Page 345 - Schaum's Outline of Differential Equations

328 AN INTRODUCTION TO DIFFERENCE EQUATIONS [CHAP. 34

34.13. Solve d n+1 = 2d n + 6n, by guessing a solution of d n = An + B, where A and B are coefficients to be
determined.
Substitution of our guess into the difference equation leads to the identity A(n + 1) + B = 2(An + B) + 6n.
Equating the coefficients of like powers of n, we have A = 2A + 6 and A + B = 2B, which implies A = B = -6. Hence
our solution becomes d n = —6n — 6.
We note that the method of "undetermined coefficients" was presented in Chapter 11 regarding differential
equations. Our guess here is the discrete variable counterpart, assuming a first degree polynomial, because the non-
homogeneous part of the equation is a first degree polynomial.

34.14. Find the general solution to the non-homogeneous difference equation d n+l = 2d n + 6n, if we know that
n
the general solution to the corresponding homogeneous equation is d n = k(2) , where k is any constant.
Because the theory of solutions for difference equations parallels that of differential equations (see Chapter 8),
the general solution to the non-homogeneous, equation is the sum of the general solution to the corresponding
homogeneous equation plus any solution to the non-homogeneous equation.
Since we are given the general solution to the homogeneous equation, and we know a particular solution to the
non-homogeneous equation (see Problem 34.13), the desired solution is d n = k(2) n - 6n- 6.

34.15. Consider the difference equation y n+2 + 6y n+1 + 9y n = 0. Use the guessing technique presented in
Problem 34.11, find the general solution.
Assuming y n=p" leads to p"(p 2 + 6p + 9) = 0 which implies p = —3, —3, a double root. We expect two
"linearly independent" solutions to the difference equations, since it is of the second-order. In fact, following the
identical case in which the characteristic equation for second-order differential equations has a double root (see
Chapter 9), we can easily verify that y n = Cj(—3)" + c 2n(—3) n indeed solves the equation and is, in fact, the general
solution.

34.16. Suppose you invest $100 on the last day of the month at an annual rate of 6%, compounded monthly. If
you invest an additional $50 on the last day of each succeeding month, how much money would have
been accrued after five years.
We will model this situation (see Chapter 2) using a difference equation.
Let y n represent the total amount of money ($) at the end of month n. Therefore y 0 = 100. Since the 6%
interest is compounded monthly, the amount of money at the of the first month is equal to the sum of y 0 and the
amount made during the first month which is 100(.06/12) = 0.50 (we divide by 12 because we are compounding
monthly). Hence, y 1 = 100 + 0.50 + 50 = 150.50 (because we add $50 at the end of each month). We note that
yi = y 0 + 0.005y 0 + 50 = (1.005)3; 0 + 50.
Building on this equation, we see that y 2 = (1.005)3^ + 50. And, in general, our difference equation becomes
y n+1 = (1.005)y n + 50, with the initial condition y 0 = 100.
We solve this difference equation by following the methods presented in the five previous problems. That is,
we first guess a homogeneous solution of the form y n = kp", where k is a constant to be determined.
Substitution of this guess into the difference equation yields kp n+l = (1.005)kp"; this implies p = 1.005. We
will solve for k after we find a solution to the non-homogeneous part of the difference equation.
Because the degree of the non-homogeneous part of our difference equation is 0 (50 is a constant), we guess
y n = C, where we must determine C.
Substitution into the difference equation implies C = (1.005)C + 50, which leads to C = -10,000.
Summing our solutions leads us to the general solution of the difference equation:

Finally, we obtain k by imposing our initial condition: y Q = 100. Letting n = 0 in (1) implies 100 = fc(1.005)
- 1000 = k- 1000; hence, k = 10,100. So (1) becomes