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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 thirdorder 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 2ndorder 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. 1storder 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 1storder DE for u(y). Unfortunately, this frequently proves fruitless anyway, since we may not be able to solve the resulting 1storder DES! 0) Convert y” + y’ + y2 = 0 into a 1storder 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 lstorder DE for u, and solve it, to obtain a lstorder
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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 Summer '09
 Ducharme

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