Page 344 - Schaum's Outline of Differential Equations

CHAP. 34] AN INTRODUCTION TO DIFFERENCE EQUATIONS 327

34.7. By recursive computations, generate the first 11 Fibonacci numbers using Problem 34.4.
We are given that / 0 = 1 and /j = 1, and f n+2 =f n+i +/«• Using this recursion formula, with n =0, we have,
2
f 2 =/j +/„ =1 + 1 = 2. We now let n = 1; this implies/ 3 =/ 2 +fi = +1=3. Continuing in this recursive way, we
have the following: / 4 = 5,/ 5 = 8,/ 6 = 13,/ 7 = 21,/ 8 = 34,/ 9 = 55 and/ 10 = 89.

34.8. Verify y n = c(4"), where c is any constant, solves the difference equation y n+1 = 4y n.
+1
Substituting our solution into the left-hand side of the difference equation, we have y n+i = c(4" ). The right-
+1
hand side becomes 4c(4") = c(4" ), which is precisely the result we obtained when we substituted our solution into
+1
the left-hand side. The equation is identically true for all n; that is, it can be written as 4c(4") = c(4" ). Hence, we
have verified our solution. We note that this solution can be considered the general solution to this linear, first-order
equation, since the equation is satisfied for any value of c.
34.9. Consider the difference equation a n+2 + 5a n+1 + 6a n = 0 with the imposed conditions; a 0 = 1, a^ = -4.
Verify that a n = 2(-3)" - (-2)" solves the equation and satisfies both conditions.
Letting n = 0 and n = 1 in a n clearly gives a 0 = 1 and a 1 = —4, hence our two subsidiary conditions are satisfied.
Substitution of a n into the difference equation gives

Thus the equation is satisfied by a n.
We note that this solution can be considered a particular solution, as opposed to the general solution, because
this equation is coupled with specific conditions.

34.10. Verify that p n = c^S)" + c 2(5)" + 3 + 4n, where c 1 and c 2 are any constants, satisfies the difference
equationp n+2 = 8p n+1 - I5p n + 32n.
Letting n, n + 1 and n + 2 into p n and substituting into the equation yields

whence, both sides simplify to 9ci(3)" + 25c 2(5)" + 11 + 4n, thereby verifying the solution.

34.11. Consider the difference equation, y n+l = -6y n. By guessing y n = p " for p ^ 0, find a solution to this equation.
Direct substitution gives p n+1 = -6p" which implies p = -6. Hence, y n = (-6)" is a solution to our difference
equation, which we can easily verify. We note that y n = k(-6) n also solves the difference equation, where k is any
constant. This can be thought of as the general solution.

34.12. Using the technique employed in the previous problem, find the general solution to
3b n+2 + 4b n+1 + b n=0.
2
Substitution of the guess b n = ff into the difference equation gives 3p n+2 + 4p n+1 + p " = p " (3p + 4p + 1) = 0,
2
which implies 3p + 4p + 1 = 0. This results in p = —, -1. So the general solution, as can easily be verified, is
where the c ; are arbitrary constants.

2
We note that 3p + 4p+ 1 = 0 is called the characteristic equation. Its roots can be treated in exactly the
same way as the characteristic equations derived from constant coefficient differential equations are handled (see
Chapter 9).