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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 Linear Recurrence Relations with Constant Coefficients, 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 Linear Recurrence Relations with Constant Coefficientsis 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, Linear Recurrence Relations with Constant Coefficientsare all solutions of equation (i).

Also, we have Linear Recurrence Relations with Constant Coefficientsare all solutions of equation (i). The sums of solutions are also solutions.

Hence, the homogeneous solutions of the difference equation are

Linear Recurrence Relations with Constant Coefficients

Case2: If the characteristics equation has repeated real roots.

If ∝1=∝2, then (A1+A2 K) Linear Recurrence Relations with Constant Coefficientsis also a solution.

If ∝1=∝2=∝3 then (A1+A2 K+A3 K2) Linear Recurrence Relations with Constant Coefficientsis also a solution.

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

          (A1+A2 K+A3 K2+......+An Kn-1) Linear Recurrence Relations with Constant Coefficients

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 =Linear Recurrence Relations with Constant Coefficients andLinear Recurrence Relations with Constant Coefficients

Therefore, the homogeneous solution of the equation is given by
yK=(C1+C2 k).Linear Recurrence Relations with Constant Coefficients

Example3: Solve the difference equation yK-yK-1-yK-2=0.

Solution: The characteristics equation is s2-s-1=0
s=Linear Recurrence Relations with Constant Coefficients

Therefore, the homogeneous solution of the equation is

Linear Recurrence Relations with Constant Coefficients

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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