Section
3.5
Undetermined
Coefficients:
Problem
4
Previous
Problem
Problem
List
Next
Problem
(1
point)
Find
a
particular
solution
to
the
differential
equation
y'
=5y’
+6y=108¢.
Yy
=
l
187
+
450
+
57
+
Solution:
The
complementary
solution
is
y.
=
¢;e?
+
cye.
Since
3
and
its
derivatives
do
not
duplicates
any
of
the
terms
in
Y.,
we
assume
a
particular
solution
of
the
form
y,
=
A+
Bt
+C
+
D?,
for
which
Yy
=B+2Ct+3Df
and
yj
=2C
+6Dt.
Substituton
in
the
original
differential
equation
yields
2C
+
6Dt
—
5(B+2Ct
+3Dr*)
+
6(A
+
Bt
+
Ct*
+
Dr’)
=
1081’
Rearranging
terms
gives
(2C
—
5B+
6A)
+
(6D
—
10C
+
6B)t
+
(6C
—
15D)*
+
6D
=
108’
We
equate
coefficients
of
like
powers
of
¢
to
get
2C-5B+6A
=
0
6D
—-10C+6B
=
0
6C—-15D
=
0
6D
=
108
with
solution
A
=
%,B
=57,C=45and
D
=
18.
Hence
a
particular
solution
is
yp
=
+57t
+457
+18°
