Solve
y
′
′
+
3
y
′
−
4
y
=
0
,
y
(
0
)
=
−
1
,
y
′
(
0
)
=
−
11
\displaystyle {y}{''}+{3}{y}'-{4}{y}={0},\quad{y}{\left({0}\right)}=-{1},\quad{y}'{\left({0}\right)}=-{11}
y
′′
+
3
y
′
−
4
y
=
0
,
y
(
0
)
=
−
1
,
y
′
(
0
)
=
−
11
y
(
t
)
\displaystyle {y}{\left({t}\right)}
y
(
t
)
=
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Question 6
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