Explain the detemaine 2.8.4 Example D100The equationYk+1 = /2yk + 2 – 3/2yk(2.193)76Difference Equationscorresponds to a hyperbola in the (yk,Yk+1) plane having asymptotes Ykand Yk+1= 0= 3. The= 1 is an1/2yk + 2. There are two fixed points, Yk = 1 and ykslopes at (1, 1) and (3, 3) are, respectively, 2 and 2/3. Therefore, ykunstable fixed point, while Yk = 3 is a stable fixed point.A geometric analysis indicates that arbitrary starting points lead finallyto the stable fixed point. See Figure 2.10.The first approximation to the solution of equation (2.193), near ykis given by the expression= 3,Yk3+ A(2/3)*,(2.194)where A is an arbitrary constant. Therefore, if we lett = A(?/3)*, z(t) = Yk,(2.195)then equation (2.193) becomes2t2z(t)z= 2(t)² + 4z(t) – 3.(2.196)We assume a solution of the form2(t) = 3+t+ A2t² + A3t³ + ....(2.197)Substitution of this expression into equation (2.197) gives, respectively, forthe left- and right-hand sides2t2z(t)z318 + 10t + 2/3(13A2 + 2)t?+ 2%(35A3 + 10A2)t² + · · · ,(2.198)and2(t)2 + 4z(t) – 3 = 18 + 10t + (10A2 + 1)t?+ (10А3 + 2A2)t3 + ...(2.199)

Name: Explain the detemaine 2.8.4 Example D100The equationYk+1 = /2yk + 2 –
Uploaded: 2024-03-20T07:06:41.000Z
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Description: Explain the detemaine 2.8.4 Example D100The equationYk+1 = /2yk + 2 – 3/2yk(2.193)76Difference Equationscorresponds to a hyperbola in the (yk,Yk+1) plane having asymptotes Ykand Yk+1= 0= 3. The= 1 is an1/2yk + 2. There are two fixed points, Yk = 1 an