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
p
(
t
)
\displaystyle {y}_{{p}}{\left({t}\right)}
y
p
(
t
)
=
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