Judicious Guessing
(or Undetermined Coefficients)
Recall that Variation of Parameters gave us a way to produce a particular solution to the nonho
mogeneous equation, given that we could solve the homogeneous equation. Here we will look at a
few shortcuts that will allow us to skip the sometimes diﬃcult integrations required by Variation of
Parameters.
±
Judicious Guessing
The goal is to ﬁnd a particular solution
ψ
(
x
)
to the equation
y
00
+
p
(
x
)
y
0
+
q
(
x
)
y
=
g
(
x
)
by guessing the form of solution, and then solving for the
undetermined coeﬃcients
in that form. For
this reason this method is also known as the method of undetermined coeﬃcients. It is not always
applicable, but it does work in the following cases. Throughout, we will assume the homogeneous
equation has constant coeﬃcients.
±
Case 1: RHS is polynomial
In this case we will assume that our equation is of the form
ay
00
+
by
0
+
cy
=
a
0
+
a
1
x
+
···
+
a
n
x
n
.
Here because the RHS is a polynomial, we will guess that the particular solution
ψ
(
x
)
is also a
polynomial. In order to make the degree of the polynomials on both sides agree, our guess will
depend on which derivatives appear on the LHS.
In this case we guess:
ψ
(
x
) =
A
0
+
A
1
x
+
···
+
A
n
x
n
if
c
6
= 0
A
0
x
+
A
1
x
2
+
···
+
A
n
x
n
+1
if
c
= 0
and
b
6
= 0
A
0
x
2
+
A
1
x
3
+
···
+
A
n
x
n
+2
if
c
= 0
and
b
= 0
To ﬁnd the undetermined coeﬃcients
A
0
,A
1
,...,A
n
, we plug in our guess and compare the coeﬃ
cients of
t
k
for each
k
.
Remark:
We included the case where
b
= 0
and
c
= 0
for completeness, but you will NEVER
NEED TO APPLY THIS CASE. Since the original equation in this case just looks like
ay
00
=
p
(
x
)
for some polynomial
p
(
x
)
, we can ﬁnd
y
just by integrating this polynomial twice.
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Example
Find a particular solution to the equation
y
00
+ 3
y
0
=
x
2
+ 6
x
+ 1
.
Solution
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 '08
 staff
 Exponential Function, Complex number, RHS

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