t
)
+
c
1
cos(2
t
) +
c
2
sin(2
t
)
,
which shows the
homogeneous
and
particular
solutions
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Second Order Linear Equations
Part 2  Nonhomogeneous
— (17/27)
CauchyEuler Equation
Review
Variation of Parameters
Motivating Example
Technique of Variation of Parameters
Main Theorem for Nonhomogeneous DE
Variation of Parameters
1
Technique of Variation of Parameters:
Consider the
nonhomogeneous problem
y
00
+
p
(
t
)
y
0
+
q
(
t
)
y
=
g
(
t
)
,
where
p
,
q
, and
g
are given continuous functions
Assume we know the
homogeneous solution
:
y
c
(
t
) =
c
1
y
1
(
t
) +
c
2
y
2
(
t
)
Try a
general solution
of the form
y
(
t
) =
u
1
(
t
)
y
1
(
t
) +
u
2
(
t
)
y
2
(
t
)
,
where the functions
u
1
and
u
2
are to be determined
Differentiating yields
y
0
(
t
) =
u
1
(
t
)
y
0
1
(
t
) +
u
2
(
t
)
y
0
2
(
t
) +
u
0
1
(
t
)
y
1
(
t
) +
u
0
2
(
t
)
y
2
(
t
)
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Second Order Linear Equations
— (18/27)
CauchyEuler Equation
Review
Variation of Parameters
Motivating Example
Technique of Variation of Parameters
Main Theorem for Nonhomogeneous DE
Variation of Parameters
2
Variation of Parameters:
As before, there must be a condition
relating
u
1
and
u
2
, so take
u
0
1
(
t
)
y
1
(
t
) +
u
0
2
(
t
)
y
2
(
t
) = 0
This simplifies the derivative of the general solution to
y
0
(
t
) =
u
1
(
t
)
y
0
1
(
t
) +
u
2
(
t
)
y
0
2
(
t
)
Differentiating again yields:
y
00
(
t
) =
u
1
(
t
)
y
00
1
(
t
) +
u
2
(
t
)
y
00
2
(
t
) +
u
0
1
(
t
)
y
0
1
(
t
) +
u
0
2
(
t
)
y
0
2
(
t
)
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Second Order Linear Equations
Part 2  Nonhomogeneous
— (19/27)
CauchyEuler Equation
Review
Variation of Parameters
Motivating Example
Technique of Variation of Parameters
Main Theorem for Nonhomogeneous DE
Variation of Parameters
3
Variation of Parameters:
We now have expressions for the
general
solution
,
y
(
t
) =
u
1
(
t
)
y
1
(
t
) +
u
2
(
t
)
y
2
(
t
), and its derivatives,
y
0
(
t
) and
y
00
(
t
), which we substitute into the nonhomogeneous problem:
y
00
+
p
(
t
)
y
0
+
q
(
t
)
y
=
g
(
t
)
,
This can be written in the form:
u
1
(
t
) [
y
00
1
(
t
) +
p
(
t
)
y
0
1
(
t
) +
q
(
t
)
y
1
(
t
)]
+
u
2
(
t
) [
y
00
2
(
t
) +
p
(
t
)
y
0
2
(
t
) +
q
(
t
)
y
2
(
t
)]
+
u
0
1
(
t
)
y
0
1
(
t
) +
u
0
2
(
t
)
y
0
2
(
t
) =
g
(
t
)
The quantities in the square brackets are
zero
, since
y
1
and
y
2
are
solutions of the
homogeneous equation
, leaving
u
0
1
(
t
)
y
0
1
(
t
) +
u
0
2
(
t
)
y
0
2
(
t
) =
g
(
t
)
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Second Order Linear Equations
— (20/27)
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CauchyEuler Equation
Review
Variation of Parameters
Motivating Example
Technique of Variation of Parameters
Main Theorem for Nonhomogeneous DE
Variation of Parameters
4
Variation of Parameters:
This gives two
linear algebraic
equations
in
u
0
1
and
u
0
2
u
0
1
(
t
)
y
1
(
t
) +
u
0
2
(
t
)
y
2
(
t
)
=
0
u
0
1
(
t
)
y
0
1
(
t
) +
u
0
2
(
t
)
y
0
2
(
t
)
=
g
(
t
)
Recall Cramer’s Rule for solving a system of two linear equations in
two unknowns, which above are the functions
u
0
1
(
t
) and
u
0
2
(
t
).
u
0
1
(
t
) =
det
0
y
2
(
t
)
g
(
t
)
y
0
2
(
t
)
det
y
1
(
t
)
y
2
(
t
)
y
0
1
(
t
)
y
0
2
(
t
)
and
u
0
2
(
t
) =
det
y
1
(
t
)
0
y
0
1
(
t
)
g
(
t
)
det
y
1
(
t
)
y
2
(
t
)
y
0
1
(
t
)
y
0
2
(
t
)
From before we recognize the denominator as the
Wronskian
:
W
[
y
1
, y
2
](
t
) = det
y
1
(
t
)
y
2
(
t
)
y
0
1
(
t
)
y
0
2
(
t
)
=
y
1
(
t
)
y
0
2
(
t
)

y
2
(
t
)
y
0
1
(
t
)
Joseph M. Mahaffy,
h
[email protected]
i
Lecture Notes – Second Order Linear Equations
Part 2  Nonhomogeneous
— (21/27)
CauchyEuler Equation
Review
Variation of Parameters
Motivating Example
Technique of Variation of Parameters
 Fall '08
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