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

Given data in matrix Jorm is 3t 20 -2 + xa -1 e-6t 2) out a = (2 7] + F 4) = (2 3t eest Eigenvalues of A are LA-Il=0 1-2-1 -7 12 = 0 -> 7-5) (++3)=0 4-1 id=5,-3 for tas is [A-SI] Vio Eigen vector ::]6) - = 0 =) - x - x2 = 0 let take (1= we get pla=-1 = (-1) Eigenvector for t= -3 is [A + ?I] Va = 0 7] (3) = 0 = 2-782=0 x2=1 we get 2,=7 -7 let take va = (3)

est pilt)= ett vi (i); $2 (t)=est (7) :. The general solution for homogeneous system is (xict) X(t)= =cidilt) + (2 dalt) xact) xit): ciest (.) +622134 (5) ciest -3t X(t)= +7cae -ci est + ca e-3t ③ Here § (t)=rest 7e-3t Ties -est e- 3t Particular integral Xp= $(t) , & '(t) F(t) dt § (t)=1 e-3t -7e-3t seat C est est e-st § - '(t) = -1 e-st 8 8 е 3+ 3t 8 8

re-st 3t (t) f(t) = } e-st 8 3t e34 engt 8 8 a '(t) Fit) = 3 test 1 allt 8 8 3 test+1-34 8 -3 te-st s &.' (t) f(t) dt = 40 a e-st + Zellt 88 I test 1 3t 8 1e-3t 24 24 .: Xp = § (t) , § -"(t) F(t) dt Xp=rest .-5t Ze-3t 2 o test 2s + Iellt 88 3 e-3t te 3t -I 24 est -1 24 e-3t Xp= t - 7 199 600 e-6t 33 (Particular integral st e-6t 600 4 33