EDITOR: B. SOMANATHAN NAIR
Since difference
equations are modifications of differential equations, we can expect similar
solutions in the case of second-order difference equations, where we have to
substitute the term an for the term et. Consider the second-order difference
equation
y(n) + ay(n‒1) + by(n‒2) = 0 (1)
The
characteristic equation of this difference equation can be written as
(C2 + aC + b)y = 0 (2)
where we have used C2 y = y(n),
and so on. The roots of this equation, as in the case of differential equations,
can be taken as α and β and depending on the values of these, we
have four possible
solutions to Eq. (2) as given below.
Solution I: Both a and b
exist and are real and unequal
In this case, both the roots are real and
unequal. That is the roots are α ± β. The solution can be written in a way
similar to that given in Eq. (4) of solution to differential equation as
y(n)
= A(α + β)n + B(α ‒ β)n (3)
Solution II: Both the roots are real and equal
In this case, only α exists, β. = 0. The solution
can be written similar to that given in Eq. (5) as
y(n) = αn(A +
Bn) (4)
Solution III: Both the roots have real and imaginary parts
The roots are α ± jβ, and the solution can be written similar to that given in Eq. (6) as
y(n) = A(α + jβ)n u(n)+ B(α ‒ jβ)n
u(n) (5)
Equation (5) may be manipulated mathematically
to yield its trigonometric form
y(n)
= rn‑(C cos nθ + D sin nθ)
(6)
where we have
used the new constants C = A+B and D = j(A−B).
Solution IV: Both the roots are imaginary and are equal to ± jb
The solution can be written in a form similar to that
given in Eq. (6). Thus, we have
y(n) = C
cos nb + D sin nb (7)
In the forthcoming blogs, we illustrate
the solutions of differential and difference equations using typical numerical
examples.
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