Solve the question in the same way as a book Plot the following curves, locate any fixed points, and obtain asymp-totic expansions to their solution as k(a) Yk+1 = y% +a², a =Yk (y%+3)3y+1(c) 2y%+1 - 5yk+1Yk + 2y% = 2,real number,(b) Ук+1 —(d) ук+1ayk+1Ykta »(е) Ук+15y -6yk+26y -8yk+3'(f) 5yk+15y+6yk+19 Example DThe equationYk+1 =1/2yk +2-3/2yk(2.193)76Difference Equationscorresponds to a hyperbola in the (yk, Yk+1) plane having asymptotes Yk = 0and yk+1 =slopes at (1, 1) and (3, 3) are, respectively, 2 and 2/3. Therefore, Yk = 1 is anunstable 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 yk = 3,is given by the expression1/2yk + 2. There are two fixed points, yk = 1 and yk = 3. TheYk = 3+ A(2/3)*,(2.194)where A is an arbitrary constant. Therefore, if we lett = A(2/3)*, 2(t) = Yk,(2.195)then equation (2.193) becomes2t2:(0): ()2(t)? + 4z(t) – 3.(2.196)=We assume a solution of the formz(t) = 3+t+ A2t? + Azt +....(2.197)Substitution of this expression into equation (2.197) gives, respectively, forthe left- and right-hand sides2z(t)z2t= 18 + 10t + 2/3(13A2 + 2)t+ 2/%(35A3 + 10A2)t² + ...,(2.198)andz(t)? + 4z(t) – 3 = 18 + 10t + (10A2 + 1)t?+ (10A3 + 2A2)t³ + ....(2.199)Equating the coefficients of corresponding powers of t gives2/3(13A2 + 2) = 10A2 + 1,(2.200)2/9(35A3 + 10A2) = 10A3 + 2A2,andA2 = 1/4, A3 =140.(2.201)Therefore,z(t) = 3+t+at2 + 1/40t + ... ,andYk = 3+ A(2/3)* +14A? (2/3)2k + 1/40A° (2/3)3k + ....(2.202)

Name: Solve the question in the same way as a book Plot the following curves
Uploaded: 2024-03-05T11:52:21.000Z
Channel: Bartleby
Description: Solve the question in the same way as a book Plot the following curves, locate any fixed points, and obtain asymp-totic expansions to their solution as k(a) Yk+1 = y% +a², a =Yk (y%+3)3y+1(c) 2y%+1 - 5yk+1Yk + 2y% = 2,real number,(b) Ук+1 —(d) ук+1ay

Question

Accepted Answer

As you can see in the figure attached there is only one fixed point and when k goes to infinity the…

5y-6yk+2 6y-8yk+3) (e) Yk+1