Question

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(1 point) Consider the linear system ;' = -5 -3): a. Find the eigenvalues and eigenvectors for the coefficient matrix. 21 = v= and 12 = V2= II b. Find the real-valued solution to the initial value problem 3yı + 2y2, y = -5yı - 3y2, yı(0) = -4, y2(0) = 10. Use t as the independent variable in your answers. yı(t) = y2(t) =

Answer 1

Now, The ☺ 3 2 -5 -3 [] given System is> } = Ay where } - Now, 1A-21) = 0 = (3-1)(-3-1) + 10 = 0. (2+3) (9-3) +10 = 0. → 22 -9 +10 = 0 → 22+1 = 0 → = Let us be the eigen Vector Corresponding to a fi Then . Ah = 20 (A-(I) = 0 2 3-i ㅋ J[]-[0] -5 ولا -3-i ..Now, (3-1)), +212 = 0 Therefore (3-1), 키봇 2. 7 59; -(3+1)9,50 If y = 2k. then yo = -13-ik Where KEIR-893 V = K. 2 (1-3) taking K=1. We get Let I be the eigen Vector Corresponding to 3=+1) AV = 3 → (A +i I) x = 3ti -3ti [-] 11 = 1 Then, 2 JE}]-[0] -5

*3-(3+0) (3+1) y, 2 Now (3+i) 9, +24 - 0 -54, +(-3+1)%=0 If y=2d. then where d ER-{o} * = -(3+i)d Then, 2 & 22=(-1) -sto] taking das me get - 2 - 1 2] 1 b) Now, The given System Can be Written As - it. 2 t G - 3 ya у 39, +292) y = -59,-392 Then the solution of this system is it 2 (3+1) where a. q are arbitary constant y, (O) = (-4) ► 24+24 = (-4) = 9+=(-2) 92 (0) = 10 7 9. (1-3) – 6:(3+1) = 10 > i (9-~) - 3 (4+5) = 10 . i (9-x) = 10-6 =4 - (-412). → G-Q=441) (ii) ** Iow. (i) +C) => 29 = 42–41) = -(+2) L()

(i) — Lij) 26 = (-2+41) G = (-1+20) Therefore, 9, (+) = 29 eit +2gelt ->y, (t) = fag (-41-2) (lost + usint) + (+2+41) (6054 Elsint > $() -t 4 lost + 8 Sint) i. Now, (t) = (1-3) 9. elit +-(3+i) sett ►g(t) --(i-3). (1 +2 1) (Cost+Usint) -(3+1)(–1+2i) (cost-isint) → $(t) =-(-5-5i). (fost + i sint) (-5+51) (cost-i'sint) → 4(1) = (5+54) (lost + isin t) + (-5i +5)/lost-Lan! + y(t) 10 lost - 10 sint Therefore, the Requided Solution is → 4,(t) - (-4604 +8 Sint) $(t) = (1 Olost - 10 Sint)