Solve
y
′
′
−
2
y
′
+
10
y
=
0
,
y
(
0
)
=
1
,
y
′
(
0
)
=
−
8
\displaystyle {y}{''}-{2}{y}'+{10}{y}={0},\quad{y}{\left({0}\right)}={1},\quad{y}'{\left({0}\right)}=-{8}
y
′′
−
2
y
′
+
10
y
=
0
,
y
(
0
)
=
1
,
y
′
(
0
)
=
−
8
y
(
t
)
\displaystyle {y}{\left({t}\right)}
y
(
t
)
=
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Question 6 Part 1 of 2
The behavior of the solutions are:
Oscillating with decreasing amplitude
Oscillating with increasing amplitude
Steady oscillation
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