Solve
y
′
′
+
4
y
′
+
20
y
=
0
,
y
(
0
)
=
−
2
,
y
′
(
0
)
=
12
\displaystyle {y}{''}+{4}{y}'+{20}{y}={0},\quad{y}{\left({0}\right)}=-{2},\quad{y}'{\left({0}\right)}={12}
y
′′
+
4
y
′
+
20
y
=
0
,
y
(
0
)
=
−
2
,
y
′
(
0
)
=
12
y
(
t
)
\displaystyle {y}{\left({t}\right)}
y
(
t
)
=
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