Question

Solve the IVP 3 [31' = [ +5] [i] [OF

Answer 1

The given IvP is 2-57 5-8 ac where x = and A= jie. X'=AX with initial Condition -X-6] - [:-] Let X=ę ert (20) be a trial solution of (). Then (1) becomes greet = Açert > (A-r In) sext=0 → (A-7 12) $ =O (since ext=0) This shows that go is an eigen value of a and § is a corresponding eigen rector. The charecteristic equation for A is 2--5 - O 5 -8-ro le. (y-2)(3+1) +25 20 po +87-16-2*+25=0 70 +679+9=0 → (70+3)=0 .. A has a repeated eigen value-3. » = -3 -3 (read repeated) roots Let be an rector of A copresponding eigen to the eigen value 3

Then 2 -5 5 - 9 = -3 22 224-522 = -321 and 534 -8X2=-322 which are equivalent to 74-72=0 '. 74-72 If x=0 then s4=c 13- [8] = CT!] where CER. > CER. Therefore eigen vectors of A corresponding to the reapeated eigenvalue -3 are of the form e e [] only linearly independent eigen vector associated with the repeated eigenvalue. get one solution X, of (1) so there are we Taking cal, as X,= ** [8] Let there be a and solution of the form: X=8te 3t, where § is a Constant vector to be determined. substituting for X in equation (1) gives (-3te3t+ e3t) = Aistest >> -35 test +fe 3t-A ste3t = 0 For the above equation to be satisfied for all t, it is necessary for coefficients

of test and eat both to be zero. From the term in e 3t we find that =0 Hence there is no solution of the system (1) of the form Cşte-3t. We need to assume that X=ęte 3t are constant vectors to bedetermined. +n rest where & in are ) we Upon substituting this expression for X in Eq. (1) , on we obtain -35te3d(standet Afsteenest) Equating coefficients of test and est, get (A +31) 5=0 (Arn Equation (2) shows that is satisfied if he is eigen rector of A corresponding to the eigen value -3 . So 5- Ti] get Therefore from (3) , we 2 - (A+3 son-1 ]*[:])-[1] -]N-H 5-8 5 -5 5

sn, 5%2=1 > n-n2=/5 Take Mk Then ny = 5 +k > where KER. n= [] - [***] - *[:-[:] x=ętest the 3t 3t + est te 3t +k + est e-3t The and term is a multiple of the first solution X= [:] ignoring the and term take the solution X₂ of (1) as so 2nd X₂ test so the general solution of the given system (1) X = 4X2 + CzXg [1] [i]+ezt ] arbitrary constants. tie. Gest te: + czest where e and C are

Initial condition given: [sco [you]= [-] Cine, come pending Therefore from (4 we get [1] = a[i] + c2[%] G + Ś C2 = 3 CI= -1 :.4+ ķez = 3 => ' = 3-0, = 3 - 61) = 4 e,=20 putting the values of a and cq in o) we get the solution of the given IvP as U1 - * [+20 te **[:] +20€* [s] -3t