +(2+(3 [:°=]

GY, 54-26 +26, -2 -30=6 and 24+2+ (z=0 ~, 5(-2)-244 +243 = 2 C = -2 OY, 24-2)+(2+63 =0 - 262 +263 = 2+10 on C₂ + (3 =4 -C2 + 3 = 6 fix) a Adding (ix) and (2) = 6 - Cy+63 = +(2+(3=4 263 =10 193=57 From (x); C2 = 4-5 -1 So, we have G = -2 C2 = -1 and (3 = 5. Using these vaus from feüi) twe have, X(+) = -2 5 pat -3 (9)**000). 3+ a X(+) - / -1022" be at -9e2t + X (+) = -1022+ + 20 C +2e-t + 10 e 34 Geat -442+ --+ 5e37

The solution of the given initial value problem (i) is X(+) 11 -1002t + 24-4710834 Geet -402t-e-t +503 +