Question

problem 6 please!

In each of Problems 1 through 12 find the general solution of the given system of equations. 13 1. X' X+ 2. x' = X + 3 13 e 1 2 COST -2t 3. x' = X+ ( 4. x' = X + ( 1 sint 4 –zet 4 5. x' = X+ t> 0 8 -4 65-7-2)*+(24) t> 0

Answer 1

94-%, so 6. + x + > 0 + 4 het 2 {! A= -4 2 ) . (1942.)x+ ( and get) = x'- Ax + g(4) txo We first find the complementary solution, when Can be found by taking gas=o.in, by solving 1 the system: &'- Ax. Now, The Characteristic polyn caeation of A is piese 0 = (-4-2)(1-2) – 4:0 14-a 2 -Ina 5) (4+1) (1+) -4 20 -) 22+50+ 4- 4 - 0 a(at5) - o ano -5. -) Now we will find the eigenvectors for each the values of a. for a=0 we'll need to solve. -4 2 (23)(1)-(3) =) -42,72x2=0 & 294 - also x2 = 224 24 het . The eigen-vector is (1) be the taking) be the eigen vector corresponding to 255 A (21) = a (si CM) - (0) & 2020 & 2n't aro ) + 2y = 0 -) :-/

ast xo(t) = er c'))iczenst (?). particular &elation to Now, the need to find the value of . The eigen - vector corresponding so das in - 2m, ies, ni (2) (taking up-1), The general solution of x'= An is *C)6,200 G6 ) +6,654/?) (1) + CC (?). " The Complementary solution is we will find a'- Art glt) ht the particular solution be of the form Xp = X(+) VCt), where g(+) = 2 vct). have, in a = A xrg (t ) A xp + g(t) -) Xehv (4) + X4 v'(4): Axcusu(t) + Ct). Now, note that x lt) is a solution of X'CH): AXC). . have, X(+) w'CH) = g(4) =) vict) XCg (t) & CA) S xc7 g(4) dt. X(+) {xer) gadt - -20-54 eest we cue Puting xp from 1 cue УР

0-54 2est por min ad;CA) deta - Se 2 - ما دراما 2 o5t šest 2 X'g(t): 1 / 24 144 ost ś 1 + 8 5t St st t + Now (S (- / + 8) de Sfestat Sxigent as the lar: (re 2 at ligt S 4 est 25 xp = X(+) 4 e + logt st 25 2erst &ttlogt -st - 90 2 25 8 et lagt 25 1st asligi + 3/5 8 : The general solution is 2(t) = C($)+ Cease + 84 +logt - is 16 + + 2 logt + 2/5 $