Find the general solution of this ODE:
d
2
y
d
t
2
+
12
d
y
d
t
+
45
y
=
4
e
−
6
t
\displaystyle \frac{{{d}^{{2}}{y}}}{{{\left.{d}{t}\right.}^{{2}}}}+{12}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}+{45}{y}={4}{e}^{{-{6}{t}}}
d
t
2
d
2
y
+
12
d
t
d
y
+
45
y
=
4
e
−
6
t
The solution will be of the form:
y
(
t
)
=
C
y
1
(
t
)
+
D
y
2
(
t
)
+
y
p
(
t
)
\displaystyle {y}{\left({t}\right)}={C}{y}_{{1}}{\left({t}\right)}+{D}{y}_{{2}}{\left({t}\right)}+{y}_{{p}}{\left({t}\right)}
y
(
t
)
=
C
y
1
(
t
)
+
D
y
2
(
t
)
+
y
p
(
t
)
so use
C
\displaystyle {C}
C
and
D
\displaystyle {D}
D
as the arbitrary constants.
y
(
t
)
=
\displaystyle {y}{\left({t}\right)}=
y
(
t
)
=
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