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Answer To: https://webwork.math.nau.edu/webwork2/CDiggans_239/ username: mka53 password: Ms XXXXXXXXXX

Robert answered on Dec 23 2021
131 Votes
CDiggans 239 mka53 WW 09 Systems
1
u
′′
+ 4u

+ 6u = 0
v = u

v

= u
′′
= −4u′ − 6u

u

= v
v

= −4v − 6u
d
dt
[
u
v
]
=
[
0 1
−6 −4
] [
u
v
]
2
u
′′ − 8u′ + 6u = 7sin(3t)
v = u

v

= u
′′
= 8u
′ − 6u+ 7sin(3t)
u

= v
v

= 8v − 6u+ 7sin(3t)
d
dt
[
u
v
]
=
[
0 1
−6 8
] [
u
v
]
+
[
0
7sin(3t)
]
[
u(1)
v(1)
]
=
[
−5
−3.5
]
1
3
t2u
′′
+ 2.5tu

+ (t2 + 6)u = −2sin(3t)
v = u

v

= u
′′
=
−2.5tu′ − (t2 + 6)u− 2sin(3t)
t2
u

= v
v

= −2.5
t
v − t
2 + 6
t2
u− 2sin(3t)
t2
d
dt
[
u
v
]
=
[
0 1
−...
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