Question

9. IfA = -1 2 - 3 4 then eAt 3 et - 2 e2t -2 et + 2 e2 t 3 et - 3 e2 t -2 et + 3 e2 t a) (5 pts) What is de At A.eAt = ? dt b) (15 pts) Solve the system Yı - Y1 + 2 Y2 y2 - 3 yı + 4 y2 Yi (0) = 1, Y2 (0) -1. Extra Credit (10 pts) Then write down a particular solution to Yı Yi + 2 Y2 + 2 Y2 - 3 yı + 4 y2 - 3 Yi (0) = 0, y2 (0) = 0.

Answer 1

gila) A - (-si). ) ent 'set_2e2t set - 3024 - 20€ +2e21 -2 et + 3824 to salutim- Here, 3et-re²t -2 et +2e24 EAR 1) 3et 3 ezt -264 +30²t dett ->0 dt 3 et 4e2t set cezt -zet t4e2t -zet +6e2t NOW, 2 A eAL (-: -3 4 3et - rezt -zet treat 3et-zert - zet + 3e21 - 3 et +2824 +6et Geel 2e1-2824_464 +6 024 -get tre2e +1264_12024 cetuse2d_ged +1264) = 3 et 4021 +4e2t :( -zet - zet treat 3 et rezt dett [using 0] 2 dt Therefore, dett dk Aedd dett - A.eft => dt dett [ ans Thus, A.eht at 6) Y = - Y, +292 -70 yz' =-34, +4Y2 4,10- | , 42(0)=-1

Lely y=eme, Salution - Differentelating 0,w.71 , we get " -- +242 y" -- ;' +2 (-39, +492) [ using a Yita ,' -69, + 892 Y," --Y/ -69, +4/9,'+y) [ using 0] Y;" = -4,'- 64, +49;' +49, y," ~3; + 29, = 0 (0) be the trial solution of ③ Then, aurilany equation of ® is given by, m²_3m +2=0 => m²-2m-m+2=0 => m(m-2)-1 (m-2)=0 3 => [m-2) (m-1)=0 m-250 '. m-1=0 -> m=1 -> m=2 so, the salution of ® is, arbitery ),(t)= 9,621 tezed; where e, s ez are 20 constant Here, Y, 10)-', Y210) = -1 Then from D, y,'(0) = -9,10) +29210) = -1 -2 .: Y;'(o)=-3 Thus, we get, y, (0) = 1, Y;'(0) = -3

so the general solution of the given differential equation ® Now Diff. ④ , word d, we get y,'(t) = 2 9,624 tezet → 6 Now, Y, (0 = 1 gives Rj: t ez a y,'(o) = =-3 gives > 29, tey=-3 Now, from ® -o 22+92-2,-'z 3-3-1 1.413-4 Then from © 2 C221-e, = 174 nie2 = 5 Thus from og Y,(t)=-4e2+ +5et From ©, y,'(4) = -802 +504 Now, from Og 2 Y2 = y;' + y, a-8e24+5 86-4824 +591 272 = -12e24 tio et :. Yz1t)a -relt +5 et isa yolt)= -4e2+5 et Yelt): -Keitt set [ ans

y = -9, +29, +2 Yz' = - 34, +442-3 410)-0, 42(0)=0 salution- Differentiating , wort ,t, we get Y," =-;' + 23 Y," = -4/' +2 (-39, +482-3) [using ® ] y," -14 is, equation of B © = -4'-64, +4.292 - 6 y," - -Y ;-cy, +4 ()' +,-2)-6 Y" -- Y; -29, +49;' +49,-8-6 y," - 34/' +241 = corresponding Lomogeneous y," - 39,' +29, = 0 Now by the first part hf of ③ g y, = 9,824 + eget teget is the palution of ①. 30, The collemonteny furetion of ® is, e, e2tter et where y&lq are arbitery write we can C.Fz constan. Now, we we find the particular integral by s'aferater methot. & ut Da Then, the particular integral of 6 is, (-14) P.Ja D2- 3D +2 1 (1) - 14 / (1-D+ --7.(1- ŽD+ b) (1)

:-7 (1+{ D - -7 I P.J. -7 so, The general solution of is, Y, CH), 2deed - 7 = y;' = 26,826 teget o Now, from 0, Y;' 10) = -7/10) +2 Y210) +2 0 +0+2 ;. 9, 10) = 2 Now , 4, (0)=0 gives, e, tex - 7= 0 => 9, +62 = 7 Now, y,'(o)= 2 gives, (8 26,+ Ⓡ2 = 2 from - © 2 2-7 24,+92-01-91 = i e,=-5 Then from o - 5+02=7 => ezz7 +5 C2 - 12 9 Then from Ⓡ, Y, (t) = -5€ 9, (+) = -5224 +12et-7

- 5e2t + 5 ylt) = -5247 +12e7-7 from og y,' s → 0 -10827 +izet from 0, 242 = y, ty, -2 3-10 ezt +1284 +5 224-1zét +7-2 272 = Y2 5 e2t + 5/5 tso, The solution of the giren non- homogeneous system is, +1284-7 "(t) = -5827 Y2(t) = -5 elt + From @ 2 y; = -1002t tizet Then from og 272 29,'+7,-2 - 1002t +12 eta 5e2t +12et_7-2 2y = -15224 +24et-9 +12 et -Ź Y2 = - 12 h ezt so, the salution of the giren hon-homogeneam system is, [Ans] Y2(t) = -12 eet +1284_o