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View Full DocumentMAT A35 Solutions to Homework 1 Due Date: Week 3. Graded Problems & solutions: 2. For xed values of r, R, consider the ODEV f 2rf + (r2 + 2 )f = 0. Verify that each function of the form f (t) = Aert cos(t) + B ert sin(t) where A, B R, satises this ODE. Find a solution of this ODE having the initial values f (0) = 1 and f (0) = 2. Solution: If f (t) = Aert cos(t) + Bert sin(t) then f (t) = (Ar + B )ert cos(t) + (A + Br)ert sin(t) and f (t) = (A(r2 2 ) + 2Br)ert cos(t) + (2Ar + B (r2 2 ))ert sin(t) Therefore (r2 + 2 )f 2rf + f = (r2 + 2 )A 2r (Ar + B ) + A(r2 2 ) + 2Br ert cos(t) ert sin(t) + (r2 + 2 )B 2r (A + Br) + 2Ar + B (r2 2 ) =0 The initial conditions clearly reduce to: 1 = f (0) = Aer0 cos( 0) + Ber0 sin( 0) = A hence A = 0 and 2 = f (0) = (Ar + B )er0 cos( 0) + (A + Br)er0 sin( 0) hence B = 2r 3. Let a population evolve according to the Logistic equation p = ap bp2 and suppose that t1 is the time at which half the limiting population is achieved. Show that p(t) = a/b . 1 + ea(tt1 ) Solution: As for the logistic equation, we can proceed as usual: dp = ap bp2 dt dp = dt ap bp2 1/a b/a dp + dp = t + C p a bp 1/a ln |p| 1/a ln |a bp| = t + C 1 p =t+C only care about p > 0 ln a a bp p = aCeat a bp p = (a bp)aCeat p= p(t) = a2 Ceat 1 baCeat a/b 1 + C eat Now we can determine the remaining constant C by means of the initial condition, as clearly the limiting population is a : b 1a a/b = p(t1 ) = 2b 1 + C eat1 solving this gives 2 = 1 + C eat1 and hence we have C = eat1 Substituting this back in to the general solution gives p(t) = a/b 1 + ea(tt1 ) ... View Full Document
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View Full DocumentMAT135 Test 3 Study Package 1
mat237 prob set 2
Lay - Solution Manual to Linear Algebra and Its Applications, 3rd Edition(aw 2002)(0201709708)(576
Lecture1
Problem Set 1 Solutions
2008 Final Exam (Online)
pp07s
chapt7x
newell_lectures16-21
pp07
homework3
B
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