Use undetermined coefficients to find the particular solution to
y
′
′
+
5
y
′
+
6
y
=
4
e
−
2
t
\displaystyle {y}{''}+{5}{y}'+{6}{y}={4}{e}^{{-{2}{t}}}
y
′′
+
5
y
′
+
6
y
=
4
e
−
2
t
Y
(
t
)
\displaystyle {Y}{\left({t}\right)}
Y
(
t
)
=
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