Question

Use the method of undetermined coefficients to find a general solution to the system x'(t) = Ax(t) + f(t), where A and f(t) are given. - 10 5 1 Ав 24 f(t) = -2 X(t)

Answer 1

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The given system is xCE) - An(t) + f(t) 30 Where A2 f(t)= L- first We find the general Solution of the homogeneous Part x'ct) = AXLE) first amume that etu, where I and a are a constant vector and a scalar respectively: to: determined 1.-13] substituting x = a to in the system, ale get ranto = a entre Further et is never zero, so we have Here AVEV (A-11) v=0 where I = 12 m.]{X:3.1:] kle find eigen values and corresponding eigen vector of A The characteristic equation is 5-1 al 2 4- (5-1) (4-8)-2.=0 20-52-47 +4-2=0 9 -9% +18 = 0 - 67-3x + 18 = 0 1(2-6)-3(2-6)=0

2-6,3. for ?,= 3 becomes J: [: 1283-(3) which gives &V, TU2=0 V2 = -20 If we choose V, 1 then 12 = -2 Therefore #() in a solution fo 2=6. 6 becomes 1 -V,+ V₂ = 0 {e 20,-2 U₂ = 0 klhich gives V, = U₂ If ule choose U=1 then 12=1 est (/) in also a solution Since the two eigen values are distinct, the general sdution of systemu 126) = Ge*(4)+cie khere, &, and C are arbitary Constant We now use the method of undetermined coefficient a particular solution of ③ to find

-10 Since f(t) = kule mume 2 vtra, where a ū a vector to determined Substituting into equation ④ we get Aa + [-] Αα - (03] ()(a) [13] Where a: 50,+ a2 = 10 2a, tua2 = 2 UX given 200, +ulz za, -ul, = no-2. 189, = 33 a, : 3819 180 from (4 we have 19 9 2149) + 492 = 2 ua2 = -2-38 9 -20 9 ya a. -519 1919 -sla

Therefore xp = 1919 -51a ña particulo Solution of ① Therefore the general solution of in 2415 6, 2" ()+63240)( 19la -519 Where, and care arbitrary constants.