Mathematical Methods for Physics and Engineering : A Comprehensive Guide

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS

and, ifkis a constant, the particular solution is equally straightforward:wn=K

for alln, providedKis chosen to satisfy

K=aK+k,

i.e.K=k(1−a)−^1. This will be sufficient unlessa=1,inwhichcaseun=u 0 +nk

is obvious by inspection.

Thus the general solution of (15.23) is

un=

{

Can+k/(1−a) a=1,

u 0 +nk a=1.

(15.24)

Ifu 0 is specified for the case ofa=1thenCmust be chosen asC=u 0 −k/(1−a),

resulting in the equivalent form

un=u 0 an+k

1 −an

1 −a

. (15.25)

We now illustrate this method with a worked example.

A house-buyer borrows capitalBfrom a bank that charges a fixed annual rate of interest

R%. If the loan is to be repaid overYyears, at what value should the fixed annual payments

P, made at the end of each year, be set?For a loan over 25 years at6%, what percentage

of the first year’s payment goes towards paying off the capital?

Letundenote the outstanding debt at the end of yearn,andwriteR/100 =r. Then the
relevant recurrence relation is

un+1=un(1 +r)−P

withu 0 =B. From (15.25) we have

un=B(1 +r)n−P

1 −(1 +r)n

1 −(1 +r)

.

AstheloanistoberepaidoverYyears,uY= 0 and thus

P=

Br(1 +r)Y

(1 +r)Y− 1

.

The first year’s interest isrBand so the fraction of the first year’s payment going
towards capital repayment is (P−rB)/P, which, using the above expression forP,isequal
to (1 +r)−Y. With the given figures, this is (only) 23%.

With only small modifications, the method just described can be adapted to

handle recurrence relations in which the constantkin (15.23) is replaced bykαn,

i.e. the relation is

un+1=aun+kαn. (15.26)

As for an inhomogeneous linear differential equation (see subsection 15.1.2), we

may try as a potential particular solution a form which resembles the term that

makes the equation inhomogeneous. Here, the presence of the termkαnindicates