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Mathematical Methods for Physics and Engineering - Matematica.NET

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HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSverify that the golden mean is equal to the larger root of the recurrence relation’scharacteristic equation.15.16 In a particular scheme <strong>for</strong> numerically modelling one-dimensional fluid flow, thesuccessive values, u n , of the solution are connected <strong>for</strong> n ≥ 1 by the differenceequationc(u n+1 − u n−1 )=d(u n+1 − 2u n + u n−1 ),where c <strong>and</strong> d are positive constants. The boundary conditions are u 0 =0<strong>and</strong>u M = 1. Find the solution to the equation, <strong>and</strong> show that successive values of u nwill have alternating signs if c>d.15.17 The first few terms of a series u n , starting with u 0 ,are1, 2, 2, 1, 6, −3. The seriesis generated by a recurrence relation of the <strong>for</strong>mu n = Pu n−2 + Qu n−4 ,where P <strong>and</strong> Q are constants. Find an expression <strong>for</strong> the general term of theseries <strong>and</strong> show that, in fact, the series consists of two interleaved series given byu 2m = 2 + 1 3 3 4m ,u 2m+1 = 7 − 1 3 3 4m ,<strong>for</strong> m =0, 1, 2,... .15.18 Find an explicit expression <strong>for</strong> the u n satisfyingu n+1 +5u n +6u n−1 =2 n ,given that u 0 = u 1 = 1. Deduce that 2 n − 26(−3) n is divisible by 5 <strong>for</strong> allnon-negative integers n.15.19 Find the general expression <strong>for</strong> the u n satisfyingu n+1 =2u n−2 − u nwith u 0 = u 1 =0<strong>and</strong>u 2 = 1, <strong>and</strong> show that they can be written in the <strong>for</strong>mu n = 1 ( )5 − 2n/2 3πn√ cos5 4 − φ ,where tan φ =2.15.20 Consider the seventh-order recurrence relationu n+7 − u n+6 − u n+5 + u n+4 − u n+3 + u n+2 + u n+1 − u n =0.Find the most general <strong>for</strong>m of its solution, <strong>and</strong> show that:(a) if only the four initial values u 0 =0,u 1 =2,u 2 =6<strong>and</strong>u 3 = 12, are specified,then the relation has one solution that cycles repeatedly through this set offour numbers;(b) but if, in addition, it is required that u 4 = 20, u 5 =30<strong>and</strong>u 6 = 42 then thesolution is unique, with u n = n(n +1).15.21 Find the general solution ofx 2 d2 ydx − x dy2 dx + y = x,given that y(1) = 1 <strong>and</strong> y(e) =2e.15.22 Find the general solution of(x +1) 2 d2 y+3(x +1)dydx2 dx + y = x2 .526

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