The differential equation
y
(
3
)
+
7
y
′
′
+
10
y
′
−
7
y
=
sin
(
x
)
\displaystyle {y}^{{{\left({3}\right)}}}+{7}{y}{''}+{10}{y}'-{7}{y}={\sin{{\left({x}\right)}}}
y
(
3
)
+
7
y
′′
+
10
y
′
−
7
y
=
sin
(
x
)
can be rewritten in the form
L
(
y
)
=
sin
(
x
)
\displaystyle {L}{\left({y}\right)}={\sin{{\left({x}\right)}}}
L
(
y
)
=
sin
(
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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