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Author Topic: TUT0602 Quiz4  (Read 6746 times)

Yuchen Cong

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TUT0602 Quiz4
« on: October 19, 2019, 05:11:25 PM »
Question:
2y''+2y+1=0
Solution:
2r2+2r+1=0
r=[-2+-((-2)2-(4)(2)(1))(1/2)]/2(2)
r=[-2+-(-4)(1/2)]/4
r=-(1/2)+-(1/2)i
λ=-(1/2), µ=1/2
y(t)=C1e-(1/2)tcos(1/2)t+C2e-(1/2)tsin(1/2)t


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