Use the method of reduction of order to find a second solution to
t
y
′
′
−
(
8
t
−
4
)
y
′
+
(
16
t
−
16
)
y
=
0
,
t
>
0
\displaystyle {t}{y}{''}-{\left({8}{t}-{4}\right)}{y}'+{\left({16}{t}-{16}\right)}{y}={0},\quad{t}\gt{0}
t
y
′′
−
(
8
t
−
4
)
y
′
+
(
16
t
−
16
)
y
=
0
,
t
>
0
Given
y
1
(
t
)
=
e
4
t
\displaystyle {y}_{{1}}{\left({t}\right)}={e}^{{{4}{t}}}
y
1
(
t
)
=
e
4
t
y
2
(
t
)
\displaystyle {y}_{{2}}{\left({t}\right)}
y
2
(
t
)
=
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Question 6
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