Solve the following linear second order homogeneous differential

Blackettyl2j 2022-03-25 Answered

Solve the following linear second order homogeneous differential equation
y 32y+916y=0, y(0)=1, y'(0)=-1

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Answers (2)

Gia Edwards
Answered 2022-03-26 Author has 12 answers
Solution:
y32y+916y=0, y(0), y(0)=1
The characteristic equation is given by
γ232γ+916=0
16γ224γ+9=0
Hence
γ=24±2424169216
=24±57657632
=2432=34
Since the root are repeated therefore the solution are
y1(t)=eγt and y2(t)=teγt
=e34 and y2(t)=te34t
Let yc be the complimentary solt
Then
yc(t)=c1y1+c2y2, where c1 and c2 constant
=c1e34t+c2te34t
=e34t(c1+c2t)
given, y(0)=1, so
y(0)=1 c1=1
y(t)=34e34t(c1+c2t)+e34tc2
y(0)=34c1+c2=1
c2=134c1
=134=74
Hence
y(t)=e34t74te34t
=e34(174t)
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Jeffrey Jordon
Answered 2022-03-31 Author has 2262 answers

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