Find the particular solution to the equation
y
′
′
+
2
y
′
+
y
=
e
−
t
ln
t
\displaystyle {y}{''}+{2}{y}'+{y}={e}^{{-{t}}}{\ln{{t}}}
y
′′
+
2
y
′
+
y
=
e
−
t
ln
t
y
p
(
t
)
=
\displaystyle {y}_{{p}}{\left({t}\right)}=
y
p
(
t
)
=
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