MATA35HW1solution - -t 1 ) . Solution: As for the logistic...

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MAT A35 — Solutions to Homework 1 Due Date : Week 3. 2. For fixed values of r, ω R , consider the ODEV f 00 - 2 rf 0 + ( r 2 + ω 2 ) f = 0. Verify that each function of the form f ( t ) = A e rt cos( ωt ) + B e rt sin( ωt ) where A, B R , satisfies this ODE. Find a solution of this ODE having the initial values f (0) = 1 and f 0 (0) = 2. Solution: If f ( t ) = Ae rt cos( ωt ) + Be rt sin( ωt ) then f 0 ( t ) = ( Ar + ) e rt cos( ωt ) + ( - + Br ) e rt sin( ωt ) and f 00 ( t ) = ( A ( r 2 - ω 2 ) + 2 Bωr ) e rt cos( ωt ) + ( - 2 Arω + B ( r 2 - ω 2 )) e rt sin( ωt ) Therefore ( r 2 + ω 2 ) f - 2 rf 0 + f 00 = ( ( r 2 + ω 2 ) A - 2 r ( Ar + ) + ( A ( r 2 - ω 2 ) + 2 Bωr )) e rt cos( ωt ) + ( ( r 2 + ω 2 ) B - 2 r ( - + Br ) + ( - 2 Arω + B ( r 2 - ω 2 ) )) e rt sin( ωt ) = 0 The initial conditions clearly reduce to: 1 = f (0) = Ae r 0 cos( ω 0) + Be r 0 sin( ω 0) = A hence A = 0 and 2 = f 0 (0) = ( Ar + ) e r 0 cos( ω 0) + ( - + Br ) e r 0 sin( ω 0) hence B = 2 - r ω 3. Let a population evolve according to the Logistic equation p 0 = ap - bp 2 and suppose that t 1 is the time at which half the limiting population is achieved. Show that p ( t ) = a/b 1 + e - a ( t
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Unformatted text preview: -t 1 ) . Solution: As for the logistic equation, we can proceed as usual: dp dt = ap-bp 2 Z dp ap-bp 2 = Z dt Z 1 /a p dp + Z b/a a-bp dp = t + C 1 /a ln | p | -1 /a ln | a-bp | = t + C 1 a ln ± ± ± ± p a-bp ± ± ± ± = t + C only care about p > p a-bp = aCe at p = ( a-bp ) aCe at p = a 2 Ce at 1-baCe at p ( t ) = a/b 1 + e Ce-at Now we can determine the remaining constant e C by means of the initial condition, as clearly the limiting population is a b : 1 2 a b = p ( t 1 ) = a/b 1 + e Ce-at 1 solving this gives 2 = 1 + e Ce-at 1 and hence we have e C = e at 1 Substituting this back in to the general solution gives p ( t ) = a/b 1 + e-a ( t-t 1 )...
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