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MAT A35 Solutions to Homework 1 Due Date : Week 3. Graded Problems & solutions: 2. For fixed values of r, R , consider the ODEV f 00- 2 rf + ( r 2 + 2 ) f = 0. Verify that each function of the form f ( t ) = A e rt cos( t ) + B e rt sin( t ) where A, B R , satisfies this ODE. Find a solution of this ODE having the initial values f (0) = 1 and f (0) = 2. Solution: If f ( t ) = Ae rt cos( t ) + Be rt sin( t ) then f ( t ) = ( Ar + B ) e rt cos( t ) + (- A + Br ) e rt sin( t ) and f 00 ( t ) = ( A ( r 2- 2 ) + 2 Br ) e rt cos( t ) + (- 2 Ar + B ( r 2- 2 )) e rt sin( t ) Therefore ( r 2 + 2 ) f- 2 rf + f 00 = ( ( r 2 + 2 ) A- 2 r ( Ar + B ) + ( A ( r 2- 2 ) + 2 Br )) e rt cos( t ) + ( ( r 2 + 2 ) B- 2 r (- A + Br ) + (- 2 Ar + B ( r 2- 2 ) )) e rt sin( t ) = 0 The initial conditions clearly reduce to: 1 = f (0) = Ae r cos( 0) + Be r sin( 0) = A hence A = 0 and 2 = f (0) = ( Ar + B ) e r cos( 0) + (- A + Br ) e r sin( 0) hence B = 2- r 3. Let a population evolve according to the Logistic equation p = 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- 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 )... View Full Document