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Consider the following equation: f(E)yk = (E – 2)²yk = 2k. Che solution to the homogeneous equation is (H) : (c1 + c2k)2*. (4.232) Now f(E) has a double root and we must apply Theorem 4.9' again. [For this ase g(E) = 1 and m = 2.] Thus, 2k-212 k22k (P) = (E – 2)–2 . 2k (4.233) %3D (1)(?)

Consider the following equation: f(E)yk = (E – 2)²yk = 2k. Che solution to the homogeneous equation is (H) : (c1 + c2k)2*. (4.232) Now f(E) has a double root and we must apply Theorem 4.9' again. [For this ase g(E) = 1 and m = 2.] Thus, 2k-212 k22k (P) = (E – 2)–2 . 2k (4.233) %3D (1)(?)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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4.6.4 Example D Consider the following equation: f(E)yk = (E – 2)²yk 2k. The solution to the homogeneous equation is (H) Yk : (c1 + c2k)2*. (4.232) Now f(E) has a double root and we must apply Theorem 4.9' again. [For this case g(E) = 1 and m = 2.] Thus, (P) 2k-2k2 k22k = (E – 2)–2 . 2k Yk (4.233) (1)(2!) 8. and the general solution of equation (4.232) is k22k Yk = (C1 + c2k)2* + 8 (4.234)
Transcribed Image Text:4.6.4 Example D Consider the following equation: f(E)yk = (E – 2)²yk 2k. The solution to the homogeneous equation is (H) Yk : (c1 + c2k)2*. (4.232) Now f(E) has a double root and we must apply Theorem 4.9' again. [For this case g(E) = 1 and m = 2.] Thus, (P) 2k-2k2 k22k = (E – 2)–2 . 2k Yk (4.233) (1)(2!) 8. and the general solution of equation (4.232) is k22k Yk = (C1 + c2k)2* + 8 (4.234)
Theorem 4.9'. Let f(a) = 0, where f(E) = (E – a)"g(E) and g(a) + 0. Then -m km ak- f(E)-'ak (4.206) g(a)m! Proof. We assume that f(r) has a zero atr=a of multiplicity m. Therefore, ak f(E)-'a* = (E – a)-"g(E)¯'a* = (E – a)* g(a) -1k -m ak - (aЕ — а)- ak-m 1 g(a) ak-m k(m) -(E – 1) ™. 1 (4.207) -m g(a) k-m -m 1 = g(a) g(a) m!
Transcribed Image Text:Theorem 4.9'. Let f(a) = 0, where f(E) = (E – a)"g(E) and g(a) + 0. Then -m km ak- f(E)-'ak (4.206) g(a)m! Proof. We assume that f(r) has a zero atr=a of multiplicity m. Therefore, ak f(E)-'a* = (E – a)-"g(E)¯'a* = (E – a)* g(a) -1k -m ak - (aЕ — а)- ak-m 1 g(a) ak-m k(m) -(E – 1) ™. 1 (4.207) -m g(a) k-m -m 1 = g(a) g(a) m!
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