Find the particular solution to
y
′
′
−
y
′
−
20
y
=
2
e
4
t
\displaystyle {y}{''}-{y}'-{20}{y}={2}{e}^{{{4}{t}}}
y
′′
−
y
′
−
20
y
=
2
e
4
t
y
p
\displaystyle {y}_{{p}}
y
p
=
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