DHARM64 MATHEMATICAL FOUNDATION OF COMPUTER SCIENCESubstituting P. 2n into the difference equation we obtain,
P 2n + 5 P 2n–1 + 6P 2n –2 = 3. 2n ;
It simplifies to 5P 2n = 3. 2n; so P = 3/5;
Hence, particular solution is
an(p) = 3/5. 2n;
(d) Consider the difference equation
an + an–1 = 2 n .3n;
Here, f(n) = 2. n. 3n ; where n is a polynomial of degree 1 followed by 3n and 3 is not a
characteristic root then the particular solution is of the form of expression (3.17), i.e.
(P 1 n + P 2 ). 3n
Substituting (P 1 n + P 2 ). 3n into the difference equation, we obtain
(P 1 n + P 2 ). 3n + (P 1 (n – 1) + P 2 ). 3n–1 = 2.n.3n;
Then after simplification we obtain
P 1 = 3/2 and P 2 = 3/8;
Hence, the particular solution is
an(p) = (3/2n + 3/8).3n;
(e) Consider the difference equation an – 2 an–1 = 3.2n ; Here f(n) is 3.2n which is of the
form βn, where β = 2. Since 2 is a characteristic root of multiplicity 1 therefore general form of
the particular solution is like expression (3.18) i.e.
P.n.2n ;
Substituting P.n.2n into the difference equation, we obtain
P.n.2n – 2P.(n – 1).2n–1 = 3.2n ;
that yields P = 6.
Hence, the particular solution is
an(p) = 6.n.2n.
(f) Consider difference equation an – 2 an–1 + an–2 = (n+ 1). 2n where, f(n) = (n + 1).2n
which is a polynomial of degree 1 followed by 2n. Since 2 is a characteristic root of multiplicity
2 so the general form of the particular solution will be like expression (3.18) i.e.
n^2 (P 1 n + P 2 ).2n;
Substituting above expression into the difference equation we obtain,
n^2 (P 1 n + P 2 ).2n – 2(n – 1)^2 (P 1 (n – 1) + P 2 ).2n–1 + (n – 2)^2 (P 1 (n – 2) + P 2 ).2n–2 = (n +1).2n;
After simplifing and comparing the coefficients of 2n and the constant term we get the
values of P 1 & P 2
Hence, particular solution will be
an(p) = n^2 (P 1 n + P 2 ).2n ;
(g) Consider the difference equation an – an–1 = 5 ; Since, 1 is the characteristic root of
the difference equation, so we can write f(n) = 5.1n. Thus the form of particular solution is
P. n. 1n or P n ;
Since expression P.n satisfies the difference equation so we obtain,
P. n. – P.(n – 1) = 5 ; or P = 5 ;
Hence, Particular solution is
an(p) = 5. n