Question

(1 point) Consider the linear system 3 2 ' = y. -5 -3 a. Find the eigenvalues and eigenvectors for the coefficient matrix. 2 = 15 and 2 V2 b. Find the real-valued solution to the initial value problem Syi ly 3y1 + 2y2, -541 – 3y2, yı(0) = 0, y2(0) = -5. Use t as the independent variable in your answers. yı(t) y2(t)

Answer 1

3 2 is given by the Characterestic equation 1+1=0. b) a Griven y at are 3 2 5 -3 The eigen values of the coefficient matrix, Az values of A. the eigen Now we will compute the eigen rectos corresponding where 2 ~ to the eigen value, (A - 11) (3-1) 4, +24₂ = 0 (-54) + (-3- i) Y2 = 0 selring for y, G₂ gives - 3+1 Aguin eigen vector corresponding to eigen value ^2=-1, is given by, (A + is) 7.0 where 52 = ( 2 ) (3+1)x+ 2x2=0. -521+(-3+2) x2 = 0 selring we get, -3. i 4= 2, -3- b) is eigen veelos corresponding eigen ralue 72= -1 Therefore the solution that we got from first eigen value and eigen rector u, منهل 2 7 (4) e -3+1 2 -3+1 2 eust tizsit -3 eust-Sint til cost-3Sit)

i e. gi (t) = 2 cost mil in dependent use them to form a general solution and g (t) = 4, 2 costite zsent. 2 Sint - 3 cost-sint Cosa 3 Sint u It) + iu (t) Since u It) and 3 (t) are linearly solutions to the differential equation. This means that we com they are both real solution: The general solution is, ý (ta a ū (t) + 2 W (t) 9,10) e + ₂ east-3 Sint Sz(t) = 41(-3 los t_sin t) + ļy (Cost-3 sent) Now using initial condition, 4,10)20, 4₂(0)=-5 e, zo. & C. (-3-0) + (1-0): -5 =) -3e1f205 4.0, 2-5 2 sint 2 12eost - 3 east-Sint we get, Hence, y (t) = -10 sint Y2 (4) 2 -5(360st -3 sint)