{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

H6_1130

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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,.... ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern