The differential equation
y
(
4
)
+
9
y
(
3
)
−
y
′
′
+
5
y
=
e
x
\displaystyle {y}^{{{\left({4}\right)}}}+{9}{y}^{{{\left({3}\right)}}}-{y}{''}+{5}{y}={e}^{{x}}
y
(
4
)
+
9
y
(
3
)
−
y
′′
+
5
y
=
e
x
can be rewritten in the form
L
(
y
)
=
e
x
\displaystyle {L}{\left({y}\right)}={e}^{{x}}
L
(
y
)
=
e
x
where
L
\displaystyle {L}
L
is a differential operator based on
D
=
d
d
x
\displaystyle {D}=\frac{{d}}{{{\left.{d}{x}\right.}}}
D
=
d
x
d
. Find
L
\displaystyle {L}
L
.
L
=
\displaystyle {L}=
L
=
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