Question

[-/2 Points] DETAILS Find the general solution of the given system. 4 - 1 2 X' = -1 4 0 X -1 04 X(t) = Submit Answer [-/2 Points] DETAILS Solve the given initial-value problem. (1 -4 -6 X' = 1 2-3 X, X(0) = 1 -2 - 2 X(t) [-/2 points) DETAILS Solve the given initial-value problem. X' = 8 -1 5 +6)x, x(0) = (-3) X(t)

Answer 1

Saluhion=0 hiven that Di- Given that Solution 4 -1 -1 2 4 o 4 X 6 -1 New coeficient. matrix 8 4 -1 A = 4 -1 -1 0 4-d 4 -1 2 4- do ad = 0 ANI O o =0 (4-0)|(4--12-0) + 1 (0-1) +244-4) (4-2) I (4-2) ² - 1-2 23 (4-1) { 16 + d²-8d-3} (4-d) { d² - od +133 8 + 564-52 d = 4 2 o & J12 d = 41 2 4, 4 ± 3 d eigen vector are, TA - NIJU Now for A = 4 26 2 O Vi V2 - V3 -1 O 0 o CS Scanned with Camscanner

-12 +2V3 = 0 -Vi = Vico こ○ D V - 2 1 for do 4 +13 eigen vector are À FAIU=O 53 -1 2 In O 1 53 0 42 . -1 O J3 43 D J341-42 +243 -4, +5342 = 0 + 4 = 5342 -YL +34 g = 0 → 41 = 5343 =h ES 1 I for-d=+4√3 CS Scanned with CamScanner

for d = 4-53 then eigen vector are A-sI) W=0 To -1 2 0 - Ľ 0 -I3 1) W2 Wz O -1 -53 0 J3 Wi Wz+ 2W3=0 wa 3 Wzzo 5W3 =0 -- -53 wz الرا - - الرا و ليا - a دارا دل Hence = لا L 1 Hence general solution of differential equation 6 18. -41 Ge 13 _X6 47 the cost 4+ +ਉਦੇ ਨੇ Z 1 2 1 -.- 2 Now again for 5 -4 - 6 2 -3 Х to-2 X 1 1 3 1 -2 manix equation The eigen value of coefficient ID 16, 1-d LA HI! -4 2-4 -3 1 -2-1 * (1-1). [42-1) 12+) + 3) +4 1 2+d) + 3) +41-(24d) +3] -6-11-zo ICS Scanned with CamScanner

- (1-2) - [4-42-3 +3) +4/- 2 - + 5) 66--1) - 71-10- 4 - 12+3 ) +4 (1-2) - (-1) 20 1ि-+) (-1) + 4-1-4).-6(1-2).50 4 (1-1) 2 ! +4 +6 त+9) ८० ---1,36 4. 7 (1-11 Now eogen. Value d= | then vector Sen (त) %3D 0 D 4 -6 0 1 1 -3 3 0 1 1 । 3 41 42 43 ५॥ 42 0 O -6 4 1. U D ।"0 OOO 43 442 4 +42 642 0 2) 242 + 343 = 0 342 0 18 V = 2 । then U1=-9- -93 g 3 प M -2 + -36 then for fox-eigen-vedor (A-d1) 120 1-30 L 4 - 2-3 -3 । -2-31 VI V2 V %3D 2 0 ICS Scanned with CamScanner

4 1 -31 a 0 -3 1 V2 0 2-31 1-30 o -1+31 V3 O =0 19-31) Vs - 4% - 6V, 50 Uit (2-3 L) U2 - 303 (1-31]V2 - 0-0 V3 = 1/2 = 1 - V3 = Vr=41+34) let T 1 +31 tente. es I 1 then vector w CUS MON 1E-1 = X Q2M (IP-H eigen 0 4 - 6 الا 1 foi 0 -3 عليا こ 1 2+31 0 T WI -2+31 T -4 - 6 اليأ 1 +31 D -3 د لیا Coo I ?E+2 17[+T -1-21 W] 6W3 = 0 (+31) W, awz Wi +(2+31) W2 - 3W1 EC (1934 W320 (+31) 2 let wz - I = W3 - 1 : الدا ti+31°) -1 +30 1 Hence CS

hence general solution of di Hesenhal equations -1 +31) Tesi + G81n3t X (1) net 3 + C2 (431) 1 more -2 피... initial condition کیا وہ Love -X ooo S Hence 3 –961-+(1134) G 36 t (2 =S 90 + 362 -96 +(1+34) ų 106 (4731) C2 = 11 IT C2 2 4+31 11 2 3C, -C B +31_-1) 4+31 3 +34 4+37: IGE 4 +31 -171 4730 C3=918 arbi and arbitrary Hence solution 18 6 1+31 Titsil f ---- -lt 4+31 + (0337 4730 } I HEUSDt -2 1 ICS

13 Again y! 1 6 . -Х -6 8 ish v t - -40)=1 ergen value of value of coefficient coethicient matrix Now -A-dt 8-1 -1 6-d s 。 (8-1) (6-1) + S 40 -14d td? +.5=0 12-141+53:30 14 – 196- 4453 14 + 1196-212 -Hence de 그 2 2 7 #21 da 14 E 1-16 2 and carrespondi Cramog ponding Mence eigen value d= 7+21 eigen vector as (A-d Is U=0 1-21 5 -1-21 42 - 0 (1-21) 4 42 (1721).4.2 85 0 (1-22) = 541- この s 42 O 511-21)-4.! 5(1-26741_.-5.42.2 (1-21)-47 42 Y2 = (1-21 42 to = I ICS Scanned with CamScanner

1 Hence 4 = 1-21 How d = 21 then correspon espording eigen vector TIAL -1 VI V2 5 -1 +21 11 o s (1+21) 5 5 Vi 5(1+21)_61+22) (1+20 Vz (1+21) U V2-0 el a vz=1421 Hence Vz 1 H2O Fence [ General solution of differential equation W 4 X (H = Gett Cosat e77 1-21 +G sinzt 1 دراه | 1720 using initial condition, -6 8 -b Land Ca = b 18 arbitro Hence polution in bitary * fett oszt + bett sinat 1-21 1+2 CS Scanner with Cam