Solve
4
y
′
′
−
17
y
′
+
15
y
=
0
,
y
(
0
)
=
6
,
y
′
(
0
)
=
12.75
\displaystyle {4}{y}{''}-{17}{y}'+{15}{y}={0},\quad{y}{\left({0}\right)}={6},\quad{y}'{\left({0}\right)}={12.75}
4
y
′′
−
17
y
′
+
15
y
=
0
,
y
(
0
)
=
6
,
y
′
(
0
)
=
12.75
y
(
t
)
\displaystyle {y}{\left({t}\right)}
y
(
t
)
=
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