# Linear Recurrence Relations with Constant Coefficients

A Recurrence Relations is called linear if its degree is one.

The general form of linear recurrence relation with constant coefficient is

C0 yn+r+C1 yn+r-1+C2 yn+r-2+⋯+Cr yn=R (n)

Where C0,C1,C2......Cn are constant and R (n) is same function of independent variable n.

A solution of a recurrence relation in any function which satisfies the given equation.

### Linear Homogeneous Recurrence Relations with Constant Coefficients:

The equation is said to be linear homogeneous difference equation if and only if R (n) = 0 and it will be of order n.

The equation is said to be linear non-homogeneous difference equation if R (n) ≠ 0.

Example1: The equation ar+3+6ar+2+12ar+1+8ar=0 is a linear non-homogeneous equation of order 3.

Example2: The equation ar+2-4ar+1+4ar= 3r + 2r is a linear non-homogeneous equation of order 2.

A linear homogeneous difference equation with constant coefficients is given by

C0 yn+C1 yn-1+C2 yn-2+⋯......+Cr yn-r=0 ....... equation (i)

Where C0,C1,C2.....Cn are constants.

The solution of the equation (i) is of the form , where ∝1 is the characteristics root and A is constant.

Substitute the values of A ∝K for yn in equation (1), we have

C0 A∝K+C1 A∝K-1+C2 A∝K-2+⋯....+Cr A∝K-r=0.......equation (ii)

After simplifying equation (ii), we have

C0r+C1r-1+C2r-2+⋯Cr=0..........equation (iii)

The equation (iii) is called the characteristics equation of the difference equation.

If ∝1 is one of the roots of the characteristics equation, then is a homogeneous solution to the difference equation.

To find the solution of the linear homogeneous difference equations, we have the four cases that are discussed as follows:

Case1: If the characteristic equation has n distinct real roots∝1, ∝2, ∝3,.......∝n.

Thus, are all solutions of equation (i).

Also, we have are all solutions of equation (i). The sums of solutions are also solutions.

Hence, the homogeneous solutions of the difference equation are Case2: If the characteristics equation has repeated real roots.

If ∝1=∝2, then (A1+A2 K) is also a solution.

If ∝1=∝2=∝3 then (A1+A2 K+A3 K2) is also a solution.

Similarly, if root ∝1 is repeated n times, then.

(A1+A2 K+A3 K2+......+An Kn-1) The solution to the homogeneous equation.

Case3: If the characteristics equation has one imaginary root.

If α+iβ is the root of the characteristics equation, then α-iβ is also the root, where α and β are real.

Thus, (α+iβ)K and (α-iβ)K are solutions of the equations. This implies

(α+iβ)K A1+α-iβ)K A2

Is also a solution to the characteristics equation, where A1 and A2 are constants which are to be determined.

Case4: If the characteristics equation has repeated imaginary roots.

When the characteristics equation has repeated imaginary roots,

(C1+C2 k) (α+iβ)K +(C3+C4 K)(α-iβ)K

Is the solution to the homogeneous equation.

Example1: Solve the difference equation ar-3ar-1+2ar-2=0.

Solution: The characteristics equation is given by

s2-3s+2=0 or (s-1)(s-2)=0
⇒ s = 1, 2

Therefore, the homogeneous solution of the equation is given by

ar=C1r+C2.2r.

Example2: Solve the difference equation 9yK+2-6yK+1+yK=0.

Solution: The characteristics equation is

9s2-6s+1=0 or (3s-1)2=0
⇒ s = and Therefore, the homogeneous solution of the equation is given by
yK=(C1+C2 k). Example3: Solve the difference equation yK-yK-1-yK-2=0.

Solution: The characteristics equation is s2-s-1=0
s= Therefore, the homogeneous solution of the equation is Example4: Solve the difference equation yK+4+4yK+3+8yK+2+8yK+1+4yK=0.

Solution: The characteristics equation is s4+4s3+8s2+8s+4=0
(s2+2s+2) (s2+2s+2)=0
s = -1±i,-1±i

Therefore, the homogeneous solution of the equation is given by

yK=(C1+C2 K)(-1+i)K+(C3 +C4 K)(-1-i)K

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