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H6_1130 - SECTION 3.7 11 Two linearly independent solutions...

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Unformatted text preview: SECTION 3.7 11. Two linearly independent solutions of the homogeneous D.B. are y,(t) - eat 2". Applying Theorem 3.7.1 with W(y1,y2)(t) - —e and y2(t) = e 5t, we obtain 3t 9(8) 3 5s .25 =f[e3(t‘°)- e2(t'3’]g(s)ds. The complete solution is then obtained by adding ole“ + czen' to Ytt). Note: since we are taking e“ under the integral the integration variable can't be t. Y(t) = -e and en SECTION 3.9 15. We must solve the three I.V.P.: (nu; + u1 - Pot, O < t < n, u1(0) = u'1(0) = O; (2) u; + u2 - Poms-t), n < t < 23;, u2(n) = “1(a), u'zpr) = u',(n); and (3) u; + u3 = o, 2:: < t, u3(21t) = u2(21':), u‘3(2a) = u’2(2:u. The conditions at n and 2:: insure the continuity of u and u' at those points. The general solutions of the D.E. are u1 =. blcost + bzsint + Pot, u2 = clcost + czsint + Puma-t), and u3 = dlcost + dzsint. The LG. and matching conditions, in order, give b1 - 0, b2 + to = 0, -l:~1 4- 1:30 = -c1 + 31%, -b2 + F0 = -c2 - F0, c:1 = d1, and c2 - 1'0 8 d2. Solving these equations we obtain t - sint , 0 s t s n u=ro (Zn-t)-38int,n<t521t - daint , Zn < t. SECTION 10.1 17. It can be shown, as in Problem 12, that A > 0. Setting A = #2, the general solution of the resulting ODE is y(a:) =c1 cospz+C2 sin ’11:, with y’(z) = —pc1 sinus + pczcos pas. Imposing the first boundary condition, we find that 62 = 0. Therefore y(z) = (:1 cos #3. The second boundary condition re- quires that c1 cos pL = 0. For a nontrivial solution, it is necessary that cospL = 0, that is, p = (2n — l)vr/ (2L), with n = l, 2,. . . . Therefore the eigenvalues are (211 — 1)2 «2 = T The corresponding eigenfunctions are given by _ cos (2n — l)1r:r 3’" ‘ 2L A" ,n=1,2,.... ,n=l,2,.... ...
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