A5.W09 - Applied Math 250 Assignment#5 Winter 2009 due...

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Unformatted text preview: Applied Math 250 Assignment #5 Winter 2009 due February 20th A/ Problem Set 3 #1, #2, #3, #8 Other recommended problems from Problem Set 3 (not for submission): #4, #6, #7 13/ 1/ 11/ 111/ Solve the IVP y” + y’ - 6y = 26‘“, y(0) = 0, y’(0) = 1. The technique we’ve developed works for third-order equations as well (as long as they’re linear, with constant coefficients). The only complications are that the characteristic polynomial will be of third degree, and we require three linearly independent solutions, so that we can accommodate three initial conditions. Find the general solutions to the following equations (hint: if you haven’t had experi- ence in factoring cubics, note that (it: — a)(:1: — b)(:c — c) = x3 + — abc, so if we can factor the constant term we should be able to identify a real root of a cubic by inspection, and then use long division to find the remaining quadratic factor). a) yI/I _ 3y” + 3y! _ y = e—t b) y’” + By” + 12y’ + 10y = 0 Unfortunately, if we encounter a 2nd-order equation which is either nonlinear or linear with nonconstant coefficients, we have little hope of solving it. Here are a few exceptions: o If 3; does not appear in the equation explicitly, then we can view the equation as a. 1st-order equation for y’, and then integrate to find y. This can be formalized by making the substitution u = 3/, so that u’ = y”. Find the general solutions to the following problems: a) y” + if = 0 b) y” + 2303/ = 0 Note: integrating y’ is a problem in (b). The best we can do is to express the result in terms of the “error function”: erf(x) = % f0”B 13“2 dt. 0 If a: does not appear in the equation explicitly, it turns out that the same . . dy dzy du du dy du substitution ma hel . Let u = —, so —- = — = —— = — y p d2: dz? d3: dy dz “113/ and then we can express the DE as a 1st-order DE for u(y). Unfortunately, this frequently proves fruitless anyway, since we may not be able to solve the resulting 1st-order DES! 0) Convert y” + y’ + y2 = 0 into a 1st-order DE for u = g5. This will be neither linear nor separable, so you won’t be able to solve it. d) Convert y” = 3/2 into a lst-order DE for u, and solve it, to obtain a lst-order DE for y(a:). We get further this time, but we still can’t finish the problem. e) Now try 3/” = 2yy'. You should find that either y = C (a constant), or i“ = y2 + K. Proceeding from this point requires three cases, for K > 0, K < 0, and K = 0; including the equilibrium solutions 3/ = C there are AM 250 — Assignment #5 Page 2 actually 4 completely different families of solutions! (There is no guarantee of u’niqueness for nonlinear second—order initial value problems.) Assume K = 0 and find one of these families. Verify that your result satisfies the DE. ...
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