Mathematical Methods for Physics and Engineering : A Comprehensive Guide

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS

both linear and homogeneous, and is satisfied by bothvn=Aλn 1 andvn=Bλn 2 ,its

general solution is

vn=Aλn 1 +Bλn 2.

If the coefficientsaandbare such that (15.30) has two equal roots, i.e.a^2 =− 4 b,

then, as in the analogous case of repeated roots for differential equations (see

subsection 15.1.1(iii)), the second term of the general solution is replaced byBnλn 1

to give

vn=(A+Bn)λn 1.

Finding a particular solution is straightforward ifkis a constant: a trivial but

adequate solution iswn=k(1−a−b)−^1 for alln. As with first-order equations,

particular solutions can be found for other simple forms ofkby trying functions

similar tokitself. Thus particular solutions for the casesk=Cnandk=Dαn

can be found by tryingwn=E+Fnandwn=Gαnrespectively.

Find the value ofu 16 if the seriesunsatisfies

un+1+4un+3un− 1 =n

forn≥ 1 ,withu 0 =1andu 1 =− 1.

We first solve the characteristic equation,

λ^2 +4λ+3=0,

to obtain the rootsλ=−1andλ=−3. Thus the complementary function is

vn=A(−1)n+B(−3)n.

In view of the form of the RHS of the original relation, we try

wn=E+Fn

as a particular solution and obtain

E+F(n+1)+4(E+Fn)+3[E+F(n−1)] =n,

yieldingF=1/8andE=1/32.
Thus the complete general solution is

un=A(−1)n+B(−3)n+

n

8

+

1

32

,

and now using the given values foru 0 andu 1 determinesAas 7/8andBas 3/32. Thus

un=

1

32

[28(−1)n+3(−3)n+4n+1].

Finally, substitutingn=16givesu 16 = 4 035 633, a value the reader may (or may not)
wish to verify by repeated application of the initial recurrence relation.