Use undetermined coefficients to find the particular solution to
y
′
′
−
2
y
′
−
15
y
=
2
e
−
t
\displaystyle {y}{''}-{2}{y}'-{15}{y}={2}{e}^{{-{t}}}
y
′′
−
2
y
′
−
15
y
=
2
e
−
t
Y
(
t
)
\displaystyle {Y}{\left({t}\right)}
Y
(
t
)
=
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