Mathematical Methods for Physics and Engineering - Matematica.NET

HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONSand, if k is a constant, the particular solution is equally straightforward: w n = Kfor all n, provided K is chosen to satisfyK = aK + k,i.e. K = k(1 − a) −1 . This will be sufficient unless a =1,inwhichcaseu n = u 0 + nkis obvious by inspection.Thus the general solution of (15.23) is{Ca n + k/(1 − a) a ≠1,u n =u 0 + nk a =1.(15.24)If u 0 is specified for the case of a ≠1thenC must be chosen as C = u 0 −k/(1−a),resulting in the equivalent formu n = u 0 a n + k 1 − an1 − a . (15.25)We now illustrate this method with a worked example.◮A house-buyer borrows capital B from a bank that charges a fixed annual rate of interestR%. If the loan is to be repaid over Y years, at what value should the fixed annual paymentsP , made at the end of each year, be set? For a loan over 25 years at 6%, what percentageof the first year’s payment goes towards paying off the capital?Let u n denote the outstanding debt at the end of year n, andwriteR/100 = r. Then therelevant recurrence relation isu n+1 = u n (1 + r) − Pwith u 0 = B. From (15.25) we haveu n = B(1 + r) n 1 − (1 + r)n− P1 − (1 + r) .AstheloanistoberepaidoverY years, u Y = 0 and thusBr(1 + r)YP =(1 + r) Y − 1 .The first year’s interest is rB and so the fraction of the first year’s payment goingtowards capital repayment is (P − rB)/P , which, using the above expression for P ,isequalto (1 + r) −Y . With the given figures, this is (only) 23%. ◭With only small modifications, the method just described can be adapted tohandle recurrence relations in which the constant k in (15.23) is replaced by kα n ,i.e. the relation isu n+1 = au n + kα n . (15.26)As for an inhomogeneous linear differential equation (see subsection 15.1.2), wemay try as a potential particular solution a form which resembles the term thatmakes the equation inhomogeneous. Here, the presence of the term kα n indicates498