Example 10-39. (Characteristic equation with real roots)
Solve the second-order difference equation
yk+2 − 3 yk+1 + 2 yk= 0.
Solution: Assume solutions of the form yk=λk, where λis a constant. This assumed
solution produces yk+1 =λk+1 and yk+2 =λk+2. Substituting these values into the
difference equation produces the equation
(λ^2 − 3 λ+ 2)λk= 0 ,
which tells us the required values for λ in order that yk=λksatisfy the difference
equation. For a nontrivial solution it is required that λ= 0. This produces the
characteristic equation
λ^2 − 3 λ+ 2 = 0.
A short cut for writing down the characteristic equation is to observe the form
of the given difference equation, when written in an operator form involving the
stepping operator E. One can quickly obtain the characteristic equation from this
operator form. For example, the given difference equation can be expressed in the
form (E^2 − 3 E+ 2)yk= 0 ,where the operator E^2 − 3 E+ 2 shows us the general form
of the characteristic equation when Eis replaced by λ. The characteristic equation
has the roots λ 1 = 2 and λ 2 = 1,and hence two linearly independent solutions are
y 1 (k) = 2k and y 2 (k) = 1k= 1
which is called a fundamental set of solutions. The general solution can be written
as a linear combination of this fundamental set and so one can write
y(k) = yk=c 1 (2)k+c 2 ,
where c 1 and c 2 are arbitrary constants. Given a set of initial conditions of the form
y 0 =Aand y 1 =B, where Aand Bare given constants, one can form the equations
y 0 =A=c 1 +c 2
y 1 =B= 2c 1 +c 2 ,
from which the constants c 1 and c 2 can be determined. Solving for these constants
produces the solution of the initial value problem which satisfies the given initial
conditions. The desired solution is unique and found to be
yk= (B−A)(2)k+ (2 A−B).
