Question

Given the non-homogeneous linear system of differential equations Xi' = -2x1 – 7x2 + 3t xz' = -X1 + 4x2 + e-6 a. Find its homogeneous solution using the eigenvalue-eigenvector approach b. Use the variation-of-parameters method to find its particular solution

Answer 1

Page :- 1) X = -2*1-7X2+3+ x = -x + 2 te-bt in matrix form. -7 x' 1-2 1 13] X + - Le e6t A = -2 -7 -1 Where x = [] (a) corresponding home homo. Settemit X'=AX we find eigenvalue, & eigenvector of A det (ArdI) = derado de 15 = (d+ 3) (d-5) = 0 So eigenvalues are d = ~, da = 5 eigen weates for [d, = -35 det land and will A-36-37) Halbe (531) C) n = 712 12 ()
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RY A de IF soi homogeneous sot is. Page: -2 eigenrector = for d₂ = 5 (47) R₂ + RARI, a loo) rive (I) (%) = 18) (A) = 12 (1) Zelt) = c, e n = 12 eigenuston, Vali -3t (1) + 5+ tal 00) 1b ) we find Now by variation particular solution of parameters. -3t 5t те we have matrix 8 = -3t e est
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Page: 31 at at 7 été det 3 = = 8e24 5 . 5t 0-1 -34 8 e 2t - е 7834 36 8 3t e 8 - St ze -st e 8 8 we have get) = 3t -67 3t 3t е ola 3t 67 4 8 8 St 7e -e 8 8 -3d 37 8 118 3 t e t e 8 - 3t e st + 7 e 8 8 Now do the integral So geht 3+ Bt + - 3 te 8 8 -3t 7 e + 8 8
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Page 4 3t te 3 t e 8 8 3t 13t 11 3 #e 3. d ] dt 8 3 3t 3 3t e t e u - 13 3 8 3 3t te ) ) 3t e 24 8 st - 5t - 3 dt +e S+ -3d e dt mloo 8 do st - 5t e dt 4 ) té 5 f 5 .st 3 3 l t e-st 25 st 3t e -st +3e yo 200
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-3t 3t Page : 5 ( 67 get 3t e е 24 24 8 -57 1e 5t 3té Yo +3é 200 So, particuler sol is. & Solgdt Xp 3t -3t e (t-1)- 5t 6-3t 8 7 e - e 24 5t е St -1lt e 3t зе 40 (+ ++)-70 88 - 6t 11- ) -ze- *(*+5) + 7 ē 88 -6+ 7 e if ( t ) – est +31++) - 88 24 -bt 4 t - 7 23 75 33 -bt ist - 4 e 2 33 75
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