MA 36600 LECTURE NOTES: FRIDAY, FEBRUARY 27
3
Then the general solution to the nonhomogeneous equation is the function
y
(
t
) =
c
1
y
1
(
t
) +
c
2
y
2
(
t
) +
Y
(
t
)
for some constants
c
1
and
c
2
. We remark that typically finding
y
1
and
y
2
is easy: when the coe
ﬃ
cients
a
(
t
),
b
(
t
), and
c
(
t
) are constant functions, one can compute
y
1
and
y
2
by considering roots of the characteristic
equation
a r
2
+
b r
+
c
= 0. We will spend the next couple of lectures discussing how to find a solution
Y
.
Note that once
one
solution
Y
is found, we can compute
all
solutions
y
.
We give a proof of the claim above. Say that
y
=
y
(
t
) is any solution to the nonhomogeneous equation,
and let
Y
=
Y
(
t
) be as in assumption (iii). Denote the di
ff
erence
u
(
t
) =
y
(
t
)
−
Y
(
t
)
.
We see that
u
=
u
(
t
) is a solution to the homogeneous equation:
a
(
t
)
u
+
b
(
t
)
u
+
c
(
t
)
u
=
a
(
t
)
y
−
Y
+
b
(
t
)
y
−
Y
+
c
(
t
)
y
−
Y
=
a
(
t
)
y
+
b
(
t
)
y
+
c
(
t
)
y
−
a
(
t
)
Y
+
b
(
t
)
Y
+
c
(
t
)
Y
=
f
(
t
)
−
f
(
t
)
= 0
.
By assumption (i), both
y
1
and
y
2
are also solutions to
a
(
t
)
u
+
b
(
t
)
u
+
c
(
t
)
u
= 0; and by assumption (ii)
these functions are linearly independent. Hence
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 Spring '09
 EdrayGoins
 Derivative, Characteristic polynomial, Nonhomogeneous Equation

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