hw4sol - 46 v.6 24. ywaq The homogeneous equation y” —...

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Unformatted text preview: 46 v.6 24. ywaq The homogeneous equation y” — 3 y’ — 10y = 0 has characteristic equation A2 —— 3)» — 10 = (A — 5)(A + 2) = 0 with zeros A] = 5 and A2 = —2. Thus, the homogeneous solution is yh = CleS‘ + C2e_2‘. Thus, the forcing term of y” —- 3y’ ~— 10y = 362’ is a solution of the homogeneous equation. Substitute yp = Ate—2' and its derivatives y; = Ae-2'(1 — 2t) y; = (—4 — 4t)Ae~2t into the equation, (—4 + 4t)Ae"2‘ — 3Ae*2‘(1 — 22) — 10Ate‘2’ : 3e~21 [(—4 + 4t) — 3(1 ~ 2:) — 10t1Ae*2’ : 3e~2t Solving for A, we get A = 3/7. Thus, y —(_-5/7)te‘2‘ is a particular solution. The homogeneous equation x” + x = 0 has x10) 2 cos I and x20) = sin! as a fundamental set of solu— tions. The Wronskian is cost sin I Wcost,si‘1t 2 . ( l ) —s1nt cost = 1. Form the solution xp = v1x1+ szz where v1 and v2 are to be determined. Indeed, —sinttan2t 1 —sint(sec2t — 1) = —secttant + sint. vi: Thus, v1: — sect —— cost. and homogeneous solution M. = (Cl +‘C2t)e“2’- The forcing term of y”+4y/ +4y = 2e‘2’ and Ate‘Z’ are solutions of the homogeneous equation, so mul— ii tiply by another factor of t and try yp = At2e—2’. The derivatives of yp are y; = 2Ae—2'(t —— t2) yg = 2Ae"2'(2t2 — 4t + 1). Substituting these in y” + 4}” + 4y = 2e‘2’, 2Ae—2‘(2t2 — 41‘ + 1) + sAe—2t + (t — t2) + 4At2e‘2t [2(2:2 — 4t + 1) + 80? — :2) + 49] A = 2 Solving for A shows that A = 1. Thus, yp = tze'z‘ is a particular solution. Secondly, ~ , cost tanzt U2 = "‘ _' 1“ = cost(sec2 t — 1) = sect « cos 1. Thus, v2 = lnlsect +tant| — sint. Inserting these results in xp = mm + vzxz, xp = (~ sect —— cost) cost + (lnlsect +tant| -— sin t) sint . ' * .2 = ~1 —cos2t+s1ntln|sect+tant| —s1nt =—~2+sintln|sect+tant| is the particular solution we seek. . The homogeneous equation y” + 4y’ + 4y = 0 has characteristic equation AZ + 4)» + 4 = (A + 2)2 = O 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 yz(t) = sin t. 3 3 1 We look for a solution of the form y" + ~t— ' - t—Zy : [—3. y(t) = mg) yl (t) + v2(t)y2(t) First, check y; (t) = t is a solution. =v tcost+v tsint. , 3 3 3 3 1() 2() y/+_’_fiy=(0)+;—(l)-—;§(t)=0. t Differe tiatin we et . . n g g Check that y2(t) = f3 is a solution. I I I - - y :11 cost+v smt—vlsmt+v2cost. 3 3 _ 3 _ 3 _ ‘ 2 y” + ;y’ ~ t—zy = (12: 5) + 7(—3t 4) ~ t—za 3) To snnplify future calculations we set 2 lzt—s __ 9t—5 _ 3t—5 vicost+v§sint=0. =0- . Calculate the Wronskian. Then y’ =: ~v1s1nt + v2 cost, and *3 I W(t, f3) = It t _4 = —4t‘3 y' ==—visint+v§cost—vlcost—vzsint. 1 "’3t . Next, Adding, we get v, _ —y2g(t) l _ . W y” + y = ~vi s1nt + 1); cost __t_3t_3 =sect+cost—1. = ~4t_3 Thus we must solve the system : lt-3 4 vi cost + v; sin 1‘ = 0, Thus, 1 —visint+v§cost=sect+cost— 1. v1: —§t”2. The solutions are NC)“, I 3’18“) I - . U2 : v1:——tant—smtcost+smt and W v'2=1+coszt—cost. I = "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 I "2 = ‘glz. v t) = —t + ~sinzcost — sink 2( 2. 2 Form Thus the solution is W = 01M + U2Y2 . 1 —2 1 2 a3 y(t) = v1(t) cost + v2(t) s1nt = *gt 1 + —§l t 1 3 = =cosz-1n(cost)+—cost+—sint~1. _ 1 2 2 - ‘5- Thus, the general solution is y(t)=C1t+'§"‘”- :m, 4. Using Definition 1.1, F(s)=/ e3’e_"dt 0 = lim (WM; T—>oo O 1- e—(s—3)t T = 1111 ~ T—->oo s—3 0 provided s > 3. 22. If y" + y : sin 4t, y(0) = 0, y’(0) = 1, then, letting Y (s) = £(y)(s), 4 S2+42 4 s2£(y)(s) — sy(0) ~ y'(0) + £(y)(s) = s2Y(s)— 1+ Y(s) = Solving for Y(s), 4 (s2+1)Y(s)=l+s2+16 sz+20 s2+16 s2+20 WW6? (s2 +1)Y(s)= Y(s) = y’ + y = e’sin3t, Y(O) '5 0’ n with £(y)(s) = Y(S), ’J‘ _ 20’ + y)(s) = s 50%") ~ 3%- : SY(S) + Y(s) = (s +1)Y(s). \./ (2M5) i '. Consider the transform pair, 3 324—9. f(t)=sin3t <—> F(s) = By Proposition 2.12, £{e‘t sin 3t}(s) :: F(s +1) _ 3 " (s +1)2 +9‘ Therefore, 3 (s+1)2+9 3 = (s+1)(s2+2s+10)' (s +1)Y(s) = Y(s) 32+16' 29. Beamse / 0’ t < 3 N) = H(t * 3V” '1 {e022 t Z 3’ the graph is: The Laplace transform is Y(s) = foo y(t)e—“dt 0 3 oo 2/ 0e”“dt+/ eo'z’e”s’dt 0 3 T _. lirn e“(s_0‘2)’ dz T~>oo 3 ate—(#02): T . e~(s 02,2" 6*3ts—O 2)] :rlinéi 5—02 + s~02 J e—3(S-0,2) 2 3 ~01. ’ y’ + 3y = t2, y(0) = —1, then, letting Y (s) 2 C(37)“), s Loxs) — M) + 3£(y)(s) = 60%) l sY(s) + 1 +3Y(s) = S Solving for Y (s), 2 (3+3)Y(s) = —3 ~1 S - 3 M) = s (s + 3) ...
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hw4sol - 46 v.6 24. ywaq The homogeneous equation y” —...

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