3.1.3 Example C By direct substitution, we can show that the equation Yk+2 – 5yk+1 + 6yk (3.27) has the solutions (2) - 2" , (1) - 3*. (3.28) Therefore, for arbitrary constants C1 and С2, Yk = C12k + c23k (3.29) is a solution to equation (3.27). Likewise, we can show by direct substitution that Yk = 1/2(k? + 3k + 5) (3.30) is a solution to the inhomogeneous equation Yk+2 5yk + 6yk = k². (3.31) Therefore, from Theorem 3.3, we conclude that c2* + c23k + 1/2(k² + 3k + 5) (3.32) Yk = is a solution to equation (3.31).

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Explain the determaine please

3.1.3
Example C
By direct substitution, we can show that the equation
Yk+2 – 5yk+1 + 6yk
(3.27)
has the solutions
Y = 2k,
(1)
(2) – 3k.
(3.28)
Therefore, for arbitrary constants c1 and c2,
Yk
C12* + c23k
(3.29)
is a solution to equation (3.27).
Likewise, we can show by direct substitution that
(Y = 1/2(k? + 3k + 5)
(3.30)
is a solution to the inhomogeneous equation
Yk+2 – 5yk + 6yk = k².
(3.31)
Therefore, from Theorem 3.3, we conclude that
c12* + c23* + 1/2(k² + 3k + 5)
(3.32)
Yk = C
is a solution to equation (3.31).
Transcribed Image Text:3.1.3 Example C By direct substitution, we can show that the equation Yk+2 – 5yk+1 + 6yk (3.27) has the solutions Y = 2k, (1) (2) – 3k. (3.28) Therefore, for arbitrary constants c1 and c2, Yk C12* + c23k (3.29) is a solution to equation (3.27). Likewise, we can show by direct substitution that (Y = 1/2(k? + 3k + 5) (3.30) is a solution to the inhomogeneous equation Yk+2 – 5yk + 6yk = k². (3.31) Therefore, from Theorem 3.3, we conclude that c12* + c23* + 1/2(k² + 3k + 5) (3.32) Yk = C is a solution to equation (3.31).
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