If y = e^(3x) cos 4x, then y₂ − 6y₁ + ky = 0, where k is:
If y = e^(3x) cos 4x, then y₂ − 6y₁ + ky = 0, where k is:
- A. 25
- B. 7
- C. −7
- D. −25
Answer: A) 25
Explanation: y′ = e^(3x)(3 cos 4x − 4 sin 4x) = e^(3x)·R, y″ = e^(3x)[(3 cos 4x − 4 sin 4x)3 + (−12 sin 4x − 16 cos 4x)]? Better: y = Re(e^(3+4i)x), characteristic roots 3±4i satisfy r²−6r+25=0, so k=25.
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