Sol.4.6 - Excrcises 4.6 :T"ERCIStrS 4.6: Variation of...

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Excrcises 4.6 :T"ERCIStrS 4.6: Variation of Pararneters :. From Exatnple 1in the text, we know that functions E1(t) :.ort and Az(t): sinl are tn'o linearly independent solutions to the corresponding horrrogeneous equatiorr, and so -rs gerreral solulion is giverr by l)nltl : c1 cos I * cz siri l. \orv we apply the method of variation of pararneters to find a particular solution to the ligirrai equa,tion. By the fbrrnula (3) in the text, Ar,(t) bas the fonn ar(t) : u r(t)sr(t) + r,2(t)y12(t). d(t\ -- (c-ost\ : - s\rt. svstem (9) becomes l rrr uN \ -- \s\rL\ -- cssr. u'r(t) cost + tr!(t) sint: 0 -ur(t) sin f f t'i (f ) cos I : sec l. \lLrltiplying the first equation by sinf and the second equation by cosri yielcls z'i (t) sin f cosf + u'r(t)sin2, : 0 -dr(t)sin f cos t + u;(t)cos2, : 1. \dding these equations together, rve obtain u'r(t) (cos2 f * sin2 t) : t or u'r(t) : 7. From the first equa,tion in the system, we can now fincl ,r(t). ar(t): _ u;tt) ttnl: -tanl. "' 'cosf So, ,\(t):-tant * "1(t) :-/tantilt:lnlcostl+ca uL(t):7 ur(t): J'dt:tIct. Since we are looking for a particular solution, we can take cz: c4: 0 ancl get ao(t) : (cost) lrr l cosf| + tsint. Thus, a general solution to the given equation is A(t) :
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Chairter 4 4. This eqrration has associatecl homogeneolls ecluation !/" - lJ : 0. The rc-rots of , associated auxiliary equation, 12 - 7: 0, are r' : :t1. Therefore, a general solutiorr this equation is yr,(t) : cteL t cz€ t.
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This note was uploaded on 03/30/2009 for the course MA diff eq taught by Professor Fenn during the Spring '09 term at N.C. State.

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Sol.4.6 - Excrcises 4.6 :T"ERCIStrS 4.6: Variation of...

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