Question

find the general solution for 14,15,16

differential equation

14. z't) 1 -5 21) =(0) 15. z'(t) = 10 0 2 1 -2 (t) 3 2 -1 16. x'(t) = -3 0 2 1 -1 0 (t) -2 -1 0

Answer 1

Que 14. solution. Given , (t) n(t) -5 ie racist) A milt) 012 (1) con M' (t) 2 - where , AE اسے - - Now, A-AI! 20 -) 2 1-1 -5 -1/ --) -) (1-0) 1-1-1) +10 - 1 + 12 +10 2 + 3i ie eigen Let values of are A, =si and A2 = -3i. ū be eigen vector corresponding to do = Bi. [A-d, I] ū then , 1-3 i 2 اکا -1-3i U (4-3i) v, +2 427 = 10 -50, +(-1-32) U2] Take , (1-3i) v, +2 U2 =0 (4 Uz = (12)

then , 교 U2 (312) , Take U,=2 then we get 1 i 2 15 11 (3-] Then 2 31-1 Now let Jy is since , U2 31-1 is an eigen vector of ty = 3i . an eigen vector of d2 = -3i. T a lie. I conjugate , 00 . U. 2 3i-1 ..t е. If the solution we get from value and the first eigen eigen vector is 71, (t) sit (27 31-1 = *Com (8 +) + i fin (3+)4 2 Cos (3t) + 2i sin(3 t) --COA( 3t) –-3 sin(3 t) + 3 icos (3t) – Sin(3) 2 Cop (3 t) 2 sin (3) - cos (3 t) -38in (3+) 360p(3 t) - Sin(36) + 2 31-1 ,]+ últ) + i Tyle) (day)

Then the is 2 Bin (37) -]+s{scombe)-timet general solution of equation r(t) = a qi(t) +2 ult) 2Con (3) = 4 -Cop(8t) - Sin 8 t) le 24 COD (3 t) + 2 (2 8in (3+) +9+362) COS (3 t) +(-4-62) Bin (8t) which is the required solution. - (จ) 15. 1 0 0 solution. Given gilt) 8(t) 2 I-2 3 2 ie. (91) Milt) A () H2 (t) 0 0 where a a 1 2 3 -2 - 2 0 0 Now, 1A-TI) 2 3 2. -1-1 + =) (1-1) [(1-1) 6-1-2) +4] - 3) 11-0) [+1 +2+4) 5) (1-1) (12+3) = 0 3) + T5 i di = h i d2 = - Bi , dg = 1 (Bay) v= ie

If eigen vector corresponding to do is up then [A-A, 1] To Bi 0 2 1-13 i -2 3 2 -1-13 18:00 1.3 :) (1-53 i) vi 2U, +(1-13 ) U2 - 203 30, +2 U2 -(1+B i) 03 Vi = 0 ax take a then, U2 = (1+Bi) va | 1920) 0 25 U U3 U3 03 take U3=2 we get, اے 0 tabi 2 If eigen vector corresponding to iz is Jž then , B 3 (dz = 7 ) ie. d 1-23 2 If the solution we get from first eigen value and eigen vector is,

ni (t) = elita ا to it 0 e 13 2 [Conc.TB +) i Ain (13 )] + 3 2.

0 2 tenlis y este) til 15 15 ) = fantes, a fuerung -COA(13 t) -13 Sin(J3 ++ips3 cop(8t) + Bindst)} 2 Cos (13 t) + i sin (13 t) 0 N3 Con (13 t) + sin (18+) ū,(t) + i úz ( 18 ay) Then the solution we get from first and second eigen values and eigen vectors are EG, ū(t) + egūscu (t) 2 EG COM(13 t)-13 Pink3 t) + C2 53 Confit3t) + sin(83t) 2. COA (18+) Sin (13 t) the one

81 (t) = 166 +13 (2) COA (134) +(62 - 134) fin (13 t) 2G Coulis t) + CZ sin (18+) If the eigen vector corresponding to alig is then, [A-d3I] U} 0 O 0 :) ใน 2 O JE 3 2 -2 5) an, -293 (3M, +2 M2 -2 m3 [3] 3) ^, = mg and m2 -Ž 2 then 15 న న mz .nz/2 113 n3 take ไว -2 then, u 2 2 then the solution we get from third eigen value & eigen vector is ña (t) = e.e 1 Azt 33

ng(t) = gel 2 Then Filt) = R (E t the general solution of equation ① is, ñ (t) + molt 2 e |(6,+ 13 (2) Con (18+)+(52 -1361) Binkis t) -czet 26 cos (13 +) +CZ fin (13 t) +2 Get which is the required solution of equation . il. 16. 3 0 2 Sdution. Given, gilt Mit) -1 -2 ie x' (E) A.MCU where , -3 2 A 1 -2 0 Now, (A-NI) s -3-1 2 0 -2 od +3-1) 61-7)) +2 (-1 +2(---)) =0 (1+d)(3+al) -2.-4 (1+0) =0 =)

n To +(3+44+02) +2 +4 +48 = 0 13 +4 +2 +74+ 6 =0 13 +212 +212 +44 +34 +6 =0 72 (x+2) +20 (4+2) +3(4+2) =0 (+2) (112 +21+3) =0 na -2 -2 + N4-12 :) =) - 2 ie Y -2 - t i - i-e. 'W 2-2 -1 + J2 i , y = -1-12 i, da If eigen vector corresponding to do is öt then, [A-d, I] Vi -2-2i 0 2 U 1 O -dzi 0 -2 1-62 i :) be (-2-F2 i) U, + 2 U3 U, - r2 i U₂ - 20,- 02 + (1 - ) i) Us vi Uz $ 03 =-62-21), 12 i Take , ie Uz civi 2+12 s uz (21. juni Tz

s then, si 11 559 ht |(24) a take u,=2 then, 2 ū. -idz (2 + 12 v) If the solution we get from the first eigen value and eigen vector is, Lt) = enit, vi (-1+F2i)t 2 -da 2 trz t fcom (12 +) ti pin£12x)} 2 di 2+dzi 2 cup (12 t) + 2 i sin(at) 12 sin(det) - Izi Cop (2 t) 201[t)f3 sir{fat)+iff.conf2 t) + 2 fin(f2+}}|

2000612+)-12 finkat)] 12 con(18+) +2 Sin (12 t) 2 cop (2 t) 122 sin (2 t) -t e 2 sin (2 t) -12 contrat) = ùilt) ti ū (t) (o ay) Then the solution we get from front and second eigen values and eigen vectors are qu, (t) + 2 Palt) ac, con (2 t) +2 pinat) 12G sin (12 t) - d2 C Cop (2 t) (26thon (21)+(-79) Rin (2 t) 22 (t) e If the eigen vector corresponding to rls in Tz then, [A-dg IJ DE =) -5 2 m, -3 O -18] -2 - 2 - ) -5m, t2n₂ n, -3 M2 M2-293 m3 5 m, s -2n, 1, 11 Take , 2 then , J3 ni M2 M₂

take n = 6 we get, 6 15 2 from the third eigen value then the solution we and eigen vector is, enzt. og Hy(t) с. е. 2t с. е 3 15 2 Then the general solution of equatio 0 is, 31 (t) = (t) + no(t) ilt) 26,Con (12 t) + 26 din 2t) 12 G8in (2 t) - 126 Cooldat) (24+(2) col 12 t) +1247-12) Ain 2 v) which in the required general solution of equation 0. - e 15 2