**preview**has

**blurred**sections. Sign up to view the full version! View Full Document

**Unformatted text preview: **MAT A35 Solutions to Homework 1
Due Date: Week 3. Graded Problems & solutions: 2. For xed values of r, ω R, consider the ODEV f 2rf + (r2 + ω 2 )f = 0. Verify that each function of the form f (t) = Aert cos(ωt) + B ert sin(ωt) 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(ωt) + Bert sin(ωt) then f (t) = (Ar + Bω )ert cos(ωt) + (Aω + Br)ert sin(ωt) and f (t) = (A(r2 ω 2 ) + 2Bωr)ert cos(ωt) + (2Arω + B (r2 ω 2 ))ert sin(ωt) Therefore (r2 + ω 2 )f 2rf + f = (r2 + ω 2 )A 2r (Ar + Bω ) + A(r2 ω 2 ) + 2Bωr ert cos(ωt) ert sin(ωt) + (r2 + ω 2 )B 2r (Aω + Br) + 2Arω + B (r2 ω 2 ) =0 The initial conditions clearly reduce to: 1 = f (0) = Aer0 cos(ω 0) + Ber0 sin(ω 0) = A hence A = 0 and 2 = f (0) = (Ar + Bω )er0 cos(ω 0) + (Aω + Br)er0 sin(ω 0) hence B =
2r ω
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