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Math 11

# Math 11 - 13 The formulae given in the text depend upon the...

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Unformatted text preview: 13. The formulae given in the text depend upon the fact that the coefﬁcient of y” is 1. We start by dividing 12. A fundamental system of solutions to the homoge— our equation by :2. neous equation is 3210‘) = cost and 3220‘) = sin t. 3 3 1 We look for a solution of the form 3/" + ~t- ' —- t—Zy = 73. y(t) = v1(t)y1(t) + v2(t)y2(t) First, check y; (t) = t is a solution. =v tcost+v tsint. , 3 3 3 3 1() 2() y/+?’_;2_y=(0)+;—(1)-—;§(t)=0. D' t' . ifferen latlng we get Check that y2(t) = f3 is a solution. ’='u’cost+v’sint—v sint+v cost. 3 3 _ 3 _ 3 _ y ‘ 2 ‘ 2 y”+ ;y’~ 72y = (12: 5)+ ;<—3t 4) ~ 72" 3) To simplify future calculations we set 2 12t‘5 __ 9t'5 _ 3t"5 vicost+v§sint=0. =0- . Calculate the Wronskian. Then y’ :2 ~v1s1nt + v2 cost, and *3 W(t, f3) = It ’ _4 = _4r3 y” z: —vi sint + v; cost — v1 cost — v2 sint. 1 ”’3t _ Next, Adding, we get 11’ _ —y2g(t) ” _ , . , 1 W y + y _ ~v1s1nt+ v2 cost ——t‘3t“3 =sect+cost—1. = ~4t‘3 Thus we must solve the system : i163 vi cost + v; sin 1‘ = 0, Thus, 1 —visint+v§cost=sect+cost— 1. v1: —§t”2. The solutions are Next, I 3’18“) I - . U2 : v1:——tant—smtcost+smt and W v'2=1+coszt—cost. , = "H3 __4t-3 Integrating we get 1 t 1 _ 4 v1(t) = ln(cos t) = 5 cos2 t — cost and Thus, 1 3 1 i 1 U2 :1 ‘glz. v t) = —t ~sinzcost — si It 2( 2. + 2 11 Form Thus the solution is M) = ‘01 3’1 + ”23’2 . 1 —2 1 2 ,3 y(t) = v1(t) cost + v2(t) s1nt = —§t t + —§t t 1 3 = =cosz-1n(cost)+—cost+—sint~1. _ 1 2 2 - —4*- t t Thus, the general solution is C C2 1 W) T 1t+ t3 4t' ...
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