Find the general solution.(D − 1)(D − 2)(D2 − 4D + 13)2y = 0.

1.	Find the general solution.(D − 1)(D − 2)(D2 − 4D + 13)2y = 0.
Answer» We have λ = 1 and λ = 2 as roots of multiplicity 1, so they contribute basic solutions e^x and e^2x. The roots of the quadratic λ²− 4λ + 13 are λ = 2 ± 3i and these conjugate roots both have multiplicity 2. Thus, this pair of conjugate roots contributes the basic solutions e^2xcos(3x), e^2x sin(3x), xe^2x cos(3x) and xe^2xsin(3x). Thus, the general solution is (1.1) y = C₁e^x + C₂e^2x + C₃e^2xcos(3x) + C₄e^2xsin(3x) + C₅xe^2x cos3x + C₆xe^2xsin(3x).

Answer»

We have λ = 1 and λ = 2 as roots of multiplicity 1, so they contribute basic solutions e^x and e^2x. The roots of the quadratic λ²− 4λ + 13 are λ = 2 ± 3i and these conjugate roots both have multiplicity 2. Thus, this pair of conjugate roots contributes the basic solutions e^2xcos(3x), e^2x sin(3x), xe^2x cos(3x) and xe^2xsin(3x). Thus, the general solution is

(1.1) y = C₁e^x + C₂e^2x + C₃e^2xcos(3x) + C₄e^2xsin(3x) + C₅xe^2x cos3x + C₆xe^2xsin(3x).

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Find the general solution.(D − 1)(D − 2)(D2 − 4D + 13)2y = 0.

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