Linear Recurrence Relations with Constant CoefficientsA Recurrence Relations is called linear if its degree is one. The general form of linear recurrence relation with constant coefficient is C_{0} y_{n+r}+C_{1} y_{n+r1}+C_{2} y_{n+r2}+⋯+C_{r} y_{n}=R (n) Where C_{0},C_{1},C_{2}......C_{n} 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 nonhomogeneous difference equation if R (n) ≠ 0. Example1: The equation a_{r+3}+6a_{r+2}+12a_{r+1}+8a_{r}=0 is a linear nonhomogeneous equation of order 3. Example2: The equation a_{r+2}4a_{r+1}+4a_{r}= 3r + 2^{r} is a linear nonhomogeneous equation of order 2. A linear homogeneous difference equation with constant coefficients is given by C_{0} y_{n}+C_{1} y_{n1}+C_{2} y_{n2}+⋯......+C_{r} y_{nr}=0 ....... equation (i) Where C_{0},C_{1},C_{2}.....C_{n} 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 y_{n} in equation (1), we have C_{0} A∝^{K}+C_{1} A∝^{K1}+C_{2} A∝^{K2}+⋯....+C_{r} A∝^{Kr}=0.......equation (ii) After simplifying equation (ii), we have C_{0} ∝^{r}+C_{1} ∝^{r1}+C_{2} ∝^{r2}+⋯C_{r}=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 (A_{1}+A_{2} K) is also a solution. If ∝_{1}=∝_{2}=∝_{3} then (A_{1}+A_{2} K+A_{3} K^{2}) is also a solution. Similarly, if root ∝_{1} is repeated n times, then. (A_{1}+A_{2} K+A_{3} K^{2}+......+A_{n} K_{n1}) 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 αiβ is also the root, where α and β are real. Thus, (α+iβ)^{K} and (αiβ)^{K} are solutions of the equations. This implies (α+iβ)^{K} A_{1}+αiβ)^{K} A_{2} Is also a solution to the characteristics equation, where A_{1} and A_{2} are constants which are to be determined. Case4: If the characteristics equation has repeated imaginary roots. When the characteristics equation has repeated imaginary roots, (C_{1}+C_{2} k) (α+iβ)^{K} +(C_{3}+C_{4} K)(αiβ)^{K} Is the solution to the homogeneous equation. Example1: Solve the difference equation a_{r3ar1+2ar2=0.} Solution: The characteristics equation is given by s^{2}3s+2=0 or (s1)(s2)=0⇒ s = 1, 2 Therefore, the homogeneous solution of the equation is given by a_{r}=C^{1}_{r}+C^{2}.2^{r}.Example2: Solve the difference equation 9y_{K+2}6y_{K+1}+y_{K}=0. Solution: The characteristics equation is 9s^{2}6s+1=0 or (3s1)^{2}=0⇒ s = and Therefore, the homogeneous solution of the equation is given by Example3: Solve the difference equation y_{K}y_{K1}y_{K2}=0. Solution: The characteristics equation is s^{2}s1=0 Therefore, the homogeneous solution of the equation is Example4: Solve the difference equation y_{K+4}+4y_{K+3}+8y_{K+2}+8y_{K+1}+4y_{K}=0. Solution: The characteristics equation is s^{4}+4s^{3}+8s^{2}+8s+4=0 Therefore, the homogeneous solution of the equation is given by y_{K}=(C_{1}+C_{2} K)(1+i)^{K}+(C_{3} +C_{4} K)(1i)^{K}
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