Mathematical Methods for Physics and Engineering - Matematica.NET

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSboth linear and homogeneous, and is satisfied by both v n = Aλ n 1 and v n = Bλ n 2 ,itsgeneral solution isv n = Aλ n 1 + Bλ n 2.If the coefficients a and b are such that (15.30) has two equal roots, i.e. a 2 = −4b,then, as in the analogous case of repeated roots for differential equations (seesubsection 15.1.1(iii)), the second term of the general solution is replaced by Bnλ n 1to givev n =(A + Bn)λ n 1.Finding a particular solution is straightforward if k is a constant: a trivial butadequate solution is w n = k(1 − a − b) −1 for all n. As with first-order equations,particular solutions can be found for other simple forms of k by trying functionssimilar to k itself. Thus particular solutions for the cases k = Cn and k = Dα ncan be found by trying w n = E + Fn and w n = Gα n respectively.◮Find the value of u 16 if the series u n satisfiesu n+1 +4u n +3u n−1 = nfor n ≥ 1, withu 0 =1and u 1 = −1.We first solve the characteristic equation,λ 2 +4λ +3=0,to obtain the roots λ = −1 andλ = −3. Thus the complementary function isv n = A(−1) n + B(−3) n .In view of the form of the RHS of the original relation, we tryas a particular solution and obtainyielding F =1/8 andE =1/32.Thus the complete general solution isw n = E + FnE + F(n +1)+4(E + Fn)+3[E + F(n − 1)] = n,u n = A(−1) n + B(−3) n + n 8 + 132 ,and now using the given values for u 0 and u 1 determines A as 7/8 andB as 3/32. Thusu n = 1 32 [28(−1)n +3(−3) n +4n +1] .Finally, substituting n =16givesu 16 = 4 035 633, a value the reader may (or may not)wish to verify by repeated application of the initial recurrence relation. ◭500